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Calculus

Calculus—by Md. Shouvik Iqbal—is an open access mathematics textbook published under CC BY-NC-SA 4.0 International License. Download the PDF for free at: https://md-shouvik-iqbal.github.io/calculus

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Calculus

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Copyright © 2024 Md. Shouvik Iqbal. This book is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.

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Calculus Md. Shouvik Iqbal

Copyright © 2024 Md. Shouvik Iqbal. This book is licensed under the Creative
Commons Attribution-NonCommercial-ShareAlike 4.0 International License
(CC BY-NC-SA 4.0)

Edition: First

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Design: Md. Shouvik Iqbal

AMS 2020 Mathematics Subject Classification: 97-01, 26A06, 97I40, 97I50

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Preface 13

I The Basis of Modern Calculus

1 Limits and Continuity 23
1.1 An Introduction to the Limit ···························································· 24
1.2 One-Sided Limits ············································································ 30
1.3 Existence of a Limit ········································································ 32
1.4 Analytic Evaluation of Limit of a Function········································· 34
1.4.1 General Properties of Limit of a Function · · · · · · · · · · · · 34
1.4.2 Direct Substitution · · · · · · · · · · · · · · · · · · · · · · · · 37
1.4.2.1 The Limit of Polynomial and Rational Functions · · · 37
1.4.2.2 The Limit of a Reciprocal · · · · · · · · · · · · · · · · 39
1.4.2.3 The Limit of Transcendental Functions · · · · · · · · · 40
1.4.2.4 The Limit of a Composite Function · · · · · · · · · · 42
1.4.3 Algebraic Manipulation · · · · · · · · · · · · · · · · · · · · · · 44
1.4.3.1 Dividing Out Technique · · · · · · · · · · · · · · · · · 46
1.4.3.2 Rationalizing Technique · · · · · · · · · · · · · · · · · 48
1.4.4 The Squeeze/Sandwich/Pinching Theorem · · · · · · · · · · · · 52
1.4.5 Trigonometric Manipulation · · · · · · · · · · · · · · · · · · · · 60
1.5 Limits Involving ∞·········································································· 68
1.5.1 Infinite Limits · · · · · · · · · · · · · · · · · · · · · · · · · · · 69
1.5.2 Limit at Infinity · · · · · · · · · · · · · · · · · · · · · · · · · · 73
1.5.3 Infinite Limits at Infinity and End Behaviors · · · · · · · · · · 75
1.5.3.1 The Limit of Power functions as x → ±∞ · · · · · · · 77
1.5.3.2 The Limit of Polynomial Functions as x → ±∞ · · · · 78
1.5.3.3 The Limit of Rational Functions as x → ±∞ · · · · · 80
1.5.3.4 The Limit of Functions Involving Radicals as x → ±∞ 88
1.5.3.5 Limits of Transcendental Functions as x → ±∞ · · · · 91
1.6 The Concept of Continuity ······························································· 94
1.6.1 Continuity at a Point · · · · · · · · · · · · · · · · · · · · · · · 94

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1.6.2 Continuity over an Interval · · · · · · · · · · · · · · · · · · · · 97
1.6.3 Properties of Continuity · · · · · · · · · · · · · · · · · · · · · · 100
1.6.3.1 Continuity of Combined Functions · · · · · · · · · · · 100
1.6.3.2 Continuity of Composite Functions · · · · · · · · · · · 102
1.6.4 The Intermediate Value Theorem (IVT) · · · · · · · · · · · · · 102
1.7 (ε − δ) Definition of Limit ································································ 104

II Differential Calculus
2 An Introduction to Differential Calculus 115
2.1 The Tangent Line Problem ······························································· 117
2.2 Definition of the Derivative of a function············································ 127
2.3 Notation for Derivatives ··································································· 134
2.4 Differentiability and Continuity························································· 136
2.5 Higher-Order Derivatives·································································· 139

3 The Derivation of Derivatives 143

3.1 Elementary Rules of Differentiation ··················································· 143
3.1.1 The Constant Rule · · · · · · · · · · · · · · · · · · · · · · · · · 144
3.1.2 The Constant Multiple Rule · · · · · · · · · · · · · · · · · · · 145
3.1.3 The Power Rule · · · · · · · · · · · · · · · · · · · · · · · · · · 147
3.1.4 The Summation Rule · · · · · · · · · · · · · · · · · · · · · · · 150
3.1.5 The Difference Rule · · · · · · · · · · · · · · · · · · · · · · · · 152
3.1.6 The Product Rule · · · · · · · · · · · · · · · · · · · · · · · · · 153
3.1.7 The Quotient Rule · · · · · · · · · · · · · · · · · · · · · · · · · 158
3.2 Differentiation Rules for Trigonometric Functions································ 161
3.2.1 Differentiation of sin(x) · · · · · · · · · · · · · · · · · · · · · · 161
3.2.2 Differentiation of cos(x) · · · · · · · · · · · · · · · · · · · · · · 164
3.2.3 Differentiation of tan(x) · · · · · · · · · · · · · · · · · · · · · · 165
3.2.4 Differentiation of cot(x) · · · · · · · · · · · · · · · · · · · · · · 167
3.2.5 Differentiation of sec(x) · · · · · · · · · · · · · · · · · · · · · · 169
3.2.6 Differentiation of csc(x) · · · · · · · · · · · · · · · · · · · · · · 171
3.3 Differentiation of Composite Functions
(The Chain Rule) ············································································ 172
3.4 Implicit Differentiation····································································· 182
3.5 Differentiation Rules for Exponential Functions ·································· 188
3.5.1 Differentiation of ex · · · · · · · · · · · · · · · · · · · · · · · · 188
3.5.2 Differentiation of ax · · · · · · · · · · · · · · · · · · · · · · · · 192
3.6 Differentiation Rules for Logarithmic Functions ·································· 195

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3.6.1 Differentiation of ln(x) · · · · · · · · · · · · · · · · · · · · · · 195
3.6.2 Differentiation of loga (x) · · · · · · · · · · · · · · · · · · · · · 198
3.7 Logarithmic Differentiation······························································· 200
3.8 Differentiation Rules for Inverse Trigonometric Functions····················· 206
3.8.1 Differentiation of sin−1 (x) · · · · · · · · · · · · · · · · · · · · · 206
3.8.2 Differentiation of cos−1 (x) · · · · · · · · · · · · · · · · · · · · 208
3.8.3 Differentiation of tan−1 (x) · · · · · · · · · · · · · · · · · · · · 210
3.8.4 Differentiation of cot−1 (x) · · · · · · · · · · · · · · · · · · · · · 212
3.8.5 Differentiation of sec−1 (x) · · · · · · · · · · · · · · · · · · · · · 214
3.8.6 Differentiation of csc−1 (x) · · · · · · · · · · · · · · · · · · · · · 216
3.9 Differentiation of General Inverse Functions ······································· 218
3.10 Summary ······················································································· 221

4 Applications of Derivatives 223

4.1 Extreme Values of a Function ··························································· 223
4.1.1 Absolute (Global) Extreme Values · · · · · · · · · · · · · · · · 224
4.1.2 Relative (Local) Extreme Values · · · · · · · · · · · · · · · · · 227
4.1.3 Critical Points · · · · · · · · · · · · · · · · · · · · · · · · · · · 228
4.1.4 Location of Absolute (Global) Extreme Values · · · · · · · · · 233
4.2 Increasing and Decreasing Functions·················································· 236
4.2.1 Increasing and Decreasing Functions · · · · · · · · · · · · · · · 236
4.2.2 The First Derivative Test · · · · · · · · · · · · · · · · · · · · · 241
4.3 Concavity and the Second Derivative Test·········································· 245
4.3.1 The Point of Inflection · · · · · · · · · · · · · · · · · · · · · · · 247
4.3.2 The Second Derivative Test · · · · · · · · · · · · · · · · · · · · 250
4.4 Rolle’s Theorem and The Mean Value Theorem ·································· 250
4.4.1 Rolle’s Theorem · · · · · · · · · · · · · · · · · · · · · · · · · · 250
4.4.2 The Mean Value Theorem (MVT) · · · · · · · · · · · · · · · · 253
4.5 Related Rates ················································································· 257
4.6 Linearization and Differentials ·························································· 262
4.6.1 Linearization · · · · · · · · · · · · · · · · · · · · · · · · · · · · 263
4.6.2 Differentials · · · · · · · · · · · · · · · · · · · · · · · · · · · · 267
4.7 Newton’s Method ············································································ 271
4.8 L’Hôpital’s Rule ·············································································· 277
0
4.8.1 Indeterminate Form · · · · · · · · · · · · · · · · · · · · · · · 278
0
∞
4.8.2 Indeterminate Form · · · · · · · · · · · · · · · · · · · · · · 282
∞
4.8.3 Indeterminate Forms 0 · ∞ and ∞ − ∞ · · · · · · · · · · · · · 284
4.8.4 Indeterminate Forms 1∞ , 00 , and ∞0 · · · · · · · · · · · · · · 289

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III Integral Calculus
5 An Introduction to Integral Calculus 295
5.1 The Area Under a Curve·································································· 297
5.1.1 The Summation ( ) Notation · · · · · · · · · · · · · · · · · · 298
P

5.1.2 Area Approximation · · · · · · · · · · · · · · · · · · · · · · · · 304

5.1.3 Generalized Area Approximation · · · · · · · · · · · · · · · · · 310
5.1.3.1 The Left-Endpoint Approximation · · · · · · · · · · · 312
5.1.3.2 The Right-Endpoint Approximation · · · · · · · · · · 313
5.1.4 Riemann Sum and the Definition of the Area Under a Curve · · 321
5.2 The Definite Integral ······································································· 323
5.2.1 Net Signed Area and the Definition of the Definite Integral · · 324
5.2.2 The Properties of the Definite Integral · · · · · · · · · · · · · · 332
5.2.2.1 The Reversal of Limits of Integration and Identical
Limits of Integration · · · · · · · · · · · · · · · · · · · 332
5.2.2.2 Integrals Involving a Sum or Difference of Functions · 334
5.2.2.3 Integrals Involving a Constant · · · · · · · · · · · · · 335
5.2.2.4 Integrals over Multiple Sub-intervals · · · · · · · · · · 336
5.2.2.5 Integrals Involving Absolute Values · · · · · · · · · · · 336
5.3 Numerical Integration ······································································ 340
5.3.1 The Midpoint Rule · · · · · · · · · · · · · · · · · · · · · · · · 340
5.3.2 The Trapezoidal Rule · · · · · · · · · · · · · · · · · · · · · · · 343

6 The Connection Between Differential and Integral Calculus 349

6.1 Antidifferentiation ··········································································· 349
6.1.1 Antiderivative · · · · · · · · · · · · · · · · · · · · · · · · · · · 350
6.1.2 Indefinite integral · · · · · · · · · · · · · · · · · · · · · · · · · 356
6.2 The Fundamental Theorem of Calculus (FTC) ··································· 364
6.2.1 The Inverse Relationship between Differentiation and Integration 370

7 Methods of Integration 377

7.1 Integration by Substitution······························································· 378
7.1.1 Substitution Rule for Definite Integrals · · · · · · · · · · · · · 391
7.2 Integration by Parts ········································································ 394
7.2.1 The LIATE Rule · · · · · · · · · · · · · · · · · · · · · · · · · · 397
7.2.2 Integration by Parts for Definite Integrals · · · · · · · · · · · · 400
7.2.3 Integration by Reduction Formulas · · · · · · · · · · · · · · · · 402
7.3 Trigonometric Integrals ···································································· 403
7.3.1 Integrals Involving Powers of sin(x) and cos(x) · · · · · · · · · 404
7.3.2 Integrals Involving Product of Powers of sin(x) and cos(x) · · · 407

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7.3.3 Integrals Involving Powers of tan(x) and sec(x) · · · · · · · · · 409
7.3.4 Integrals Involving Product of Powers of tan(x) and sec(x) · · · 411
7.4 Trigonometric Substitution······························································· 412
√
7.4.1 Integrals Involving a2 − x2 · · · · · · · · · · · · · · · · · · · 413
√
7.4.2 Integrals Involving a2 + x2 · · · · · · · · · · · · · · · · · · · 419
√
7.4.3 Integrals Involving x2 − a2 · · · · · · · · · · · · · · · · · · · 425
7.5 Partial Fractions ············································································· 431
7.5.1 The denominator Q(x) is a product of linear factors · · · · · · 433
7.5.2 The denominator Q(x) contains repeated product of linear factors435
7.5.3 The denominator Q(x) contains irreducible quadratic factors · · 438
7.5.4 The denominator Q(x) contains repeated irreducible quadratic
factors · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · 441
7.6 Improper Integrals··········································································· 443
7.6.1 Infinite Intervals · · · · · · · · · · · · · · · · · · · · · · · · · · 443
7.6.2 Discontinuous Integrands · · · · · · · · · · · · · · · · · · · · · 446
7.7 Table of Integrals ············································································ 450
7.7.1 Elementary Integrals · · · · · · · · · · · · · · · · · · · · · · · · 451
7.7.2 Trigonometric Integrals · · · · · · · · · · · · · · · · · · · · · · 452
7.7.3 Inverse Trigonometric Integrals · · · · · · · · · · · · · · · · · · 453
7.7.4 Reduction Formula for Trigonometric Functions · · · · · · · · · 453
7.7.5 Integrals of the Form a2 − x2 , a > 0 · · · · · · · · · · · · · · · 454
7.7.6 Integrals of the Form x2 − a2 , a > 0 · · · · · · · · · · · · · · · 455
7.7.7 Integrals of the Form a2 + x2 , a > 0 · · · · · · · · · · · · · · · 455
7.7.8 Integrals of the Form ax ± b, a 6= 0, b > 0 · · · · · · · · · · · 456
7.7.9 Integrals with Exponential and Trigonometric Functions · · · · 457
7.7.10 Integrals with Exponential and Logarithmic Functions · · · · · 457
7.7.11 Miscellaneous Integral Forms · · · · · · · · · · · · · · · · · · · 457

8 Applications of Integration 459

8.1 Average Value of a Function ····························································· 459
8.1.1 The Mean Value Theorem for Integrals · · · · · · · · · · · · · 462
8.2 Area of a Region Between Two Curves··············································· 464
8.2.1 Area of a Region between Two Curves with respect to y · · · · 466
8.3 Volume: The Slicing Method ···························································· 470
8.3.1 The Definition of the Volume of a Solid · · · · · · · · · · · · · 470
8.3.2 The Disk Method · · · · · · · · · · · · · · · · · · · · · · · · · 474
8.3.3 The Washer Method · · · · · · · · · · · · · · · · · · · · · · · · 478
8.4 Volume: The Method of Cylindrical Shells ········································· 481
8.5 Arc Length ····················································································· 487

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8.6 The Area of a Surface of Revolution ·················································· 491
8.6.1 The Generalized Area Formula for a Surface of Revolution · · · 495
8.7 Work ····························································································· 501

Index 507

License 513

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Preface

“The only constant in life is change”

Indeed! Everything in life is changing except for the state of being changed itself.
From the smallest particles to the largest of galaxies and beyond, everything is
in a state of flux—a continuous change. Calculus—as we know it today—is the
mathematics of this very change. More precisely, it is the mathematics to deal
with variables that are continuously changing and can thus be used to under-
stand the changing nature around us and to answer the fundamental questions
within our minds.

Unfortunately, in today’s world, the fundamental questions and ideas behind

calculus have tragically been reduced to not much more than an intimidating
academic obstacle. The subject of deep exploration, requiring years and years to
fully comprehend, has now been relegated to the status of a mere requirement—
a box to be ticked. The emphasis is no longer put on understanding the deeper
meaning behind the ideas; instead, it seems like it has turned to rote memo-
rization of formulas and techniques to reach the correct answer, all in the hope
of getting a good score, seen as currency for future opportunities. Should one
fail to do so within a certain time period, their prospects—at least in society’s
eyes—are in jeopardy!

Calculus—at least, in my eyes—is not just all about that. I view it as a pro-
found inspiration where philosophical ideas intersect with that of mathematical
reasoning to produce something nothing short of elegant. What inspired this
book was my desire to share this elegance I found within calculus. The origin

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of which can be traced back to the collection of notes I made during my own
journey of learning the subject years ago. This, I believe, ensures that the topics
that fascinated me as a learner will resonate just as deeply with readers. I’ve
made every effort to present the material in a clear and accessible language,
without sacrificing the eloquence of tone. Likewise, I tried my best to explain
the reasoning behind each concept without excluding any contents covered in
any other formal textbook. That is, this book covers all the topics typically in-
cluded in any introductory calculus text, but it distinguishes itself in how those
ideas are explained. Therefore, the reader may find it comfortable to use it as a
main text, especially when learning the concepts.

The Structure of the Book.

The book is divided into three parts to discuss two major branches of calculus:
differential calculus and integral calculus. The first part serves as a foundation
for the contents that follow and consists of the opening chapter. The second
part focuses on differential calculus and includes chapters two, three, and four.
The third part focuses on integral calculus, covering chapters five through eight.
A detailed purpose and direction of each chapter is discussed below.

• Chapter 1. The Concept of Limits and Continuity.

The purpose of this chapter is to introduce the foundational concept of limit and continuity
of a function in a way that demonstrates its essence. This chapter is organized in a way
that covers all the necessary information about limits and continuity at a foundational level.
The chapter begins with a thought-provoking observation and analysis to set the reader
in the correct mindset to understand the topics that follow. After giving the reader an
informal definition of the limit of a function, the chapter begins to discuss and define what
is referred to as one-sided limits and the conditions for a limit to exist. Following that, the
chapter dives into a long discussion of evaluating the limit of a function—showing various
methods and techniques. Then the limits involving infinity are discussed in a similarly
detailed fashion. Following that, the concept of continuity of a function is discussed along
with their associated properties. At the end, the chapter concludes by discussing the formal
(ε − δ) definition of limits along with its execution.

• Chapter 2. An Introduction to Differential Calculus.

The purpose of this chapter, as the name suggests, is to introduce a subfield of calculus—
differential calculus. Understanding the ideas covered in this chapter is essential for com-
prehending the chapters that follow. The chapter begins with a brief overview of differential
calculus, leading to a key question related to the problem of the tangent line, which is cen-
tral to understanding the concept of the derivative. The chapter then discusses the tangent
line problem in detail—defining the slope of the tangent line and the instantaneous rate of
change. From this, the concept of the derivative is introduced, defined in terms of the slope
of the tangent line. After discussing the derivative of a function with several examples,
the chapter discusses various notations used to represent derivatives. Finally, the chapter
concludes by addressing two closely related concepts: differentiability and continuity.

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• Chapter 3. The Derivation of Derivatives.
The purpose of this chapter is to derive the derivatives of various functions in order to avoid
reapplying the definition of the derivative repeatedly. The chapter begins with the section
“Elementary Rules of Differentiation,” which covers several fundamental rules: Constant
Rule, Constant Multiple Rule, Power Rule, Summation Rule, Difference Rule, Product Rule,
and Quotient Rule. These rules serve as the foundation for differentiating a wide range of
functions. The next section, “Differentiation Rules for Trigonometric Functions,” focuses
on the derivatives of the six trigonometric functions along with their examples. Following
this, the section “Differentiation of Composite Functions” introduces the Chain Rule and
explores its application through numerous examples. Understanding the Chain Rule is cru-
cial for comprehending the following topics. Next, in the section “Implicit Differentiation,”
the process of differentiating implicitly defined functions is discussed, along with multi-
ple examples to illustrate the method. The section “Differentiation Rules for Exponential
Functions” derives and discusses the derivatives of exponential functions with various ex-
amples. The subsequent section “Differentiation Rules for Logarithmic Functions,” derives
and discusses the derivatives of logarithmic functions with various examples. Following this,
the section on “Logarithmic Differentiation” explains a technique for simplifying functions
using logarithmic properties before differentiating, which proves beneficial in various cases.
The next section, “Differentiation Rules for Inverse Trigonometric Functions,” derives the
derivatives of the six inverse trigonometric functions and provides various examples. Follow-
ing that, the section “Differentiation of General Inverse Functions,” extends the discussion
to cover general inverse functions. Finally, the chapter concludes with a summary of the
key derivatives for quick reference.

• Chapter 4. Applications of Derivatives.

The purpose of this chapter is to show how the concept of derivative of a function, as
discussed previously, can be applied to solve a variety of mathematical problems. The
chapter begins with the section “Extreme Values of a Function,” which covers topics such
as absolute (global) extreme values, relative (local) extreme values, critical points, and the
location of absolute (global) extreme values. The next section, “Increasing and Decreasing
Functions,” discusses the functions that are increasing or decreasing, followed by an intro-
duction to the First Derivative Test for determining intervals of increase or decrease. In
the section “Concavity and the Second Derivative Test,” the chapter discusses the Point of
Inflection and the Second Derivative Test as a tool to analyze the concavity of functions
and identify points where the concavity changes. Next, the section “Rolle’s Theorem and
The Mean Value Theorem” presents two important results in differential calculus: Rolle’s
Theorem and the Mean Value Theorem (MVT), along with their applications. Following
that, the section “Related Rates” discusses problems of related rates and shows how the
derivatives of multiple variables can be used to solve practical problems with various ex-
amples. In “Linearization and Differentials,” the concept of linearization is introduced to
show how functions can be approximated near a given point using derivatives and is fol-
lowed by a discussion of differentials. Then the section “Newton’s Method” introduces a
numerical method to approximate the roots of a function using derivatives. At the end, the
chapter concludes with “L’Hôpital’s Rule,” which provides a method for evaluating limits
of indeterminate forms using derivatives.

• Chapter 5. An Introduction to Integral Calculus.

The purpose of this chapter is to introduce a subfield of calculus—Integral calculus. It
begins with a discussion of the concept of area and its historical methods of calculation,
leading into a discussion of the section “area under a curve.” The topics discussed in that
section include the summation notation, area approximation, generalization of area ap-
proximation, Riemann sums, and the formal definition of the area under a curve in terms
of Riemann sums. Following that, the chapter discusses the net signed area in terms of
Riemann sums, and introduces the definite integral as a formal way of representing net

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signed area. The properties of definite integrals are then discussed through various exam-
ples. Finally, the chapter concludes with a discussion of numerical integration, specifically
the Midpoint Rule and the Trapezoidal Rule, which are methods for approximating definite
integrals.

• Chapter 6. The Connection Between Differential and Integral Calculus.

The purpose of this chapter is to establish the connection between two subfields of calculus
through the Fundamental Theorem of Calculus (FTC). The goal is to show that differen-
tiation and integration are essentially the inverse operations of each other in a way that
one can be undone by applying the other. To set up this inverse relationship, the chapter
discusses the fundamental topics such as antiderivatives and indefinite integrals and then
introduces the Fundamental Theorem of Calculus. Finally, the chapter concludes with a
discussion of the inverse relationship between differentiation and integration.

• Chapter 7. Methods of Integration.

The purpose of this chapter is to discuss various methods of integration. It begins with the
section “Integration by Substitution,” which covers the u-substitution rule through various
examples for both definite and indefinite integrals. Next is the section “Integration by
Parts,” which discusses the integration of products of functions and introduces the LIATE
rule, integration by parts for definite integrals, and reduction formulas. Following this, the
“Trigonometric Integrals” section discusses methods for integrating functions involving the
powers of sine, cosine, tangent, and cotangent functions, as well as the products of these
functions. After that comes the “Trigonometric Substitution” section, which discusses
techniques for integrating functions of a specific form with the help of trigonometry. The
chapter then moves on to “Partial Fractions” to discuss integrating rational functions. Next,
the section on “Improper Integrals” discusses integrating functions that either become
infinite within an interval or have intervals extending to infinity. Finally, the chapter
concludes with a comprehensive table of indefinite integrals as a reference.

• Chapter 8. Applications of Integration.

The purpose of this chapter is to show how integration can be used to solve a range of
mathematical problems. It starts with the section “Average Value of a Function” which
introduces the concept of average value of a function and the mean value theorem for in-
tegrals with examples. The next section “Area of a Region Between Two Curves” explains
how to find the area of a region enclosed by two curves with examples. The section “Vol-
ume: The Slicing Method” discusses how to find the volume of a three-dimensional solid
of revolution. It begins by defining the volume of such solids using integrals and then dis-
cusses the disk and washer methods with examples. The following section “Volume: The
Method of Cylindrical Shells” presents another method to calculate the volume of a solid
of revolution with examples. The section “Arc Length” explains how to precisely calculate
the length of a curve using the formula for arc length. The section “The Area of a Surface
of Revolution” shows how to calculate the area of the surface of solids of revolution with
examples. Finally, the chapter concludes with the section “Work” which discusses the work
done by a variable force with examples.

Prerequisite.

The true study of calculus necessitates a certain level of mathematical knowl-

edge, as well as some awareness of its historical context. Although this book
is not filled with just mathematically rigorous statements and proofs, it does,
however, assume that the readers have a basic understanding of the precalculus
topics such as algebra, geometry, and trigonometry. This is because these topics

Copyright © 2024 Md. Shouvik Iqbal. This book is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License (CC BY-NC-SA 4.0)
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are not covered in detail within this book, except when necessary for clarity.
While prior exposure to introductory calculus concepts, such as limits, may be
beneficial, it is not required in this text. What is required is a solid grasp of the
precalculus contents to follow along comfortably.

From my experience, there are three category of potential readers: (1) those
who have a strong grasp of the precalculus topics, (2) those with a basic under-
standing of the precalculus topics, and (3) those with little to no knowledge of
the precalculus topics and may struggle comprehending statements such as “Let
a be a real number defined over the domain of f (x)” or “for all x, there exist y
such that this and that is true” or something along those lines.

Readers in the first category can proceed through the book without additional
resources. Those in the second category are encouraged to keep a relevant math-
ematical textbook that covers the precalculus topics or use online resources as
needed. However, the third category, lacking a foundation in precalculus, may
find the book difficult to follow and are generally advised not to begin without
first mastering the necessary background, though this choice is ultimately up to
them.

So, the bottom line is that a firm command of precalculus topics is crucial to
fully engage with the material, as these topics are not explicitly covered in the
book.

About the Author.

Born and raised in Bangladesh, I have completed both my secondary (SSC)

and higher secondary (HSC) education under the Board of Intermediate and
Secondary Education in Bangladesh. I am currently looking forward to my
undergraduate studies. This book began as a collection of notes when I was 16,
shortly after my SSC exam and before my admission to intermediate college, and
is now being published at the age of 19. Writing this book has been a central
focus throughout my academic years. In addition to this, I have a strong interest
in academic research in mathematics (as of now, mainly number theory) and
philosophy and have authored several research articles and notes. I also have a
longstanding engagement with creative disciplines such as painting, music, and
literature, for which I have earned awards at both national and international

—Md. Shouvik Iqbal

Dhaka, Bangladesh
30 October 2024

The figure in the cover of this book was taken from the following photograph of
the statue of Issac Newton, located in the chapel of Trinity College, Cambridge.

Figure 1: Statue of Isaac Newton

Photograph © Andrew Dunn, 8 September 2004.

Website: http://www.andrewdunnphoto.com.
Visit https://commons.wikimedia.org/wiki/File:StatueOfIsaacNewton.jpg
for more details.

Copyright © 2024 Md. Shouvik Iqbal. This book is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License (CC BY-NC-SA 4.0)
Download for free at https://md-shouvik-iqbal.github.io/calculus
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Download for free at https://md-shouvik-iqbal.github.io/calculus
Part I

The Basis of Modern

Calculus

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Download for free at https://md-shouvik-iqbal.github.io/calculus
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Chapter 1

Limits and Continuity

In order to travel a distance and reach its end, one must first have
to cover half of that distance. From there, to reach the end, the re-
maining half must be covered, and then again the remaining half, and
again the half of what remains, and the process continues endlessly.
Each step brings closer and closer to the endpoint, but never exactly
reaches it, as there’s always another halfway point to be traveled to
reach the endpoint. Thus, the conclusion left to be drawn is that the
endpoint can never be exactly reached.

This seemingly absurd observation above appears to have been first made by
Zeno of Elea, a pre-Socratic Greek philosopher. It is a paradox suggesting that
a moving object can certainly approach the endpoint but never exactly reach it.
The underlying idea is similar to that of the mathematical concept referred to
as the “limit.” The notion of such in mathematics stands as such an ineffable
construct that, without its existence, the entire field itself might not have devel-
oped in the form we see it today. For example, calculus is a fundamental branch
of mathematics, and among the significant topics in calculus lie derivatives and
integrals, both of which can be thought of as a unique representation of limits.
In this chapter, we aim to focus on exploring limits, their definitions, properties,
associated theorems, and tackling problems related to limits. Thus, we start
with the introduction of the concept of limit.

§ 1.1 An Introduction to the Limit

The concept of limit is fundamental in modern calculus as it formalizes the
informal idea of a variable “approaching a fixed value,” which arises in many
contexts. Simply put, a limit provides a precise and rigorous way to describe
how a function behaves as the input gets closer to a specific point. It captures
not only the function’s behavior near a point but also how it behaves when it is
arbitrarily close to that point. This allows us to analyze situations where direct
calculation either fails or leads to ambiguous results by—informally speaking—
considering the value predicted by the surrounding points.

Quite frankly, it is hard—if not totally impractical—to try to understand the

concept of a limit without first realizing the idea of approaching a value. It’s an
informal term frequently used in situations such as the following.

Consider a hypothetical bouncing ball “b” with the following property “P.”

Property P: When b is dropped from a certain height on a surface, it rebounds

up to exactly 1 half of its previous height.

So, we use the symbol “b(P)” to denote a bouncing ball with the property P.

Say, we drop the ball b(P) from a height of 1 unit. When it is dropped from
that height, it bounces back up 0.5 units on the first bounce, then 0.25 units,
0.125 units, and so on. So, the following question arises naturally,

Question: How many times will the ball bounce before it comes to rest on the
ground, if it ever does?

Let us logically think about this problem. The ball was released from a height
of 1 unit. So, the initial height of that ball is 1 unit. Then, after each bounce,
1
the new height of the ball is half, i.e., times the previous height due to the
2
property P. This follows that,
1 1
after bounce = 1, the new height = 1 × = units
|{z} 2 2
previous height

1
with no loss of physical properties such as energy in a way that if dropped from a height
x
of x units, it rebounds back up to units after 1 bounce.
2

1 1 1
after bounce = 2, the new height = × = 2 units
2
|{z} 2 2
previous height

1 1 1 1
after bounce = 3, the new height = × × = 3 units
2 2 2 2
| {z }
previous height

1 1 1 1 1
after bounce = 4, the new height = × × × = 4 units
2 2 2 2 2
| {z }
previous height
..
.
previous height
}| { z
1 1 1 1 1 1
after bounce = n, the new height = × × × ··· × × = n units
2 2 2 2 2 2
| {z }
n

1
Notice that as the number of bounces n increases, the height n tends to ap-
2
proach a value of 0, as shown in the table below.
Table 1.1: Table

n 0 1 2 3 4 5 ··· ∞
1
1 0.5 0.25 0.125 0.0625 0.03125 ··· 0
2n
Therefore, we say,

1
as n approaches ∞, approaches 0
2n
1
where, n is the number of bounces and n is the height of the ball b(P) after
2
each bounce. This can symbolically be written as,
1
as n → ∞, →0
2n
by letting the symbol “→” to mean “approaching,” just like the symbol “=” is
used to mean “equal.”

1
Since, the height n gets closer and closer to a value 0 as the number of bounces
2
n increases without a bound, therefore, the value 0 can be thought of as the
1
ultimate destination—the limit—for the height n to ultimately reach, as n
2
approaches infinity. We express this idea symbolically as follows,
1
lim n = 0
n→∞ 2

1
In other words, n → 0, as n → ∞. Thus, we have found our answer to the
2
question above that—hypothetically—the ball b(P) will increasingly approach
the ground to settle down as the number of bounces grows without a bound.

Now, notice in the table above that for each bounce n, there exists a corre-
1
sponding new height n . Therefore, the table above can be modified into a
2
1
height function f of each bounce n, that is, f (n) = n . Notice that n in the
2
1
function f (n) = n is restricted to be a positive integer as the number of bounces
2
can’t be negative or fractional. The function, however, maps these positive in-
tegers to real numbers as an output.

It is, by no means, essential to do this to understand the concept of approaching

a value. Functions can be defined for inputs beyond positive integers for the
concept of limits to be still applicable. To demonstrate, consider the same
function but with the variable x instead of n, indicating the function is defined
on real numbers, as follows.
1
f (x) = x
2
1
Visually, it can be seen that f (x) = x → 0 as x → ∞ in the following figure.
2
y

1.5

0.5

x
0.5 1 1.5 2 2.5

Figure 1.1

1
Notice that the domain (simply, the input) of f (n) = n is a subset of the
2
1 1
domain of f (x) = x , therefore if the limit of f (x) = x is 0 as the input ap-
2 2
1
proaches infinity, then the limit of f (n) = n automatically becomes 0 for the
2

same.

Now, what if the input x approaches a finite value, say x = 1? What value does
1
the function f (x) = x approach? It can be observed—with no effort—from the
2
graph shown below that when x → 1, f (x) → 0.5.

1.5

0.5

x
0.5 1 1.5 2 2.5

Figure 1.2

It may seem absurd to even consider the concept of approaching a value in this
1
case. Because anyone can evaluate the function f (x) = x at the point x = 1
2
1
and get the answer that f (1) = 1 = 0.5. However, the idea of limits really
2
becomes necessary when a function is not defined at a point. For example,
consider the following function,

x2 − 42
f (x) =
x−4
What is the value of the function at the point x = 4? Evaluating the function
at x = 4, results in the following,

x2 − 42
f (x) =
x−4
42 − 42
=
4−4
0
=
0
This can be interpreted as the measure of nothing is broken down into the
measure of nothing, which is nothing but absurdity. However, by examining
the behavior of the function f (x) in the vicinity of x = 4, we can predict by

estimating the value of f (x) at x = 4. To do this, we approximate f (x) using

values close to x = 4, albeit not precisely equal to it. These values may either
be slightly less than 4 or slightly greater than 4. We start by estimating values
of f (x) using values of x less than 4.

x2 − 42
x<4 f (x) =
x−4
3.9 7.9
3.99 7.99
3.999 7.999
3.9999 7.9999

The table above indicates that as the values of x are getting closer and closer
to the point x = 4, the values of f (x) are getting closer and closer to the point
f (x) = 8. Now, we estimate f (x) using values of x greater than 4.

x2 − 42
x>4 f (x) =
x−4
4.1 8.1
4.01 8.01
4.001 8.001
4.0001 8.0001

Similar to the previous table, this table above also indicates that as x is ap-
proaching 4, f (x) is approaching the value 8.

This can also be seen visually as follows.

2 4 6 8

Figure 1.3

Thus, it may be sufficiently logical to conclude that the limit of the function
x2 − 42
f (x) = , as x gets closer and closer to the value 4, is the value 8.
x−4

We can generalize the idea behind the limit and define the limit of a function as
follows. Note that this definition is informal and the formal definition is given
at the end of this chapter.

Definition 1.1.1. (Informal Definition of Limit)

Let f (x) be a function defined on an open interval containing a, except
possibly at the point a. If f (x) gets arbitrarily close to the real number
L, as x gets arbitrarily close to the point a, then the limit of f (x) as x
approaches a, is L. Symbolically,

lim f (x) = L
x→a

Note that x does not necessarily have to be defined at the point a. In fact, when
examining the limit of f (x) as x approaches a, the value of x at a is not even
taken into account. The sole concern is how f (x) behaves in the vicinity of a.

By now, if it isn’t already obvious, it should be pointed out that as x approaches

a particular value, it can approach from two directions on the number line. It
can either approach that value through values less than that or through values
greater than that particular value, as done previously in the tables above.

In general terms, if x approaches a finite value a from values less than a, it

appears visually to be approaching from the left side of a on the number line. If
x approaches a from values greater than a, it appears visually to be approaching
from the right side of a on the number line. Thus, this prompts us to study one-
sided limits as outlined in the next section.

The End of Section 1.1

An Introduction to the Limit

§ 1.2 One-Sided Limits

A one-sided limit refers to the value that a function approaches as the input
approaches a particular point from only one side—either from the left or from
the right. For example, consider we have an arbitrary function y = f (x), as
shown below.
y

x
a

Figure 1.4

Here, as x approaches a, the function f (x) approaches a limit L. If x approaches

a through values less than a, it is visually seen that x approaches a from the left
side of a on the number line, as indicated by the blue shaded region the figure
above. Symbolically, we denote this as x → a− .

On the other hand, if x approaches a from values greater than a, it is visually

seen that x approaches a from the right side of a on the number line, as indicated
by the red shaded region in the figure above. Symbolically, this is expressed as
x → a+ .

Thus, the informal definition of this one-sided limit can be given as,

Definition 1.2.1. (One-sided Limits)

1. Limit from the left: Let f (x) be a function defined over an open in-
terval (b, a) and let L be a real number. If f (x) → L as x → a, where
x < a, then the limit of f (x) is L as x approaches a from the left.
Symbolically,
lim− f (x) = L
x→a

2. Limit from the right: Let f (x) be a function defined over an open
interval (a, b) and let L be a real number. If f (x) → L as x → a,
where a < x, then the limit of f (x) is L as x approaches a from the
right. Symbolically,
lim+ f (x) = L
x→a

At first glance, defining one-sided limits might seem trivial. Yet, it turns out
to be a crucial clarification. This is because not every function approaches the
same limit as its input approaches a single value. For example, consider the
x
graph of the function f (x) = . As x → 0, what value does y = f (x) get
|x|
closer to?
y

x
0

Figure 1.5

The answer is not obvious because if x approaches 0 from the left side of 0, it
seems like f (x) approaches the value −1. However, if x approaches 0 from the
right side of 0, it seems like f (x) approaches the value 1. Therefore, the overall
limit of f (x) as x → 0, does not agree on any single limit and thus does not

exist, even though side-wise limits do exist, that is,

x x
lim− = −1 and lim+ =1
x→0 |x| x→0 |x|
.
Therefore, the situation at hand gives rise to the study of the existence of a limit
of a function, as outlined below.

The End of Section 1.2

One-Sided Limits

§ 1.3 Existence of a Limit

It follows from the previous section’s discussion that a limit L of a function only
exists if both the right-sided limit and the left-sided limit yield the same value
L. If not, then the limit of the function does not exist, as the function doesn’t
approach any specific value. Thus, we have the following theorem.

Theorem 1.3.1. (Existence of a Limit) Let f (x) be a function and let a and
L be real numbers. Then,

lim f (x) = L if and only if lim f (x) = L and lim+ f (x) = L

x→a x→a− x→a

This theorem implies,

if, lim f (x) 6= L
x→a

then, lim f (x) 6= L or, lim f (x) 6= L or, both

x→a+ x→a−

While violating Theorem 1.3.1 indeed leads to the absence of a function’s limit,
it’s important to note that this isn’t the sole factor causing the existence of
a limit to fail. Various other reasons can also contribute to the failure of a
function’s limit to exist. For example, consider the function below,

1
f (x) =
(x − 4)2
What value does f (x) approach as x → 4? To determine, we look at the graph
1
of f (x) = , as shown below.
(x − 4)2

x
2 4 6 8

Figure 1.6

1
It is evident that as x → 4 from both sides, the function f (x) = grows
(x − 4)2
1
with a bound. That is, f (x) = → ∞. Since ∞ is not a number,
(x − 4)2
1
therefore the limit of f (x) = as x → 4, does not exists.
(x − 4)2

One other reason for which the limit of a function fails to exist is oscillating
behavior. For example, consider the following function,

1
f (x) = sin
x

1
What value does f (x) = sin approach as x → 0?
x
1 1
As x → 0, → ∞ and sin oscillate between -1 and 1 infinitely often.
x x
1
Since f (x) = sin does not approach a fixed value, therefore the limit of
x
1
f (x) = sin as x → 0, does not exist.
x

In general, the following are the most common reasons for which the limit of a
function fails to exist.

Remark 1.3.1.

1. The limit of f (x) differs between the left and right sides, as x → a.

2. f (x) → ±∞, as x → a.

3. f (x) shows oscillating behavior as x → a.

The End of Section 1.3

Existence of a Limit

§ 1.4 Analytic Evaluation of Limit of a Function

So far, we have discussed the concept of limits and used two indirect method for
their evaluation. The first one is a numerical method, where we determine the
limit by numerically analyzing the function near a given point, often by a table
of values. The second one is a graphical method, relying on the visual inspec-
tion of the graph of a function. However, relying solely on these methods proves
inadequate for evaluating the limits of a function. Therefore, in this section, we
introduce several theorems regarding limits, enabling the analytical evaluation
of the limit of a function.

We begin by discussing a few essential properties of the limit of a function.

Chap. 1 / Sec. 1.4 / Subsec. 1.4.1 : General Properties of Limit of a Function

The following are some of the fundamental properties of the limit of a function,
commonly referred to as the “limit laws.” By leveraging these laws alongside
previously discussed methods, we can evaluate the limit of various kinds of
algebraic functions. The proofs for these limit laws are omitted in this section.

Theorem 1.4.1. (Limit Laws)

If L, M, a, and c are real numbers, and

lim f (x) = L and lim g (x) = M

x→a x→a

then,

1. lim c = c
x→a

2. lim x = a
x→a

3. lim c · f (x) = c · lim f (x) = c · L
x→a x→a

4. lim f (x) + g (x) = lim f (x) + lim g (x) = L + M
x→a x→a x→a

5. lim f (x) − g (x) = lim f (x) − lim g (x) = L − M
x→a x→a x→a

6. lim f (x) · g (x) = lim f (x) · lim g (x) = L · M
x→a x→a x→a

f (x)
lim f (x) L
7. lim = x→a
= where, g(x) 6= 0
x→a g (x) lim g (x) M
x→a
n n
8. lim f (x) = lim f (x) = Ln
x→a x→a
√
9. lim lim f (x) = where, n is a positive integer.
p q n
n
f (x) = n L,
x→a x→a

If n is even, then lim f (x) > 0 is assumed.

x→a

Acknowledging these limit laws, we now begin to use them to evaluate the limit
of a function.

I Example 1.4.1. If f (x) = 33, what value does f (x) approaches as x → 2?

Solution. From Law 1, we know that

lim c = c
x→a

This implies,
lim 33 = 33
x→2

Therefore, as x → 2, the function f (x) → 33.

I Example 1.4.2. If f (x) = x, what value does f (x) approaches as x → 2?

Solution. From Law 2, we know that

lim x = a
x→a

This implies,
lim x = 2
x→2

Therefore, as x → 2, the function f (x) → 2.

I Example 1.4.3. If lim f (x) = 4, then evaluate lim πf (x).

x→2 x→2

Solution. Since π is a constant, therefore by Law 3, we write,

lim πf (x) = π · lim f (x)

x→2 x→2

=π·4
= 4π

Therefore, the limit of πf (x), as x → 2 is 4π.

π
I Example 1.4.4. If lim f (x) = and lim g (x) = 4, then evaluate
x→e 2 x→e

πf (x) + g (x)
lim
x→e f (x) − f (x) g (x)

Solution. Since e and π are constants, therefore by the combination of Law 3, 4,

5, 6, and 7, we proceed,

πf (x) + g (x)
lim πf (x) + g (x)
lim = x→e
x→e f (x) − f (x) g (x)
lim f (x) − f (x) g (x)
x→e
lim (π · f (x)) + lim g (x)
x→e x→e
=
lim f (x) − lim (f (x) g (x))
x→e x→e
π · lim f (x) + lim g (x)
x→e x→e
=
lim f (x) − lim f (x) · lim g (x)
x→e x→e x→e
π
π· +4
= π 2π
− ·4
2 2
2
π +8
= 2
π − 4π
2
2
π +8 2
·

=
2
−3π

π2 + 8
=−
3π
π2 + 8

πf (x) + g (x) π
Therefore, lim =− , given that lim f (x) = and
x→e f (x) − f (x) g (x) 3π x→e 2
lim g (x) = 4.
x→e

5
I Example 1.4.5. If lim Z (X) = 5, then evaluate lim Z (X)
X→4 X→4

Solution. By Law 8, we write,

5 5
lim Z (X) = lim Z (X)
X→4 X→4
5
=5
= 3125
5
Therefore, the limit of lim Z (X) is 3125, if lim Z (X) = 5.
X→4 X→4

Chap. 1 / Sec. 1.4 / Subsec. 1.4.2 : Direct Substitution

Direct substitution is one of the simplest methods for evaluating limits. The core
idea is to directly substitute the value of x approaching a specific point into the
function. If this substitution yields a finite number without any indeterminate
0 ∞
forms such as , , etc. then the limit is simply that value.
0 ∞

More specifically, for a function f (x) continuous at x = a, the limit of f (x) as

x → a is f (a), given that f (a) exists and is finite. That is,

lim f (x) = f (a)

x→a

· · · / Subsec. 1.4.2 / Segment 1.4.2.1 : The Limit of Polynomial and Rational

Functions
We will now use the limit laws to determine the limit of polynomial functions.
For example, take the polynomial function

P (x) = x5 + 2x4 + 4x3 + 7x + 9

What value does P (x) approach as x → 5? To determine, we can use the limit
laws to evaluate the limit of P (x), as follows.

lim P (x) = lim x5 + 2x4 + 4x3 + 7x + 9

x→5 x→5

= lim x5 + lim 2x4 + lim 4x3 + lim (7x) + lim (9)

x→5 x→5 x→5 x→5 x→5

= lim x + 2 · lim x + 4 · lim x + 7 · lim (x) + lim (9)

5 4 3

x→5 x→5 x→5 x→5 x→5
5 4 3
=5 +2·5 +4·5 +7·5+9
= 4919

From the example above, it can be seen that the limit of the polynomial function
P (x) is easily found by direct substitution. That is, if P (x) is a polynomial
function, then the limit of it, as x → a, is P (a). This is due to the fact that
polynomial functions are one of the simplest types of functions and are defined
for all real numbers. Thus, lim P (x) in the example above, can be evaluated
x→5
directly as,

lim P (x) = lim x5 + 2x4 + 4x3 + 7x + 9

x→5 x→5

= 5 + 2 · 54 + 4 · 53 + 7 · 5 + 9
5

= 4919

Therefore, we have the following theorem.

Theorem 1.4.2. (The Limit of Polynomial Functions) If P (x) is a polynomial

function and a is a real number, then,

lim P (x) = P (a)

x→a

The theorem above further suggests that if P (x) and Q(x) are polynomial func-
P (x) P (a)
tions such that Q(x) 6= 0, then the limit of , as x → a, is . Thus, we
Q(x) Q(a)
have the following theorem.

P (x)
Theorem 1.4.3. (The Limit of Rational Functions) If R(x) = is a
Q(x)
rational function, and a is a real number such that Q(x) 6= 0, then,

P (x) P (a)
lim R(x) = lim =
x→a x→a Q(x) Q(a)

A few examples are provided below.

x3 + 3x
I Example 1.4.6. Evaluate lim .
x→1 x2 + 1

Solution. From the theorem above, we write,

x3 + 3x 13 + 3 · 1
lim =
x→1 x2 + 1 12 + 1
1+3
=
1+1
4
=
2
=2

x3 + 3x
Therefore, lim =2
x→1 x2 + 1

x1 + x2 + x3 + x4 + x5
I Example 1.4.7. If → L as x → 1, then what is the
x6 + x7 + x8 + x9 + x10
value of L?

Solution. From the theorem above, we write,

1
x + x2 + x3 + x4 + x5 11 + 12 + 13 + 14 + 15

lim =
x→1 x6 + x7 + x8 + x9 + x10 16 + 17 + 18 + 19 + 110
1+1+1+1+1
=
1+1+1+1+1
5
=
5
=1

x1 + x2 + x3 + x4 + x5
Therefore, 6 → 1 as x → 1.
x + x7 + x8 + x9 + x10

· · · / Subsec. 1.4.2 / Segment 1.4.2.2 : The Limit of a Reciprocal

The limit of a reciprocal refers to the limit of a function that is the reciprocal
of another function as x approaches a certain value. The key idea is to find the
1
limit of as x → a. To do this, we need to consider the behavior of f (x)
f (x)
itself. This is summarized below.

Theorem 1.4.4. (The Limit of a Reciprocal) If lim f (x) = L such that

x→a
L 6= 0, then,
1 1
lim =
x→a f (x) L

A few examples are shown below.

1
I Example 1.4.8. Evaluate lim .
x→2 2x
Solution. From the theorem above, we get,
1 1
lim x = 2
x→2 2 2
1
=
4
1 1
Therefore, lim x =
x→2 2 4

1
I Example 1.4.9. If f (x) = x3 , then what value does approaches as x →
|x3 |
−π?
Solution. From the theorem above, we get,
1 1
lim 3
=
x→−π |x | |−π 3 |
1
= 3
π
1 1
Therefore, 3 approaches 3 , as x → −π.
|x | π

· · · / Subsec. 1.4.2 / Segment 1.4.2.3 : The Limit of Transcendental Functions

Transcendental functions are a category of functions that are not algebraic and
include functions such as exponential functions, logarithmic functions, trigono-
metric functions, their inverses, etc. For transcendental functions, the limit of
a function as x → a can be found by directly substituting the value of a into
the function, provided that the function is continuous2 at x = a and it does not
0 ∞
result in an indeterminate form such as or . This is summarized below.
0 ∞
2
discussed later in this chapter

Theorem 1.4.5. (The Limit of Transcendental Functions) Let a be a real

number defined in the domain of the following functions, then,

1. lim bx = ba ,
x→a

2. lim ln (x) = ln (a)

x→a

3. lim logb (x) = logb (a)

x→a

Let a be defined in the domain of the following trigonometric functions;

then,

1. lim sin (x) = sin (a) 4. lim cot (x) = cot (a)
x→a x→a

2. lim cos (x) = cos (a) 5. lim sec (x) = sec (a)
x→a x→a

3. lim tan (x) = tan (a) 6. lim csc (x) = csc (a)
x→a x→a

Let a be defined in the domain of the following inverse trigonometric func-

tions; then,

1. lim sin−1 (x) = sin−1 (a) 4. lim cot−1 (x) = cot−1 (a)
x→a x→a

5. lim sec−1 (x) = sec−1 (a)

2. lim cos−1 (x) = cos−1 (a) x→a
x→a
6. lim csc−1 (x) = csc−1 (a)
3. lim tan−1 (x) = tan−1 (a)
x→a
x→a

A few examples of the theorem above are provided below.

I Example 1.4.10. Evaluate lim ex

x→1

Solution. From the theorem above, we get,

lim ex = e1
x→1

Therefore, lim ex = 0
x→1

I Example 1.4.11. Evaluate lim ln(x)

x→π

Solution. From the theorem above, we get,

lim ln(x) = ln(π)

x→π

Therefore, lim ln(x) = ln(π).

x→π

I Example 1.4.12. Evaluate lim sin−1 (x)

x→1

Solution. From the theorem above, we get,

lim sin−1 (x) = sin−1 (1)

x→1
π
=
2
π
Therefore, lim sin−1 (x) = .
x→1 2

· · · / Subsec. 1.4.2 / Segment 1.4.2.4 : The Limit of a Composite Function

The limit of a composite function refers to the limit of a function that is the com-
position of two or more functions. The key idea is that the limit of a composite
function can be evaluated by first determining the limit of the inner function
and then applying the outer function to that limit. This process relies heavily
on the continuity3 of the outer function f at the limit of the inner function g(x).
This is summarized below.

Theorem 1.4.6. (The Limit of a Composite Function) If f (x) and g(x) are
functions such that lim g (x) = L and lim f (x) = f (L), then,
x→a x→L

lim f g (x) = f lim g (x) = f (L)
x→a x→a

A few examples of the theorem above are provided below.

I Example 1.4.13. Evaluate lim e x

x→∞
3
discussed later in this chapter

Solution. Here,
1
f (x) = ex and g(x) =
x
1
First, find the limit of the inner function g(x) = as x → ∞, as shown below.
x
1
=0 lim
x→∞ x

Now, apply the outer function f (x) = ex to the limit of the inner function, that
1
is, lim = 0, as shown below.
x→∞ x
e0 = 1
lim 1
Therefore, lim e x = ex→∞ x = 1.
1

x→∞

√
I Example 1.4.14. Evaluate lim x2 + π
x→1
Solution. Here,
√
f (x) = x and g(x) = x2 + π
First, find the limit of the inner function g(x) = x2 + π as x → 1,
lim x2 + π = 1 + π
x→1
√
Now, apply the outer function f (x) = x to the limit of the inner function.
That is,
√
1 + π ≈ 2.035
√
Therefore, lim x2 + π = lim (x2 + π) ≈ 2.035.
q

x→1 x→1

I Example 1.4.15. Evaluate lim ln(x2 )

x→1
Solution. Here,
f (x) = ln(x) and g(x) = x2
First, find the limit of the inner function g(x) = x2 as x → 1,
lim x2 = 1
x→1

Now, apply the outer function f (x) = ln(x) to the limit of the inner fnction.
That is,
ln(1) = 0
Therefore, lim ln(x2 ) = ln(lim x2 ) = 0.
x→1 x→1

Chap. 1 / Sec. 1.4 / Subsec. 1.4.3 : Algebraic Manipulation

So far, we’ve evaluated the limit of a function by direct substitution using the
limit laws. However, situations occur where applying the limit laws directly does
not evaluate the limit. For example, consider the following function,
√
x−1
f (x) =
x−1
√
x−1
What is the limit of the function f (x) as x → 1? That is, lim =? By
x→1 x − 1
direct substitution, we get,
√ √
x−1 1−1
lim =
x→1 x − 1 1−1
1−1
=
1−1
0
=
0
This is absurd. In this case, direct substitution fails since the function f (x)
becomes undefined at x = 1. Although we can deal with the limit by numerical
or graphical methods, a more accurate and efficient way to do that is to find a
different function g(x) such that it equals f (x) for all values of x except at the
point x = 1 in a way that √direct substitution is applicable for g(x). For example,
x−1 1
for the function f (x) = a function g(x) = √ can be found such
√ x − 1 x + 1
x−1 1
that =√ for all x 6= 1. Their equivalence can be seen in the figure
x−1 x+1
below,

2 2

1 1

1 2 3 1 2 3

(a) (b)

Figure 1.7

The only difference between these two functions is that f (x) is not defined at
the point x = 1, whereas g(x) is defined at that point. Thus, f (x) = g(x) for

all points except x = 1. Because the concept of a limit focuses solely on the
neighboring points of a given point, ignoring the specific behavior at that point
itself, the limits of both functions are equal, as the functions happen to be equal
everywhere except at x = 1. This allows us to write,
√
x−1 1
lim = lim √
x→1 x − 1 x→1 x+1
1
=√
1+1
1
=
1+1
1
=
2
√
x−1 1
Therefore, lim = . The following theorem presents the generalization
x→1 x − 1 2
of this method to evaluate the limit of a function.

Theorem 1.4.7. Let a be a real number and f (x) = g(x) be functions defined
on an open interval containing a, except at the point x = a. If

lim g (x) = g (a)

x→a

exists, then
lim f (x) = lim g (x) = g (a)
x→a x→a

1
Note that in the previous example, we found the function g(x) = √ as a
√ x + 1
x−1
function that equals f (x) = for all x except x = 1. Yet, the method by
x−1
which we accomplished this remains undiscussed. Relying on guesswork alone
1
is insufficient. In fact, the expression √ was actually derived from the ex-
√ x+1
x−1
pression algebraically. Thus, it becomes necessary to be aware of certain
x−1
algebraic techniques associated with finding the limit of a function analytically.
We will talk about two such important techniques here. One is referred com-
monly as the “dividing out technique” and the other one is as the “rationalizing
technique.”

· · · / Subsec. 1.4.3 / Segment 1.4.3.1 : Dividing Out Technique

The Dividing Out Technique is an essential algebraic technique used to evaluate
0
limits when direct substitution results in an indeterminate form, such as . The
0
idea behind this technique is to simplify the expression by factoring the numer-
ator and the denominator and then canceling out the common factors. After
this simplification, we can substitute the limiting value again, often resolving
the indeterminate form. For example, consider the same function as discussed
in the prior section,
x2 − 42
f (x) =
x−4
In Section 1.1, we have confirmedly concluded, through numerical and graphical
x2 − 42
methods, that the limit of f (x) = , as x → 4, is 8. However, we now
x−4
evaluate this limit by the analytical method as follows.

x2 − 42
For that, first notice that the numerator of the expression can be fac-
x−4
(x + 4) · (x − 4)
tored into (x + 4) · (x − 4) resulting . Now, notice that the
x−4
numerator and the denominator have a common factor of (x − 4), thus they get
canceled out and we’re left with a new function g (x) = x + 4, that is equivalent
x2 − 42
to f (x) = for all x but x = 4. Thus, by the theorem above, we write,
x−4
x2 − 42
lim = lim (x + 4)
x→4 x − 4 x→4

=4+4
=8

x2 − 42
Which matches the previous answer that the limit of f (x) = , as x → 4,
x−4
is 8.
The dividing out technique thus can be generalized as follows.

Remark 1.4.1.

1. Factor numerator and denominator.

2. Cancel out any common factors.

A few examples of the remark above are provided below.

x2 − 9
I Example 1.4.16. Evaluate lim
x→3 x − 3

Solution. Substituting x = 3 directly into the function yields an indeterminate

form,
32 − 9 0
=
3−3 0
Now, to solve the problem notice that the numerator x2 − 9 can be factored into
(x + 3)(x − 3). Thus,

x2 − 9 (x−3)
(x + 3)
=
x−3 (x−3)

=x+3 for x 6= 3

Now, substituting x = 3 into x + 3 yields 3 + 3 = 6. Therefore,

x2 − 9
lim = lim (x + 3) = 6
x→3 x − 3 x→3

x3 − 1
I Example 1.4.17. Evaluate lim
x→1 x − 1

Solution. Substitution x = 1 directly into the function yields an indeterminate

0
form of . So, to solve the problem notice that the numerator x3 − 1 can be
0
factored into (x − 1)(x2 + x + 1). Thus,

x3 − 1 2
(x−1)(x + x + 1)
lim = lim

x→1 x − 1 (x− 1)
x→1

= lim (x2 + x + 1)
x→1

x3 − 1
Therefore, lim =3
x→1 x − 1

x3 − 6x2 + 11x − 6
I Example 1.4.18. Evaluate lim
x→1 x−1

Solution. Substitution x = 1 directly into the function yields division by zero,

which is not defined. So, to solve the problem notice that the numerator
x3 − 6x2 + 11x − 6 can be factored into (x − 1)(x − 2)(x − 3). Thus,

x3 − 6x2 + 11x − 6 (x−1)(x

− 2)(x − 3)
lim = lim

x→1 x−1 x→1 (x−1)

= lim (x − 2)(x − 3)
x→1

x3 − 6x2 + 11x − 6
Therefore, lim =2
x→1 x−1

· · · / Subsec. 1.4.3 / Segment 1.4.3.2 : Rationalizing Technique

The rationalizing technique is a technique used for simplifying limits that involve
radicals and is used when direct substitution results in an indeterminate form,
0
such as . By multiplying the expression by the conjugate of the radical, we can
0
eliminate the radical, simplify the expression, and solve the limit. For example,
consider the previous function,
√
x−1
f (x) =
x−1
√
x−1 1
Previously, we’ve seen that → , as x → 1. It was determined by an
x−1 2
1 1
equivalent function g (x) = √ . It is now shown how the expression √
x√+1 x+1
x−1
is derived from the expression .
x−1
√
x−1
To do that, notice that the numerator of the expression contains a
x−1
radical. In order to remove it, multiply the numerator and denominator by the
√
conjugate of x − 1, that is,
√ √ √
x−1 x−1 x+1
·1= ·√
x−1 x−1 x+1
√ √
( x + 1) · ( x − 1)
= √ (1)
(x − 1) · ( x + 1)
√ 2
Notice that the numerator of equation (1) can be simplified into ( x) − 12 ,
by the identity (a + b) · (a − b) = a2 − b2 . Which further simplifies to x − 1.

Therefore we get, √
x−1 (x − 1)
= √
x−1 (x − 1) ( x + 1)
Notice that we now have a common factor in the numerator and denominator
1
which can be canceled out and be left with an expression √ , that is defined
x+1
1
as g(x) = √ . This is precisely the function g(x) that we’d previously used
x+1 √
x−1
to express its equivalence to f (x) = for all x but x = 1.
x−1

Now taking the limit as x → 1, yields the same result as before. That is,
√
x−1 1
lim = lim √
x→1 x − 1 x→1 x+1
1
=√
1+1
1
=
1+1
1
=
2
The rationalizing technique now can be generalized as follows.

Remark 1.4.2.

1. Multiply both the numerator and denominator by the conjugate of

the numerator.

2. Cancel out any common factors.

A few examples are provided below.

√ √
1+x− 1−x
I Example 1.4.19. Evaluate lim
x→0 x
Solution. Substituting x = 0 directly into the function yields an indeterminate
form, √ √
1+0− 1−0 0
=
0 0
Now, to solve the problem we multiply the numerator and denominator by the
√ √
conjugate of 1 + x − 1 − x, and simplify as follows.
√ √ √ √ √ √
1+x− 1−x 1+x− 1−x 1+x+ 1−x
lim = lim √ √
x→0 x x→0 x 1+x+ 1−x

(1 + x) − (1 − x)
= lim √ √
x→0 x 1+x+ 1−x
2x
= lim √ √
x 1+x+ 1−x
x→0

2
= lim √ √
x→0 1+x+ 1−x
2
= √ √
1+0+ 1−0
=1
√ √
1+x− 1−x
Therefore, lim =1
x→0 x

√ √
3 + 7x − 3 + 5x
I Example 1.4.20. Evaluate lim
x→0 x
Solution. Substituting x = 0 directly into the function yields an indeterminate
form, √ √
3+7·0− 3+5·0 0
=
0 0
Now, to solve the problem we multiply the numerator and denominator by the
√ √
conjugate of 3 + 7x − 3 + 5x, and simplify as follows.
√ √ √ √ √ √
3 + 7x − 3 + 5x 3 + 7x − 3 + 5x 3 + 7x + 3 + 5x
lim = lim √ √
x→0 x x→0 x 3 + 7x + 3 + 5x
(3 + 7x) − (3 + 5x)
= lim √ √
x→0 x 3 + 7x + 3 + 5x
7x − 5x
= lim √ √
x→0 x 3 + 7x + 3 + 5x
2
x
= lim √ √
x 3 + 7x + 3 + 5x
x→0

2
= lim √ √
x→0 3 + 7x + 3 + 5x
2
=√ √
3+7·0+ 3+5·0
2
= √
2 3
1
=√
3
√ √
3 + 7x − 3 + 5x 1
Therefore, lim =√
x→0 x 3

x
I Example 1.4.21. Evaluate lim √
x→0 1+x−1
Solution. Substituting x = 0 directly into the function yields an indeterminate
form,
0 0
√ =
1+0−1 0
Now, to solve the problem we multiply the numerator and denominator by the
√ √
conjugate of 1 + x − 1, which is 1 + x + 1, and simplify as follows.
√
x x 1+x+1
lim √ = lim √ √
x→0 1 + x − 1 x→0 1+x−1 1+x+1
√
x 1+x+1
= lim
x→0
√1+x−1
x 1+x+1
= lim

x→0
√ x
= lim 1+x+1
x→0
√
= 1+0+1
=2
x
Therefore, lim √ =2
x→0 1+x−1

√ √
a + bx − a + dx
I Example 1.4.22. Evaluate lim
x→0 x
Solution. Substituting x = 0 directly into the function yields an indeterminate
form, √ √
a+b·0− a+d·0 0
=
0 0
Now, to solve the problem we multiply the numerator and denominator by the
√ √ √ √
conjugate of a + bx − a + dx, which is a + bx + a + dx, and simplify as
follows.
√ √ √ √ √ √
a + bx − a + dx a + bx − a + dx a + bx + a + dx
lim = lim √ √
x→0 x x→0 x a + bx + a + dx
(a + bx) − (a + dx)
= lim √ √
x→0 x a + bx + a + dx
bx − dx
= lim √ √
x→0 x a + bx + a + dx

x (b − d)
√ = lim √

a + bx + a + dx x
x→0

b−d
= lim √ √
x→0 a + bx + a + dx
b−d
=√ √
a+0+ a+0
b−d
= √
2 a
√ √
a + bx − a + dx b−d
Therefore, lim = √
x→0 x 2 a

Remark 1.4.3.
√ √
a + bx − a + dx b−d
lim = √
x→0 x 2 a
√ √
a + bx − a − dx b+d
lim = √
x→0 x 2 a

Chap. 1 / Sec. 1.4 / Subsec. 1.4.4 : The Squeeze/Sandwich/Pinching Theo-

rem
Up to this point, we have discussed techniques applicable to various functions
for evaluating the limit of a function. Yet, these approaches alone often fall
short in evaluating limits analytically. Thus, in this part of the section, we’ll
be introduced to a new technique often referred to as the “Squeeze Theorem”,
synonymous the “Sandwich Theorem” or occasionally the “Pinching Theorem”.

The squeeze theorem might seem convoluted, however, the idea behind is quite
simple and is synonymous with the following analogy.

Imagine a race competition among three cat breeds: Bengal, Turkish Angora,
and Ragdoll, each completing to showcase their speed. While they share a
comparable running pace on average, however the Bengal cat stands out as
particularly athletic due to its natural traits. Therefore, the premise is that the
Bengal cat outpaces the Turkish Angora, which in turn outpaces the Ragdoll.
Thus, the hierarchical speed order among them can be shown by the following

inequality.

Bengal Cat ≥ Turkish Angora ≥ Ragdoll (1)

Unfortunately, on the official attempt, the Turkish Angora couldn’t make it to

the race due to some unforeseen circumstances, meanwhile the Bengle cat im-
pressed everyone with an impressive speed of 32 kilometers per hour. However,
to everyone’s surprise, the well-trained Ragdoll cat was able to keep pace with
the Bengal cat, matching its exact speed during the official race. So, by the
premise, it’s reasonable to conclude that had the Turkish Angora participated,
it would have also run at the same speed. The rationale behind this conclu-
sion is due to the premise that the Turkish Angora cannot surpass the Bengal
cat in speed, nor can it lag behind the Ragdoll. If both cat breeds have equal
speeds, it follows that the Turkish Angora—being trapped in the middle of the
inequality (1)—must also share the same speed. Therefore, it is concluded that
if the Turkish Angora had been in the competition, it would also have run at 32
kilometers per hour.

Similar to the analogy above, the Squeeze Theorem is based on the idea that
a function f (x) is squeezed between two other functions h(x) and g(x) near a
point x = a. If the functions h(x) and g(x) have the same limit L at the point
x = a, then the function f (x) being trapped between h(x) and g(x) must also
have the same limit L, at the point x = a. This can be better understood by
looking at the figure below.

Figure 1.8

Now, this is interesting. Because, no matter how complicated f (x) is, if h(x) and

g(x) are relatively simple functions and their limits are equal at a point x = a
where f (x) is squeezed between them near that point, then evaluating the limit
of f (x) at that point becomes trivial, as it doesn’t require explicit evaluation of
the limit of f (x).

Thus, we have the Squeeze Theorem as follows.

Theorem 1.4.8. (The Squeeze Theorem)

If h(x) ≤ f (x) ≤ g(x) for all x over an open interval containing a, except
possibly at a itself, and if the following is true,

lim h (x) = lim g (x) = L

x→a x→a

then,
lim f (x) = L
x→a

This theorem allows us to analytically evaluate functions that were not supposed
to be evaluated by other methods discussed before. For example, consider the
following function,
sin (x)
f (x) =
x
What is the limit of f (x), as x → 0?

If we proceed with the previous methods, we get,

sin (x) sin (0)

lim =
x→0 x 0
0
=
0
Which is absurd. However, by the squeeze theorem, we can evaluate this limit
analytically. Before that, we must recall that,

1
the area of triangle = (base × height)
2
1
the area of a sector = r2 θ
2
where, r is the radius and θ is the angle.

With that, we draw the following figure of a unit circle to evaluate the limit
sin (x)
lim . Note that, since it is a unit circle, therefore the radius r = 1.
x→0 x
y

B’
1

x
x
O A A’
1

Figure 1.9

In the figure, we see that, for ∆AOB,

opposite side
sin (x) =
hypotenuse
opposite side
=
1
= opposite side
∴ opposite side = sin (x)

For ∆A0 OB 0 ,
opposite side
tan (x) =
adjacent side
opposite side
=
1

= opposite side
∴ opposite side = tan (x)

Now, referring to Figure 1.9, it’s evident that,

area of ∆OA0 B ≤ area of sector OA0 B ≤ area of ∆OA0 B 0 (1)

where,
1 1 sin (x)
area of ∆OA0 B =
(base × height) = · 1 · sin (x) =
2 2 2
1 1 x
area of sector OA0 B = r2 · θ = (1)2 · x =
2 2 2
1 1 tan (x)
area of ∆OA0 B 0 = (base × height) = · 1 · tan (x) =
2 2 2
Thus, inequality (1) can be written as,

sin (x) x tan (x)

≤ ≤
2 2 2
sin (x)
Now, dividing it by yields,
2

sin (x)

x tan (x)
2 2 2

≤ ≤
sin (x)

sin (x) sin (x)
2

2 2
x tan (x)
1≤ ≤
sin (x) sin (x)
sin (x)
x cos (x)
1≤ ≤
sin (x) sin (x)
x 1
1≤ ≤
sin (x) cos (x)
Now, by taking the reciprocals, we get,

sin (x)
1≥ ≥ cos (x)
x
Now, taking the limit as x → 0, we get,

sin (x)

lim 1 ≥ lim ≥ lim cos (x)
x→0 x→0 x x→0

sin (x)
1 ≥ lim ≥1
x→0 x
Therefore, by the Squeeze Theorem, we get,

sin (x)
=1 lim
x→0 x
The limit above is crucial because it frequently aids in evaluating other limits of
functions using the Squeeze Theorem. Therefore, we have the following remark.

Remark 1.4.4. (An Important Limit)

sin (x)
lim =1
x→0 x

A few examples of the Squeeze Theorem are provided below.

x
I Example 1.4.23. Evaluate lim
x→0 sin(x)

sin (x)
Solution. We know that lim = 1, therefore,
x→0 x
x 1
lim = lim
x→0 sin(x) x→0 sin(x)
x
1
=
sin(x)
lim
x→0 x
1
=
1
x
Therefore, lim =1
x→0 sin(x)

Remark 1.4.5.
sin (x) x
lim = lim =1
x→0 x x→0 sin(x)

tan(x)
I Example 1.4.24. Evaluate lim
x→0 x

sin(x)
Solution. We know that lim = 1 and lim cos(x) = 1, therefore,
x→0 x x→0

sin(x)
tan(x) cos(x)
lim = lim
x→0 x x→0 x
sin(x) 1
= lim ·
x→0 cos(x) x
1 sin(x)
= lim ·
x→0 cos(x) x
1 sin(x)
= lim · lim
x→0 cos(x) x→0 x
=1
tan(x)
Therefore, lim =1
x→0 x

x
I Example 1.4.25. Evaluate lim
x→0 tan(x)

tan(x)
Solution. We know that lim = 1, therefore,
x→0 x
x 1
lim = lim
x→0 tan(x) x→0 tan(x)
x
1
=
tan(x)
lim
x→0 x
=1
x
Therefore, lim =1
x→0 tan(x)

Remark 1.4.6.
tan(x) x
lim = lim =1
x→0 x x→0 tan(x)

cos(x) − 1
I Example 1.4.26. Evaluate lim
x→0 x
sin(x)
Solution. Since lim = 1, therefore,
x→0 x
cos(x) − 1 cos(x) − 1 cos(x) + 1

lim = lim ·
x→0 x x→0 x cos(x) + 1

cos2 (x) − 1
= lim
x→0 x · (cos(x) + 1)

−sin2 (x)
= lim
x→0 x · (cos(x) + 1)

sin(x) sin(x)

= − lim ·
x→0 x cos(x) + 1

0
= −1 ·
1+1
=0
cos(x) − 1
Therefore, lim .
x→0 x

Remark 1.4.7.
cos(x) − 1
lim =0
x→0 x

sin(x)
I Example 1.4.27. Evaluate lim
x→∞ x
Solution. We know that for all x, the sine function oscillates between −1 to 1.
Therefore,

−1 ≤ sin(x) ≤ 1
1 sin(x) 1
≤ ≤ −
x x x
1 sin(x) 1
− lim ≤ lim ≤ lim
x→∞ x x→∞ x x→∞ x
sin(x)
0 ≤ lim ≤0
x→∞ x
As a result, by the Squeeze Theorem,
sin(x)
lim =0
x→∞ x

cos(x)
I Example 1.4.28. Evaluate lim
x→∞ x
Solution. We know that for all x, the sine function oscillates between −1 to 1.
Therefore,

−1 ≤ cos(x) ≤ 1

1 cos(x) 1
≤ − ≤
x x x
1 cos(x) 1
− lim ≤ lim ≤ lim
x→∞ x x→∞ x x→∞ x
cos(x)
0 ≤ lim ≤0
x→∞ x
As a result, by the Squeeze Theorem,

cos(x)
lim =0
x→∞ x

Remark 1.4.8.
sin(x) cos(x)
lim = lim =0
x→∞ x x→∞ x

Chap. 1 / Sec. 1.4 / Subsec. 1.4.5 : Trigonometric Manipulation

In addition to the technique discussed above, trigonometric manipulation is a
crucial technique in solving limit problems involving trigonometric functions.
It often requires the use of specific trigonometric identities, algebraic simpli-
fications, and limit theorems to simplify expressions that would otherwise be
challenging to evaluate directly. The goal is to transform the original expression
into a form where the limit can be easily computed, often by eliminating inde-
0
terminate forms such as . The following trigonometric identities are commonly
0
used.

1. Reciprocal Identities

1 1
sin(x) = csc(x) =
csc(x) sin(x)
1 1
cos(x) = sec(x) =
sec(x) cos(x)
1 1
tan(x) = cot(x) =
cot(x) tan(x)

2. Tangent and Cotangent Identities

sin(x) cos(x)
tan(x) = cot(x) =
cos(x) sin(x)

3. Pythagorean Identities

sin2 (x) + cos2 (x) = 1 1 + cot2 (x) = csc2 (x)

1 + tan2 (x) = sec2 (x)

4. Reduction Formulas

sin(−x) = − sin(x) csc(−x) = − csc(x)

cos(−x) = cos(x) sec(−x) = sec(x)

tan(−x) = − tan(x) cot(−x) = − cot(x)

5. Cofunction Identities
π π
sin − x = cos(x) csc − x = sec(x)
2 2
π π
cos − x = sin(x) sec − x = csc(x)
2 2
π π
tan − x = cot(x) cot − x = tan(x)
2 2

6. Sum and Difference Formulas

sin(x ± y) = sin(x) cos(y) ± cos(x) sin(y)

cos(x ± y) = cos(x) cos(y) ∓ sin(x) sin(y)

tan(x) ± tan(y)
tan(x ± y) =
1 ∓ tan(x) tan(y)
7. Double Angle Formulas

sin(2x) = 2 sin(x) cos(x)

cos(2x) = cos2 (x) − sin2 (x) = 2cos2 (x) − 1 = 1 − 2sin2 (x)

2 tan(x)
tan(2x) =
1 − tan2 (x)
8. Power Reducing Formulas
1 − cos(2x)
sin2 (x) =
2

1 + cos(2x)
cos2 (x) =
2
1 − cos(2x)
tan2 (x) =
1 + cos(2x)
9. Sum to Product Formulas

x+y x−y
sin(x) + sin(y) = 2 sin cos
2 2

x+y x−y
sin(x) − sin(y) = 2 cos sin
2 2

x+y x−y
cos(x) + cos(y) = 2 cos cos
2 2

x+y x−y
cos(x) − cos(y) = −2 sin sin
2 2
10. Product to Sum Formulas
1
sin(x) sin(y) = [cos(x − y) − cos(x + y)]
2
1
cos(x) cos(y) = [cos(x − y) + cos(x + y)]
2
1
sin(x) cos(y) = [sin(x + y) + sin(x − y)]
2
1
cos(x) sin(y) = [sin(x + y) − sin(x − y)]
2
A few examples using the identities and formulas above are provided below.

1 − sin(x)
I Example 1.4.29. Evaluate limπ
x→ 2 cos(x)
Solution. In order to evaluate the limit, we first simplify it as follows.
1 − sin(x) 1 − sin(x) 1 + sin(x)
limπ = limπ ·
x→ 2 cos(x) x→ 2 cos(x) 1 + sin(x)
(1 + sin(x))(1 − sin(x))
= limπ
x→ 2 (1 + sin(x)) cos(x)
1 − sin2 (x)
= limπ
x→ 2 (1 + sin(x)) cos(x)

Since, sin2 (x) + cos2 (x) = 1, therefore, we write,

1 − sin2 (x) cos2 (x)
limπ = limπ
x→ 2 (1 + sin(x)) cos(x) x→ 2 (1 + sin(x)) cos(x)

cos(x)
= limπ
x→ 2 1 + sin(x)
π
cos
= 2
π
1 + sin
2
0
=
1+1
=0

1 − sin(x)
Therefore, limπ =0
x→ 2 cos(x)

I Example 1.4.30. Evaluate lim (csc(x) − cot(x))

x→0

Solution. In order to evaluate the limit, we simplify it as follows,

cos(x)

1
lim (csc(x) − cot(x)) = lim −
x→0 x→0 sin(x) sin(x)
1 − cos(x)
= lim
x→0 sin(x)
Now, it is known, from the double angle identity, that cos(2x) = 1 − 2 sin2 (x),
therefore,
2 x

cos(x) = 1 − 2 sin
2
2 x
cos(x) − 1 = −2 sin
2
2 x
1 − cos(x) = 2 sin
2
x x
Similarly, it is know, from the double angle identity, that sin(2x) = 2 sin cos ,
2 2
therefore,
x x
sin(x) = 2 sin cos
2 2
This implies that,
x
1 − cos(x) 2sin

2

lim = lim x 2 x

x→0 sin(x) x→0
2 sin
cos
x 2 2
sin
= lim 2x
x→0
cos
2

x
= lim tan
x→0
2
0
= tan
2
=0

Therefore, lim (csc(x) − cot(x)) = 0

x→0

1 1
I Example 1.4.31. Evaluate lim −
x→0 sin(x) tan(x)
Solution. In order to evaluate the limit, we first simplify it as follows.
cos(x)

1 1 1
lim − = lim −
x→0 sin(x) tan(x) x→0 sin(x) sin(x)
1 − cos(x)
= lim
x→0 sin(x)
x
Now, since cos(2x) = 1 − 2 sin2 (x), therefore, cos(x) = 1 − 2 sin2 . Likewise,
x x2
since sin(2x) = 2 sin(x) cos(x), therefore, sin(x) = 2 sin cos . Thus,
2 2
2 x

1 − cos(x) 1 − 1 − 2sin
lim = lim x2
sin(x)
x
x→0 x→0
2 sin cos
x 2
x
2sin2

= lim x 2 x
x→0
2 sin
cos
xx 2
sin
= lim 2x
x→0
cos
x 2
= lim tan
x→0
2
0
= tan
2
=0

1 1
Therefore, lim − =0
x→0 sin(x) tan(x)

Some limit problems necessitate a combination of algebraic and trigonometric

manipulations to solve them. This is shown in the following examples.

1 − cos(x)
I Example 1.4.32. Evaluate lim
x→0 x
Solution. Since it is known, from the double angle identity, that cos(2x) = 1 −
2sin2 (x), therefore,
2 x

cos(x) = 1 − 2 sin
2
2 x
cos(x) − 1 = −2 sin
2
2 x
1 − cos(x) = 2 sin
2
This implies that,
x
2
1 − cos(x) 2sin
lim = lim 2
x→0 x x→0 x
sin(x)
Now, we do the following simplification to apply the standard limit lim =
x→0 x
1. That is,
2 x

1 − cos(x) 2sin
lim = lim 2
x→0 x x→0 x
x
2sin2
= lim 2 ·x
x→0 x x
x
2sin2
= lim 2 ·x
x→0 x2
x
2sin2
= lim 2 ·x
x→0 x2
·4
4
x
2sin2
= lim 22 · x
x→0 x2
·4
2
x
2sin2 x
= lim 22 ·
x→0 x2 4
2
 
x
 sin2  x
= lim
2 ·  22 ·
 
x→0  x2  4
2

x 2

 sin 2  x
= lim  2   ·2
x→0  x
2
 2
x
 sin 2  x
= lim  2  · lim
x→0  x  x→0 2

2
x 1
Now, as x → 0, → 0, since x is multiplied by a constant of . Therefore, by
2 2
sin (x)
the the standard limit lim = 1, we get the following,
x→0 x
 2  2
x x
 sin 2  x  sin 2  x
lim  2  · lim = xlim  2  · xlim
x→0  x  x→0 2 2
→0  x  →0 2
|2 {z }
2 2 0
| {z }
1

=1·0
=0

1 − cos(x)
Therefore, lim =0
x→0 x

cos(2x) − cos(3x)
I Example 1.4.33. Evaluate lim
x→0 x2
Solution. Using the Sum to Product
Formula

x+y x−y
cos(x) − cos(y) = −2 sin sin , we write the following,
2 2

2x + 3x 2x − 3x
−2 sin sin
cos(2x) − cos(3x) 2 2
lim = lim
x→0 x2 x→0
x2
5x x
2 sin sin
2 2
= lim
x→0 x2
sin (x)
Now, we do the following simplification to apply the standard limit lim =
x→0 x

1.

5x
 
 sin 2
x
5x   sin 2
 
x
2 ·  · x · 
5x x 5x 2 2
2 sin sin

2 2 2 2
lim 2
= lim
x→0 x x2 x→0

5x
sin
x
sin
2
2
· 2 · 5x · x
5x x 2 2
2
= lim 2
2
x→0
x
5x
 
 sin 2
x
sin 2
= lim  2 · 2 · 5x  · 1
x

x→0 5x 4  x2

2 2

5x

 sin 2
x
sin 2
= lim  · 2 · 5x · 1 
x

x→0  5x 2 x2 
2 2

5x x 
 sin 2 sin
2 · 5
= lim 
 · x

x→0 5x 2
2 2

5x
sin
x
sin
= lim
2
· lim 2 ·5
x→0 5x x→0 x 2
2 2
5x x
Now, as x → 0, both → 0 and → 0, since they are multiplied by a constant
2 2
5 1 sin (x)
of and , respectively. Therefore, by the the standard limit lim = 1,
2 2 x→0 x
we write,

5x 5x
sin sin
x x
sin sin
lim
2
· lim 2 · 5 = lim 2
· lim 2 ·5
x→0 5x x→0 x 2 5x
→0 5x x
→0 x 2
2 2
2 2 2 2
5
=1·1·
2
5
=
2
cos(2x) − cos(3x) 5
Therefore, lim 2
=
x→0 x 2

The End of Section 1.4

Analytic Evaluation of Limit of a Function

§ 1.5 Limits Involving ∞

Limits of functions can involve infinity and they can do so in two common ways.
First, the function f (x) can increase or decrease without a bound as its input
x approaches a finite value, these limits are referred to as “Infinite Limits.”
Second, the input x can increase or decrease without bound, these are referred
to as “Limits at Infinity.” The concepts behind these two are rather distinct.
For example, consider the function we’d seen before.

1
f (x) =
(x − 4)2
The graph of f (x) is shown below.
y

x
2 4 6 8

Figure 1.10

Based on that, consider the following limits,

1 1
1. lim =∞ 2. lim =0
x→4 (x − 4)2 x→∞ (x − 4)2

In the first limit, when the input x approaches a finite value of 4 from both
sides, the function f (x) increases without a bound and thus the limit does not
exist.

On the other hand, in the second limit, as x increases without a bound, f (x)
approaches a finite value of 0, and thus the limit, in this case, does exist.

The first limit is an example of an “infinite limit,” whereas the second one is of
“limit at infinity.”

In general, with an infinite limit, the dependent variable approaches infinity as

the independent variable approaches a finite value. With a limit at infinity,
the dependent variable approaches a finite value as the independent variable
approaches infinity.

Chap. 1 / Sec. 1.5 / Subsec. 1.5.1 : Infinite Limits

Given our understanding of infinite limits, we have the following definition.

Definition 1.5.1. (Infinite Limits) Let f (x) be a function defined for all x in
an open interval containing a. Then,

1. If f (x) increases without a bound as x → a, where, x 6= a, then we

write,
lim f (x) = +∞
x→a

2. If f (x) decreases without a bound as x → a, where, x 6= a, then we

write,
lim f (x) = −∞
x→a

In both of these cases, the limit does not exist.

Note that, since the symbol ∞ does not represents a number, therefore the limits
lim f (x) = +∞ and lim f (x) = −∞ does not imply that the limit exists, but
x→a x→a
rather denotes the behavior of the limit (whether the function is growing in
a positive or negative direction). Definition above can be elaborated in detail
through the use of one-sided limits as follows.

Definition 1.5.2. (One-sided Infinite Limits)

1. Infinite limits from the left: Let f (x) be a function defined for all x
near a. Then,

(a) If f (x) increases without a bound as x → a, where, x < a, then

lim f (x) = +∞
x→a−

(b) If f (x) decreases without a bound as x → a, where, x < a, then

lim f (x) = −∞
x→a−

2. Infinite limits from the right: Let f (x) be a function defined for all
x near a. Then,

(a) If f (x) increases without a bound as x → a, where, x > a, then

lim f (x) = +∞
x→a+

(b) If f (x) decreases without a bound as x → a, where, x > a, then

lim f (x) = −∞
x→a+

Consider the following figures to better understand the definition above.

y y

x
a

(a) (b)

y y

x
a

Figure 1.11

Figure 1.11(a) shows the limit lim− f (x) = +∞, Figure 1.11(b) shows the limit
x→a
lim− f (x) = −∞, Figure 1.11(c) shows the limit lim+ f (x) = +∞, lastly, Fig-
x→a x→a
ure 1.11(d) shows the limit lim+ f (x) = −∞.
x→a

Notice that, as x → a, the function f (x) approaches a vertical line at x = a in all

of the infinite limits shown in Figure 1.11. The vertical line at the point x = a is
called a vertical asymptote. It is a line that is approached by the function f (x)
when x → a.

Definition 1.5.3. (Vertical Asymptote)

If f (x) approaches positive infinity or negative infinity as x → a from the
right or the left side, then the line x = a is called a vertical asymptote of
f (x).

We now began to evaluate functions that have an infinite limit. However, before
a
that, notice that, if we have a fraction, , where a 6= 0 and b 6= 0, then
b
a
1. if b → 0, then → ∞.
b

a
2. if b → ∞, then → 0.
b

x2 + 1
I Example 1.5.1. Evaluate lim+
x→0 x
x2 + 1
Solution. Notice that as x → 0+ , the denominator of the function ap-
x
proaches 0, whereas the numerator (x2 + 1) approaches the value 1. Therefore,
x2 + 1
lim+ =∞
x→0 x
2
x +1
Therefore, lim+ =∞
x→0 x

I Example 1.5.2. Evaluate lim+ cot (x).

x→0
Solution. We know that,
cos (x)
lim+ cot (x) = lim+
x→0 x→0 sin (x)
We know further that as x → 0+ , cos (x) → 1 and sin (x) → 0, where, sin(x) > 0.
Therefore,
cos (x)
lim+ cot (x) = lim+
x→0 x→0 sin (x)
= +∞
Therefore, lim+ cot (x) = +∞
x→0

I Example 1.5.3. Evaluate lim− cot (x).

x→0
Solution. We know that,
cos (x)
lim− cot (x) = lim−
x→0 x→0 sin (x)
We know further that as x → 0− , and sin (x) → 0 and cos (x) → 1, where,
sin(x) < 0. Therefore,
cos (x)
lim− cot (x) = lim−
x→0 x→0 sin (x)
= −∞
Therefore, lim− cot (x) = −∞
x→0

Chap. 1 / Sec. 1.5 / Subsec. 1.5.2 : Limit at Infinity

The limit at infinity suggests that the limit of a function will have a finite value
as its input approaches positive or negative infinity. For example, consider the
following function,
1
f (x) = π +
x
If x → +∞, what value does f (x) approach? To determine, notice that as
1
x → +∞, the term → 0 in the function above. Therefore,
x

1 1
lim π + = lim π + lim
x→+∞ x x→+∞ x→+∞ x

=π+0
=π
1 1
Similarly, if x → −∞, the term → 0, that is, lim = 0. Thus, we write
x x→−∞ x
the following,

1 1
lim π+ = lim π + lim
x→−∞ x x→−∞ x→−∞ x

=π+0
=π

The result can be seen visually in the following figure.

-4 -2 2 4

Figure 1.12

In general, we have the following definition of the limit at infinity.

Definition 1.5.4. (Limit at Infinity) If f (x) becomes arbitrarily close to L

for all sufficiently large x > 0, then

lim f (x) = L
x→+∞

If f (x) becomes arbitrarily close to L for all sufficiently large x < 0, then

lim f (x) = L
x→−∞

Notice that if f (x) → L, as x → ±∞, the graph of f (x) approaches the hori-
zontal line y = L. The horizontal line y = L is called the horizontal asymptote
of f (x).

Definition 1.5.5. (Horizontal Asymptote)

If lim f (x) = L or lim f (x) = L, then the line y = L is called a
x→+∞ x→−∞
horizontal asymptote of f (x).

1
For example, since lim π+ = π, therefore the line y = π is the horizontal
x→∞ x
1
asymptote for the function f (x) = π + , as it was shown in the Figure above.
x

ex + 1
I Example 1.5.4. Evaluate lim
x→∞ x
Solution. In order to evaluate the limit, we do the following simplification,
ex + 1 ex 1
lim = lim +
x→∞ x x→∞ x x
1
= lim e +
x→∞ x
1
Now, as x → ∞, the term → 0. This implies that,
x
1 1
lim e + = lim e + lim
x→∞ x x→∞ x→∞ x

=e+0
=e

ex + 1
Therefore, lim =e
x→∞ x

x + x2 + x3 + x4 + x5
I Example 1.5.5. Evaluate lim
x→∞ x5
Solution. In order to evaluate the limit, we do the following simplification,

x + x2 + x3 + x4 + x5 x x2 x3 x4 x5
lim = lim + + + +
x→∞ x5 x→∞ x5 x5 x5 x5 x5
1 1 1 1
= lim 4 + 3 + 2 + + 1
x→∞ x x x x
Now, as x → ∞, the all term except the last one approach 0 in the limit above,
therefore, we write,
1 1 1 1 1 1 1 1
lim + + + + 1 = lim + lim + lim + lim + lim 1
x→∞ x4 x3 x2 x x→∞ x4 x→∞ x3 x→∞ x2 x→∞ x x→∞

=0+0+0+0+1
=1

x + x2 + x3 + x4 + x5
Therefore, lim =1
x→∞ x5

Chap. 1 / Sec. 1.5 / Subsec. 1.5.3 : Infinite Limits at Infinity and End Behav-
iors
If a function f (x) increases or decreases without a bound as its input x increases
or decreases without a bound, then we have an infinite limit at infinity. For
example, consider the following function,

f (x) = x3

In the given function, f (x) becomes arbitrarily large as x becomes arbitrarily

large, that is, as x → +∞, the function f (x) → +∞. Similarly, as x → −∞,
the function f (x) → −∞. This can be seen in the following figure.

-2 -1 1 2

-1

-2

Figure 1.13

In general, we have the following definition.

Definition 1.5.6. (Infinite Limits at Infinity) If f (x) increases without a

bound as x → +∞ or x → −∞, then

lim f (x) = ∞ and lim f (x) = ∞

x→+∞ x→−∞

If f (x) decreases without a bound as x → +∞ or x → −∞, then

lim f (x) = −∞ and lim f (x) = −∞

x→+∞ x→−∞

Having discussed the idea of the infinite limits at infinite, it is now crucial to
discuss the end behavior of a function. End behavior refers to how a function
f (x) behaves as its input x increases or decreases indefinitely. The function
can show various behaviors at its endpoints. The following are among the most
common ones.

1. f (x) approaches a horizontal asymptotes y = L.

2. f (x) → +∞ or f (x) → −∞

3. Neither 1 nor 2.

· · · / Subsec. 1.5.3 / Segment 1.5.3.1 : The Limit of Power functions as x →

±∞
Consider the following power function, where, n is a positive integer.

f (x) = xn

Now, if n is even, then f (x) → +∞, as x → ±∞. In contrast, if n is odd, then

f (x) → +∞, as x → +∞ and f (x) → −∞, as x → −∞. More generally, we
have the following remark.

Remark 1.5.1.
(
−∞, n = 1, 3, 5, · · ·
lim xn = ∞ n = 1, 2, 3, · · · and lim xn =
x→∞ x→−∞ +∞, n = 2, 4, 6, · · ·

Now consider the reciprocal of the power function, where n is a positive integer.
1
f (x) = = x−n
xn
For the reciprocal, as x → ±∞, f (x) → 0, regardless of whether n is even or
odd.

I Example 1.5.6. Analyze the end behavior of f (x) = x3 .

Solution. Here, n = 3, therefore n is odd. That’s why,

1. As x → +∞, f (x) → +∞

2. As x → −∞, f (x) → −∞

I Example 1.5.7. Analyze the end behavior of f (x) = x4 .

Solution. Here, n = 4, therefore n is even. That’s why,

1. As x → +∞, f (x) → +∞

2. As x → −∞, f (x) → +∞

1
I Example 1.5.8. Analyze the end behavior of f (x) = .
x3
1
Solution. Here, f (x) = is a reciprocal of x3 . That’s why,
x3
1. As x → +∞, f (x) → 0

2. As x → −∞, f (x) → 0

· · · / Subsec. 1.5.3 / Segment 1.5.3.2 : The Limit of Polynomial Functions as

x → ±∞
Now, using the power function, we can determine the behavior of any polynomial
function as x → ±∞. For that, consider the polynomial function,

P (x) = an xn + an−1 xn−1 + · · · + a1 x + a0

where, an 6= 0 and n ≥ 1. Now, factoring P (x) yields,

P (x) = an xn + an−1 xn−1 + · · · + a1 x + a0

n
an−1 xn−1

n an x a1 x a0
=x + + ··· + n + n
xn xn x x
n n−1

x x x 1
= xn an n + an−1 n + · · · + a1 n + a0 n
x x x x

n n−n n−1−n 1−n 1
= x an x + an−1 x + · · · + a1 x + a0 n
x
an−1 a1 a0
= xn an + + · · · + n−1 + n
x x x
As x → ±∞, all the terms inside the parentheses approach 0, except the first
term. Therefore,

lim an xn + an−1 xn−1 + · · · + a1 x + a0 = lim an xn

x→±∞ x→±∞

This implies that as x → ±∞, the behavior of the polynomial P (x) is determined
by the term with the highest power of x, that is, an xn . Therefore, we have the
following remark.

Remark 1.5.2. The end behavior of a polynomial function P (x) resembles

the end behavior of its highest degree term.

A few examples are shown below.

I Example 1.5.9. If P (x) = 2x4 − 3x3 + 5x2 − 7x + 11, evaluate lim P (x).
x→±∞

Solution. Here, P (x) = 2x4 − 3x3 + 5x2 − 7x + 11

1. As x → +∞, the highest degree term dominates, that is,
lim P (x) = lim 2x4
x→+∞ x→+∞

=∞

2. As x → −∞, the highest degree term again dominates, that is,

lim P (x) = lim 2x4
x→−∞ x→−∞

=∞
since the degree is even.

I Example 1.5.10. If P (x) = −3x5 + 5x4 − 7x3 + 11x2 − 13x + 17, evaluate
lim P (x)
x→±∞

Solution. Here, P (x) = −3x5 + 5x4 − 7x3 + 11x2 − 13x + 17

1. As x → +∞, the highest degree term dominates, that is,
lim P (x) = lim −3x5

x→+∞ x→+∞

= −∞

2. As x → −∞, the highest degree term again dominates, that is,

lim P (x) = lim −3x5

x→−∞ x→−∞

=∞
since the degree is odd and the leading coefficient is negative.

· · · / Subsec. 1.5.3 / Segment 1.5.3.3 : The Limit of Rational Functions as x →

±∞
As of yet, we’ve explored the end behavior of power and polynomial functions
with relative ease. However, the end behavior of rational functions isn’t as
straightforward as seen previously. For example, consider the following rational
function,
x2 − 1
f (x) = 3
x −1
Here, as x → ∞, both the numerator and denominator approaches ∞, which
doesn’t make sense. We can, however, simplify this to analyze the end behavior
as follows.

Since the degree of the numerator is less than that of the denominator, therefore
we modify the expression by dividing every term in both the numerator and
denominator by the greatest power of x in the denominator. By doing so, we’ll
be able to identify the term in the overall expression that dominates the behavior
of the function for large values of x. That is,

x2 − 1
2
x −1 3
lim 3 = lim 3x
x→∞ x − 1 x→∞ x − 1

x3
x2 1
3
− 3
= lim x3 x
x→∞ x 1
−
x3 x3
1 1
− 3
= lim x x
x→∞ 1
1− 3
x
1 1
lim − lim
x→∞ x x→∞ x3
=
1
1 − lim
x→∞ x3
0−0
=
1−0
=0

x2 − 1
This can be verified by looking at the graph of f (x) = 3 as x → ∞ as
x −1
follows.

-2 -1 1 2

-1

-2

Figure 1.14

x2 − 1
Notice that, since lim 3 = 0, therefore f (x) has an horizontal asymptote
x→∞ x − 1
at y = 0.

Now, we shall consider a function where the degree of the numerator is equal to
that of the denominator and x → ∞. That is,
9x + π
f (x) =
3x + e
Since π and e are constants, therefore as x → ∞, both the numerator and de-
nominator approach ∞. To deal with this absurdity, we divide each term in both
the numerator and denominator by the highest power of x in the denominator
as before. That is,
9x + π
f (x) =
3x + e
9x + π
= x
3x + e
x
9x π
+
= x x
3x e
+
x x
π
9+
= x
e
3+
x

π
9+
∴ f (x) = x
e
3+
x
Now, taking the limit of f (x) as x → ∞, we get,
π
9+
lim x = 9+0
x→∞ e 3+0
3+
x
9
=
3
=3
9x + π
Note that, since lim = 3, therefore f (x) has a horizontal asymptote at
3x + e x→∞
y = 3. This can be seen in the following figure.

5 3

-10 -5 5 10

Figure 1.15

Now, having explored rational functions with numerators and denominators of

equal degree, as well as those where the numerator’s degree is less than that
of the denominator, we now focus on rational functions where the numerator’s
degree exceeds the degree of the denominator. For that, consider the function
below,
x2
f (x) =
1+x
As x → ∞, both the numerator and denominator approaches ∞. To deal with
that, we again divide each term in both the numerator and denominator by the
highest power of x in the denominator, as before. That is,
x2
f (x) =
1+x
x2
= x
1
+1
x

x
=
1
1+
x
x
∴ f (x) =
1
1+
x
Now, if we take the limit of f (x) as x → ±∞, then the denominator of f (x)
approaches the value 1, and the numerator approaches +∞ as x → +∞ and
−∞ as x → −∞. Thus, the limit is,
x
lim = ±∞
x→±∞ 1
1+
x
This presents a special case. Previously, we observed that rational functions
tend to approach a horizontal asymptote as x → ±∞. However, in this case,
the limit doesn’t approach a horizontal asymptote as x → ±∞. Nevertheless, it
does approach a linear asymptote, which becomes apparent when seen visually
as follows.

-10 -5 5 10

-5

-10

Figure 1.16

Looking at the figure above, it appears that the function f (x) resembles a linear
function as x → ±∞. To determine the linear function that f (x) is approaching,
we do the long division on f (x) and obtain the following.
x2
f (x) =
1+x
1
by long division
=x−1+
x+1
Now, we take the limit as x → ±∞,

1 1
lim x − 1 + = lim (x − 1) + lim
x→±∞ x+1 x→±∞ x→±∞ x + 1

1
Here, as x → ±∞, the term → 0, therefore,
x+1

1
lim x − 1 + = lim (x − 1)
x→±∞ x+1 x→±∞

This suggests that the function f (x) behaves similarly to the linear function
x − 1, as x → ±∞. For this reason, the graph of f (x) approaches the graph
of linear function x − 1, as x → ±∞. Thus, the line x − 1 is an asymptote of
the function f (x) as x → ±∞. Such an asymptote is called slant asymptote or
oblique asymptote. Oblique asymptotes occur when the numerator has a degree
that is exactly one more than that of the denominator.

In general, when a linear asymptote is neither vertical nor horizontal, it is called

an oblique asymptote or slant asymptote. If the distance between a line mx + b
and a function f (x) approaches 0, as x increases or decreases indefinitely, that
is,
lim f (x) − (mx + b) = 0
x→±∞
then the line mx + b is called the oblique asymptote.

Summarizing these function’s behavior, we have the following remark.

Remark 1.5.3.
Let P (x) and Q(x) be polynomial functions such that

P (x)
R (x) =
Q (x)
an xn + an−1 xn−1 + · · · + a1 x + a0
=
bm xm + bm−1 xm−1 + · · · + b1 x + b0
where, an 6= 0, bm 6= 0. Now,

1. If the degrees of the numerator n and denominator m are equal (i.e.,

an
n = m), then R(x) has a horizontal asymptote at y = , as x →
bm
±∞.

2. If the degree of the numerator n is less than the denominator m (i.e.,

n < m), then R(x) has a horizontal asymptote at y = 0, as x → ±∞.

3. If the degree of the numerator n is greater than the denominator m

(i.e., n > m), then R(x) does not have neither vertical nor horizontal
asymptote, as x → ±∞.

A few examples are shown below.

2x3 + 3x2 + 5x + 7
I Example 1.5.11. Evaluate lim
x→±∞ 11x2 + 13x + 17

Solution. Since the degree of the numerator (that is, 3) is greater than the degree
of the denominator (that is, 2), we divide the numerator and denominator by
xn , where n is the greatest power in the denominator (that is, 2).

2x3 + 3x2 + 5x + 7
2x3 + 3x2 + 5x + 7 x2
lim 2
= lim 2
x→±∞ 11x + 13x + 17 x→±∞ 11x + 13x + 17

x2
2x3 3x2 5x 7
2
+ 2 + 2+ 2
= lim x x x x
x→±∞ 11x2 13x 17
+ 2 + 2
x2 x x
5 7
2x + 3 + + 2
= lim x x
x→±∞ 13 17
11 + + 2
x x
= ±∞

2x3 + 3x2 + 5x + 7
Therefore, as x → +∞, the function f (x) = → +∞. Simi-
11x2 + 13x + 17
2x3 + 3x2 + 5x + 7
larly, as x → −∞, the function f (x) = → −∞.
11x2 + 13x + 17

πx3 + e
I Example 1.5.12. Evaluate lim
x→±∞ ex2 + π

πx3 + e
3
πx + e x2
lim = lim
x→±∞ ex2 + π x→±∞ ex2 + π

x2
πx3 e
2
+ 2
= lim x 2 x
x→±∞ ex π
+
x2 x2

e
πx +
= lim x2
x→±∞ π
e+ 2
x
= ±∞

πx3 + e
Therefore, as x → +∞, the function f (x) = → +∞. Similarly, as
ex2 + π
πx3 + e
x → −∞, the function f (x) = → −∞.
ex2 + π

3x3 + 2x2
I Example 1.5.13. Evaluate lim
x→±∞ 9x3 + 6

Solution. Since the degree of the numerator (that is, 3) is equal to the degree of
the denominator (that is also 3), we divide the numerator and denominator by
xn , where n is the greatest power in the denominator.

3x3 + 2x2
3x3 + 2x2 x3
lim 3
= lim 3
x→±∞ 9x + 6 x→±∞ 9x + 6

x3
3x3 2x2
3
+ 3
= lim x 3 x
x→±∞ 9x 6
3
+ 3
x x
2
3+
= lim x
x→±∞ 6
9+ 3
x
3
=
9
3x3 + 2x2 3
Therefore, as x → +∞, the function f (x) = → . Similarly, as
9x3 + 6 9
3x3 + 2x2 3
x → −∞, the function f (x) = 3
→ .
9x + 6 9

2eπx2 + x
I Example 1.5.14. Evaluate lim
x→±∞ x2 + 2x

Solution. Since the degree of the numerator (2) is equal to the degree of the
denominator (2), divide the numerator and denominator by xn , where n is the

greatest power in the denominator. That is,

2eπx2 + x
2
2eπx + x x2
lim = lim 2
x→±∞ x2 + 2x x→±∞ x + 2x

x2
2eπx2 x
+
= lim x2 x2
2
x→±∞ x 2x
2
+ 2
x x
1
2eπ +
= lim x
x→±∞ 2
1+
x
= 2eπ

2eπx2 + x
Therefore, as x → +∞, the function f (x) = → 2eπ. Similarly, as
x2 + 2x
2eπx2 + x
x → −∞, the function f (x) = → 2eπ.
x2 + 2x

3x3 + x
I Example 1.5.15. Evaluate lim
x→±∞ 5x5 + 2x2

Solution. Since the degree of the numerator (3) is less than the degree of the
denominator (5), we divide the numerator and denominator by xn , where n is
the greatest power in the denominator. That is,

3x3 + x
3
3x + x x5
lim = lim
5
x→±∞ 5x + 2x 2 x→±∞ 5x + 2x2
5

x5
3
3x x
5
+ 5
= lim x5 x
x→±∞ 5x 2x2
+
x5 x5
3 1
2
+ 4
= lim x x
x→±∞ 2
5+ 3
x
0
=
5
=0

3x3 + x
Therefore, as x → +∞, the function f (x) = → 0. Similarly, as
5x5 + 2x2
3x3 + x
x → −∞, the function f (x) = → 0.
5x5 + 2x2

x3 + 2x2 + 3x + 4
I Example 1.5.16. Evaluate lim 5
x→±∞ x + 2x4 + 3x3 + 4x2 + 5x + 6

x3 + 2x2 + 3x + 4
x3 + 2x2 + 3x + 4 x5
lim 5 = lim 5
4 3 2
x→±∞ x + 2x + 3x + 4x + 5x + 6 x→±∞ x + 2x + 3x + 4x2 + 5x + 6
4 3

x5
3
x 2x2 3x 4
5
+ 5
+ 5+ 5
= lim 5 x
4
x
3
x
2
x
x→±∞ x 2x 3x 4x 5x 6
5
+ 5 + 5 + 5 + 5+ 5
x x x x x x
1 2 3 4
+ + +
= lim x2 x3 x4 x5
x→±∞ 2 3 4 5 6
1+ + 2 + 3 + 4 + 5
x x x x x
0
=
1
=0

x3 + 2x2 + 3x + 4
Therefore, as x → +∞, the function f (x) = →
x5 + 2x4 + 3x3 + 4x2 + 5x + 6
x3 + 2x2 + 3x + 4
0. Similarly, as x → −∞, the function f (x) = 5 →
x + 2x4 + 3x3 + 4x2 + 5x + 6
0.

· · · / Subsec. 1.5.3 / Segment 1.5.3.4 : The Limit of Functions Involving Rad-

icals as x → ±∞
Previously, we’ve seen how to deal with the limit of rational functions as x
increases or decreases indefinitely. However, sometimes these functions can con-
tain a radical. For example, consider the same function we previously examined

as x → ∞, but this time the numerator contains a radical.

√
9x + π
f (x) =
3x + e
Now, to evaluate the limit of this function as x → ∞, we divide the numerator
and denominator by the highest power of x. That is,
√
√ 9x + π
9x + π
lim = lim x
x→∞ 3x + e x→∞ 3x + e

√ x
9x + π
= lim x
e
x→∞
3+
x
Now, a problem arises due to the presence of a radical in the numerator. To
address this issue, notice that x ≥ 0 as x → +∞, therefore
√
x = |x| = x2

. Therefore, it is totally possible for us to evaluate the limit in the following

manner.
√ √
9x + π 9x + π
√
lim x = lim x2
x→∞ e x→∞ e
3+ 3+
x r x
9x + π
x2
= lim e
x→∞
3+
r x
9 π
+ 2
x x
= lim e
x→∞
3+
√ x
0+0
=
3+0
=0

This same strategy can be used to evaluate the limit of the following function.
Consider the function below.
10x2 + x
f (x) = √
2x2 + 8

What is the limit of f (x) as x → +∞ and x → −∞?

We will use the previous strategy. Divide the numerator and denominator by
the highest power of x in the denominator. The highest power of x in the
denominator of f (x) is x2 . However, since x2 is under a square root, therefore
√
we need to divide both the numerator and denominator by x2 . That is,

10x2 + x
√
10x2 + x 2
lim √ = lim √ x
x→+∞ 2x2 + 8 x→+∞ 2x2 + 8
√
x2
10x2 + x
√
2
= lim r x
x→+∞ 2x2 + 8
x2
2
10x x
√ +√
2 x2
= lim rx
x→+∞ 2x2 8
+
x2 x2
√
Since x2 = |x| = x, for all x ≥ 0, therefore,

10x2 x 10x2 x
√ +√ +
2 x2 = lim r x
lim rx x
x→+∞ 2 x→+∞ 2
2x 8 2x 8
2
+ 2 2
+ 2
x x x x
10x + 1
= lim r
x→+∞ 8
2+ 2
x
+∞ + 1
= √
2+0
= +∞

Therefore, f (x) → +∞, as x → +∞.

√
Let us now consider x → −∞, where x2 = |x| = −x, for all x < 0,
10x2 + x
√
10x2 + x 2
lim √ = lim √ x
x→−∞ 2x2 + 8 x→−∞ 2x2 + 8
√
x2

10x2 + x
= lim r −x
x→−∞ 2x2 + 8
x2
2
10x x
+
= lim r −x −x
x→−∞ 2
2x 8
2
+ 2
x x
−10x − 1
= lim r
x→−∞ 8
2+ 2
x
− (−∞) − 1
= √
2+0
= +∞

Therefore, f (x) → +∞, as x → ±∞.

· · · / Subsec. 1.5.3 / Segment 1.5.3.5 : Limits of Transcendental Functions as

x → ±∞
Transcendental functions, such as exponential, logarithmic, and trigonometric
functions, behave in distinct ways as x → ∞ or x → −∞.

Remark 1.5.4. (Exponential Functions) Let f (x) = ax , where a > 1, then,

lim ax = ∞ and lim ax = 0

x→∞ x→−∞

If 0 < a < 1, then,

lim ax = 0 and lim ax = ∞

x→∞ x→−∞

A few examples are provided below.

I Example 1.5.17. Evaluate lim ex

x→∞

Solution. Since e > 1, therefore, in this case, exponential functions increase

without a bound as x → ∞. Thus,

lim ex = ∞
x→∞

I Example 1.5.18. Evaluate lim π x

x→−∞

Solution. Since π > 1, therefore π x approach zero as x → −∞. Thus,

lim π x = 0
x→−∞

I Example 1.5.19. Evaluate lim e−x

x→∞
Solution. Since lim e = ∞, therefore, x
x→∞

1 1
lim =
x→∞ ex lim ex
x→∞

Therefore, lim e−x = 0

x→∞

Remark 1.5.5. (Logarithmic Functions) Let f (x) = loga (x), where a > 1,
then,
lim loga (x) = ∞ and lim+ loga (x) = ∞
x→∞ x→0

If 0 < a < 1, then,

lim loga (x) = −∞ and lim loga (x) = ∞

x→∞ x→0+

A few examples are provided below.

I Example 1.5.20. Evaluate lim ln(x)

x→∞

Solution. Since ln(x) = loge (x) and e > 1, therefore, the natural logarithmic
function grows slowly but unbounded as x → ∞. Therefore,

lim ln(x) = ∞
x→∞

I Example 1.5.21. Evaluate lim+ ln(x)

x→0

Solution. Since ln(x) = loge (x) and e > 1, therefore, the natural logarithmic
function approaches negative infinity as x approaches 0 from the positive side.
Thus,
lim+ ln(x) = −∞
x→0

Remark 1.5.6. (Trigonometric Functions) The behavior of six trigonometric

functions as x → ±∞ are as follows.

1. Sine Function: As x → ±∞, f (x) = sin(x) does not approach a single

value; instead, it oscillates between −1 and 1 indefinitely. Therefore,

lim sin(x) does not exist

x→±∞

2. Cosine Function: As x → ±∞, f (x) = cos(x) does not approach

a single value; instead, it oscillates between −1 and 1 indefinitely.
Therefore,
lim cos(x) does not exist
x→±∞

3. Tangent Function: The tangent function f (x) = tan(x) has vertical

(2n + 1)π
asymptotes at x = , where n ∈ Z. However, between
2
these asymptotes, tan(x) increases or decreases without a bound.
Therefore,
lim tan(x) does not exist
x→±∞

4. Cotangent Function: The Cotangent function f (x) = cot(x) has ver-

tical asymptotes at x = nπ, where n ∈ Z. However, between these
asymptotes, cot(x) increases or decreases without a bound. There-
fore,
lim cot(x) does not exist
x→±∞

5. Secant Function: The Secant function f (x) = sec(x) has vertical

(2n + 1)π
asymptotes at x = , where n ∈ Z. However, between these
2

asymptotes, sec(x) increases or decreases without a bound. There-

fore,
lim sec(x) does not exist
x→±∞

6. Cosecant Function: The Cosecant function f (x) = csc(x) has vertical

asymptotes at x = nπ, where n ∈ Z. However, between these asymp-
totes, csc(x) increases or decreases without a bound. Therefore,

lim csc(x) does not exist

x→±∞

The End of Section 1.5

Limits Involving ∞

§ 1.6 The Concept of Continuity

The concept of continuity is one of the most important concepts in the whole
of mathematics. It describes the behavior of a function. A function f (x) is said
to be continuous if its graph has no interruptions and maintains a smoothness
across every point. An analogy can be drawn to railway tracks, where an unin-
terrupted and smooth track is essential for safe travel. Likewise, for a function
to be considered well-behaved, it must demonstrate continuity across all points.

In informal terms, a function is considered continuous if its graph can be drawn

without lifting the pen from the page.

We begin by discussing the concept of continuity for a function f (x) at a specific

point x = a. Following that, we’ll delve into functions that exhibit continuity
across an interval.

Chap. 1 / Sec. 1.6 / Subsec. 1.6.1 : Continuity at a Point

A function f (x) is continuous at a point x = a, if it does not have any discontinu-
ities. Discontinuities are points at which a function does not behave “smoothly”
or continuously. Discontinuities are classified into three types: removable dis-
continuities, jump discontinuities, and infinite discontinuities. Each has distinct
characteristics that impact how functions behave at a certain point.

1. Removable Discontinuities: A removable discontinuity occurs when

a function has a “hole” at a certain point, but can be made continuous
by appropriately redefining the function at that point. Essentially, the
limit of the function as it approach the point of discontinuity exists, but
the function’s value either is not defined or differs from this limit. For
example, consider the following figures.

Figure 1.17

Here, the function f (x) is not defined at the point x = a, therefore f (x) is
discontinuous at the point x = a. A similarity can be seen in the following
figure.

Figure 1.18

Here, the value of the function f (x) at the point x = a differ from the limit
of f (x) as x → a, therefore f (x) is discontinuous at the point x = a.

2. Jump Discontinuities: A jump discontinuity occurs when there is a

sudden “jump” in the value of the function at a certain point, causing the
left-sided and the right-sided limit to differ. The function’s value jumps
from one level to another without passing through any intermediate values.
For example, consider the figure below.

Figure 1.19

Here, lim− f (x) 6= lim+ f (x), therefore the limit of f (x) as x → a does
x→a x→a
not exist.

3. Infinite Discontinuities: An infinite discontinuity occurs when the value

of the function f (x) approaches infinity as x approaches the point of dis-
continuity. Rational functions often have discontinuities at points where
the denominator is zero, resulting in vertical asymptotes. For example,
consider the figure below.

Figure 1.20

Here, as x → a, the limit of the function f (x) grows indefinitely, therefore

the limit of f (x) as x → a has an infinite discontinuity.

Having seen a few descriptions of discontinuities, we now summarize to define

the continuity of a function at a point as follows.

Definition 1.6.1. (Continuity at a Point) A function f (x) is continuous at a

point a, if

1. f (a) is defined.

2. lim f (x) exists.

x→a

3. lim f (x) = f (a)

x→a

If a function does not satisfy any of the specified criteria at the point a, then
the function is not continuous at that point.

Chap. 1 / Sec. 1.6 / Subsec. 1.6.2 : Continuity over an Interval

When we talk about a function being continuous over an interval, we are saying
that the function behaves predictably and without interruption across all points
within that interval. In other words, the graph of the function can be traced
without lifting the pen. In order to fully understand continuity over an interval,
it’s essential to first recognize the type of intervals.

1. Open intervals: An interval (a, b) is called open if it includes all points

between a and b, but not the endpoints a and b themselves. A function is
continuous on an open interval if it satisfies the conditions of continuity at
a point, across all the interior points.

2. Closed intervals: An interval [a, b] is called closed if it includes all points

between a and b, as well as the endpoints a and b themselves. When
defining continuity at a point, we rely on two-sided limits, which consider
both directions around a point. However, at the endpoints of a closed
interval, two-sided limits aren’t possible since there’s no interval extending
beyond that boundary. To handle this, we define continuity at an endpoint
using a one-sided limit. In other words, a function is continuous at an
endpoint if the one-sided limit exists and equals the function’s value at
that point.

3. Half-open intervals: If we combine the open and the closed intervals

together, we get intervals such as (a, b] or [a, b). These intervals are semi-
closed, that is, they include only one of the endpoints and not the other.
Continuity over a half-open interval requires that the function be contin-
uous at all interior points as well as at the endpoint that is included.

To understand this concept visually, consider the following figure.

a b

Figure 1.21

In the figure above, the function f (x) is continuous over the open interval (a, b)
and at the left endpoint over the closed interval [a, b], because

lim f (x) = f (a)

x→a+

However, the function f (x) is not continuous at the right endpoint over the
closed interval [a, b], because

lim f (x) 6= f (b)

x→b−

In general, we have the following definition for continuity at endpoints.

Definition 1.6.2. (Continuity at Endpoints)

1. Left-continuous: A function f (x) is continuous from the left at the

point a, if
lim− f (x) = f (a)
x→a
.

2. Right-continuous: A function f (x) is continuous from the right at

the point a, if
lim+ f (x) = f (a)
x→a
.

Now, based on definition above, the continuity of a function over a closed interval
[a, b] is defined as follows.

Definition 1.6.3. (Continuity over an Closed Interval)

A function f (x) is continuous over a closed interval [a, b], if the following
is true:

1. f (x) is continuous over an open interval (a, b).

2. lim+ f (x) = f (a)

x→a

3. lim− f (x) = f (b)

x→b

To fully grasp the concept of continuity over an interval, consider the following
example.

I Example 1.6.1. Consider the figure below to identify points where f (x) has
discontinuities within the interval [0, 8].

f(x)

x
1 2 3 4 5 6 7 8

Figure 1.22

Solution. Using the definition of continuity of a function, we get the following.

1. f (x) is discontinuous at x = 1 because f (1) is not defined.

2. f (x) is discontinuous at x = 2 because lim− f (x) 6= lim+ f (x)

x→2 x→2

3. f (x) is discontinuous at x = 5 because lim f (x) = ∞

x→3

4. f (x) is discontinuous at x = 7 because lim f (x) 6= f (4)

x→4

Note that the discontinuities of f (x) at points x = 1, 7 are called removable

discontinuity as they can be removed by redefining the function at these points.
The discontinuity at x = 2 is called jump discontinuity. Whereas, the disconti-
nuity at x = 5 is called infinite discontinuity.

Chap. 1 / Sec. 1.6 / Subsec. 1.6.3 : Properties of Continuity

In the previous subsection, we discussed the concept of a function’s continuity.
However, when determining the continuity of a complex function, it is often
helpful to break the function down into simpler functions. The idea is that if
the complex function can be expressed in terms of simpler functions that are
continuous, then the entire function will also be continuous. For example, if
the complex function is written as a sum, difference, product, quotient (with a
non-zero denominator), or composition of continuous functions, then the whole
function remains continuous. This approach simplifies the analysis by allowing
us to assess continuity through its simpler components, rather than tackling the
entire function at once. We discuss this matter in details in the segments below.

· · · / Subsec. 1.6.3 / Segment 1.6.3.1 : Continuity of Combined Functions

Functions can be combined in various ways, such as through addition, subtrac-
tion, multiplication, division, or composition. Understanding the continuity of
these combined functions is essential because it allows us to determine how the
continuity of individual functions influences the continuity of the overall func-
tion. Fortunately, there are clear rules that dictate when combined functions
remain continuous, based on the continuity of their component functions as
summarized below.

Theorem 1.6.1. (Continuity of Combined Functions)

If c is a real number and n is a positive integer, and both functions f (x)
and g(x) are continuous at the point a, then the functions,

1. c · f (x)

2. f (x) + g(x)

3. f (x) − g(x)

4. f (x) · g(x)
f (x)
5. , where, g(x) 6= 0
g(x)
are also continuous at the point a.

The following types of functions are continuous at every point within their re-
spective domains.

Remark 1.6.1. (Continuous Functions)

The functions below are continuous at every point within their defined
domains.

1. Polynomial Functions, that is, P (x) = an xn +an−1 xn−1 +· · ·+a1 x1 +a0

P (x)
2. Rational Functions, that is, R(x) = , where, P (x) and Q(x) are
Q(x)
both polynomials and Q(x) 6= 0.
√
3. Radical Functions, that is, f (x) = n
x; provided that it is defined

4. Trigonometric Functions, that is, sin(x), cos(x), tan(x), cot(x), sec(x),

csc(x)

5. Inverse Trigonometric Functions, that is, sin−1 (x), cos−1 (x), tan−1 (x),
cot−1 (x), sec−1 (x), csc−1 (x).

6. Exponential and Logarithmic Functions, that is, f (x) = ax , f (x) = ex ,

f (x) = ln(x), f (x) = logb (x)

· · · / Subsec. 1.6.3 / Segment 1.6.3.2 : Continuity of Composite Functions

A composite function f (g(x)) is a result of applying the function g to the input
x and then applying the function f to the output of g(x). To determine whether
the composite function is continuous, we rely on the following theorem.

Theorem 1.6.2. (Continuity of a Composite Function) If g(x) is continuous

at the point a and f is continuous at the point g(a), then the composite
function, f (g(x)) is also continuous at the point a.

The theorem above suggests that for f (g(x)) to be continuous at a point a, both
the inner function g(x) must be continuous at a, as well as the outer function f
must be continuous at the value that g(x) takes at a, that is, g(a). For example,
consider f (g(x)) = sin(x2 ), where f (x) = sin(x) and g(x) = x2 . The function
g(x) = x2 is continuous on its entire domain, and the function f (x) = sin(x)
is also continuous everywhere. Therefore, the function f (g(x)) = sin(x2 ) is also
continuous everywhere.

Chap. 1 / Sec. 1.6 / Subsec. 1.6.4 : The Intermediate Value Theorem (IVT)
The Intermediate Value Theorem (IVT) is a fundamental result in Calculus. It
formalizes the intuitive idea that if a function is continuous on a closed interval,
and if it takes on two distinct values at the endpoints of that interval, then it
must take on every value between those two endpoints at some point within
the interval. The idea is indeed intuitive and to get a gist of it, consider the
following analogy.

Suppose we’re hiking up a hill, starting at point A on the ground and heading
upwards to point B, which is at a higher elevation. The path we’re taking is
smooth, without any sudden drops or jumps. As we walk up, we inevitably pass
through every height between A and B. For instance, if we start at 0 meters
and end at 100 meters, then there must be a point where we have been at 50
meters. This is because the path is continuous, so we can’t skip any heights or
jump straight from 0 to 100 meters. This is the core idea of the Intermediate
Value Theorem.

Suppose now the hill’s height is represented by a continuous function, f (x), and

let two points a and b on the x-axis represent the start and end of the walk,
respectively. If f (a) = 0 and f (b) = 100, then there must be a point c between
a and b such that the function’s value f (c) = 50. This is guaranteed by the
continuity of the function.

Following the same logic as in the analogy above, we can state the following
theorem.

Theorem 1.6.3. (The Intermediate Value Theorem (IVT))

If f (x) is a continuous function defined on the interval [a, b] and L is any
number between f (a) and f (b). Then, there exists at least one point c ∈
(a, b) such that,
f (c) = L

The Intermediate Value Theorem (IVT) can be used in many clever ways to
ensure the existence of solutions within a certain interval. A few of them are
shown below.

I Example 1.6.2. Prove that the function f (x) = x3 − x − 2 has a solution on

the interval [1, 2].

Solution. Notice that f (x) = x3 − x − 2 is a polynomial, and therefore it is con-

tinuous everywhere. As a result, it is also continuous on the interval [1, 2].

Now, evaluating f (x) for x = 1 yields: f (1) = 13 − 1 − 2 = −2 Similarly, f (x)

for x = 2 yields: f (2) = 23 − 2 − 2 = 4

Since f (1) = −2 and f (2) = 4, therefore, according to the Intermediate Value

Theorem, there exist a point c ∈ [1, 2] where f (x) changed its sign from negative
to positive, resulting 0. In other words, there must be at least one point c ∈ [a, b]
such that f (c) = 0. Thus, we have proved that the function f (x) = x3 − x − 2
has a solution at the point c.

I Example 1.6.3. Prove that the function f (x) = x3 − 10 has a solution on the
interval [2, 2.2].

Solution. The function f (x) = x3 − 10 is a polynomial, and therefore it is con-

tinuous everywhere. That’s why we evaluate f (x) at the interval’s endpoints to
identify a point where the function may have changed its sign.

f (2) = 23 − 10 = −2
f (2.2) = 2.23 − 10 = 0.648

Since f (2) = −2 and f (2.2) = 0.648, therefore, by the Intermediate Value The-
orem, there must exist a point c ∈ (2, 2.2) such that f (c) = 0, because f (x) is
continuous and crosses the x-axis between x = 2 and x = 2.2.

The End of Section 1.6

The Concept of Continuity

§ 1.7 (ε − δ) Definition of Limit

The (ε − δ) definition of a limit serves as a pivotal shift in Calculus from intu-
itive reasoning to formal mathematics. The early development of calculus was
marked by controversies and ambiguities regarding the fundamental principles
upon which it was built. The intuitive idea that a function f (x) approaches
some value L as its input x approaches some value a is always suggestive, but
lacks a clear definition of vague words such as “approach.”

Indeed, what does it mean for a function to “approach” a value? It may be

intuitively clear, but this intuition is not what to be called mathematics.

To understand why, suppose you are the mathematician who devised the concept
of limits. Intuitively, what you’ve written in the following is pretty clear.

lim f (x) = L
x→a

That is, as x gets close to a, the function f (x) gets close to L.

Now, the question arises: how do you define what it means to be “close,” as
you have used this term in your definition of limits? Specifically, what does it
mathematically mean for some variable x to be “close” to a number a? After
pondering, you may figure out that the distance between x and a is simply x−a.
Since distance cannot be negative, therefore you use the absolute value of x − a,

that is, |x−a|. So, “closeness” is, in this case, if the distance |x−a| is kept small.

So, this implies that the function f (x) is close to a value L if the distance be-
tween f (x) and L, that is, |f (x) − L| is kept small by keeping the distance
between x and a sufficiently small.

The issue, however, is not resolved. Because, you may have clearly defined what
“closeness” implies in this case, but it was done so by leveraging the vagueness
of the term “smallness.” That is, “smallness” is a relative term, so “how small
is really small?” is a question to be answered now.

To address the issue, you can assign a fixed real number to indicate the exact
amount of “smallness.” Say, ε ∈ R is such a real number. So, |f (x) − L| < ε
implies that the distance between f (x) and L is less than the chosen “small”
real number ε, no matter how small. In other words, f (x) is within ε-distance
of L.

You can similarly assign a fixed value to clearly define the “smallness” for the
distance between x and a. Let δ ∈ R be such a real number. So, |x − a| < δ
implies that the distance between x and a is less than the chosen “small” real
number δ, no matter how small.

So, after precisely defining the ambiguous words, you can now ensure that if
|x − a| < δ, then |f (x) − L| < ε, where ε is the chosen value. However, you soon
realize that since x 6= a for the reason that the function may not even be defined
at x = a, therefore, the distance between x and a can never be zero, that is,
|x − a| > 0. So, this implies that

if 0 < |x − a| < δ, then |f (x) − L| < ε

In other words, to keep the distance between f (x) and L less than ε, i.e.,
|f (x) − L| < ε; you would have to keep the distance between x and a less
than δ and greater than 0, i.e., 0 < |x − a| < δ. Notice that the magnitude of
δ depends on the function f (x) itself. For very steep functions, you would need
a much smaller δ to keep f (x) within a given ε. So, this implies that for every
chosen real number ε > 0 (no matter how small), there exists a corresponding
δ > 0, but δ will vary based on the behavior of the function.

So, putting everything together, we write the following

lim f (x) = L
x→a

to mean that the limit of f (x), as x approaches a value a, is L; if the following

property holds:

for every real number ε > 0, there exist a corresponding real number δ > 0
such that for all real number x, if 0 < |x − a| < δ, then |f (x) − L| < ε.

All the details of the (ε − δ) definition might seem overwhelming at once. There-
fore, to better illustrate this concept, we will consider a visual approach. For
that, consider the following figure.

x
a

Figure 1.23

From the figure above, it is evident that f (x) → L as x → a from both sides.
Thus, the limit lim f (x) = L may informally be defined as “the limit of f (x),
x→a
as x approaches the value a, is the value L.”

This definition, however, is vague. For example, what is meant by “f (x) ap-
proaches the value L” or “x approaches the value a”? If this implies that the
value of f (x) gets closer and closer to L whenever x gets closer and closer to a,
then how close is really “close”? How close is close enough to conclude that a
limit is indeed approaching a certain value?

We deal with the idea of closeness by introducing the idea of error tolerance.
That is, if we assign a value of error tolerance below which the distance be-
tween the function f (x) and the value L must be kept, then we can keep x close

enough to a in order for the distance between f (x) and L to be less than that
given error tolerance. The key idea is that the distance between f (x) and L can
be minimized to be smaller than any specified level of error tolerance by letting
x sufficiently close to a.

We let ε be a small positive real number as a value of error tolerance. Since the
error tolerance ε is the value less than which the distance between f (x) and L
must be, therefore the symbolic form of this is |f (x) − L| < ε. The absolute
value is used to show that distance cannot be a negative quantity. Now, we
derive the range within which f (x) must lie as follows,

|f (x) − L| < ε
−ε < f (x) − L < ε
−ε + L < f (x) − L + L < ε + L
L − ε < f (x) < L + ε

This implies that f (x) must lie inside an open interval (L − ε, L + ε), as shown
in the Figure 1.24.

Now, we let δ be the value less than which the distance between x and a must
be kept in order for |f (x) − L| to be less than the given error tolerance ε. The
symbolic form of this is |x − a| < δ. Based on that, we derive the range for x as
follows,

|x − a| < δ
−δ < x − a < δ
−δ + a < x − a + a < δ + a
a−δ <x<a+δ

This implies that in order for the distance between f (x) and L be less than a
given error tolerance ε, x must be kept within an open interval (a − δ, a + δ),
where, x 6= a, as shown in the following figure.

f(x)
L

x
a
x

Figure 1.24

Since, x 6= a, therefore |x − a| > 0, and suggests the following,

since, 0 < |x − a| and,

|x − a| < δ
∴ 0 < |x − a| < δ

As a result, a rigorous definition of the limit of a function can be given as follows.

Definition 1.7.1. (The Formal Definition of a Limit)

Let f (x) be a function defined over an open interval containing the point
a (except possibly at a), and let L be a real number. Then the meaning of
the limit
lim f (x) = L
x→a

is that for all every chosen ε > 0, there exists a corresponding δ > 0 such
that

if 0 < |x − a| < δ, then |f (x) − L| < ε

A few examples are provided below.

I Example 1.7.1. Consider the function f (x) = 2x − 5. It is intuitively clear that

f (x) → 1 as x → 3. That is,

lim (2x − 5) = 1
x→3

Now, what is the value less than which the distance between x and 3 must be
kept in order for |f (x) − 1| to be less than ε = 0.0001?

Solution. The value less than which the distance between x and 3 must be kept
in order for |f (x) − 1| to be less than 0.0001, is the value δ. That is, |x − 3| < δ.

We know that, for each ε > 0, there is a δ > 0 such that if 0 < |x − a| < δ, then
|f (x) − L| < ε.

Since, f (x) = 2x − 5 and L = 1, therefore,

|f (x) − L| < ε = |(2x − 5) − 1| < 0.0001

= |2x − 6| < 0.0001
= |2 (x − 3)| < 0.0001
= 2 |x − 3| < 0.0001
0.0001
= |x − 3| <
2
0.0001
This implies that δ = = 0.00005 is the value less than which the distance
2
between x and 3 must be kept in order for |f (x) − 1| to be less than 0.0001.

I Example 1.7.2. Verify the result derived for δ in Example 1.7.1.

Solution. In example 1.7.1, it was found that δ = 0.00005. This implies that
|x − 3| < 0.00005. Based on that, we derive,

|f (x) − L| = |(2x − 5) − 1|
= |2x − 6|
= 2 |x − 3|
< 2 (0.00005) = 0.0001 = ε

This matches the error tolerance ε given in Example 1.7.1. This completes the
verification.

I Example 1.7.3. Prove that lim (3x + 1) = 7

x→2

Solution. We can prove that lim (3x + 1) = 7, by showing that for any chosen
x→2
ε > 0, there exists a δ > 0 such that if 0 < |x − 2| < δ, then |(3x + 1) − 7| < ε.

Since f (x) = 3x + 1 and L = 7, therefore,

|f (x) − L| < ε = |(3x + 1) − 7| < ε

= |3x − 6| < ε
= 3 |x − 2| < ε
ε
= |x − 2| <
3
ε ε
This implies that if 0 < |x−2| < , then |(3x+1)−7| < ε. Therefore, δ = .
3 3

I Example 1.7.4. Prove that lim x2 = 1

x→1

Solution. We can prove that lim x2 = 1, by showing that for any chosen ε > 0,
x→1
there exists a δ > 0 such that if 0 < |x − 1| < δ, then |x2 − 1| < ε.

Since f (x) = x2 and L = 1, therefore,

|f (x) − L| < ε = x2 − 1 < ε

= −ε < x2 − 1 < ε
= −ε + 1 < x2 < ε + 1
√ √
= 1−ε<x< 1+ε

Now, since a = 1, therefore,

|x − a| < δ = |x − 1| < δ
= −δ < x − 1 < δ
=1−δ <x<1+δ

This implies that, either,

√
1−δ = 1−ε
√
δ =1− 1−ε

or,
√
1+δ = 1+ε

√
δ= 1+ε−1
√ √
Thus, we choose the smallest distance δ = min{1 − 1 − ε, 1 + ε − 1}.

√ √
This implies that if 0 < |x − 1| < min{1 − 1 − ε, 1 + ε − 1}, then |x2 − 1| <
ε.

√
I Example 1.7.5. Prove that lim x=2
x→4
√
Solution. We can prove that lim x = 2, by showing that for any chosen ε > 0,
√ x→4
there exists a δ > 0 such that if 0 < |x − 4| < δ, then | x − 2| < ε.

√
Since f (x) = x and L = 2, therefore,
√
|f (x) − L| < ε = x − 2 < ε
√
= −ε < x − 2 < ε
√
=2−ε< x<2+ε
= (2 − ε)2 < x < (2 + ε)2

Now, since a = 4, therefore,

|x − a| < δ = |x − 4| < δ
= −δ < x − 4 < δ
=4−δ <x<4+δ

This implies that, either,

4 − δ = (2 − ε)2
δ = 4 − (2 − ε)2

or,

4 + δ = (2 + ε)2
δ = (2 + ε)2 − 4

Thus, we choose the smallest distance δ = min{4 − (2 − ε)2 , (2 + ε)2 − 4}.

√
This implies that if 0 < |x − 4| < min{4 − (2 − ε)2 , (2 + ε)2 − 4}, then | x − 2| <
ε.

The End of Section 1.7

(ε − δ) Definition of a Limit

Differential Calculus

113

Copyright © 2024 Md. Shouvik Iqbal. This book is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License (CC BY-NC-SA 4.0)
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Copyright © 2024 Md. Shouvik Iqbal. This book is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License (CC BY-NC-SA 4.0)
Download for free at https://md-shouvik-iqbal.github.io/calculus
Chapter 2

An Introduction to Differential
Calculus

Differential calculus is a subfield of calculus that studies the rates at which quan-
tities change at any given instant. In modern terms, it primarily addresses the
question: how does a function react to a change in its input variable? For exam-
ple, consider the function f (x). We know that when the independent variable
x changes, the dependent variable f (x) also changes correspondingly, but the
question that differential calculus essentially addresses is more about the “how”
than the “how much.” That is, it explores the nature of this change—to what
sensitivity does f (x) react to a change in x? What’s the “rate1 ” at which f (x)
changes reacting to a change in x, no matter how small?

Finding the rate of change of f (x) may seem straightforward: measure the
amount x changes and compare it with the corresponding change in f (x). For
example, if x changed from the point x = a to x = a + h, where h > 0 is
an increment, then the change in x is the difference (a + h) − a = h and the
corresponding change in f (x) for an increment h is the difference between f (a+h)
and f (a), that is, f (a + h) − f (a). Now, by comparing these two changes, we
get,

f (a + h) − f (a)
(1)
h
1
A rate is a specific type of ratio that compares two quantities of different units. It describes
how one quantity changes relative to another.

115

This quotient above represents the “rate” at which f (x) changes as x changes
from x = a to x = a+h, and is called the difference quotient . So, for any change
f (x + h) − f (x)
h in x in general, f (x) changes at a rate of . For example, if
h
f (x) = x, then as x changes from x = a to x = a + h, the change in f (x) is,

f (a + h) − f (a) = (a + h) − a
=h

Therefore, the rate at which f (x) changes yields,

f (a + h) − f (a) h
=
h h
=1

This is because the function is f (x) = x, therefore the rate of change of f (x)
matches with the rate of change of x, resulting in a ratio of 1. It is also noticeable
when the function is graphed, as shown below.

Figure 2.1

It is worth mentioning here is that, when graphed, the rate of change given by
the difference quotient is essentially the same as the slope of the straight line.
This is because we know that
vertical change
slope, m =
horizontal change
f (a + h) − f (a)
=
(a + h) − a

f (a + h) − f (a)
=
h
Therefore, in general, the slope of a line on a graph represents the rate of change
between two variables being graphed. In other words, it represents the rate at
which the dependent variable changes reacting to a change in the independent
variable.

With that in mind, let us now pose a “philosophical question” with the purpose
to understand the tool through which differential calculus examines the rate of
change of a function—namely, the derivative.

Philosophical Question: At what rate does f (x) change when there’s no change
in x whatsoever?

This question is absurd! Why wouldn’t it be? The dependent variable f (x) only
changes when the independent variable x changes. If the change in x is 0, then
0
the change in f (x) is also 0, and thus the rate at which f (x) changes is... ?
0
2 0
Can there be such a real thing as ?
0

This absurd philosophical question is in fact equivalent to asking: what is the

rate at which f (x) changes at a point single point, say x = a. This question is
equivalently absurd because if there’s no change in x at the point x = a, then
basically x changed from x = a to x = a + 0, and the difference h between them
0
is essentially 0, resulting again . But why do we need to consider the rate of
0
change of a function at a single point a?

This question can best be answered through problems that have perplexed
solvers for centuries. One such problem is the tangent line problem. Under-
standing which can allow us to define the aforementioned heart of differential
calculus: the derivative.

§ 2.1 The Tangent Line Problem

Given the graph of a function f (x) and a point P = (a, f (a)) on the graph of
f (x) as shown below, can a line be drawn such that it is tangent to the graph
of f (x) at the point P ?
2
within the real number system.

Figure 2.2

First of all, what does it mean for a line to be “tangent” to the graph of a function
at a point? The term “tangent” originated from the Latin word “tangere,”
meaning “to touch”; so the term “tangent” is indicative of touching something.
However, defining this precisely isn’t as straightforward as it might appear. The
definition of a tangent line is simple for common curves such as a circle. For
example, a line is tangent to a circle if it meets the circle at exactly one and
only point, known as the point of tangency, as shown below.

Figure 2.3

This intuitive definition might suggest that a line is tangent to a curve at a point
if it “meets” the curve at that point. However, this definition only applies to
simple curves like the one shown above and falls short for more complex curves.
To understand why, consider the following figure.

Figure 2.4

Here, the line “meets” the curve at exactly one point, but it goes without saying
that it does not align with our previous understanding of a tangent line. There-
fore, a revised definition would be that a line is tangent to a curve at a point if
it “just touches” the curve at that point. This revised definition emphasizes the
idea of a line gradually nearing the curve until it makes contact.

Sadly, however, this revised definition is also not sufficient for the reason depicted
below.

Figure 2.5

Here, the tangent line “just touches” the curve at only one point P , but it also
crosses by intersecting the curve at that point. That is, a part of the curve lies
above the tangent line while another part falls beneath it. Thus, it violates the
current definition of a tangent line. Therefore, a revised definition would be that

a line is tangent to a curve at a point if it either “just touches” or “intersects”

the curve at exactly one point. However, even with this definition, the situation
at hand is unsolved. Consider the figure below as an example,

Figure 2.6

Here, the tangent line not only intersects the curve at point P , but also at an-
other point ahead of P . This is again a violation of the current revised definition
as it requires the point of intersection to be exactly one. Therefore, the defini-
tion can again be revised such that a line is tangent to a curve at a point if it,
“up close,” just touches or intersects the curve at that point. But unfortunately,
even this definition eventually falls short.

This is frustrating. There’s something inherently weird about defining a tangent

line. This is because, in general, a tangent line requires no more than a single
point for its definition (else it would become a secant line). However, to define a
line, at least two points are required. Therefore, we’re facing an seeming para-
dox: how can a tangent line be defined using only a single point?

That is, in order to define a line, at least two points are required. But if two
points are used, then the line can no longer be defined as a tangent line—it
becomes a secant line! A secant line to a curve is one that intersects the curve
at two different points. It “cuts” the curve at least at two distinct points, as the
term “secant” itself originated from the Latin word “secare,” meaning “to cut.”

It is at this point we may come to realize why the absurd philosophical question
was posed earlier.

To understand why, first recall that in Analytic Geometry a line can be defined
using a single point if its slope is known. For example, recall the point-slope
form of a line for a given point (x1 , y1 ) and a slope m,

y − y1 = m(x − x1 ) (1)

Now, if the slope m of the tangent line can somehow be determined, it then
becomes possible to define the tangent line using only a single point by the
equation (1). However, the core problem remains as the slope or the rate of
0
change of two quantities at a single point becomes , as previously explained.
0

Historically, efforts were made to calculate the slope of the tangent line to the
curve at a single point by introducing a new quantity known as the “infinitesi-
mal.” To oversimplify, it was a quantity that was infinitely small but not exactly
equal to 0. More specifically, it is a positive real number less than every positive
real number. For example, consider the figure below,

Figure 2.7

In order to calculate the slope of the tangent line to the curve of f (x) at the
point P = (a, f (a)), a small increment dx is made to x = a, resulting in a point
Q = (a + dx, f (a + dx)) and then to draw a secant line P Q. Now, let dx be
an infinitesimal increment. Since dx is an infinitesimal increment and infinitely
small, therefore the point Q is infinitely closer to the point P in a way that
they merge into one single point, becoming indistinguishable from each other,
and thus, the slope of the secant line P Q is infinitely closer to the slope of the

tangent line to the curve at the point P in a way that it is indistinguishably the
same.

The unfortunate truth, however, is that the formulation above is—without a

doubt—an invalid one within the framework of the real number system. There
exists no such positive number that is less than every positive real number in
the real number system. It can be proven. For example, if dx is a positive
number smaller than any other positive number, where dx > 0. Then consider
dx
a real number r > 1 that divides dx, resulting > 0. Thus, we’ve found a
r
dx
new positive real number > 0 that contradicts the assumption that dx > 0
r
dx
is the smallest positive real number because dx > > 0. This process can be
r
repeated infinitely, disproving the existence of a positive number less than every
positive real number.

In the historical development of calculus, mathematicians did indeed struggle

with the concept of infinitesimals, leading some to criticize them by remark-
ing, “Ghosts of departed Quantities.” The accurate formulation of this concept
was not possible until the rigorous formulation of the limit came into existence,
namely the (ε − δ) definition of limits.

In modern calculus, we can simply calculate the slope of the tangent line by a
limiting process where the point Q = (a + dx, f (a + dx)) approaches the point
P = (a, f (a)) along the curve, i.e., the difference between Q = (a+dx, f (a+dx))
and P = (a, f (a)) approaches 0 as a function as dx → 0, given that dx 6= 0.

More specifically, consider calculating the slope of the tangent line to the curve
at point P , as shown in the figure below.

Figure 2.8

The slope can be determined by a limiting process using the slope of the secant
line. To do this, let Q = (a + h, f (a + h)) be a second point on the curve that
is close to P = (a, f (a)), where h > 0. Now, we draw the secant line P Q that
intersects the points P and Q, as follows.

Figure 2.9

The slope of the line P Q is,

vertical change
msec =
horizontal change
f (a + h) − f (a)
=
h
Since h is the difference between x = a and x = a + h, therefore taking the limit3
3
assuming the limit exists

of the slope as h → 0 causes the point Q to approach the point P by sliding

along the curve of f (x). The process is shown below.

Figure 2.10

Thus, the slope of the tangent line is defined as

mtan = lim msec
h→0
f (a + h) − f (a)
= lim
h→0 h
In general, the slope of the tangent line to the graph of f (x) at a point is defined
using the slope of the secant line through a limiting process. Note that since
the slope of a line is a measure of the rate of change between two points on
that line, therefore the slope of the tangent line represents the “instantaneous
rate of change.” This is because the tangent line touches the curve at only one
point, thus it represents the instantaneous rate of change at that precise instant.

Summarizing the facts discussed so far, we have the following definition,

Definition 2.1.1. (The Slope of a Tangent Line and The Instantaneous Rate
of Change)

The slope of the curve y = f (x) at a single point a equals the slope of the
tangent line or the instantaneous rate of change of the curve at that point.
Which is defined as,
f (a + h) − f (a)
mtan = lim
h→0 h
provided that the limit exists, where h > 0 is a real number.

Similarly, an alternative definition can be given when the point a + h in Figure

2.10 is considered as a movable point x, as shown in the following figure.

Figure 2.11

This time, Q approaches P if the moving point x approaches a. Thus, an

alternative definition of the slope of the tangent line mtan yields,

Definition 2.1.2. (The Slope of a Tangent Line and The Instantaneous Rate
of Change)

The slope of the curve y = f (x) at a single point a equals the slope of the
tangent line or the instantaneous rate of change of the curve at that point.
Which is defined as,

f (a) − f (x)
mtan = lim
x→a x−a
provided that the limit exists, where x 6= a.

Since we can now determine the slope of the tangent line mtan at the point
P = (a, f (a)) on the curve, we can define the tangent line using the point-slope
form of a line as given by the equation 1 above. That is,

y − y1 = m(x − x1 )
y − f (a) = mtan (x − a)
= mtan (x − a) + f (a)

f (a + h) − f (a)
= lim (x − a) + f (a)
h→0 h
Summarizing our discussion, we finally have the long-awaited definition of a line
that is tangent to a curve at a single point as follows.

Definition 2.1.3. (Tangent Line) A line that is tangent to the graph of f (x)
at the point P = (a, f (a)) is given by,

y = mtan (x − a) + f (a)

where,
f (a + h) − f (a)
mtan = lim
h→0 h
provided that the limit exists.

This is the generalized equation of a line that is tangent to the graph of f (x)
at the point a. Using this definition, the tangent line can be determined to the
graph of any function at any point. For demonstration, consider the following
example.

I Example 2.1.1. What is the tangent line to the graph of f (x) = x2 at the point
(2, 4).

Solution. Since we’ve to find the tangent line at the point (2, 4), therefore we
first have to determine its slope at that point. From Definition 2.1.1, we know
that the slope of the tangent line is given by,
f (a + h) − f (a)
mtan = lim
h→0 h
Therefore, the slope of the tangent line at the point (2, 4) yields,
f (2 + h) − f (2)
mtan = lim
h→0 h
2
(2 + h) − 22
= lim
h→0 h
2 + 2 · 2 · h + h2 − 22
2
= lim
h→0 h
2
4h + h
= lim
h→0 h
4h h2
= lim +
h→0 h h

= lim 4 + h
h→0

Since mtan = 4, therefore the tangent line from Definition 2.1.3 yields,

y = mtan (x − a) + f (a)
= 4(x − 2) + 4
= 4x − 8 + 4
= 4x − 4

Therefore, 4x − 4 is the line that is tangent to the graph of f (x) = x2 at the

point (2, 4).

Having discussed the tangent line problem and its resolution, we now move to
discuss the derivative of a function in the following section and see how it is
connected to the tangent line problem.

The End of Section 2.1

The Tangent Line Problem

§ 2.2 Definition of the Derivative of a function

We have, so far, calculated the slope of the tangent line to the curve at a fixed
point. However, if that fixed point itself is not fixed and changes along the
curve, then the tangent line also changes accordingly, resulting in changes to its
slope for every new change. As a result, the slope of the tangent line to the
curve of f (x) becomes a new function on its own, dependent on the point to be
evaluated. This new function has a special name called the “derivative” of f (x).
Notice how this new function is termed as a derivative “of ” some other function.
This is because the derivative is a function that is “derived” from some other
function f (x).

The derived function or the derivative can be expressed symbolically. For ex-
ample, if f (x) represents a function, then we let f 0 (x) represent its derivative.
Although many other notations exist and we will discuss them later on, but for
now, we let f 0 (x) (read as f prime of x) denote the derivative of a function f (x).

f (a + h) − f (a)
Since lim represents the slope of the tangent line at a single
h→0 h
point a, therefore,
f (x + h) − f (x)
lim
h→0 h
represents the slope of the tangent line at a varying point x, which is precisely
said to be the derivative and thus denoted by f 0 (x). Notice how the limit
f (x + h) − f (x)
f 0 (x) = lim can only be evaluated to find the slope of the tan-
h→0 h
gent line when x is specified, and thus the derivative f 0 (x) is indeed a function
of x.

Summarizing the facts discussed, we have the following definition of the deriva-
tive of a function.

Definition 2.2.1. (The Derivative of a Function) The derivative f 0 (x) of f (x)

at the point x is defined as

f (x + h) − f (x)
f 0 (x) = lim
h→0 h
provided that the limit exists.

The derivative of a function can be thought of as a tool that helps us find the
slope of the tangent line, i.e., the instantaneous rate of change of a function f (x)
at any given point. So, the derivative of a function may informally be said as
follows.

“The derivative of a function is a mathematical tool used to measure the rate at

which the function f (x) changes with respect to a change in its input x at any
given instant.”

Think of it in this way: a curve does not maintain a constant slope at every
point—it changes as you move along it. To determine the slope of the curve at
a specific point, a tangent line is drawn to the curve at that point through a
limiting process. The slope of this tangent line represents the curve’s slope at
that exact location. By calculating the slope at various points along the curve,
using a variable x that moves across all points on the curve, we generate a new
function called the “derivative.” This derivative describes how the slope varies
along the curve, representing the curve’s instantaneous rate of change at each
point.

The definition of the derivative of a function can alternatively be defined as

follows.

Definition 2.2.2. (The Derivative of a Function) The derivative of f (x) at

the point x is given by

f (x) − f (a)
f 0 (x) = lim
x→a x−a
provided that the limit exists.

To thoroughly understand the concept behind derivatives, let us “differentiate,”

that is, find the derivative of the function f (x) = x2 .

In order to do this, notice that

for x =⇒ f (x) = x2
for x+h =⇒ f (x + h) = (x + h)2
= x2 + 2xh + h2

So, the change in y and x yields the following,

f (x + h) − f (x) = x2 + 2xh + h2 − |{z}

x2 =⇒ 2xh + h2

| {z }
f (x)
f (x+h)

(x + h) − x =⇒ h

Now, by comparing the changes as in the difference quotient, we get,

f (x + h) − f (x) 2xh + h2
=
h h
2xh h2
= +
h h
= 2x + h

f (x + h) − f (x)
Since the derivative is defined as lim , therefore
h→0 h
f (x + h) − f (x)
lim = lim (2x + h)
h→0 h h→0

= 2x + lim h
h→0

= 2x

This implies that the derivative of f (x) = x2 is f 0 (x) = 2x—but what does it
really mean? The function f (x) = x2 defines a specific relationship between
the input x and the output f (x), resulting in a parabola visually. However, its
derivative, f 0 (x) = 2x, explains how this relationship changes at every point
along the curve. Specifically, it shows how the slope (i.e., the rate of change)
of the parabola varies at different values of x. Basically, while f (x) = x2 repre-
sents a parabolic curve that changes its slope continuously at different points,
its derivative, f 0 (x) = 2x, tells us the exact rate of this change at those points.
For example, at x = 1, the slope of f (x) = x2 is f 0 (1) = 2 · 1 = 2. At x = 2,
the slope of f (x) = x2 is f 0 (2) = 2 · 2 = 4. At x = 3, the slope of f (x) = x2 is
f 0 (2) = 2 · 3 = 6.

This is shown geometrically below.

f(x)

x
1 2 3 4

f’(x)

x
1 2 3 4

Figure 2.12

The figures above illustrate that the graph of f (x) = x2 becomes progressively
steeper as x increases. This suggests that the slope, or rate of change, also grows
with every increase of x. As a result, the graph of the derivative f 0 (x) = 2x of

the function f (x) = x2 also increases at a constant rate of 2 as x increases.

A few examples to derive the derivative of a function are provided below.

I Example 2.2.1. If f (x) = 5x − 5, then derive the derivative of f (x).

Solution. By the definition of derivative, we get,

f (x + h) − f (x) [5(x + h) − 5] − (5x − 5)
lim = lim
h→0 h h→0 h
5x + 5h − 5 −5x
+ 5
= lim

h→0 h
5h
= lim
h→0 h
= lim 5
h→0

Therefore, the derivative of f (x) = 5x − 5 is f 0 (x) = 5.

√
I Example 2.2.2. If f (x) = x, then derive the derivative of f (x).

Solution. By the definition of derivative, we get,

√ √
f (x + h) − f (x) x+h− x
lim = lim
h→0 h h→0
√ h
√ √ √
x+h− x x+h+ x
= lim ·√ √
h→0 h x+h+ x
√ 2 √ 2
x + h − ( x)
= lim √ √
h→0 h x+h+ x
x+ h −
x
= lim √

√
h x+h+ x
h→0

1
= lim √ √
h→0 x+h+ x
1
=√ √
x+0+ x
1
= √
2 x
√ 1
Therefore, the derivative of f (x) = x is f 0 (x) = √ .
2 x

I Example 2.2.3. If f (x) = x3 , then derive the derivative of f (x).

Solution. By the definition of derivative, we get,

f (x + h) − f (x) (x + h)3 − x3
lim = lim
h→0 h h→0 h
x + 3x2 h + 3xh2 + h3 − x3
3
= lim
h→0 h
3x + 3xh + h2
h 2
= lim
h→0 h

= lim 3x + 3xh + h2
2

h→0

= lim 3x2 + lim 3xh + lim h2

h→0 h→0 h→0
2
= 3x + 0 + 0
= 3x2

Therefore, the derivative of f (x) = x3 is f 0 (x) = 3x2 .

1
I Example 2.2.4. If f (x) = , then derive the derivative of f (x).
x2
Solution. By the definition of derivative, we get,
1 1
2 − x2
f (x + h) − f (x) (x + h)
lim = lim
h→0 h h→0
h
1 1 1
= lim · −
h→0 h (x + h)2 x2
!
1 x2 − (x + h)2
= lim ·
h→0 h x2 (x + h)2
!
1 x2 − x2 − 2xh − h2
= lim ·
h→0 h x2 (x + h)2
−2xh − h2

1
= lim ·
h→0 h x2 (x + h)2
1 h (−2x − h)
= lim ·
h x2 (x + h)2
h→0

−2x − h
= lim
h→0 x2 (x + h)2

−2x − 0
=
x2 (x + 0)2
−2x
= 2 2
x ·x
2
=− 3
x
1 2
Therefore, the derivative of f (x) = 2
is f 0 (x) = − 3 .
x x

The End of Section 2.2

Definition of the Derivative of a function

§ 2.3 Notation for Derivatives

In the previous section, the concept of the derivative of a function was intro-
duced, and the notation f 0 (x) was used to represent it. It is important to note,
however, that several notations exist for the derivative, each rooted in its his-
torical developments. These different notations are not merely a reflection of
tradition but also offer practical advantages, as will become evident in their
dy
varied applications. One such convenient notation is the Leibniz notation , ,
dx
named after one of the co-founders of calculus: Gottfried Wilhelm Leibniz. To
understand this, consider the following graph of a function that changed from
x to x + ∆x, resulting in a change from f (x) to f (x + ∆x), where ∆x > 0 is a
small increment.

Figure 2.13

The slope of the secant line that passes through the points P and Q (the yellow
colored line) is given by the difference quotient,
f (x + ∆x) − f (x)
∆x
Since the change in y is often denoted as ∆y = f (x + ∆x) − f (x), therefore, the
slope of the secant line yields,
f (x + ∆x) − f (x) ∆y
=
∆x ∆x
Since the slope of the tangent line at a movable point x (i.e., the derivative) is
f (x + ∆x) − f (x)
defined as lim , therefore
∆x→0 ∆x
f (x + ∆x) − f (x)
derivative, f 0 (x) = lim
∆x→0 ∆x
∆y
= lim
∆x→0 ∆x
dy
=
dx
dy ∆y
Here, is simply a single notation to represent the limit lim is NOT a
dx ∆→0 ∆x
dy
ratio of dy over dx. The notation is read as “the derivative of y with respect
dx
to x,” or simply, “dee y dee x.” Note that it is never read as “dee y divided by
dy
dee x” or “the ratio of dee y and dee x.” The notation is not yet a quotient
dx

of two quantities dy and dx because the meaning of dy and dx have not yet
dy ∆y
been defined other than the fact that represents the limit lim . We will
dx ∆x→0 ∆x
later consider giving dy and dx a meaning to set up a real quotient between them.

Two other equivalent forms of the Leibniz notation are as follows,

df (x) d
and f (x)
dx dx
d
The second notation “ ( )” is thought of as an operator applied to a function
dx
f (x) to perform differentiation, yielding its derivative f 0 (x). This implies,
d
f (x) = f 0 (x)
dx
In addition to these notations, there exist a few others as listed below.
dy df (x) d
f 0 (x) = y 0 = ẏ = = = f (x) = Df (x) = Dx f (x)
dx dx dx
If it’s shown that the derivative is evaluated at any point a, then the notations
above are modified as follows.

dy d
f 0 (a) f (x)
dx x=a dx x=a
Considering all the different notations for derivatives at once may be overwhelm-
ing. However, it’s important to note that in most cases, the Leibniz notation
with its variants and the f 0 (x) notation introduced earlier are the ones most
commonly used. The notation f 0 (x) is named after the Italian mathematician
Joseph-Louis Lagrange.

The End of Section 2.3

Notation for Derivatives

§ 2.4 Differentiability and Continuity

In calculus, differentiability and continuity are two fundamental and related con-
cepts, yet they are not interchangeable. If a function is continuous at a point
then, informally, it does not have any sudden “jump” or “hole” at that point.
On the other hand, differentiability indicates that the function has a well-defined
tangent line (i.e., a derivative in broader terms) at that point.

Not every continuous function has a derivative at every point. When a function
f (x) has a derivative f 0 (x) at any given point x, the function is said to be dif-
ferentiable at that point. Similarly, when f (x) is differentiable at every point
over an open interval I 4 , it is said to be differentiable over that interval I.

Differentiability of a function at a point can be defined as follows.

Definition 2.4.1. (Differentiability) Let a be a point in the domain of the

function f (x). The function f (x) is said to be differentiable at the point a
if the following limit exists,

f (a + h) − f (a)
f 0 (a) = lim
h→0 h
This limit, if it exists, is called the derivative of f (x) at the point a.

A crucial result in the derivative of a function f (x) lies in the fact that the
differentiability of f (x) implies its continuity. That is, if a function f (x) is
differentiable at the point x = a, i.e., if it has a derivative at the point x = a,
then that function is necessarily continuous at that point x = a.

Theorem 2.4.1. (Differentiability Implies Continuity) If f (x) is differentiable

at a point a, then it must be continuous at that point.

Proof. Assume f (x) is differentiable at a point x = a. If so, then the following

limit exists,
f (x) − f (a)
f 0 (a) = lim
x→a x−a
Now, to show that f (x) is not only differentiable but also continuous at x = a,
we need to show the following,

lim f (x) = f (a)

x→a

In order to do this, consider the following.

f (x) = f (x)
0
z }| {
= f (x) − f (a) + f (a)
x−a
= f (x) − f (a) · + f (a)
x−a
4
finite or infinite

f (x) − f (a)
∴ f (x) = · (x − a) + f (a)
x−a
Now, we take the limit as x → a on both sides of the equation above and simplify
as follows.

f (x) − f (a)
lim f (x) = lim · (x − a) + f (a)
x→a x→a x−a
f (x) − f (a)
= lim · lim (x − a) + lim f (a)
x→a x−a x→a
| {z } x→a | {z }
0 f (a)
f (x) − f (a)
= lim +f (a)
x→a
| x−a }
{z
0

= f (a)
∴ lim f (x) = f (a)
x→a

This implies that f (x) is also continuous at the point x = a. This completes the
proof.
The theorem above states that if f (x) is differentiable at a point x = a, then it
necessarily is continuous at that point x = a. This logically follows that if f (x)
is not continuous at x = a, then it is necessarily not differentiable at x = a and
the following theorem states just that.

Theorem 2.4.2. (Not Continuous Implies Not Differentiable) If f (x) is not

continuous at a point a, then f (x) is also not differentiable at that point.

Notice carefully that if f (x) is continuous at a given point, then it may or may
not be differentiable at that point. Conversely, if f (x) is differentiable at a given
point, then it is necessarily continuous at that point.

In practice, generally, a function f (x) is not differentiable at a point a, if one of

the following conditions occurs.

Remark 2.4.1.

1. f (x) is not continuous at a.

2. f (x) has a sharp corner at a.

3. f (x) has a vertical tangent line at a.

Consider the following example for a better insight.

I Example 2.4.1. Determine the differentiability and continuity of f (x) = |x| at

the point x = 0.

Solution. The function f (x) = |x| is continuous at the point x = 0 because

lim |x| = |0| = 0

x→0

The function, however, is not differentiable at x = 0, because the graph of f (x)

has a sharp cornet at x = 0. For this reason, the left-side and the right-side
limits in the definition of the derivative of f (x) = |x| are not equal. That is,
|0 + h| − |0| |h| −h
f 0 (0− ) = lim− = lim− = lim− = −1
h→0 h h→0 h h→0 h
|0 + h| − |0| |h| +h
f 0 (0+ ) = lim+ = lim+ = lim+ = +1
h→0 h h→0 h h→0 h
Since f 0 (0− ) 6= f 0 (0+ ), therefore, the derivative does not exist at x = 0 as well.
As a result, the function is continuous but not differentiable at x = 0.

The End of Section 2.4

Differentiability and Continuity

§ 2.5 Higher-Order Derivatives

From the previous sections, we know that if f (x) is a differentiable function,
then f 0 (x) represents its derivative. Notice that the derivative f 0 (x) itself is a
function that can be further differentiated if the derivative exists. Therefore,
the derivative of the derivative of the function f (x) is a second-order deriva-
tive
ofthe function f (x). It is denoted by f (x), as it can be thought of as
00
0
f 0 (x) = f 00 (x). The process of taking derivatives can be continued endlessly,
resulting in third, fourth, fifth, etc. order derivatives each time (provided that
the function is differentiable each time). This notion can be expressed by the
following notations.

f 0 (x) = f 00 (x) = f 000 (x) = f (4) (x) = · · · = f (n) (x)

y 0 = y 00 = y 000 = y (4) = · · · = y (n)
dy d2 y d3 y d4 y dn y
= 2 = 3 = 4 = ··· = n
dx dx dx dx dx

d2 y
The second-order Leibniz notation can be viewed in terms of the first no-
dx2
tation, that is,

d2 y d dy
=
dx2 dx dx
to mean that the operation of differentiation is applied to the first derivative of
the function y = f (x).

dy d2 y
I Example 2.5.1. If f (x) = x3 and = 3x2 , then what is 2 ?
dx dx
dy
Solution. The derivative of = 3x2 and the second-order derivative of f (x) =
dx
x3 is,

d2 y 3(x + h)2 − 3x2

= lim
dx2 h→0 h
3(x + 2xh + h2 ) − 3x2
2
= lim
h→0 h
3x + 6xh + 3h2 −
2
3x2
= lim

h→0 h
h (6x + 3h)
= lim

h→0 h

= lim (6x + 3h)
h→0

= 6x + lim 3h
h→0

= 6x

d2 y
This implies that if f (x) = x , then 2 = 6x. 3

dx

dy d2 y d3 y
I Example 2.5.2. If f (x) = x3 , = 3x2 , and 2 = 6x, then what is 3 ?
dx dx dx
d2 y
Solution. The derivative of 2 = 6x and the third-order derivative of f (x) = x3
dx
is,

d3 y 6(x + h) − 6x
= lim
dx3 h→0 h
6x + 6h −6x
= lim

h→0 h

6h
= lim
h→0 h
= lim 6
h→0

d3 y
This implies that if f (x) = x3 , then = 6.
dx3

2
dy 2 d y d3 y
I Example 2.5.3. If f (x) = x , = 3x , 2 = 6x, and 3 = 6, then what is
3
dx dx dx
4
d y
?
dx4
d3 y
Solution. The derivative of 3 = 6 and the fourth-order derivative of f (x) = x3
dx
is,

d4 y 6−6
4
= lim
dx h→0 h
0
= lim
h→0 h

= lim 0
h→0

d4 y
This implies that if f (x) = x3 , then = 0.
dx4

The End of Section 2.5

Higher-Order Derivatives

Copyright © 2024 Md. Shouvik Iqbal. This book is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License (CC BY-NC-SA 4.0)
Download for free at https://md-shouvik-iqbal.github.io/calculus
Copyright © 2024 Md. Shouvik Iqbal. This book is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License (CC BY-NC-SA 4.0)
Download for free at https://md-shouvik-iqbal.github.io/calculus
Chapter 3

The Derivation of Derivatives

In the previous chapter, we discussed the concept of the derivative of a func-

tion and explored some examples to calculate them. Manually calculating these
derivatives using the formal limit definition can be quite tedious. While the
limit definition of derivative gives an exact method for computation, it is always
better to have some shortcuts or general formulas to avoid reapplying the same
limit process every time. These shortcuts are what we refer to as the differenti-
ation rules.

The key idea is that once we derive the rules from the limit definition for common
types of functions, we no longer need to perform complex limit calculations
repeatedly, over and over again. Thus, we now begin to derive the differentiation
rules for the most common types of functions in the following sections.

§ 3.1 Elementary Rules of Differentiation

In this section, we cover the fundamental differentiation rules that are widely
used. These elementary rules allow us to calculate derivatives without requiring
the limit definition and combine them to tackle more complex problems where
limit-based methods would be difficult.

143

Chap. 3 / Sec. 3.1 / Subsec. 3.1.1 : The Constant Rule

A function whose output is the same for all input is said to be a constant
function. For example, the function f (x) = c, where c is any real number, is a
constant function. Since the output of a constant function is always constant,
therefore the graph of the constant function is a horizontal line across all input
values, as shown below.

Figure 3.1

Since the slope of a horizontal line is 0, therefore the derivative of a constant

d
function is simply 0. That is, if f (x) = c, where c ∈ R, then f 0 (x) = (c) = 0.
dx
As a result, we have the following theorem.

Theorem 3.1.1. (The Constant Rule) If c ∈ R, then

d
(c) = 0
dx

Proof. If f (x) = c, where c ∈ R, then f (x + h) = c. Now, by the definition of

the derivative of a function, we write,
d f (x + h) − f (x)
(c) = lim
dx h→0 h
c−c
= lim
h→0 h
0
= lim
h→0 h

= lim 0
h→0

This completes the proof.

A few examples of the theorem above are presented below.

I Example 3.1.1. If f (x) = 27, then what is the derivative of f (x)?

Solution. The derivative of f (x) = 27 is as follows,
d
f 0 (x) = f (x)
dx
d
= (27)
dx
=0
∴ If f (x) = 27, then f 0 (x) = 0.

I Example 3.1.2. If f (x) = π, then what is f 0 (x)?

Solution. Note that π is a constant, therefore the derivative of f (x) = π is as
follows,
d
f 0 (x) = (π)
dx
=0
∴ If f (x) = π, then f 0 (x) = 0.

I Example 3.1.3. If f (x) = π e , then f 0 (x) =?

Solution. The derivative of f (x) = π e is as follows,
d
f 0 (x) = (π e )
dx
=0
∴ If f (x) = π e , then f 0 (x) = 0.

Chap. 3 / Sec. 3.1 / Subsec. 3.1.2 : The Constant Multiple Rule

This rule is obvious. The derivative of a constant multiplied by a function is
equal to the constant multiplied by the derivative of that function. That is,

Theorem 3.1.2. (The Constant Multiple Rule) If c ∈ R and f (x) is a differ-

entiable function, then
d d
c · f (x) = c · f (x)
dx dx

Proof. We will prove this directly.

d f (x + h) − f (x)
(c · f (x)) = lim
dx h→0 h
c · f (x + h) − c · f (x)
= lim
h→0
h
f (x + h) − f (x)
= lim c ·
h→0 h

f (x + h) − f (x)
= lim (c) · lim
h→0 h→0 h
f (x + h) − f (x)
= c · lim
h→0 h
0
= c · f (x)

This completes the proof.

A few examples of the theorem above are presented below.

x2
I Example 3.1.4. If f (x) = x2 , then what is the derivative of , where e is a
e
d
constant and x2 = 2x?
dx
Solution. Applying Theorem 3.1.2, we get,

d x2 d 1 2
= ·x
dx e dx e
1 d 2
= · x
e dx
1
= · 2x
e
2x
=
e
x2 2x
Therefore, the derivative of the function f (x) = is f 0 (x) = .
e e

Chap. 3 / Sec. 3.1 / Subsec. 3.1.3 : The Power Rule

Functions of the form f (x) = xn , where n ∈ N, are called the power functions.
Notice that if n = 1, then f (x) = xn simplifies to f (x) = x1 = x, representing a
straight line with a slope of 1. Therefore, the derivative of the function f (x) = x
is 1, as shown below.

Figure 3.2

Theorem 3.1.3. (Derivative of Identity Function) If f (x) = x is a differen-

tiable function, then
d
(x) = 1
dx

However, for n > 1, there isn’t a specific general case such as the one men-
tioned above. For example, how are the derivatives calculated for functions such
as f (x) = x2 or f (x) = x3 ? Currently, we can only apply the definition of
derivatives to these functions to determine their derivatives. For example, to
determine the derivative of f (x) = x2 , we proceed with the following approach.

If f (x) = x2 , then f (x + h) = (x + h)2 and it is known that,

(a + b)2 = a2 + 2ab + b2

Therefore, we apply the definition of derivative to determine the derivative of

f (x) = x2 as follows,

d f (x + h) − f (x)
f (x) = lim
dx h→0 h

(x + h)2 − x2
= lim
h→0 h
x + 2xh + h2 − x2
2
= lim
h→0 h
h(2x + h)
= lim

h→0 h

= lim (2x + h)
h→0

= 2x

Now, if f (x) = x3 , then f (x + h) = (x + h)3 and it is known that,

(a + b)3 = a3 + 3a2 b + 3ab2 + b3

Therefore, we similarly apply the definition of a derivative to determine the

derivative of f (x) = x3 as follows,
d f (x + h) − f (x)
f (x) = lim
dx h→0 h
(x + h) − x3
3
= lim
h→0 h
x + 3x2 h + 3xh2 + h3 − x3
3
= lim
h→0 h
h(3x + 3xh + h2 )
2
= lim

h→0 h

= lim (3x + 3xh + h2 )
2
h→0
2
= 3x

A pattern may be visible. That is, if f (x) = x2 , then f 0 (x) = 2x. Similarly, if
f (x) = x3 , then f 0 (x) = 3x2 . Indeed, this pattern suggests that if we have a
function f (x) defined as x raised to the power of n, where n is a positive integer,
then its derivative equals n multiplied by x raised to the power of n − 1. This
observation can be proven.

Theorem 3.1.4. (The Power Rule) If f (x) = xn is a differentiable function,

where n ∈ N, then
d n
(x ) = nxn−1
dx

Before presenting the proof of the theorem above, we first need to review the
Binomial Theorem.

We know the following,

(x + h)2 = x2 + 2xh + h2
(x + h)3 = x3 + 3x3 h + 3xh2 + h3
(x + h)4 = x4 + 4x3 h + 6x2 h2 + 4xh3 + h4
(x + h)5 = x5 + 5x4 h + 10x3 h2 + 10x2 h3 + 5xh4 + h5

Likewise, the general expansion of a binomial expression raised to nth power is

the following,
n(n − 1)xn−2
(x + h)n = xn + nxn−1 · (h) + · (h)2 + · · · + (h)n
2
Having this at hand, we now begin to prove Theorem 3.1.4 as follows,

Proof. We will prove this theorem directly.

d n f (x + h) − f (x)
x = lim
dx h→0 h
(x + h) − xn
n
= lim
h→0
h
n(n − 1)
xn + nxn−1 · (h) + (xn−2 ) · (h)2 + · · · + (h)n − xn
2
= lim
h→0 h
n−2

n n(n − 1)x
x
+h nxn−1 + · (h) + · · · + (h)n−1 − n
x
2
= lim
h→0 h

n−2
n(n − 1)x
= lim nxn−1 + · h + · · · + hn−1
h→0 2
n(n − 1)xn−2
= nxn−1 + · 0 + · · · + 0n−1
2
= nxn−1 + 0 + · · · + 0
= nxn−1

This completes the proof.

A few examples of the theorem above are presented below.

I Example 3.1.5. If f (x) = x999 , then f 0 (x) =?

Solution. Applying the Power Rule, we get,

d
x999 = 999x999−1

dx

= 999x998

Therefore, f 0 (x) = 999x998 .

dy
I Example 3.1.6. If f (x) = x1+2+3+4+5+6+7+8+9 , then what is ?
dx
Solution. Since 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45, therefore, applying the
Power Rule,
dy d
x45

=
dx dx
= 45x45−1
= 45x44

dy
Therefore, if f (x) = x1+2+3+4+5+6+7+8+9 , = 45x44 .
dx

Chap. 3 / Sec. 3.1 / Subsec. 3.1.4 : The Summation Rule

When examining a variety of functions, it’s often the case that they are formu-
lated as combinations of simpler functions. Therefore, it is important for us to
have a systematic method for finding the derivative of these combined functions.
One such combined function is obtained through the mathematical operation of
addition. That is, if f (x) and g(x) are two functions that can be easily dif-
ferentiated, then differentiation of their sum f (x) + g(x) follows the theorem
below.

Theorem 3.1.5. (The Summation Rule) If f (x) and g(x) are both differen-
tiable functions, then
d d d
f (x) + g(x) = f (x) + g(x)
dx dx dx

Proof. We will prove this directly.

d f (x + h) + g(x + h) − f (x) + g(x)
f (x) + g(x) = lim
dx h→0 h
f (x + h) + g(x + h) − f (x) − g(x)
= lim
h→0 h

f (x + h) − f (x) g(x + h) − g(x)
= lim +
h→0 h h
f (x + h) − f (x) g(x + h) − g(x)
= lim + lim
h→0 h h→0 h
0 0
= f (x) + g (x)
d d
= f (x) + g(x)
dx dx
This completes the proof.
This summation rule can be extended for more than two differentiable func-
tions such as f1 (x), f2 (x), f3 (x), · · · , fn−1 , fn (x), yielding a generalized summa-
tion rule as follows.
d d d d
f1 (x) + f2 (x) + · · · + fn (x) = f1 (x) + f2 (x) + · · · + fn (x)
dx dx dx dx

A few examples of the theorem above are presented below.

I Example 3.1.7. What is the derivative of f (x) = 2x2 + 3x3 + 5x5 + 7x7 ?

Solution. Since the derivatives of 2x2 , 3x3 , 5x5 , 7x7 are 4x, 9x2 , 25x4 , 49x6 ,
respectively, by the Power Rule. Therefore, the derivative of f (x) = 2x2 + 3x3 +
5x5 + 7x7 yields,
d d 2 d d d 7
2x2 + 3x3 + 5x5 + 7x7 = 2x + 3x3 + 5x5 +

7x
dx dx dx dx dx
d 2 d 3 d d 7
= 2x + 3x + 5x5 + 7x
dx dx dx dx
= 4x + 9x + 25x + 49x6
2 4

d
Therefore,

2x2 + 3x3 + 5x5 + 7x7 = 4x + 9x2 + 25x4 + 49x6
dx

d 2 d 3 d
I Example 3.1.8. If x = 2x and x = 3x2 , then what is x(x + x2 )?
dx dx dx
d
Solution. First, notice that, since x(x + x2 ) = x2 + x3 , therefore x(x + x2 ) =
dx
d 2
x + x3 . Thus,
dx
d 2 d 2 d 3
x + x3 = x + x
dx dx dx

= 2x + 3x2
= x(2 + 3x)

Therefore, the derivative of f (x) = x x + x2 is f 0 (x) = x(3x + 2).

Chap. 3 / Sec. 3.1 / Subsec. 3.1.5 : The Difference Rule

Similar to the Summation Rule, the Difference Rule is also used to find the
derivative of combined functions that involve the mathematical operation of
subtraction. If f (x) and g(x) are differentiable functions, then differentiation of
f (x) − g(x) follows the theorem below.

Theorem 3.1.6. (The Difference Rule) If f (x) and g(x) are both differen-
tiable functions, then
d d d
f (x) − g(x) = f (x) − g(x)
dx dx dx

Proof. We shall prove this directly as follows.

d f (x + h) − g(x + h) − f (x) − g(x)
f (x) − g(x) = lim
dx h→0 h
f (x + h) − g(x + h) − f (x) + g(x)
= lim
h→0
h
f (x + h) − f (x) g(x + h) − g(x)
= lim −
h→0 h h
f (x + h) − f (x) g(x + h) − g(x)
= lim − lim
h→0 h h→0 h
0 0
= f (x) − g (x)
d d
= f (x) − g(x)
dx dx
This completes the proof.
The difference rule can also be extended for more than two differentiable func-
tions such as f1 (x), f2 (x), f3 (x), · · · , fn−1 , fn (x), yielding a generalized difference
rule as follows.
d d d d
f1 (x) − f2 (x) − · · · − fn (x) = f1 (x) − f2 (x) − · · · − fn (x)
dx dx dx dx

A few examples of the theorem above are presented below.

I Example 3.1.9. What is the derivative of f (x) = x1 − x2 − x3 − x4 − x5 ?

Solution. Since the derivatives of x1 , x2 , x3 , x4 , and x5 are 1, 2x, 3x2 , 4x3 ,

and 5x4 , respectively, by the Power Rule. Therefore, the derivative of f (x) =
x1 − x2 − x3 − x4 − x5 yields,
d d 1 d d d d
x1 − x2 − x3 − x4 − x5 = x − x2 − x3 − x4 − x5

dx dx dx dx dx dx
1 2 3 4
= 1 − 2x − 3x − 4x − 5x

d
Therefore,

x1 − x2 − x3 − x4 − x5 = 1 − 2x1 − 3x2 − 4x3 − 5x4
dx

I Example 3.1.10. What is the derivative of

f (x) = ex1 − πxx1+1 − exxx1+1+1 − πxxxx1+1+1+1 − exxxxx1+1+1+1+1 ?

Solution. The function f (x) = ex1 − πxx1+1 − exxx1+1+1 − πxxxx1+1+1+1 −

exxxxx1+1+1+1+1 can be simplified as

f (x) = ex1 − πx3 − ex5 − πx7 − ex9

where, e and π are constant. Now, by the Difference Rule, Constant Multiple
Rule, and Power Rule; we get,
d d 1 d d d d
ex1 − πx3 − ex5 − πx7 − ex9 = ex − πx3 − ex5 − πx7 − ex9

dx dx dx dx dx dx
d 1 d 3 d 5 d 7 d
= e x − π x − e x − π x − e x9
dx dx dx dx dx
2 4 6 8
= e − 3πx − 5ex − 7πx − 9ex

Therefore, the derivative of f (x) = ex1 − πxx1+1 − exxx1+1+1 − πxxxx1+1+1+1 −

exxxxx1+1+1+1+1 is f 0 (x) = e − 3πx2 − 5ex4 − 7πx6 − 9ex8 .

Chap. 3 / Sec. 3.1 / Subsec. 3.1.6 : The Product Rule

In the previous combined functions, we’ve seen that the derivative of a sum of
functions equals the sum of their derivatives, and the derivative of a difference

of functions equals the difference of their derivatives. This might lead to an

assumption that the derivative of a product of functions equals the product of
their derivatives, but this is not the case this time. For example, if f (x) = x2 and
g(x) = x3 , then the combined function yields H(x) = f (x) · g(x) = x2 · x3 = x5 .
From the Power Rule, we know that the derivative of H(x) = x5 is H 0 (x) = 5x4 .
However, if the derivative of a product of functions were to be the product of
their derivatives, then the derivative of the combined function H(x) = x2 · x3
would’ve been H 0 (x) = 2x · 3x2 = 6x3 , which is invalid.

To understand the logic behind it, we can take a visual approach to explore the
underlying factors as follows.

Assume that f (x) and g(x) are both differentiable functions. Then the product
of these functions, f (x) · g(x) can be interpreted as an area of a rectangle as
shown below.

Figure 3.3

Now, if x changes by an increment of ∆x, then the corresponding changes in

f (x) and g(x) are as follows.

∆f = f (x + ∆x) − f (x) and ∆g = g(x + ∆x) − g(x)

This is shown in the following figure.

Figure 3.4

Now, the product f (x) + ∆f · g(x) + ∆g represents the area of the large
rectangle. And the change in the area f (x) · g(x) of the initial rectangle, that is,

Figure 3.5

is the following,

∆ f (x) · g(x) = f (x) + ∆f · g(x) + ∆g − f (x) · g(x)

= f (x) · g(x) + f (x) · ∆g + ∆f · g(x) + ∆f · ∆g − f (x) · g(x)

= ·
f (x) g(x) + f (x) · ∆g + ∆f · g(x) + ∆f · ∆g − ·
f (x) g(x)

= f (x) · ∆g + ∆f · g(x) + ∆f · ∆g (1)

Now to get the rate of the change, we compare the quantities by dividing equa-
tion (1) with ∆x, that is,

∆ f (x) · g(x) f (x) · ∆g + ∆f · g(x) + ∆f · ∆g
=
∆x ∆x

f (x) · ∆g ∆f · g(x) ∆f · ∆g
= + +
∆x ∆x ∆x
∆g ∆f ∆g
= f (x) · + g(x) · + ∆f ·
∆x ∆x ∆x
dy ∆y
Now, since the derivative is defined as = lim , there we write the
dx ∆x→0 ∆x
following to get the instantaneous rate of change (i.e., the derivative) of the
product f (x) · g(x),

d
= (f (x) · g(x))
dx
∆ f (x) · g(x)
= lim
∆x→0
∆x
∆g ∆f ∆g
= lim f (x) · + g(x) · + ∆f ·
∆x→0 ∆x ∆x ∆x

∆g ∆f ∆g
= f (x) · lim + g(x) · lim + lim ∆f · lim
∆x→0 ∆x ∆x→0 ∆x ∆x→0 ∆x→0 ∆x

∆g ∆f ∆g
= f (x) · lim + g(x) · lim + lim f (x + ∆x) − f (x) · lim
∆x→0 ∆x ∆x→0 ∆x ∆x→0 ∆x→0 ∆x
d d d
= f (x) · g(x) + g(x) · f (x) + 0 · g(x)
dx dx dx
d d
= f (x) · g(x) + g(x) · f (x)
dx dx
This implies that the derivative of the product of two functions f (x) and g(x)
d
is not merely the product of their derivatives, that is, f (x) · g(x) 6=
dx
d d d d d
f (x) · g(x), but rather, f (x) · g(x) = f (x) · g(x) + g(x) · f (x).
dx dx dx dx dx

Therefore, we have the following theorem.

Theorem 3.1.7. (The Product Rule) If f (x) and g(x) are both differentiable
functions, then
d d d
f (x) · g(x) = f (x) · g(x) + g(x) · f (x)
dx dx dx

Proof. We will prove this directly,

d
f (x) · g(x)
dx
f (x + h) · g(x + h) − f (x) · g(x)
= lim
h→0 h

0
z }| {
f (x + h) · g(x + h) −f (x + h) · g(x) + f (x + h) · g(x) −f (x) · g(x)
= lim
h→0
h
f (x + h) · g(x + h) − f (x + h) · g(x) f (x + h) · g(x) − f (x) · g(x)
= lim +
h→0 h h

g(x + h) − g(x) f (x + h) − f (x)
= lim f (x + h) · + g(x) ·
h→0 h h

g(x + h) − g(x) f (x + h) − f (x)
= lim f (x + h) · + lim g(x) ·
h→0 h h→0 h
g(x + h) − g(x) f (x + h) − f (x)
= lim f (x + h) · lim + lim g(x) · lim
h→0 h→0 h h→0 h→0 h
0 0
= f (x) · g (x) + g(x) · f (x)
This completes the proof.
Compared to the theorems previously discussed, this theorem seems to be quite
convoluted. Therefore, it is often helpful to simplify it in the following equivalent
manner.

Remark 3.1.1. If f (x) = u and g(x) = v are both differentiable functions,

then
d d d
u·v =u· v +v· u
dx dx dx

A few examples are provided below to demonstrate the theorem above.

I Example 3.1.11. What is the derivative of f (x) = ex(x + 1)?

Solution. If we let u = ex and v = x + 1, then by the Product Rule,
d d d
ex(x + 1) = ex · (x + 1) + (x + 1) · ex
dx dx dx
= ex + (x + 1) · e
= ex + ex + e
= 2ex + e
= e(2x + 1)
Therefore, the derivative of f (x) = ex(x + 1) is f 0 (x) = e(2x + 1).

d
I Example 3.1.12. Verify ex(x + 1) = e(2x + 1) in the previous example
dx
without using the Product Rule.

Solution. Notice that ex(x + 1) = ex2 + ex, therefore, by the Power Rule, we get,
d d 2 d
ex2 + ex =

ex + ex
dx dx dx
d 2 d
=e x +e x
dx dx
= 2xe + e
= e(2x + 1)

d
This confirms that ex(x + 1) = e(2x + 1)
dx

Chap. 3 / Sec. 3.1 / Subsec. 3.1.7 : The Quotient Rule

Similar to the Product Rule, the derivative of a quotient of functions does not
equal the quotient of their derivatives. It is also clear to see, for example, if
f (x) x3
f (x) = x3 and g(x) = x2 , then the quotient yields H(x) = = 2 = x.
g(x) x
Therefore, the derivative of H(x) yields, H 0 (x) = 1. Which is not equal to the
3x2 3x
quotient of the function’s derivatives, that is, H 0 (x) 6= = .
2x 2

To differentiate a quotient of functions, the Quotient Rule can be derived from

f (x)
the Product Rule. For example, consider the quotient function q(x) = .
g(x)
This implies that f (x) = g(x) · q(x). From the Product Rule, we know that,
d d
f (x) = g(x) · q(x)
dx dx
d d
= q(x) · g(x) + g(x) · q(x)
dx dx
d
Solving for q(x) yields,
dx
d d
d f (x) − q(x) · g(x)
q(x) = dx dx (1)
dx g(x)
f (x)
Since q(x) = , therefore substituting it in equation (1) yields,
g(x)
d d
d f (x) − g(x) · q(x)
q(x) = dx dx
dx g(x)

d d f (x)
f (x) − g(x) ·
dx dx g(x)
=
g(x)
d d f (x)
f (x) − g(x) ·
dx dx g(x) g(x)
= ·
g(x) g(x)
| {z }
1

d d f (x)
g(x) · f (x) − g(x) ·
dx dx g(x)
=
g(x) · g(x)

d d f (x)
g(x) f (x) − g(x) · g(x)

dx dx g(x)

=
(g(x))2
d d
g(x) f (x) − f (x) · g(x)
= dx dx
2
(g(x))
d d
d f (x) g(x) f (x) − f (x) · g(x)
∴ = dx dx
dx g(x) 2
(g(x))
f (x)
This implies that the derivative of the quotient function q(x) = is
g(x)
d d
·g(x) f (x) − f (x) · g(x)
dx
2
dx . As a result, we have the following theorem.
(g(x))

Theorem 3.1.8. (The Quotient Rule) If f (x) and g(x) are both differentiable
functions, then
d d
d

f (x)
g(x) · f (x) − f (x) · g(x)
= dx dx
dx g(x)
2
g(x)

Proof. We will prove this directly.

f (x + h) f (x)
−
d f (x) g(x + h) g(x)
= lim
dx g(x) h→0 h
f (x + h) · g(x) − f (x) · g(x + h)
g(x + h) · g(x)
= lim
h→0 h
f (x + h) · g(x) − f (x) · g(x + h)
= lim
h→0 h · g(x) · g(x + h)

0
z }| {
f (x + h) · g(x) −f (x) · g(x) + f (x) · g(x) −f (x) · g(x + h)
= lim
h→0 h · g(x) · g(x + h)
f (x + h) · g(x) − f (x) · g(x) + f (x) · g(x) − f (x) · g(x + h)
= lim h
h→0 g(x) · g(x + h)
f (x + h) · g(x) − f (x) · g(x) f (x) · g(x) − f (x) · g(x + h)
+
= lim h h
h→0 g(x) · g(x + h)

f (x + h) − f (x) g(x + h) − g(x)
g(x) · − f (x) ·
h h
= lim
h→0 g(x) · g(x + h)

f (x + h) − f (x) g(x + h) − g(x)
lim g(x) · − lim f (x) ·
h→0 h h→0 h
=
lim g(x) · g(x + h)
h→0
f (x + h) − f (x) g(x + h) − g(x)
lim g(x) · lim − lim f (x) · lim
= h→0 h→0 h h→0 h→0 h
lim g(x) · lim g(x + h)
h→0 h→0
d d
g(x) · f (x) − f (x) · g(x)
= dx dx
2
(g(x))
This completes the proof.
Similar to the Product Rule, the Quotient Rule also may appear convoluted,
thus it is often helpful to simplify it in the following equivalent manner.

Remark 3.1.2. (The Quotient Rule) If f (x) = u and g(x) = v are both
differentiable functions, then
d d
d u v · dx (u) − u · dx (v)
=
dx v (v)2

A few examples are provided below to demonstrate the theorem above.

ex
I Example 3.1.13. What is the derivative of f (x) = ?
x
Solution. Let u = ex and v = x, then by the Quotient Rule,
d d
d ex x · dx ex − ex · dx x
=
dx x x2

dx dx
x·e − ex ·
= dx dx
x2
ex − ex
=
x2
0
= 2
x
=0
ex
Therefore, the derivative of f (x) = is 0
x

I Example 3.1.14. Verify the result in the preceding example without using the
Quotient Rule.
ex
Solution. Notice that f (x) = = e. This suggests, by the constant rule, that
x
d
(e) = 0
dx
ex
Therefore, the derivative of f (x) = is 0. This completes the verification.
x

The End of Section 3.1

Elementary Rules of Differentiation

§ 3.2 Differentiation Rules for Trigonometric Functions

In this section, the derivatives of six trigonometric functions are derived. First,
the derivatives of the sine and cosine functions are derived by the limit definition
of derivative, which are then used to derive the derivatives of the remaining four
trigonometric functions. Throughout this process, it is assumed that the variable
x in sin(x), cos(x), tan(x), cot(x), sec(x), and csc(x) is measured in radians.

Chap. 3 / Sec. 3.2 / Subsec. 3.2.1 : Differentiation of sin(x)

Among the other trigonometric functions, the sine function is one of the few that
we will often encounter. This is because this function typically describes oscilla-
tions, waves, and circular motion, making it fundamental in both mathematics
and physics. Therefore, we have the following theorem.

Theorem 3.2.1. (The Derivative of the Sine Function) If f (x) = sin(x), then

d
sin(x) = cos(x)
dx

Prior to proving the theorem above, we first recall that if f (x) = sin(x), then
f (x + h) is given by the angle sum identity1 , that is,

f (x + h) = sin(x + h)
= sin(x) · cos(h) + cos(x) · sin(h)

and consider the following two limits that we had previously evaluated.

sin (x) cos (x) − 1

lim =1 and lim =0
x→0 x x→0 x
Proof. We will prove this directly.

f (x + h) − f (x)
f 0 (x) = lim
h→0 h
sin(x + h) − sin(x)
= lim
h→0
h
sin(x) · cos(h) + cos(x) · sin(h) − sin(x)
= lim
h→0 h
sin(x) · cos(h) − sin(x) + cos(x) · sin(h)
= lim
h→0
h hi
sin(x) · (cos(h) − 1) + cos(x) · sin(h)
= lim
h→0 h
h i
sin(x) · (cos(h) − 1)

cos(x) · sin(h) 
= lim  +
h→0 h h
cos(h) − 1 sin(h)

= lim sin(x) · + lim cos(x) ·
h→0 h h→0 h
cos(h) − 1 sin(h)

= sin(x) · lim + cos(x) · lim
h→0 h h→0 h
| {z } | {z }
the limit is 0 the limit is 1

= sin(x) · 0 + cos(x) · 1
= cos(x)

This completes the proof.

1
sin (α + β) = sin (α) cos (β) + cos (α) sin (β)

A few examples are provided below to demonstrate the theorem above.

I Example 3.2.1. Differentiate f (x) = x sin(x).

Solution. Since f (x) is a product of two functions, namely x and sin(x), therefore
the Product Rule will be used as outlined below.
d d
f (x) = x sin(x)
dx dx
d d
=x· sin(x) + sin(x) x
dx dx
= x · cos(x) + sin(x) · 1
= x cos(x) + sin(x)

Therefore, the derivative of f (x) = x sin(x) is f 0 (x) = x cos(x) + sin(x).

sin(x) + 1
I Example 3.2.2. Differentiate f (x) =
sin(x) − 1
Solution. Since f (x) is a quotient of two functions sin(x) + 1 and sin(x) − 1,
therefore the Quotient Rule will be used as outlined below.

d sin(x) + 1

d
f (x) =
dx dx sin(x) − 1
d d
(sin (x) − 1) · (sin (x) + 1) − (sin (x) + 1) · (sin (x) − 1)
= dx dx
(sin (x) − 1)2
d d d d
(sin (x) − 1) · sin (x) + (1) − (sin (x) + 1) · sin (x) − (1)
= dx dx dx dx
(sin (x) − 1)2
(sin (x) − 1) · cos (x) + 0 − (sin (x) + 1) · cos (x) − 0
=
(sin (x) − 1)2
(sin (x) − 1) · cos (x) − (sin (x) + 1) · cos (x)
=
(sin (x) − 1)2
sin (x) cos (x) − cos (x) − sin (x) cos (x) − cos (x)
=
(sin (x) − 1)2
− cos (x) − cos (x)
=
(sin (x) − 1)2
2 cos (x)
=−
(sin (x) − 1)2

sin(x) + 1 2 cos (x)

Therefore, the derivative of f (x) = is f 0 (x) = − .
sin(x) − 1 (sin (x) − 1)2

Now that the derivative of sin(x) is determined, we move to find the derivative
of cos(x).

Chap. 3 / Sec. 3.2 / Subsec. 3.2.2 : Differentiation of cos(x)

In the previous section, it is observed that the derivative of the sine function is
the cosine function. Now, it will be shown, surprisingly, that the derivative of
the cosine function is the negative of the sine function. That is,

Theorem 3.2.2. (The Derivative of the Cosine Function) If f (x) = cos(x),

then
d
cos(x) = − sin(x)
dx

Now, prior to proving the theorem above by the definition of derivative, it is

important to recall that if f (x) = cos(x), then f (x + h) is given by the angle
sum identity2 as follows,

f (x + h) = cos(x + h)
= cos(x) · cos(h) − sin(x) · sin(h)

Proof. We will prove this directly.

d cos(x + h) − cos(x)
cos(x) = lim
dx h→0
h
cos(x) · cos(h) − sin(x) · sin(h) − cos(x)
= lim
h→0 h
cos(x) · cos(h) − cos(x) − sin(x) · sin(h)
= lim
h→0
h h i
cos(x) · cos(h) − 1 − sin(x) · sin(h)
= lim
h→0 h
[cos(x) · (cos(h) − 1)] sin(x) · sin(h)

= lim −
h→0 h h
cos(h) − 1 sin(h)

= lim cos(x) · − lim sin(x) ·
h→0 h h→0 h
2
cos (α + β) = cos (α) cos (β) − sin (α) sin (β)

cos(h) − 1 sin(h)

= cos(x) · lim − sin(x) · lim
h→0 h h→0 h
| {z } | {z }
the limit is 0 the limit is 1

= cos(x) · 0 − sin(x) · 1
= − sin(x)

This completes the proof.

A few examples are provided below to demonstrate the theorem above.

I Example 3.2.3. Differentiate f (x) = coscos(0) (x)

Solution. Since cos(0) = 1, therefore,

d d
coscos(0) (x) = cos1 (x)
dx dx
= − sin(x)

Therefore, the derivative of f (x) = coscos(0) (x) is f 0 (x) = − sin(x).

I Example 3.2.4. Differentiate f (x) = x2 cos(x)

Solution. Since f (x) is a product of two functions x2 and cos(x), thus the Product
Rule will be used as follows.
d d 2
f (x) = x cos(x)
dx dx
d d 2
2
=x · cos(x) + cos(x) · x
dx
dx
= x2 · − sin(x) + cos(x) · 2x
= −x2 sin(x) + 2x cos(x)

= x 2 cos(x) − x sin(x)

Therefore, the derivative of f (x) = x2 cos(x) is f 0 (x) = x 2 cos(x)−x sin(x) .

Chap. 3 / Sec. 3.2 / Subsec. 3.2.3 : Differentiation of tan(x)

Now that we’ve obtained the derivatives of sin(x) and cos(x), we can use them
to derive the derivatives of other trigonometric functions. In order to derive the

derivative of tan(x), recall that

sin(x)
tan(x) =
cos(x)

This implies that,

sin(x)

d d
tan(x) =
dx dx cos(x)
d d
cos(x) · sin(x) − sin(x) · cos(x)
= dx dx
cos2 (x)
cos(x) · cos(x) − sin(x) · (− sin(x))
=
cos2 (x)
cos2 (x) + sin2 (x)
=
cos2 (x)
1
=
cos2 (x)
= sec2 (x)

This implies that the derivative of the tangent function is the secant function
squared.

Theorem 3.2.3. (The Derivative of the Tangent Function) If f (x) = tan(x),

then
d
tan(x) = sec2 (x)
dx

A few examples are provided below to demonstrate the theorem above.

I Example 3.2.5. Differentiate f (x) = tan(x) + x.

Solution. By the Summation Rule, we get,

d d
f (x) = tan(x) + x
dx dx
d d
= tan(x) + (x)
dx dx
= sec (x) + 1
2

Therefore, the derivative of f (x) = tan(x) + x is f 0 (x) = sec2 (x) + 1.

sin2 (x)
I Example 3.2.6. Differentiate f (x) =
cos(x)
sin2 (x)
Solution. Since can be simplified as follows,
cos(x)
sin2 (x) sin(x)
= · sin(x)
cos(x) cos(x)
= tan(x) · sin(x)

Therefore, by the Product Rule,

d d d
(tan(x) · sin(x)) = tan(x) · sin(x) + sin(x) · tan(x)
dx dx dx
= tan(x) · cos(x) + sin(x) · sec2 (x)
sin(x)
= cos(x)
·
+ sin(x) · sec2 (x)
cos(x)

= sin(x)(1 + sec2 (x))

sin2 (x)
Therefore, the derivative of f (x) = is f 0 (x) = sin(x)(1 + sec2 (x)).
cos(x)

Chap. 3 / Sec. 3.2 / Subsec. 3.2.4 : Differentiation of cot(x)

Following a similar method as before, the derivative of cot(x) is determined. For
that, recall that
cos(x)
cot(x) =
sin(x)
With that in mind, we proceed as follows,
d cos(x)

d
cot(x) =
dx dx sin(x)
d d
sin(x) · cos(x) − cos(x) · sin(x)
= dx dx
sin2 (x)
sin(x) · (− sin(x)) − cos(x) · cos(x)
=
sin2 (x)
− sin2 (x) − cos2 (x)
=
sin2 (x)
(−1) · sin2 (x) + cos2 (x)

=
sin2 (x)

(−1) · (1)
=
sin2 (x)
1
=− 2
sin (x)
= − csc2 (x)

This implies that the derivative of the cotangent function is the negative of the
cosecant function squared. That is,

Theorem 3.2.4. (The Derivative of the Cotangent Function) If f (x) =

cot(x), then
d
cot(x) = − csc2 (x)
dx

A few examples are provided below to demonstrate the theorem above.

I Example 3.2.7. Differentiate f (x) = 2x + 2 cot(x)

Solution. By the Summation Rule, we get,

d d d
(2x + 2 cot(x)) = 2x + 2 cot(x)
dx dx dx
dx d
= 2 + 2 cot(x)
dx dx
2
= 2 − 2csc (x)

Therefore, the derivative of f (x) = 2x + 2 cot(x) is f 0 (x) = 2 − 2csc2 (x).

cos2 (x)
I Example 3.2.8. Differentiate f (x) =
e sin(x)
cos2 (x)
Solution. Since can be simplified as follows,
e sin(x)

cos2 (x) 1 cos(x)

= · · cos(x)
e sin(x) e sin(x)
1
= cot(x) · cos(x)
e
Therefore,

d 1 1 d
cot(x) · cos(x) = · (cot(x) · cos(x))
dx e e dx

1 d d
= cot(x) · cos(x) + cos(x) · cot(x)
e dx dx
1
cot(x) · (− sin(x)) + cos(x) · (−csc2 (x))

=
e
1
− sin(x) cot(x) − cos(x)csc2 (x)

=
e
− cos(x) − cos(x)csc2 (x)
sin(x)
sin(x)

=

e
− cos(x) − cos(x)csc2 (x)
=
e
cos2 (x) − cos(x) − cos(x)csc2 (x)
Therefore, the derivative of f (x) = is f 0 (x) =
e sin(x) e

Chap. 3 / Sec. 3.2 / Subsec. 3.2.5 : Differentiation of sec(x)

To derive the derivative of f (x) = sec(x), notice that since sec(x) can be written
1
as , therefore, by the Quotient Rule,
cos(x)

d d 1
sec(x) =
dx dx cos(x)
d d
cos(x) · (1) − 1 · cos(x)
= dx dx
cos2 (x)

cos(x) · 0 − 1 · − sin(x)
=
cos2 (x)
sin(x)
=
cos2 (x)
1 sin(x)
= ·
cos(x) cos(x)
= sec(x) · tan(x)

This implies that the derivative of the secant function is the secant function
multiplied by the tangent function. That is,

Theorem 3.2.5. (The Derivative of the Secant Function) If f (x) = sec(x),

then
d
sec(x) = sec(x) tan(x)
dx

A few examples are provided below to demonstrate the theorem above.

I Example 3.2.9. Differentiate f (x) = x2 cos(0) − 2 cos(0) · sec(x)

Solution. Since cos(0) = 1, therefore, f (x) is simplified into the following,

x2 cos(0) − 2 cos(0) · sec(x) = x2 − 2 sec(x)

Therefore,
d d 2 d
x2 − 2 sec(x) = x − 2 sec(x)

dx dx dx
= 2x − 2 sec(x) tan(x)

Therefore, the derivative of f (x) = x2 cos(0) − 2 cos(0) · sec(x) is 2x−2 sec(x) tan(x).

I Example 3.2.10. Differentiate f (x) = sin(x) sec(x)

Solution. By the Product Rule,

d d d
(sin(x) sec(x)) = sin(x) · sec(x) + sec(x) · sin(x)
dx dx dx
= sin(x) · sec(x) tan(x) + sec(x) · cos(x)
1 sin(x) 1
= sin(x) · · + cos(x)
·

cos(x) cos(x) cos(x)

sin2 (x)
= +1
cos2 (x)
sin2 (x) + cos2 (x)
=
cos2 (x)
1
=
cos2 (x)
= sec2 (x)

Therefore, the derivative of f (x) = sin(x) sec(x) is f 0 (x) = sec2 (x).

Chap. 3 / Sec. 3.2 / Subsec. 3.2.6 : Differentiation of csc(x)

To derive the derivative of f (x) = csc(x), notice that csc(x) can be written as
1
, therefore, by the Quotient Rule,
sin(x)

d d 1
csc(x) =
dx dx sin(x)
d d
sin(x) · (1) − 1 · sin(x)
= dx dx
sin2 (x)
sin(x) · 0 − 1 · cos(x)
=
sin2 (x)
− cos(x)
=
sin2 (x)
1 cos(x)
=− ·
sin(x) sin(x)
= − csc(x) · cot(x)

This implies that the derivative of the cosecant function is the negative of the
cosecant function multiplied by the cotangent function. That is,

Theorem 3.2.6. (The Derivative of the Cosecant Function) If f (x) = csc(x),

then
d
csc(x) = − csc(x) cot(x)
dx

A few examples are provided below to demonstrate the theorem above.

I Example 3.2.11. Differentiate f (x) = πx − csc(x)

Solution. By the difference rule,

d d d
(πx − csc(x)) = π x − csc(x)
dx dx dx
= π + csc(x) cot(x)

Therefore, the derivative of f (x) = πx − csc(x) is f 0 (x) = π + csc(x) cot(x).

I Example 3.2.12. Differentiate f (x) = csc2 (x)

Solution. By the Product Rule, we get,

d d
csc2 (x) = (csc(x) · csc(x))
dx dx
d d
= csc(x) · csc(x) + csc(x) · csc(x)
dx dx
= csc(x) · (− csc(x) cot(x)) + csc(x) · (− csc(x) cot(x))
= −csc2 (x) cot(x) − csc2 (x) cot(x)
= −2csc2 (x) cot(x)

Therefore, the derivative of f (x) = csc2 (x) is f 0 (x) = −2csc2 (x) cot(x).

The End of Section 3.2

Differentiation Rules for Trigonometric Functions

§ 3.3 Differentiation of Composite Functions

(The Chain Rule)
The differentiation rules covered so far are useful for finding the derivatives of
many basic functions. However, these rules alone are often not enough when
we encounter more complex functions that are frequently used in practice, such
as composite functions. To handle such demanding cases, we need some handy
tools such as the Chain Rule, which allow us to differentiate functions that are
a composition of others. Take the following function as an example,

(1)
3
y = x4 + 14

To differentiate this function using the rules discussed so far, we must first need
to expand it and then apply the appropriate differentiation rules. For example,
since,

(a + b)3 = a3 + 3a2 b + 3ab2 + b3 (2)

Therefore, applying (2) to (1) yields,

y = x12 + 42x8 + 588x4 + 2744 (3)

We can now differentiate (3) using the Power and Summation Rule. That is,
d d
x12 + 42x8 + 588x4 + 2744

(y) =
dx dx

d
x12 + 42x8 + 588x4 + 2744

=
dx
d d d d
x12 + 42x8 + 588x4 +

= (2744)
dx dx dx dx
= 12x11 + 336x7 + 2352x3 (4)

Thus, we’ve derived the derivative of y = x4 + 14 to be 12x11 + 336x7 +

2352x3 . Now, can the same approach be taken to find the derivative of y =
? The answer is quite obviously no. Therefore, let us begin
235711131719
x4 + 14
to develop a new method to tackle such situations.

Notice that the function y = x4 + 14 is a composite function. That is, it

is composed of two different functions in a way that if we let a variable u to

stand for the expression x4 + 14, then y = x4 + 14 becomes y = u3 , where
3

u = x4 + 14. This implies that y is now a function of u (as y = u3 ), and u is in

turn a function of x (as u = x4 + 14).

Since, y is a function of u, therefore the derivative (i.e., the rate of change) of y

with respect to u is,
dy d 3
= (u )
du du
= 3u2

Now, since u in turn is a function of x, therefore the derivative (i.e., the rate of
change) of u with respect to x is,

du d
x4 + 14

=
dx dx
d d
x4 +

= (14)
dx dx
= 4x3 + 0
= 4x3

Now, notice carefully that,

If the output y changes 3u2 times as fast as the input u, since y = u3

And, if the output u changes 4x3 times as fast as the input x, since y = x4 + 14

Then, the output y changes 3u2 × 4x3 times as fast as x, for u = x4 + 14

In notations,
dy
= 3u2 if
du
du
and if = 4x3
dx
dy dy du
then = 3u2 × 4x3

= ×
dx du dx

Therefore, the derivative of y = x4 + 14 with respect to x is,

2
3u2 · 4x3 = 3 x4 + 14 · 4x3
2
= 12x3 x4 + 14

Which is indeed equivalent to equation (4) when expanded.

Now, we try to find the derivative of the challenging function y = x4 + 14

235711131719

using the method described above. For that, let u = x4 + 14, then y =
u235711131719 . Since y is a function of u, therefore, it is differentiated with respect
to u,
dy d
u235711131719

=
du du
= 235711131719u235711131719−1
= 235711131719u235711131718

Since u = x4 + 14 is a function of x, therefore, it is differentiated with respect

to x,
du d
x4 + 14

=
dx dx
d d
x4 +

= (14)
dx dx
= 4x3 + 0
= 4x3

Now, the derivative of y with respect to x (not u) happens to be:

dy dy du
= ×
dx du dx
= 235711131719u235711131718 × 4x3
235711131718
= 235711131719 x4 + 14 · 4x3

235711131718
= 942844526876x3 x4 + 14

Summarizing the discussion, we have the following theorem to differentiate com-

posite functions.

Theorem 3.3.1. (The Chain Rule) If y = f (u) is a differentiable function of

u and u = g(x) is a differentiable function of x, then the derivative of the
composite function y = f (g(x)) with respect to x is

dy dy du
= ·
dx du dx
or, equivalently,

(f ◦ g)0 (x) = f 0 (g (x)) · g 0 (x)

y 0 = f 0 (u) · u0

We here provide an intuitive argument of the Chain Rule. Note that it isn’t
sufficiently rigorous to be accepted as a complete proof of the theorem.
Intuitive Argument. Since u is a function of x, therefore any change ∆x in x pro-
duces a corresponding change ∆u in u. This, in turn, produces a corresponding
change ∆y in y, as y is a function of u. Now, consider the following,
dy ∆y
= lim by the definition of derivative
dx ∆x→0 ∆x
∆y
= lim ·1
∆x→0 ∆x

∆y ∆u
= lim ·
∆x→0 ∆x ∆u

∆y ∆u
= lim ·
∆x→0 ∆u ∆x

∆y ∆u
= lim · lim
∆x→0 ∆u ∆x→0 ∆x

Since, differentiability implies continuity, therefore ∆u → 0 as ∆x → 0.

∆y ∆u
= lim · lim
∆u→0 ∆u ∆x→0 ∆x
dy du
= ·
du dx
This completes the argument intuitively.

Remark 3.3.1. When applyingthe Chain

rule, it is often helpful to think of
the composite function y = f g(x) as having two parts. One is the inner
part u = g(x) and the other is the outer part y = f (u). This means that
to differentiate a composite function, we:

1. Differentiate the outer function y = f (u) with respect to the inner

function u = g(x).

2. Multiply the result by the derivative of the inner function u = g(x)

with respect to x.

To better understand the inner and outer function analogy, consider the following
functions.
sin(x2 ) and sin2 (x)
Although the inner and outer parts of these functions may resemble one another,
they are fundamentally different. The inner part of the first function is u = x2 ,
and the outer part of it is y = sin(u). Whereas the inner part of the second
function is u = sin(x), and the outer part of it is y = u2 , since sin2 (x) =
2
sin(x) .

I Example 3.3.1. Differentiate f (x) = sin(x2 )

Solution. For the function sin(x2 ), the inner part u is x2 and the outer part y is
sin(u).
Since the outer part y = sin(u) is a function of u, therefore it is differentiated
with respect to u,
dy d
= sin(u)
du du
= cos(u)
Similarly, the inner part u = x2 is a function of x, therefore it is differentiated
with respect to x,
du d 2
= (x )
dx dx
= 2x
Now, by the Chain Rule, we differentiate y with respect to x as follows,
dy dy du
= ·
dx du dx

= cos(u) · 2x
= cos(x2 ) · 2x
= 2x cos(x2 )

Therefore, the derivative of f (x) = sin(x2 ) is f 0 (x) = 2x cos(x2 ).

I Example 3.3.2. Differentiate f (x) = sin2 (x)

2
Solution. For the function sin (x) = sin(x) , the inner part u is sin(x) and
2

the outer part y is u2 .

Since the outer part y = u2 is a function of u, therefore it is differentiated with
respect of u,
dy d 2
= u
du du
= 2u

Similarly, the inner part u = sin(x) is a function of x, therefore it is differentiated

with respect to x,
du d
= sin(x)
dx dx
= cos(x)

Now, by the Chain Rule, we differentiate y with respect to x as follows,

dy dy du
= ·
dx du dx
= 2u · cos(x)
= 2 sin(x) cos(x)

Therefore, the derivative of f (x) = sin2 (x) is f 0 (x) = 2 sin(x) cos(x).

Author's Interruption. In practicality, it is often unnecessary to write out all

the details of the Chain Rule. Instead, one can directly apply the Chain
Rule without explicitly mentioning the inner and outer function.

To demonstrate this, we apply the Chain Rule directly to the function shown in
Example 3.3.1, as follows,
dy d
sin x2

=
dx dx

d
= cos x2 · x2

dx
= cos x · 2x
2

= 2x cos x2

That is, the outermost function is first differentiated and then simply multiplied
by the derivative of the inner function.

√
I Example 3.3.3. Differentiate y = x2 + x
√
Solution. For the function y = x2 + x, the inner and outer parts are u = x2 + x
√
and y = u, respectively. Therefore, by the Chain Rule,

dy dy du
= ·
dx du dx
d √ d 2

= u · x +x
du dx
1
= √ · 2x + 1
2 u
2x + 1
= √
2 u
2x + 1
= √
2 x2 + x
√ 2x + 1
Therefore, the derivative of f (x) = x2 + x is f 0 (x) = √ .
2 x2 + x

2
x2

I Example 3.3.4. Differentiate y =
x+1
2 2
x
Solution. For the function y = , the inner and outer parts are u =
x+1
x2
and u2 , respectively. Therefore, by the Chain Rule,
x+1
dy dy du
= ·
dx du dx 2
d 2
d x
= u ·
du dx x + 1
d d
(x + 1) · x2 − x2 · (x + 1)
= 2u · dx dx
(x + 1)2

(x + 1) · 2x − x2 · 1
= 2u ·
(x + 1)2
2x (x + 1) − x2
= 2u ·
(x + 1)2
2
x 2x (x + 1) − x2
=2 ·
x+1 (x + 1)2
2x2 2x2 + 2x − x2
= ·
x+1 (x + 1)2

2x2 2x2 + 2x − x2
=
(x + 1)3
4x4 + 4x3 − 2x4
=
(x + 1)3
2x3 (2x + 2 − x)
=
(x + 1)3
2x3 (x + 2)
=
(x + 1)3
2 2
x 2x3 (x + 2)
Therefore, the derivative of f (x) = is f 0 (x) = .
x+1 (x + 1)3
2 2
x x4
Note that if the equation were to be simplified as = prior to
x+1 (x + 1)2
differentiating it solely using the Quotient Rule, the result would’ve remained
the same.

Since the Chain Rule is one of the most important and frequently encountered
rule among the differentiation rules, therefore a few additional examples are
given below.

I Example 3.3.5. Differentiate y = sin (x)

Solution. The derivative of f (x) = sin (x) is,

dp
f 0 (x) = sin (x)
dx
d
sin (x) 2
1
=
dx
1 d
= sin (x)− 2 · sin (x)
1

2 dx
1 1
= · · cos (x)
2 sin (x) 12

cos (x)
= p
2 sin (x)

cos (x)
Therefore, the derivative of f (x) = sin (x) is f 0 (x) = p .
p

2 sin (x)

I Example 3.3.6. Differentiate y = sin cos x2

Solution. Applying the Chain Rule,

dy d
sin cos x2

=
dx dx
d
= cos x2 · cos x2

dx
d
= cos x2 · − sin x2 · x2

dx
= − cos x · sin x · 2x
2 2

= −2x cos x2 sin x2

Therefore, the derivative of f (x) = sin cos x2 is f 0 (x) = −2x cos x2 sin x2 .

I Example 3.3.7. Differentiate y = sin cos tan x2 .

Solution. Applying the Chain Rule,

dy d
sin cos tan x2

=
dx dx
d
= cos cos tan x2 cos tan x2

·
dx
d
= cos cos tan x2 · − sin tan x2 · tan x2

dx
d
= cos cos tan x2 · − sin tan x2 · sec2 x2 · x2

dx
= − cos cos tan x 2
· sin tan x 2
· sec x · 2x
2 2

= −2x · cos cos tan x2 · sin tan x2 · sec2 x2

Therefore, the derivative of f (x) = sin cos tan x2 is f 0 (x) = −2x·cos cos tan x2

·
sin tan x2 · sec2 x2 .

I Example 3.3.8. Differentiate y = sin cos tan cot sec csc x2

Solution. Applying the Chain Rule,

dy d
= sin cos tan cot sec csc x2

dx dx
d
= cos cos tan cot sec csc x2 cos tan cot sec csc x2

·
dx
= cos cos tan cot sec csc x 2
· − sin tan cot sec csc x2

d
tan cot sec csc x2

·
dx
= cos cos tan cot sec csc x2 · − sin tan cot sec csc x2

d
· sec2 cot sec csc x2 cot sec csc x2

·
dx
= cos cos tan cot sec csc x2 · − sin tan cot sec csc x2

d
· sec2 cot sec csc x2 · −csc2 sec csc x2 sec csc x2

·
dx
= cos cos tan cot sec csc x 2
· − sin tan cot sec csc x2

· sec2 cot sec csc x2 · −csc2 sec csc x2 · sec csc x2 tan csc x2

d
csc x2

·
dx
= cos cos tan cot sec csc x2 · − sin tan cot sec csc x2

· sec2 cot sec csc x2 · −csc2 sec csc x2 · sec csc x2 tan csc x2

d
· − csc x2 cot x2 · x2

dx
= cos cos tan cot sec csc x2 · − sin tan cot sec csc x2

· sec2 cot sec csc x2 · −csc2 sec csc x2 · sec csc x2 tan csc x2

· − csc x2 cot x2 · 2x

= − 2x cos cos tan cot sec csc x2 sin tan cot sec csc x2

sec2 cot sec csc x2 csc sec csc x2 sec csc x2 tan csc x2
2

csc x2 cot x2

Therefore, the derivative of f (x) = sin cos tan cot sec csc x2 is f 0 (x) =

−2x cos cos tan cot sec csc x2 · sin tan cot sec csc x2

· sec2 cot sec csc x2 · csc2 sec csc x2 · sec csc x2 · tan csc x2 ·

csc x2 cot x2 .

The End of Section 3.3

Differentiation of Composite Functions

(The Chain Rule)

§ 3.4 Implicit Differentiation

As of yet, all the functions that have been differentiated were of the form,

y = f (x)

Where, y is a function defined explicitly of x, as it is expressed entirely in terms

of x. However, times do occur when y is not expressed explicitly in terms of x,
while still being a function of x. For example, consider the following equation,

xy − 1 = 0

The equation above defines y implicitly as a function of x. This is because y can

be expressed in terms of x if we rearrange the equation as follows.

xy − 1 = 0
xy = 1
1
y=
x
It is now evident—after rearrangement of the equation—that y is indeed a
function of x, even though it was initially defined implicitly by the equation
xy − 1 = 0. Such a function, where y is defined indirectly by an equation, is
called an implicit function of x.

In general, if a relationship between x and y is given by an equation of the form,

F (x, y) = 0 (1)

where F is a function of both x and y, and y is not explicitly solved in terms of

x, then y is said to be an implicit function of x.

In contrast to our intuition, an implicit equation may not always define a single
function. When the graph of an implicit equation fails the vertical line test, the
equation defines more than one function. For example, consider the following
equation,
x2 + y 2 = 1

The graph of this equation is the graph of a unit circle that fails the vertical line
test. Solving this for y yields,
p
y = ± 1 − x2
√ √
where, y = + 1 − x2 represents the upper semicircle and y = − 1 − x2 repre-
sents the lower semicircle of the circle x2 + y 2 = 1.

With that in mind, notice now that some equations defining implicit functions
y are notoriously difficult—if not impossible—to express in terms of x explicitly.
And, so far, we have seen that in order to differentiate a function, it needs to be
in the form of y = f (x). So, how such implicitly defined functions be differenti-
ated? Is it even possible?

The answer is positive. In general, it is not required to explicitly solve an

equation for y in terms of x to differentiate implicitly defined functions. For
example, consider the previous function,

xy − 1 = 0
xy = 1 (2)

The typical approach to differentiating this equation is to first rewrite the equa-
1
tion as y = and then to differentiate y with respect to x, as shown below.
x

dy d 1
=
dx dx x
d −1
= x
dx
= −1 · x−1−1
= −x−2
1
=− 2
x
Now, instead of this typical method, if we consider y as a differentiable function
of x, we can differentiate equation (2) directly as follows.
d d
xy = (1)
dx dx
d d
x · (y) + y · (x) = 0
dx dx
dy
x· +y =0 (3)
dx

dy
Now, solving for algebraically yields,
dx
dy
x· = −y
dx
dy y
=−
dx x
y
Thus, the derivative of the implicit function xy = 1 is − . Notice that the
x
derivative, in this case, is expressed in terms of both x and y, instead of just
x. So, to evaluate the derivative at a specific point, both values of x and y are
required.

Additionally, notice how the Chain Rule is used in equation (3). That is, when
differentiating terms with respect to x that contain only the variable x, the pro-
cess follows the usual method. However, when we encounter terms involving
variables other than x, the Chain Rule is applied there.

The derivative obtained here is indeed equivalent to the derivative obtained by

1
the typical method if y = is substituted, as shown below.
x
dy y
=−
dx x
1
1 1 1
= −x = − · = − 2
x x x x
The method by which the derivative of xy = 1 is obtained without having to
solve for y in terms of x, is called implicit differentiation. Implicit differentia-
tion is a technique used to find the derivative of functions when the relationship
between their variables are not explicitly given in the form y = f (x). Instead,
the variables are interconnected by an equation such as F (x, y) = 0. In such
cases, y is treated as a function of x and the rules of differentiation are applied
to both sides of the equation, differentiating y implicitly.

In general, implicit differentiation has the following steps.

Remark 3.4.1.

1. Differentiate both sides of the equation with respect to x.

dy
2. Collect all terms involving on one side of the equation and move
dx

all the remaining terms to the other side.

dy
3. Factor out when necessary.
dx
dy
4. Solve for .
dx

With that in mind, we will now consider a few examples to understand the
implicit differentiation.

I Example 3.4.1. Differentiate x2 + y 3 = 5.

Solution. By implicit differentiation

x2 + y 3 = 5
d d
x2 + y 3 =

(5)
dx dx
d 2 d
x + y3 = 0
dx dx
2 dy
2x + 3y · =0
dx
dy
3y 2 · = −2x
dx
dy 2x
=− 2
dx 3y
dy 2x
Therefore, the derivative of x2 + y 3 = 5 is = − 2.
dx 3y

I Example 3.4.2. Differentiate x2 + xy + y 3 = x.

Solution. By implicit differentiation

d d
x2 + xy + y 3 =

(x)
dx dx
d 2 d d dx
x + xy + y 3 =
dx dx
dx dx
dy dx dy
2x + x · +y· + 3y 2 · =1
dx dx dx
dy dy
2x + x · + y + 3y 2 · =1
dx dx
dy dy
x· + 3y 2 · = 1 − 2x − y
dx dx

dy
x + 3y 2 = 1 − 2x − y

dx
dy 1 − 2x − y
=
dx x + 3y 2
dy 1 − 2x − y
Therefore, the derivative of x2 + xy + y 3 = x is = .
dx x + 3y 2

I Example 3.4.3. Differentiate sin(x) = tan(y)

Solution. By implicit differentiation,

sin(x) = tan(y)
dy dy
sin(x) = tan(y)
dx dx
dy
cos(x) = sec2 (y) ·
dx
dy cos(x)
=
dx sec2 (y)

dy cos(x)
Therefore, the derivative of sin(x) = tan(y) is = .
dx sec2 (y)

I Example 3.4.4. Differentiate ln(x) + ln(y) = x + y

Solution. By implicit differentiation,

d d
(ln(x) + ln(y)) = (x + y)
dx dx
d d dx dy
ln(x) + ln(y) = +
dx dx dx dx
1 1 dy dy
+ · =1+
x y dx dx
1 dy dy 1
· − =1−
y dx dx x

dy 1 1
−1 =1−
dx y x
1
dy 1 −
= x
dx 1
−1
y

x−1
= x
1−y
y
x−1 y
= ·
x 1−y
y(x − 1)
=
x(1 − y)

dy y(x − 1)
Therefore, the derivative of ln(x) + ln(y) = x + y is = .
dx x(1 − y)

r
√
q
I Example 3.4.5. Differentiate y =
p
x+ x+ x + x + ···

r
√
q
I Example 3.4.6. Since y = x + x + · · ·, therefore, it can be
p
x+ x+
simplified as follows,
s r 2
√
q
y2 =  x+ x + x + x + · · ·
r
√
q
2
y = x + x + x + x + ···
| {z }
y

y2 = x + y

Now, implicitly differentiating the equation above, we get,

d d
y2 =

(x + y)
dx dx
dy dy
2y =1+
dx dx
dy dy
2y − =1
dx dx
dy
(2y − 1) = 1
dx
dy 1
=
dx 2y − 1
1
= r
√
q p
2 x + x + x + x + ··· − 1

r
√
q
Therefore, the derivative of y = x + x + x + x + · · · is
p

dy 1
= r .
dx q p √
2 x + x + x + x + ··· − 1

The End of Section 3.4

Implicit Differentiation

§ 3.5 Differentiation Rules for Exponential Functions

Functions of the form f (x) = ax , where a is a constant and x is a variable
exponent, are called exponential functions. These functions play a crucial role
in describing growth and decay processes for the fact that they grow or shrink
at rates proportional to their current value. This is the reason they are ideal
for modeling real-world phenomena such as population growth, radioactive de-
cay, and interest calculations. In this section, we will derive the derivatives of
exponential functions.

Chap. 3 / Sec. 3.5 / Subsec. 3.5.1 : Differentiation of ex

In section 3.2, we saw that the derivative of sin(x) is cos(x). Similarly, the
derivative of cos(x) is − sin(x). Now, if − sin(x) is again differentiated, it turns
to − cos(x). Similarly, if − cos(x) is again differentiated, the derivative yields
sin(x) and the pattern continues!

Instances such as the one above are indeed interesting, especially, when consider-
ing the connection between the rates of change of the function and its derivative.
However, what’s even more interesting is the existence of a function where the
rate at which it changes at a point is equivalent to the value of the function itself
at that point!

It may sound unbelievable but it is indicated sufficiently by functions such as

f (x) = 2x . For that let us differentiate the function f (x) = 2x by the definition
of a derivative.
d x 2x+h − 2x
(2 ) = lim
dx h→0 h
2 · 2h − 2x
x
= lim
h→0 h

2h − 1

= lim 2 · x
h→0 h
h
2 − 1
= 2x · lim
h→0 h
h
2 −1
Now, we evaluate the limit lim numerically by the following table as
h→0 h
h approaches to 0,
Table 3.1: Table

2h − 1
h
h
−0.1 0.66967...
−0.01 0.69075...
−0.001 0.69290...
+0.001 0.69338...
+0.01 0.69555...
+0.1 0.71773...

The table above suggests that as h → 0 from both sides of 0, the expression
2h − 1
→ k such that 0.69290... < k < 0.69338.... This implies that
h
d x
(2 ) = k · 2x
dx
This further implies,
d x
(2 ) ∝ 2x
dx
This is indeed fascinating that there exists a function f (x) = 2x where its
derivative is proportional to itself. However, this does not answer our search
for a function whose derivative is equal to itself. Therefore, let us generalize
the case by letting f (x) = ax this time, where a ∈ R. If f (x) = ax , then the
d x
derivative (a ) is,
dx
d x ax+h − ax
(a ) = lim
dx h→0 h
a · ah − ax
x
= lim
h→0 h
h
a −1
= lim a ·
x
h→0 h
h
a −1
= a · lim
x
h→0 h

ah − 1

To evaluate the limit lim , we create the following table for different
h→0 h
values of a, where h ∈ R is a small increment less than 1 and greater than 0.
Table 3.2: Table

ah − 1
a
h
2 0.69314…
3 1.09861…
4 1.38630…
5 1.60945…

Now, a clue is found! From the table above, it is visible that between a = 2 and
ah − 1
a = 3, there exists a value of a for which the limit lim yields 1. And why
h→0 h
ah − 1
are we interested lim resulting in 1? Because then the derivative of ax
h→0 h
(whatever the value of a might be), would be equal to itself. Thus, let us first
ah − 1 ah − 1
search the value of a for which lim = 1. To do that, let ≈ 1, for
h→0 h h
a small h ∈ R. This implies, for a small h ∈ R,

ah − 1
≈1
h
ah − 1 ≈ h
ah ≈ h + 1
√
h
a≈ 1+h
1
a ≈ (1 + h) h

We’ve found the value of a to be approximately equal to (1 + h) h , for a small

h ∈ R. However, for an exact value of a, we let h → 0, that is, the limit,

lim (1 + h) h = a
1

h→0

The numerical value of (1 + h) h as h → 0, is approximated by the following

table,
Table 3.3: Table
1
h (1 + h) h
−0.1 2.86796 · · ·
−0.01 2.73198 · · ·

−0.001 2.71963 · · ·
+0.001 2.71691 · · ·
+0.01 2.70480 · · ·
+0.1 2.59373 · · ·

In general, the numerical value of (1 + h) h as h → 0 turns out to be 2.718281828459....

This implies that,

d
(2.718281828459...)x = (2.718281828459...)x
dx
Let us denote this real number 2.718281828459... as the variable e, that is,
e = 2.718281828459..., implying that,
d x
(e ) = ex
dx
As a result, we have found a special function that remains unchanged when
differentiated. That is,

Theorem 3.5.1. (The Derivative of the Exponential Function with Base e)

If f (x) = ex , where e ∈ R, then

d x
e = ex
dx

A few examples are provided below to demonstrate the theorem above.

I Example 3.5.1. Differentiate f (x) = e−x .

d x
Solution. Since it is known that e = ex , therefore only the Chain Rule needs
dx
to be further applied in order to differentiate f (x) = e−x , as shown below.
d −x d
e = e−x · (−x)
dx dx
d
= e−x · − (x)
dx
−x
= e · −1
= −e−x
Thus, the derivative of f (x) = e−x is f 0 (x) = −e−x .

The result of the example above is of such frequency that it should have its own
remark, as given below.

Remark 3.5.1. If f (x) = e−x , where e ∈ R, then

d −x
e = −e−x
dx

I Example 3.5.2. Differentiate f (x) = esin(x)

Solution. The derivative of ex is itself and the derivative of sin(x) is cos(x).
Therefore,
d sin(x) sin(x) d
e =e · sin (x)
dx dx
= esin(x) cos (x)

Thus, the derivative of f (x) = esin(x) is f 0 (x) = −esin(x) cos (x).

I Example 3.5.3. Differentiate f (x) = ee .

Solution. Applying the Chain Rule,

d ex x d x
e = ee · (e )
dx dx
x
= ee · ex
x +x
= ee

Thus, the derivative of f (x) = ee is f 0 (x) = ee .

x x +x

Chap. 3 / Sec. 3.5 / Subsec. 3.5.2 : Differentiation of ax

Now that we have found a function where its derivative matches the function
itself, we now try to find the derivative of what was previously unexplored:
f (x) = ax . It is known from the discussion above, that
h
d x a −1
(a ) = a · lim
x
dx h→0 h
h
a −1
Now, to solve the limit lim , we use a property of natural logarithms,
h→0 h
stating that

eln(x) = x

Letting a yields eln(a) = a. Therefore, substituting a = eln(a) into the limit

hx =
a −1
lim yields,
h→0 h
ah − 1
= lim
h→0 h
eln(a) − 1
h
= lim
h→0 h
ln(a)h
e −1
= lim (1)
h→0 h
Now, to evaluate the limit in equation (1), we use the following limit,
h
e −1
lim =1
h→0 h
k
To apply this limit, let k = ln(a)h. If k = ln(a)h, then h = . Now, since
ln(a)
h → 0, therefore k → 0, and thus,
eln(a)h − 1
= lim
h→0 h
ek − 1
= lim k
k→0
ln(a)
k
e −1
= lim · ln(a)
k→0
| {z }k
the limit is 1

= 1 · ln(a)
h
a −1
Thus, it can be shown that lim = ln(a) and therefore,
h→0 h
h
d x a −1
(a ) = a · lim
x
dx h→0 h
= ax · ln(a)

Therefore, we have the following theorem.

Theorem 3.5.2. (The Derivative of the Exponential Function with Base a)

If f (x) = ax , where a > 0, then

d x
(a ) = ax · ln(a)
dx

A few examples are provided below to demonstrate the theorem above.

I Example 3.5.4. Differentiate f (x) = 10x .

Solution. Letting a = 10 yields,

d
(10x ) = 10x · ln(10)
dx
Therefore, the derivative of f (x) = 10x is f 0 (x) = 10x ln(10)

I Example 3.5.5. Differentiate f (x) = (0.36)x .

Solution. By the theorem above,

d d
2(0.36)x = 2 · (0.36)x
dx dx
= 2 · (0.36)x · ln (0.36)
= 2 ln (0.36) (0.36)x

Therefore, the derivative of f (x) = (0.36)x is f 0 (x) = 2 ln (0.36) (0.36)x .

I Example 3.5.6. Differentiate f (x) = ln π ln(e) · π x .

Solution. By the theorem above,

d d
ln π ln(e) · π x = ln π ln(e) ·

(π x )
dx dx
= ln π · π ln (π)
1
x

= π x ln2 (π)

Therefore, the derivative of f (x) = ln π ln(e) · π x is f 0 (x) = π x ln2 (π).

I Example 3.5.7. Differentiate f (x) = xa ax .

Solution. By the theorem above,

d a x d x d a
(x a ) = xa · (a ) + ax · (x )
dx dx dx
= xa · ax ln (a) + ax · axa−1

axa
= x · a ln (a) + a ·
a x x
x
a
= x a ln (a) +
a x
x
a
Therefore, the derivative of f (x) = xa ax is f 0 (x) = xa ax ln (a) + .
x

The End of Section 3.5

Differentiation Rules for Exponential Functions

§ 3.6 Differentiation Rules for Logarithmic Functions

Logarithmic functions are the inverse of exponential functions. An exponential
function takes the form y = ax , where a is a positive constant, a logarithmic func-
tion reverses the process in a way that answers the question—to what exponent
a must be raised to obtain y? In this section, we will derive the differentiation
rules for the logarithmic functions.

Chap. 3 / Sec. 3.6 / Subsec. 3.6.1 : Differentiation of ln(x)

A natural logarithm ln(x) is a logarithm with a base e, that is, loge (x) = ln(x).
To differentiate natural logarithms, the inverse property of logarithms and ex-
ponentials and the Chain Rule are used, along with the implicit differentiation.

We know that for all x > 0,

x = eln(x)

Differentiating both sides of the equation with respect to x yields,

d d ln(x)
(x) = e
dx dx
d
1 = eln(x) · ln(x)
dx
Now, divide both sides of the equation by eln(x) . Then
d 1
ln(x) = ln(x)
dx e
1
=
x

1
This implies that the derivative of f (x) = ln(x) is f 0 (x) = , for all x > 0.
x
Therefore, we have the following theorem.

Theorem 3.6.1. (The Derivative of the Natural Logarithm) If f (x) = ln(x),

where x > 0, then
d 1
ln(x) =
dx x

d 1
Note that since the domain of ln(x) is (0, ∞), therefore ln(x) = is true for
dx x
all values of x > 0. These values can be extended by considering the function
ln |x|, instead of ln(x). Because, by the definition of absolute value, we know,

ln(x) if x > 0
ln |x| =
ln(−x) if x < 0
This implies that ln |x| is defined for all x 6= 0. Thus, for x > 0, we have
d d
ln |x| = ln(x)
dx dx
1
=
x
And, for x < 0, we have,
d d
ln |x| = ln (−x)
dx dx
1 d
= · (−x)
−x dx
1 d
= · − (x)
−x dx
1
= · −1
−x
1
=
x
Thus, we have the following theorem.

Theorem 3.6.2. (The Derivative of the Natural Logarithm) If f (x) = ln |x|,

where x 6= 0, then
d 1
ln |x| =
dx x

Furthermore, if there exists a function f (x) 6= 0, then the derivative of ln |f (x)|,

happens to be the following.
d 1 d
ln |f (x)| = · f (x)
dx f (x) dx

1
= · f 0 (x)
f (x)
f 0 (x)
=
f (x)

Thus, we have the following theorem.

d
Theorem 3.6.3. If ln |f (x)|, where f (x) 6= 0, then
dx
d f 0 (x)
ln |f (x)| =
dx f (x)

A few examples are provided below to demonstrate the theorem above.

I Example 3.6.1. Differentiate f (x) = x ln |x|.

Solution.
d d d
x ln |x| = x · ln |x| + ln |x| · (x)
dx dx dx
1
= x · + ln |x| · 1
x
= 1 + ln |x|

Therefore, the derivative of f (x) = x ln |x| is f 0 (x) = 1 + ln |x|.

I Example 3.6.2. Differentiate f (x) = ln |sec (x)|.

Solution.
d 1 d
ln |sec (x)| = · sec (x)
dx sec (x) dx
1
= · sec (x) tan (x)
sec (x)
= tan (x)

Therefore, the derivative of f (x) = ln |sec (x)| is f 0 (x) = tan (x).

Chap. 3 / Sec. 3.6 / Subsec. 3.6.2 : Differentiation of loga (x)

In order to differentiate the common logarithms notice that for all a > 1 and
a 6= 0,

y = loga (x) ⇔ x = ay

Now, differentiating both sides of the equation x = ay with respect to x yields,

d d y
(x) = (a )
dx dx
dy
1 = ay ln (a) ·
dx
Now, divide both sides of the equation by a ln (a). Then
y

dy 1
= y
dx a ln (a)
1
=
x ln (a)
Therefore, we have the following theorem.

Theorem 3.6.4. (The Derivative of the Common Logarithm) If f (x) =

loga (x), where a > 0 and a 6= 1, then

d 1
loga (x) =
dx x ln (a)

For all x > 0.

Similar to the natural logarithm, the values of the domain of loga (x) can be
extended by considering loga |x|. As a result, we have the following theorem.

Theorem 3.6.5. (The Derivative of the Common Logarithm) If f (x) =

loga |x|, where a > 0 and a 6= 1, then

d 1
loga |x| =
dx x ln (a)

For all x 6= 0.

Similar to the previous subsection, if there exists a function f (x) 6= 0, then the
derivative of loga |f (x)| happens to be the following,
d 1 d
loga |f (x)| = · f (x)
dx f (x) ln(a) dx

1
= · f 0 (x)
f (x) ln(a)
f 0 (x)
=
f (x) ln(a)
Thus, we have the following theorem.

d
Theorem 3.6.6. If log |f (x)|, where f (x) 6= 0, then
dx a
d f 0 (x)
loga |f (x)| =
dx f (x) ln(a)

A few examples are provided below to demonstrate the theorem above.

I Example 3.6.3. Differentiate f (x) = log7 x7 .

Solution. By the theorem above,

d 1 d
log7 x7 = 7 x7

·
dx x ln (7) dx
1
= 7 · 7x6
x ln (7)
7x6
= 7
x ln (7)
x6 1
=7· 7 ·
x ln (7)
1
= 7 · x−1 ·
ln (7)
7
=
x ln (7)
7
Therefore, the derivative of f (x) = log7 x7 is f 0 (x) = .

x ln (7)

I Example 3.6.4. Differentiate f (x) = logπ (ln(x))

Solution. By the theorem above,

d 1 d
logπ (ln(x)) = · ln(x)
dx ln(x) · ln(π) dx
1 1
= ·
ln(x) ln(π) x

1
=
x ln(π) ln(x)
1
Therefore, the derivative of f (x) = logπ (ln(x)) is f 0 (x) = .
x ln(π) ln(x)

The End of Section 3.6

Differentiation Rules for Logarithmic Functions

§ 3.7 Logarithmic Differentiation

So far, we have discussed several differentiation rules and techniques for different
types of functions. However, when a function involves a product or quotient
of multiple terms, or when variables appear in complex exponents, the typical
differentiation methods can become cumbersome. In such cases, it is often useful
to simplify the function before differentiating. A powerful technique for this
simplification is to use the properties of logarithms. The basic idea is to apply
the natural logarithm to both sides of the equation, simplify the expression using
logarithmic properties, and then differentiate implicitly. For example, consider
the following function,
ex+1 x
f (x) =
x+1
It would be necessary for us to use the the Quotient Rule, the Product Rule,
and the chain rule to differentiate the function above. This process would be
quite tedious, and simplifying afterward would add additional work and com-
plexity. However, by applying the properties of logarithms, the function can be
differentiated with relative ease. But, before that, it is important to review a
few properties of natural logarithms, as shown below.

1. ln(ab) = ln(a) + ln(b) 3. ln ab = b ln (a)

1
4. ln = − ln(a)
a
2. ln = ln(a) − ln(b) a
b
Now, using the properties of natural logarithms shown above, the function can
be rewritten as follows:
x+1
e x
ln f (x) = ln
x+1
= ln ex+1 x − ln (x + 1)

= ln ex+1 + ln (x) − ln (x + 1)

= (x + 1) ln (e) + ln (x) − ln (x + 1)
= (x + 1) + ln (x) − ln (x + 1)

Now, differentiating both sides of the equation above yields the following,
d d
ln f (x) = (x + 1) + ln (x) − ln (x + 1)
dx dx
d d d
= (x + 1) + ln (x) − ln (x + 1)
dx dx dx
f 0 (x) 1 1
=1+ −
f (x) x x+1

0 1 1
f (x) = 1 + − · f (x)
x x+1
x+1
1 1 e x
= 1+ − ·
x x+1 x+1
x+1
x (x + 1) + (x + 1) − x e x
= ·
x (x + 1) x+1
2 x+1
x +x+1 e x
= ·
x (x + 1) x+1
2
x+1
x +x+1 e
=
(x + 1)2
0 ex+1 x2 + ex+1 x + ex+1
∴ f (x) =
(x + 1)2

ex+1 x
Therefore, the derivative f 0 (x) of the function f (x) = is
x+1
ex+1 x2 + ex+1 x + ex+1
f 0 (x) = .
(x + 1)2

Using the logarithmic differentiation, the proof for the general Power Rule can
be given as follows.

Theorem 3.7.1. (The General Power Rule) If f (x) = xr , where r ∈ R, then

d r
x = rxr−1
dx

Proof. Let y = xr , where r ∈ R, then,

y = xr
ln (y) = ln (xr )

ln (y) = r · ln (x)
1 dy 1
· =r·
y dx x
dy y
=r·
dx x
Since y = xr , therefore,
dy xr
=r·
dx x
= r · xr−1

This completes the proof.

A few examples are provided below to demonstrate the general Power Rule given
above.

√
I Example 3.7.1. If f (x) = x = x 2 , then
1

d √ d 1
( x) = (x 2 )
dx dx
1 1
= x 2 −1
2
1 1
= x− 2
2
1
= √
2 x
√ 1
Therefore, the derivative of f (x) = x = x 2 is f 0 (x) = √ .
1

2 x

Remark 3.7.1.
d √ 1
( x) = √
dx 2 x

xr+1
I Example 3.7.2. If f (x) = , where r ∈ R, then,
r+1
r+1
d x 1 d
xr+1

= ·
dx r + 1 r + 1 dx
1
· r + 1 · xr+1−1

=
r+1
1
= · r+ 1 · xr

r+1

= xr

xr+1
Therefore, the derivative of f (x) = is f 0 (x) = xr
r+1

Remark 3.7.2.
xr+1

d
= xr
dx r+1
where, r ∈ R.

A few more examples are provided below to demonstrate the logarithmic differ-
entiation.

I Example 3.7.3. Differentiate f (x) = xx

Solution. By the properties of logarithms, we write

ln f (x) = ln (xx )
= x · ln (x)

Differentiating both sides yield,

d d
ln f (x) = (x · ln (x))
dx dx
f 0 (x) d dx
=x· ln (x) + ln (x) ·
f (x) dx dx
1
= x · + ln (x) · 1
x
f 0 (x)
= 1 + ln (x)
f (x)
f 0 (x) = (1 + ln (x)) f (x)
= (1 + ln (x)) xx
= xx + xx ln (x)

Therefore, the derivative of f (x) = xx is f 0 (x) = xx + xx ln (x).

√ √
I Example 3.7.4. Differentiate f (x) =
x
x

Solution. By the properties of logarithms, we write

√ √
ln f (x) = ln
x
x
√ √
= x · ln x

Differentiating both sides yield,

d d √ √
ln f (x) = x · ln x
dx dx
f 0 (x) √ d √ √ d√
= x· ln x + ln x · x
f (x) dx dx
√ 1 d√ √ 1
= x· √ x + ln x · √
x dx 2 x
1 √ 1
= √ + ln x · √
2 x 2 x
0 √
f (x) 1
= √ 1 + ln x

f (x) 2 x
√
0 1 + ln ( x)
f (x) = √ · f (x)
2 x
√
1 + ln ( x) √ √x
= √ · x
2 x
√ √x √ √x √
x + x ln ( x)
= √
2 x
√ √x √ √x √
√ √
x + x ln ( x)
Therefore, the derivative of f (x) = is f (x) = .
x 0
x √
2 x

I Example 3.7.5. Differentiate xy = y x

Solution. By the properties of logarithms, we write

xy = y x
ln (xy ) = ln (y x )
y · ln(x) = x · ln(y)

Differentiating both sides yield,

d d
(y · ln(x)) = (x · ln(y))
dx dx
d dy d dx
y· ln(x) + ln(x) · =x· ln(y) + ln(y) ·
dx dx dx dx

1 dy 1 dy
y· + ln(x) · =x· + ln(y) · 1
x dx y dx
y dy x dy
+ ln(x) = + ln(y)
x dx y dx
dy x dy y
ln(x) − = ln(y) −
dx y dx x

dy x y
ln(x) − = ln(y) −
dx y x
y
dy ln(y) −
= x
x
dx ln(x) −
y
x ln(y) − y
= x
y ln(x) − x
y
x ln(y) − y y
= ·
x y ln(x) − x
y (x ln(y) − y)
=
x (y ln(x) − x)
y (x ln(y) − y)
Therefore, differentiating xy = y x , we get f 0 (x) = .
x (y ln(x) − x)

I Example 3.7.6. Differentiate ln(xy) = x + y

Solution. By the properties of logarithms, we write

ln(xy) = x + y
ln(x) + ln(y) = x + y

Differentiating both sides yield,

d d
(ln(x) + ln(y)) = (x + y)
dx dx
d d dx dy
ln(x) + ln(y) = +
dx dx dx dx
1 1 dy dy
+ =1+
x y dx dx
1 dy dy 1
− =1−
y dx dx x

dy 1 1
−1 =1−
dx y x

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Page 206 Differentiation Rules for Inverse Trigonometric Functions Chapter 3

1
dy 1−
= x
dx 1
−1
y
x−1
= x
1−y
y
x−1 y
= ·
x 1−y
y (x − 1)
=
x (1 − y)
y (x − 1)
Therefore, differentiating ln(xy) = x + y, we get f 0 (x) = .
x (1 − y)

The End of Section 3.7

Logarithmic Differentiation

§ 3.8 Differentiation Rules for Inverse Trigonometric

Functions
Previously, we derived the differentiation rules for the six trigonometric func-
tions. We now derive the differentiation rules for their inverse functions which
have various applications across numerous fields.

Chap. 3 / Sec. 3.8 / Subsec. 3.8.1 : Differentiation of sin−1 (x)

The derivative of y = sin−1 (x), where −1 ≤ x ≤ 1, is obtained by implicitly
differentiating both sides of x = sin(y), with respect to x. That is,

x = sin (y)
d d
(x) = sin (y)
dx dx
dy
1 = cos (y) ·
dx
cos · dy
(y)
1
dx
=
cos (y) cos
(y)

dy 1
= (1)
dx cos (y)

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Chapter 3 Differentiation Rules for Inverse Trigonometric Functions Page 207

Now, from the Pythagorean identity, it is known that sin2 (θ) + cos2 (θ) = 1.
Letting θ = y yields,
sin2 (y) + cos2 (y) = 1
cos2 (y) = 1 − sin2 (y)
q
cos(y) = 1 − sin2 (y)
Now, continuing with equation (1),
d 1
(y) =
dx cos (y)
1
=p
1 − sin2 (y)
1
=√
1 − x2
d 1
sin−1 (x) = √
dx 1 − x2
1
Therefore, the derivative of f (x) = sin−1 (x) is f 0 (x) = √ .
1 − x2

Theorem 3.8.1. (The Derivative of the Inverse Sine Function) If f (x) =

sin−1 (x), then
d 1
sin−1 (x) = √
dx 1 − x2
for −1 < x < 1

A few examples are provided below to demonstrate the theorem above.

I Example 3.8.1. Differentiate f (x) = sin−1 (x)

Solution. Applying the Chain rule, we write,

dy d
sin−1 (x)
2
=
dx dx
d
= 2sin−1 (x) · sin−1 (x)
dx
1
= 2sin−1 (x) · √
1 − x2
−1
2sin (x)
= √
1 − x2
2sin−1 (x)
Therefore, the derivative of f (x) = sin−1 (x) is f 0 (x) = √
2

1 − x2

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Page 208 Differentiation Rules for Inverse Trigonometric Functions Chapter 3

sin (x)
I Example 3.8.2. Differentiate f (x) =
sin−1 (x)
Solution. Applying the Quotient Rule, we write,

sin (x)

dy d
=
dx dx sin−1 (x)
d d
sin−1 (x) · sin (x) − sin (x) · sin−1 (x)
= dx dx
sin (x)
−1
2

1
sin−1 (x) · cos (x) − sin (x) · √
1 − x2
=
sin−1 (x)
2

sin (x)
cos (x) sin−1 (x) − √
1 − x2
=
sin−1 (x)
2
√
1 − x2 cos (x) sin−1 (x) sin (x)
√ −√
1 − x2 1 − x2
=
sin−1 (x)
2
√
1 − x2 cos (x) sin−1 (x) − sin (x)
√
1 − x2
=
sin−1 (x)
2
√
1 − x2 cos (x) sin−1 (x) − sin (x)
= √
1 − x2 sin−1 (x)
2

√
1 − x2 cos (x) sin−1 (x) − sin (x)
Therefore, f (x) = 0
√ is the derivative of
1 − x2 sin−1 (x)
2
sin (x)
f (x) =
sin−1 (x)

Chap. 3 / Sec. 3.8 / Subsec. 3.8.2 : Differentiation of cos−1 (x)

The derivative of y = cos−1 (x), where −1 ≤ x ≤ 1, is obtained by differentiating
both sides of x = cos(y) implicitly with respect to x. That is,

x = cos (y)
dx d
= cos (y)
dx dx

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Chapter 3 Differentiation Rules for Inverse Trigonometric Functions Page 209

dy
1 = − sin (y) ·
dx
−sin · dy
(y)
1
dx
=
− sin (y) − sin (y)

dy 1
= (2)
dx − sin (y)

Now, from the Pythagorean identity, it is known that sin2 (θ) + cos2 (θ) = 1.
Letting θ = y yields,

sin2 (y) + cos2 (y) = 1

sin2 (y) = 1 − cos2 (y)
sin (y) = 1 − cos2 (y)
p

Now, continuing with equation (2),

d 1
(y) = −
dx sin(y)
1
= −p
1 − cos2 (y)
d 1
cos−1 (x) = − √
dx 1 − x2
1
Therefore, the derivative of f (x) = cos−1 (x) is f 0 (x) = − √ .
1 − x2

Theorem 3.8.2. (The Derivative of the Cosine Function) If f (x) = cos−1 (x),
then
d 1
cos−1 (x) = − √
dx 1 − x2
for −1 < x < 1

A few examples are provided below to demonstrate the theorem above.

I Example 3.8.3. Differentiate f (x) = cos−1 x2 + 1

Solution. Applying the Chain rule,

dy d
cos−1 x2 + 1

=
dx dx
1 d
x2 + 1

= −q ·
dx
1 − (x2 )2

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Page 210 Differentiation Rules for Inverse Trigonometric Functions Chapter 3

1
= −√ · 2x
1 − x4
2x
= −√
1 − x4
2x
Therefore, the derivative of f (x) = cos−1 x2 + 1 is f 0 (x) = − √ .

1 − x4

I Example 3.8.4. Differentiate f (x) = cos−1 (sin(x))

Solution. By the Chain Rule,

dy d
= cos−1 (sin(x))
dx dx
1 d
= −q · sin(x)
2 dx
1 − (sin(x))
1
= −p · cos(x)
1 − (1 − cos2 (x))
cos(x)
= −p
cos2 (x)
cos(x)

=−

cos(x)

= −1

Therefore, the derivative of f (x) = cos−1 (sin(x)) is f 0 (x) = −1.

Chap. 3 / Sec. 3.8 / Subsec. 3.8.3 : Differentiation of tan−1 (x)

The derivative of y = tan−1 (x) is obtained by differentiating both sides of x =
tan(y) implicitly with respect to x. That is,

x = tan (y)
d d
(x) = tan (y)
dx dx
dy
1 = sec2 (y) ·
dx
dy
1 sec
2(y) ·
dx

=
sec2 (y) sec
2(y)

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Chapter 3 Differentiation Rules for Inverse Trigonometric Functions Page 211

dy 1
= (3)
dx sec (y)
2

Now, from the Pythagorean identity, we know that tan2 (θ) + 1 = sec2 (θ).
Letting θ = y yields,
sec2 (y) = 1 + tan2 (y)

Now, continuing with equation (3),

d 1
(y) =
dx sec (y)
2

1
=
1 + tan2 (y)
d 1
tan−1 (x) =
dx 1 + x2
1
Therefore, the derivative of f (x) = tan−1 (x) is f 0 (x) =
1 + x2

Theorem 3.8.3. (The Derivative of the Inverse Tangent Function) If f (x) =

tan−1 (x), then
d 1
tan−1 (x) =
dx 1 + x2
for −∞ < x < ∞

A few examples are provided below to demonstrate the theorem above.

I Example 3.8.5. Differentiate f (x) = x2 + tan−1 (x)

Solution. Applying the elementary rules,

dy d
x2 + tan−1 (x)

=
dx dx
d 2 d
= x + tan−1 (x)
dx dx
1
= 2x +
1 + x2
1
Therefore, the derivative of f (x) = x2 + tan−1 (x) is f 0 (x) = 2x +
1 + x2

I Example 3.8.6. Differentiate f (x) = tan−1 (tan (x))

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Page 212 Differentiation Rules for Inverse Trigonometric Functions Chapter 3

Solution. Applying the Chain rule,

dy d
= tan−1 (tan (x))
dx dx
1 d
= · tan (x)
1 + tan (x) dx
2

1
= · sec2 (x)
1 + tan2 (x)
sec2 (x)
=
1 + tan2 (x)
sec
2 (x)

=

sec
2 (x)

Therefore, the derivative of f (x) = tan−1 (tan (x)) is f 0 (x) = 1

Chap. 3 / Sec. 3.8 / Subsec. 3.8.4 : Differentiation of cot−1 (x)

The derivative of y = cot−1 (x) is obtained by differentiating both sides of x =
cot(y) implicitly with respect to x. That is,

d d
(x) = cot (y)
dx dx
dy
1 = −csc2 (y) ·
dx
dy
2
−csc (y) ·

1
dx
2
= 2
−csc (y) −csc
(y)

dy 1
=− 2 (4)
dx csc (y)

Now, from the Pythagorean identity, we know that 1+cot2 (θ) = csc2 (θ). Letting
θ = y yields,

csc2 (y) = 1 + cot2 (y)

csc2 (y) = 1 + x2

Now, continuing with equation (4),

d 1
(y) = − 2
dx csc (y)

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Chapter 3 Differentiation Rules for Inverse Trigonometric Functions Page 213

d 1
cot−1 (x) = −
dx 1 + x2
1
Therefore, the derivative of f (x) = cot−1 (x) is f 0 (x) = −
1 + x2

Theorem 3.8.4. (The Derivative of the Inverse Cotangent Function) If

f (x) = cot−1 (x), then

d 1
cot−1 (x) = −
dx 1 + x2
for −∞ < x < ∞

A few examples are provided below to demonstrate the theorem above.

I Example 3.8.7. Differentiate f (x) = cot−1 (ex )

Solution. By the Chain Rule,
dy d
= cot−1 (ex )
dx dx
1 d
=− 2 · ex
1 + (ex ) dx
ex
=−
1 + e2x
ex
Therefore, the derivative of f (x) = cot−1 (ex ) is f 0 (x) = − .
1 + e2x

I Example 3.8.8. Differentiate f (x) = cot (x) cot−1 (x)

Solution. Applying the Product Rule,
dy d
cot (x) cot−1 (x)

=
dx dx
d d
= cot (x) · cot−1 (x) + cot−1 (x) · cot (x)
dx dx
1
= cot (x) · − cot −1 2

+ (x) · −csc (x)
1 + x2
cot (x)
=− − cot−1 (x) csc2 (x)
1 + x2
cot (x)
Therefore, f 0 (x) = − − cot−1 (x) csc2 (x) is the derivative of
1 + x2
f (x) = cot (x) cot−1 (x)

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Page 214 Differentiation Rules for Inverse Trigonometric Functions Chapter 3

Chap. 3 / Sec. 3.8 / Subsec. 3.8.5 : Differentiation of sec−1 (x)

The derivative of y = sec−1 (x), is obtained by differentiating both sides of
x = sec(y) implicitly with respect to x, where x ≤ −1 or x ≥ 1. That is,
d d
(x) = sec (y)
dx dx
dy
1 = sec (y) · tan (y) ·
dx
( dy
sec ((tan
((y) (((y) ·
(
1 (
dx
=
sec (y) tan (y) sec (y) tan((
(y)
( (
(( ( (
dy 1
= (5)
dx sec (y) tan (y)
Now, from the Pythagorean identity, we know that 1 + tan2 (θ) = sec2 (θ).
Letting θ = y yields,
1 + tan2 (y) = sec2 (y)
tan2 (y) = sec2 (y) − 1
tan (y) = ± sec2 (y) − 1
p

tan (y) = ± x2 − 1
p

continuing with equation (5),

d 1
sec−1 (x) =
dx sec (y) tan (y)
1
=± √
sec (y) · x2 − 1
1
=± √
x · x2 − 1
Now, in order to determine the sign of the derivative above, we need to consider
two cases. That is,
π
1. If x ≥ 1 in y = sec−1 (x), then 0 ≤ y < and thus tan(y) ≥ 0. In this
√ 2
case, tan (y) = x2 − 1
π
2. If x ≤ 1 in y = sec−1 (x), then ≤ y < π and thus tan(y) ≤ 0. In this
√ 2
case, tan (y) = − x2 − 1
This implies that the derivative of the inverse secant function is,
 1
 √ if x > 1
d  2−1
sec (x) =
−1 x x
1
dx − √
 if x < −1
x x2 − 1

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Chapter 3 Differentiation Rules for Inverse Trigonometric Functions Page 215

The derivative of y = sec−1 (x) above is quite confusing. However, we can simply
get around this confusion by using the absolute value function. That is, if we let
|x| > 1, then we can simplify the derivative of inverse secant as the following.
d 1
sec−1 (x) = √
dx |x| x2 − 1
1
Therefore, the derivative of f (x) = sec−1 (x) is f 0 (x) = √ , where
|x| x2 − 1
|x| > 1.

Theorem 3.8.5. (The Derivative of the Inverse Secant Function) If f (x) =

sec−1 (x), then
d 1
sec−1 (x) = √
dx |x| x2 − 1
for |x| > 1

A few examples are provided below to demonstrate the theorem above.

I Example 3.8.9. Differentiate f (x) = sec(x) + sec−1 (x)

Solution. By the elementary rules,

d d d
sec(x) + sec−1 (x) = sec(x) + sec−1 (x)

dx dx dx
1
= sec(x) tan(x) + √
|x| x2 − 1
Therefore, the derivative of f (x) = sec(x) + sec−1 (x) is f 0 (x) = sec(x) tan(x) +
1
√
|x| x2 − 1

I Example 3.8.10. Differentiate f (x) = sec−1 (xe )

Solution. By the Chain Rule,

d 1 d
sec−1 (xe ) = q · xe
dx dx
|xe | (xe )2 − 1
1
= √ exe−1
|x | x2e − 1
e

exe−1
=q √
(xe )2 x2e − 1

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Page 216 Differentiation Rules for Inverse Trigonometric Functions Chapter 3

e
= √
2e
x x −1
e
Therefore, the derivative of f (x) = sec−1 (xe ) is f 0 (x) = √ .
x x2e − 1

Chap. 3 / Sec. 3.8 / Subsec. 3.8.6 : Differentiation of csc−1 (x)

To differentiate y = csc−1 (x), first notice that if y = csc−1 (x), then x = csc(y),
where x ≤ −1 or x ≥ 1. Therefore,

csc(y) = x
1
=x
sin(y)
1
sin(y) =
x
Now, differentiating both sides of the equation above implicitly, we get,

d d 1
sin(y) =
dx dx x
dy 1
cos(y) · =− 2
dx x
dy 1
=− 2 (1)
dx x cos(y)

Now, from the Pythagorean identity, we know that sin2 (θ)+cos2 (θ) = 1. Letting
1
θ = y and sin(y) = yields,
x
sin2 (y) + cos2 (y) = 1
2
1
+ cos2 (y) = 1
x
1
cos2 (y) = 1 −
x2
2
x −1
= 2
rx
x2 − 1
cos(y) = 2
√ x
x2 − 1
= √
2
√ x
x2 − 1
=
|x|

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Chapter 3 Differentiation Rules for Inverse Trigonometric Functions Page 217

√
x2 − 1
∴ cos(y) =
|x|
√
x2 − 1
Now, substituting cos(y) = into equation (1) yields the following,
|x|
dy 1
=− √
dx x2 − 1
x2 ·
|x|
1
= − 2√
x x2 − 1
|x|
|x|
=− √
x2 x2 − 1
|x| 1
Since = , therefore,
x2 |x|
dy |x|
=− √
dx x2 x2 − 1
1
=− √
|x| x2 − 1
1
Therefore, the derivative of f (x) = csc−1 (x) is f 0 (x) = − √
|x| x2 − 1

Theorem 3.8.6. (The Derivative of the Inverse Cosecant Function) If f (x) =

csc−1 (x), then
d 1
csc−1 (x) = − √
dx |x| x2 − 1
for |x| > 1

A few examples are provided below to demonstrate the theorem above.

I Example 3.8.11. Differentiate f (x) = ex − csc−1 (x)

Solution. By the elementary rules,

d x d x d
e − csc−1 (x) = e − csc−1 (x)

dx dx dx
x 1
=e + √
|x| x2 − 1
1
= ex + √ √
x2 x2 − 1

1
= ex + √
x x2 − 1
1
Therefore, the derivative of f (x) = ex − csc−1 (x) is f 0 (x) = ex + √ .
x x2 − 1

I Example 3.8.12. Differentiate f (x) = csc−1 (csc−1 (x))

Solution. By the Chain Rule,

1
=√ q √ q
x (csc (x)) x − 1 (csc−1 (x))2 − 1
2 −1 2 2

1
= √ q
xcsc (x) x − 1 (csc−1 (x))2 − 1
−1 2

Therefore, the derivative of f (x) = csc−1 (csc−1 (x)) is

1
f 0 (x) = √ q .
−1 2 −1 2
xcsc (x) x − 1 (csc (x)) − 1

The End of Section 3.8

Differentiation Rules for Inverse Trigonometric Function

§ 3.9 Differentiation of General Inverse Functions

Let y = f (x) be a function with an inverse x = f −1 (y) and we want to find the
derivative of x = f −1 (y) with respect to y. In order to do that, consider the
following,

y = f (x)
 

= f f −1 (f (x))
| {z }
x

= f f −1 (f (x))

Now, differentiating both sides of the equation above, by the Chain Rule, with
respect to y; we get,
dy d
= f (f −1 (f (x)))
dy dy
d −1
1 = f 0 (f −1 (f (x))) · f (f (x))
dy
1 d −1
= f (f (x))
f 0 (f −1 (f (x))) dy
Since x = f −1 (f (x)), therefore,
d −1 1
f (f (x)) = 0 −1
dy f (f (f (x)))
dx 1
∴ = 0
dy f (x)
1
=
dy
dx
dx 1 dy
This implies that = , given that 6= 0. Therefore, we have the following
dy dy dx
dx
theorem.

Theorem 3.9.1. (The Derivative of General Inverse Functions) Let y = f (x)

a function that is differentiable on an interval I. If y = f (x) has an inverse
function x = f −1 (x), then the derivative of x = f −1 (x) is,

dx 1
=
dy dy
dx
dy
given that = f 0 (x) 6= 0
dx

A few examples to demonstrate the theorem above are provided below.

I Example 3.9.1. Differentiate f (x) = sin−1 (x)

Solution. Since y = sin−1 (x), therefore x = sin(y). Therefore,

d
sin(y) = cos(y)
dy

q
= 1 − sin2 (y)
p
= 1 − x2

Since,
dx 1
=
dy dy
dx
p 1
1 − x2 =
dy
dx
dy 1
∴ =√
dx 1 − x2
1
Therefore, sin−1 (x) = √ .
1 − x2

I Example 3.9.2. Differentiate f (x) = cos−1 (x)

Solution. Since y = cos−1 (x), therefore x = cos(y). Therefore,

d
cos(y) = − sin(y)
dy
= − 1 − cos2 (y)
p
p
= − 1 − x2

Since,
dx 1
=
dy dy
dx
p 1
− 1 − x2 =
dy
dx
dy 1
∴ = −√
dx 1 − x2
1
Therefore, cos−1 (x) = − √ .
1 − x2

The End of Section 3.9

Differentiation of General Inverse Functions

§ 3.10 Summary
Having covered the derivative of necessary functions, a few of them are sum-
marized for a quick reference. Note that the derivatives summarized below are
absolutely crucial to have on your mind for a smooth understanding of the chap-
ters that follow. However, mindless memorization is strongly discouraged, even
though arguments for being helpful may be made. It is suggested to rather
play 3 with them until they’re unforgettable. Take it as learning to type on a
keyboard. While many of us with decent typing speeds may not have memorized
each individual location of each key, enough mistakes have been made over time
to recall where they are efficiently.

3
for example, ask why the derivative of sin(x) cannot be equal to itself, while the derivative
of e can be; or notice the pattern that the derivative of sin−1 (x) and cos−1 (x), tan−1 (x) and
x

cot−1 (x), sec−1 (x) and csc−1 (x) are almost the same except for a sign in front of them; etc.

d d
1. (c) = 0 12. cos(x) = − sin(x)
dx dx
dx d
2. =1 13. tan(x) = sec2 (x)
dx dx
d n d
3. x = nxn−1 14. cot(x) = − csc2 (x)
dx dx
n+1 d
4.
d x
= xn 15. sec(x) = sec(x) tan(x)
dx n + 1 dx
d
d√ 1 16. csc(x) = − csc(x) cot(x)
5. x= √ dx
dx 2 x
d 1
d x 17. sin−1 (x) = √
6. e = ex dx 1 − x2
dx
d 1
d −x 18. cos−1 (x) = − √
7. e = −e−x dx 1 − x2
dx
d 1
8.
d x
a = ax · ln(a) 19. tan−1 (x) =
dx dx 1 + x2
d 1
9.
d
ln(x) =
d
loge (x) =
1 20. cot−1 (x) = −
dx dx x dx 1 + x2
d 1
10.
d
loga (x) =
1 21. sec−1 (x) = √
dx x ln (a) dx |x| x2 − 1
d 1
11.
d
sin(x) = cos(x) 22. csc−1 (x) = − √
dx dx |x| x2 − 1

Applications of Derivatives

In the previous chapters, we discussed the concept of derivatives and their deriva-
tion. Now, the focus turns to discussing how these ideas can be applied in
practice. The aim is to explore how derivatives can serve as more than just a
computational tool and to reveal themselves as the key to solve and understand
a wide range of mathematical questions.

§ 4.1 Extreme Values of a Function

Extreme values in mathematics refer to the highest and lowest values that a
function can attain, formally termed as extremum, indicating either maximum or
minimum values of a function. In the plural form, extremum becomes extrema,
similarly maximum and minimum become maxima and minima, respectively.
These extreme values are categorized into two types: absolute and relative. An
absolute maximum or minimum is the highest or lowest value that a function
attains globally over a given interval. In contrast, a relative maximum or mini-
mum is the highest or lowest value that a function attains locally within a specific
neighborhood of points in the interval. Since absolute maximum or minimum
values occur globally, therefore absolute maximum or minimum is also said to be
the global maximum or minimum values. Similarly, since relative maximum or
minimum values occur locally, therefore relative maximum or minimum is also
said to be the local maximum or minimum values. With that in mind, we start
our discussion by discussing the absolute extreme values as follows.

223

Chap. 4 / Sec. 4.1 / Subsec. 4.1.1 : Absolute (Global) Extreme Values

It may seem intuitive to state that a function reaches its absolute maximum or
minimum value, but how can this idea be defined precisely? What does it mean
for a function to attain its absolute maximum or minimum values? To define it
precisely, consider that if a function attains its absolute maximum value, then all
other values of that function must be less than or equal to this maximum, as it is
globally the highest value. Similarly, if a function attains its absolute minimum
value, then all other values must be greater than or equal to this minimum, as
it is globally the lowest value. By this idea, we have the following definition of
absolute maximum and minimum values of a function.

Definition 4.1.1. (Absolute Extreme Values) Let f (x) be a function defined

on an interval I and let a point c ∈ I, then

1. If f (c) ≥ f (x) for all x in I, then f (c) is the absolute maximum of

f (x) over I.

2. If f (c) ≤ f (x) for all x in I, then f (c) is the absolute minimum of

f (x) over I.

As suggested by the definition above, the existence of absolute maximum or

minimum values of a function depends both on the continuity of the function
and the choice of its interval. To illustrate why the choice of the interval matters,
consider the continuous function f (x) = x2 . Depending on the choice of interval,
the existence of absolute maximum or minimum values changes. For example,
over the entire domain of (−∞, +∞), the function f (x) = x2 has an absolute
minimum value at the point x = 0, but no absolute maximum value, as shown
below.

f(x)

Figure 4.1

However, the same cannot be said for the interval (0, a]. Over the interval (0, a],
the function f (x) = x2 has an absolute maximum value of f (a) at the point
x = a, but no absolute minimum value, as shown below.

f(x)

x
a

Figure 4.2

Similarly, over the closed interval [0, a], the function f (x) = x2 has both absolute
maximum and minimum values of f (0) and f (a) at the points x = 0 and x = a,
respectively. This is shown below.

f(x)

x
a

Figure 4.3

Likewise, over the open interval (0, a), the function f (x) = x2 neither has an
absolute maximum nor minimum value, as shown below.

f(x)

x
a

Figure 4.4

This suggests that in order for a function to have any absolute maximum or min-
imum values over an interval I, the interval I must be bounded, provided that
the function is continuous over I. As a result, we have the following theorem.

Theorem 4.1.1. (Extreme Value Theorem) Let f (x) be a function satisfying

the following conditions.

1. f (x) is continuous over an interval I.

2. The interval I is closed.

Then, there exists an absolute maximum and minimum value of the func-
tion f (x) over the interval I.

The theorem above is the existence theorem, which provides conditions for abso-
lute maximum or minimum values of a function. However, it does not provide a
method for finding these values. To determine how to find them, it is necessary
to understand the concept of “critical points,” which in turn requires knowledge
of the relative extreme values of a function. As a result, we now discuss the
relative extreme values.

Chap. 4 / Sec. 4.1 / Subsec. 4.1.2 : Relative (Local) Extreme Values

To understand the concept of Relative or Local Extreme (maximum or mini-
mum) values, consider the figure of a continuous function below.

a b c d e

Figure 4.5

At which points does the function attain its absolute maximum and minimum
values? Clearly, the function attains its absolute maximum at x = b and its
absolute minimum at x = e. However, at the points c and d, the function also
shows some characteristics of being a maximum or minimum. Specifically, at
these points, the function is a maximum or minimum relative to the neighboring
points. Thus, at the point d, the function has a relative maximum value. Simi-
larly, at the point c, the function has a relative minimum value.

Note that relative maximum or minimum values of a function occur at inte-

rior points of an interval rather than at its endpoints. Therefore, we have the
following definition of relative extreme values.

Definition 4.1.2. (Relative Extreme Values) Let f (x) be a function defined

on an interval I and let c ∈ I, then,

1. If f (c) ≥ f (x) for all x near c, then f (c) is the relative maximum of
f (x) over I.

2. If f (c) ≤ f (x) for all x near c, then f (c) is the relative minimum of
f (x) over I.

Now that we have discussed the relative extreme values, let’s now move on to
critical points. These points are essential for finding the absolute extreme values
of a function.

Chap. 4 / Sec. 4.1 / Subsec. 4.1.3 : Critical Points

To understand the concept of critical points, consider again the graph of a con-
tinuous function as shown below.

a b c d e

Figure 4.6

At which points does the function attain its relative maximum and minimum
values? As before, it is visible that the function has a relative maximum value
at the points x = b and x = d. Likewise, it has a relative minimum value at
the points x = c. Now, notice how the function only has its relative maximum
and minimum values at points where the derivative is either zero (at the point
x = b) or fails to exist (at the point x = d).

Based on that observation, we have the following theorem.

Theorem 4.1.2. (Relative/Local Extreme Value Theorem)

If f (x) has a relative maximum or minimum value at the point c and is
differentiable at c, then
f 0 (c) = 0

Proof. To prove that f 0 (c) = 0 at a relative extremum, we show that neither

f 0 (c) cannot be solely positive nor be solely negative at a relative extremum, and
thus f 0 (c) must be equal to 0. For that, consider the definition of the derivative
of a function,
f (x) − f (c)
f 0 (c) = lim
x→c x−c
We assume that the limit exists, and therefore the right-side and the left-side
limits also exist at x = c and are equal to f 0 (c). Now, separating these limits
yields the following.
f (x) − f (c) f (x) − f (c)
f 0 (c) = lim+ and f 0 (c) = lim−
x→c x−a x→c x−c
Now, we consider the following cases.

Case 1. f (x) has a relative maximum at the point x = c.

This implies, from the definition of relative maximum, that for x near c, f (c) ≥
f (x) or f (x) − f (c) ≤ 0. Now,

As x → c+ : As x → c+ , then x > c for all x near c. This implies that x − c > 0.

Therefore,
negative or zero
z }| {
f (x) − f (c)
≤0
x − }c
| {z
positive
negative or zero
z }| {
f (x) − f (c)
lim− ≤0
x→c x − }c
| {z
positive

As x → c− : As x → c− , then x < c for all x near c. This implies that x − c < 0.

Therefore,
negative or zero
z }| {
f (x) − f (c)
≥0
x − c
| {z }
negative
negative or zero
z }| {
f (x) − f (c)
lim ≥0
x→c− x − c
| {z }
negative

f (x) − f (c)
This implies that the limit f 0 (c) = lim must be 0.
x→c x−c

Case 2. f (x) has a relative minimum at the point x = c.

This implies, from the definition of relative minimum, that for x near c, f (c) ≤
f (x) or 0 ≤ f (x) − f (c). Now,

As x → c+ : As x → c+ , then x > c for all x near c. This implies that x − c > 0.

Therefore,
positive or zero
z }| {
f (x) − f (c)
≥0
x − c
| {z }
positive
positive or zero
z }| {
f (x) − f (c)
lim ≥0
x→c− x − c
| {z }
positive

As x → c− : As x → c− , then x < c for all x near c. This implies that x − c < 0.

Therefore,
positive or zero
z }| {
f (x) − f (c)
≤0
x − c
| {z }
negative
positive or zero
z }| {
f (x) − f (c)
lim ≤0
x→c− x − c
| {z }
negative

f (x) − f (c)
This again implies that the limit f 0 (c) = lim must be 0. This
x→c x−c
completes the proof.

The relative maximum or minimum values of a function can occur not only at
points where f 0 (c) = 0, but also at points where the derivative does not exist.
Taking these conditions into account, we define such points as “Critical Points.”

Definition 4.1.3. (Critical Point) Let f (x) be a function defined over an

interval [a, b] and a point c ∈ (a, b). Now, c is said to be a Critical Point
of f (x), if either one of the following condition is satisfied:

1. f 0 (c) = 0

2. f 0 (c) = undefined

The definition above implies that if a function has a relative maximum or min-
imum value at a point c, then c must be a Critical Point. However, a Critical
Point at c does not guarantee a relative maximum or minimum value at that
point.

This is because it is possible that f 0 (c) = 0, but there are no relative maximum
or minimum values of the function at that point, as shown below.

f(x)

x
a

Figure 4.7

Similarly, it is possible that f 0 (c) = undefined, yet there are no relative maximum
or minimum values of the function at that point, as shown below.

f(x)

x
a

Figure 4.8

This observation indicates that Critical Points are not necessarily where the rel-
ative maximum or minimum values of a function occur; rather, they are potential
locations—a candidate—where these extrema might be found.

I Example 4.1.1. Find the Critical Points of f (x) = 10x2 − 5x2

Solution. The derivative of f (x) = 10x2 − 5x2 is,

dy d
10x2 − 5x2

=
dx dx
= 20x − 5

Since f 0 (x) = 20x − 5 is a linear function, therefore it is defined for all x ∈ R.

Therefore, to find the Critical Points, we set f 0 (x) = 0, as follows.

f 0 (x) = 0
20x − 5 = 0
5
x=
20
5
Therefore, the only Critical Point of the function f (x) = 10x2 −5x2 is x =
20

I Example 4.1.2. Find the Critical Points of f (x) = x3 − x

Solution. The derivative of f (x) = x3 − x is,

dy d
x3 − x

=
dx dx
= 3x2 − 1

Since f 0 (x) = 3x2 −1 is a quadratic function, therefore it is defined for all x ∈ R.

Therefore, to find the Critical Points, set f 0 (x) = 0, as follows.

f 0 (x) = 0
3x2 − 1 = 0
r
1
x=±
3
r
1
Therefore, the Critical Points of the function f (x) = x − x are x = + 3
and
r 3
1
x=±− , respectively.
3

Having discussed Critical Points and their identification, we now return to the
topic that was left unexplored: locating absolute extreme values.

Chap. 4 / Sec. 4.1 / Subsec. 4.1.4 : Location of Absolute (Global) Extreme Val-
ues
The Extreme Value Theorem guarantees the existence of absolute maximum or
minimum values under certain conditions. However, it does not provide any
method to determine them. Nevertheless, with an understanding of relative
extreme values, the following assumptions can be made.

1. Absolute extreme values can occur at the endpoints of a closed interval.

2. Absolute extreme values can occur at the relative extreme values.

Recall that the relative extreme values occur at the Critical Points. Thus, with
the knowledge of locating such Critical Points, the absolute extreme values can
now be located with almost no effort.

As a result, the assumption above can be turned into a general guideline for
locating absolute extreme values as follows.

Let f (x) be a function defined over a closed interval [a, b]. Then, to locate
the absolute maximum or minimum values of f (x), consider the following
steps:

1. Find the Critical Points of f (x)

2. Evaluate f (x) for each Critical Point found in Step 01.

3. Evaluate f (x) for the endpoints of [a, b]

4. The greatest of the evaluated values is the maximum and the smallest
of the evaluated values is the minimum value of f (x).

A few examples are presented below to demonstrate the guidelines above.

I Example 4.1.3. Find the absolute maximum and minimum values of the func-
tion f (x) = x4 − 2x over [−2, 2]

Solution. In order to find the absolute maximum and minimum values of the
function f (x) = x4 −2x over [−2, 2], the Critical Points must first be determined.
In order to determine the Critical Points, the function f (x) = x4 − 2x needs to
be differentiated as follows.
d
f 0 (x) = x4 − 2x

dx
= 4x3 − 2

Now, the derivative f 0 (x) = 4x3 − 2 is defined for all x ∈ R. Therefore, to find
the Critical Points, let f 0 (x) = 0 as follows,

f 0 (x) = 0
4x3 − 2 = 0
r
3 1
x=
2
r
1
Therefore, the Critical Point of the function f (x) = x4 − 2x is x = 3
.
2

This Critical Point, along with the endpoints [−2, 2] are the possible points
where the function has its absolute maximum and minimum values. Therefore,
solving the function at those points yields,

Left-endpoint: f (−2) = (−2)4 − 2 (−2) = 20

r ! r !4 r !
3 1 3 1 1
Critical Point: f = −2 3 ≈ −1.191
2 2 2

Right-endpoint: f (2) = (2)4 − 2 (2) = 12

Comparing these values, it is evident that −1.191 < 12 < 20. Thus, the func-
tion attains its maximum value at x = 20 and minimum value at x ≈ −1.191.

I Example 4.1.4. Find the absolute maximum and minimum values of the func-
tion f (x) = 2x 3 − 4x over [−1, 1]
2

Solution. First, differentiate f (x) to find the Critical Points, as follows.

0 d 2
f (x) = 2x − 4x
3
dx
2 1
= 2 · x− 3 − 4
3
4 1
= · 1 −4
3 x3
4
= 1 − 4
3x 3 !
1
1 − 3x 3
=4 1
3x 3

Now, when x = 0.037, the function f (x) has a Critical Point. Because at
x = 0.037 the derivative of f (x) yields f 0 (0.037) = 0. Similarly, when x = 0,
the function f (x) has a Critical Point. Because at x = 0, the derivative of f (x)
yields f 0 (0) = division by zero = undefined. Therefore, the Critical Points of
the function f (x) = 2x 3 − 4x are x = 0, 0.037.
2

The Critical Points along with the endpoints [−1, 1] are the possible points for
maximum and minimum values. Therefore, solving the function at those points
yields,

Left-endpoint: f (−1) = 2(−1) 3 − 4 (−1) = 6

Critical Point 1: f (0) = 2(0) 3 − 4 (0) = 0

Critical Point 2: f (0.037) = 2(0.037) 3 − 4 (0.037) = 0.074

Right-endpoint: f (1) = 2(1) 3 − 4 (1) = −2

By comparing these values, it is evident that −2 < 0 < 0.074 < 6. There-
fore, the function attains its maximum value at x = −1 and minimum value at
x = 1.

The End of Section 4.1

Extreme Values of a Function

§ 4.2 Increasing and Decreasing Functions

It is intuitively clear from the graph of a function whether it is increasing,
decreasing, or constant. For example, consider the graph of a function below.

Increasing Constant Decreasing

a b c d

Figure 4.9

In the figure above, the function is increasing over the interval [a, b], constant
over [b, c], and then decreasing over [c, d]. That is, when x increases over the
interval [a, b], the function moves up. Over the interval [b, c], the function re-
mains the same. And lastly, it moves down, as x increases over the interval [c, d].

This intuitive understanding, however, is not mathematically sufficient for a

number of reasons and can be made mathematically rigorous as discussed below.

Chap. 4 / Sec. 4.2 / Subsec. 4.2.1 : Increasing and Decreasing Functions

To rigorously define an increasing and decreasing function, consider any two
points x1 , x2 ∈ [a, b] where x1 < x2 . If a function is increasing over an interval,

then f (x1 ) < f (x2 ), as shown in Figure 4.10(a). In contrast, if a function is

decreasing over an interval, then f (x1 ) > f (x2 ), as shown in Figure 4.10(b).
And, if a function is constant over an interval, then f (x1 ) = f (x2 ), as shown in
Figure 4.10(c).

(a) (b) (c)

Figure 4.10

Summarizing the discussion, we have the following definition of an increasing,

decreasing, and constant function.

Definition 4.2.1. (Increasing, Decreasing, and Constant Functions)

Let f (x) be a function defined over an interval I and let two points x1 , x2 ∈
I. Then

1. If f (x1 ) < f (x2 ) for x1 < x2 , then f (x) is increasing over I.

2. If f (x1 ) > f (x2 ) for x1 < x2 , then f (x) is decreasing over I.

3. If f (x1 ) = f (x2 ) for x1 < x2 , then f (x) is constant over I.

Since the precise definitions of increasing, decreasing, and constant functions

have been established, we now shift our attention to identify intervals where the
function shows such behaviors. This identification is done through the use of
the derivative, as it gives the slopes of tangent lines. When the derivative is
positive on an interval, the tangent lines on that interval have positive slopes.
Therefore, the function is increasing on that interval, as shown below.

Figure 4.11

Similarly, when the derivative is negative on an interval, the tangent lines on

that interval have negative slopes. Therefore, the function is decreasing on that
interval, as shown below.

Figure 4.12

Likewise, when the derivative is zero, the tangent line on that interval represents
a horizontal line, suggesting the function is constant on that interval.

With this intuition, we have the following theorem to identify intervals on which
a function is increasing, decreasing, or constant.

Theorem 4.2.1. (Increasing and Decreasing Test)

Let f (x) be a continuous function over the interval [a, b] and differentiable
over the interval (a, b). Then,

1. If f 0 (x) > 0 for all x ∈ (a, b), then f (x) is increasing over [a, b].

2. If f 0 (x) < 0 for all x ∈ (a, b), then f (x) is decreasing over [a, b].

3. If f 0 (x) = 0 for all x ∈ (a, b), then f (x) is constant over [a, b].

10
I Example 4.2.1. Find the intervals on which the function f (x) = x3 − 10x is
3
increasing and decreasing.

Solution. It is known from the theorem above that a function f (x) is increasing
if its derivative f 0 (x) is greater than 0, and decreasing if its derivative f 0 (x)
is less than 0. Therefore, the first task would be to differentiate the function
10
f (x) = x3 − 10x, as follows.
3

0 d 10 3
f (x) = x − 10x
dx 3

d 10 3 d
= x − (10x)
dx 3 dx
10
= ·3x2 − 10
3

= 10x2 − 10
= 10 x2 − 1

Now, since f (x) is increasing if f 0 (x) = 10 x2 − 1 > 0 and decreasing if f 0 (x) =

10 x2 − 1 < 0, therefore, we first need to evaluate f 0 (x) = 0 and the result

would be the Critical Points of f (x).

f 0 (x) = 0
10 x2 − 1 = 0

x2 − 1 = 0
x2 = 1
√
x=± 1
x = ±1

The Critical Points are x = 1 and x = −1. The Critical Points divide the real
line into the following intervals:

(−∞, −1) (−1, 1) (1, ∞)

Now, for the first interval, choose a value of x such that it lies within that
interval. Say, x = −2 ∈ (−∞, −1) is chosen as a test value. Now, evaluate
f 0 (x) = 10 x2 − 1 for x = −2. If the result yields a positive value, then f (x)

is increasing within that interval. So,

f 0 (−2) = 10 (−2)2 − 1
= 30 > 0

Since f 0 (x) > 0 over the interval (−∞, −1), therefore f (x) is increasing over
(−∞, −1).

Similarly, let x = 0 ∈ (−1, 1) as a test value for the second interval. Then,

f 0 (0) = 10 02 − 1

= −10 < 0

Since f 0 (x) < 0 over the interval (−1, 1), therefore f (x) is decreasing over (−1, 1).

Likewise, let x = 2 ∈ (1, ∞) as a test value for the third interval. Then,

f 0 (2) = 10 22 − 1

= 30 > 0

Since f 0 (x) > 0 over the interval (1, ∞), therefore f (x) is increasing over (1, ∞).

In conclusion, the function f (x) is increasing on (−∞, −1) and (1, ∞), and de-
creasing on (−1, 1).

As previously discussed, if a function f (x) has a relative (local) maximum or

minimum value at c, then c must be a Critical Point of f (x). However, not all
Critical Points correspond to relative maximum or minimum values. As a result,
it is necessary to have a method—referred to as the “First Derivative Test”—to
determine whether a function f (x) has a relative maximum or minimum at a
given Critical Point. In order to do so, we need to have a firm understanding
of the concept of increasing/decreasing functions as already discussed above.
Therefore, let us now begin our discussion of the First Derivative Test as follows.

Chap. 4 / Sec. 4.2 / Subsec. 4.2.2 : The First Derivative Test

With an understanding of the concepts of increasing, decreasing, and constant
functions, as well as the methods to determine such, it is now possible to apply
this knowledge to identify local extreme values of a function. In order to do
that, consider the following graph of a function,

f(x)

f(c)

x
a c b

Figure 4.13

In the graph above, the function f (x) has a Critical Point at x = c, as f 0 (c) = 0.
Now, observe that f (x) is increasing over the interval (a, c) and decreasing over
the interval (c, b). Which suggests that f 0 (x) > 0 over the interval (a, c) and
f 0 (x) < 0 over (c, b). The derivative f 0 (x) changes its sign from positive to
negative at the Critical Point x = c, where f 0 (c) = 0. As a result, the function
f 0 (x) has a relative maximum at the Critical Point x = c.

Similarly, for the relative minimum values, consider the following graph of a
function.

f(x)

f(c)

x
a c b

Figure 4.14

Notice that f (x) is decreasing over the interval (a, c) and increasing over the
interval (c, b). This suggests that f 0 (x) < 0 over the interval (a, c) and f 0 (x) > 0
over (c, b). The derivative f 0 (x) changes its sign from negative to positive at
the Critical Point x = c, where f 0 (c) = 0. As a result, the function f 0 (x) has a
relative minimum at the Critical Point x = c.

From the observation above, it can be noted that when a function f (x) has a
relative maximum or minimum value at a point, the derivative f 0 (x) changes its
sign at that point. This idea is crucial for identifying the relative maximum or
minimum values of a function.

Recall that Critical Points are points where a function might have a relative
maximum or minimum, but is not guaranteed. The derivative f 0 (x) does not
always change sign at these points. Therefore, examining whether the derivative
changes sign at Critical Points can serve as a test to identify if they correspond
to any relative maximum or minimum values of the function. Thus, we have the
following theorem to identify relative extreme values of a function.

Theorem 4.2.2. (The First Derivative Test)

Let f (x) be a continuous function over an interval I containing a Critical
Point x = c. If f (x) is differentiable over that interval (except possibly at
point c), then

1. If the sign of f 0 (x) changes from positive (−) to negative (+) at c,

then f (c) is a relative maximum.

2. If the sign of f 0 (x) changes from negative (+) to positive (−) at c,

then f (c) is a relative minimum

3. If the sign of f 0 (x) remains the same from both sides of c, then f (c)
is neither a relative maximum nor a relative minimum.

A few examples to demonstrate the theorem above are shown below.

I Example 4.2.2. Identify the relative maximum and minimum values of the
10
function f (x) = x3 − 10x.
3
Solution. To identify the relative extreme values of the function, we need to know
the intervals on which the function is increasing and decreasing. From Exam-
10 3
ple 4.2.1, it was known that the function f (x) = x − 10x is increasing on
3
(−∞, −1), decreasing on (−1, 1), and then again increasing on (1, ∞).

In Example 4.2.1, it has been observed through the test points that f 0 (x) changes
its sign from positive to negative as x goes through the Critical Point x = −1.
Thus, f (x) has a relative maximum value at x = −1. Similarly, since f 0 (x)
changes its sign from negative to positive as x goes through the Critical Point
x = 1, therefore f (x) has a relative minimum value at x = 1.

I Example 4.2.3. Identify the relative maximum and minimum values of the
x2 + 4x x2 + 4x + 6
function f (x) =
4
Solution. To identify the relative extreme values of the function, the first step is
to find its derivative to locate the Critical Points. Then, by analyzing whether
the derivative changes sign at these Critical Points, the relative extreme values
can be determined. Thus, the derivative of f (x) is as follows,
2
2 !
d x + 4x x + 4x + 6
f 0 (x) =
dx 4

1 2
d 2
2
d 2

= · x + 4x · x + 4x + 6 + x + 4x + 6 · x + 4x
4 dx dx
1
= · x2 + 4x · (2x + 4) + x2 + 4x + 6 · (2x + 4)

4

1
· 2x3 + 12x2 + 16x + 2x3 + 12x2 + 28x + 24

=
4
1
= · 4x3 + 24x2 + 44x + 24

4
= x3 + 6x2 + 11x + 6
= (x + 1) (x + 2) (x + 3)

Now, since f (x) is continuous at every point, therefore the f 0 (x) = 0 are the
only Critical Points. Letting f 0 (x) = 0 yields,

(x + 1) (x + 2) (x + 3) = 0
x = −1, −2, −3

These Critical Points divide the real number line into the following intervals.

(−∞, −3) (−3, −2) (−2, −1) (−1, ∞)

Now, based on these intervals, we analyze the function to check if it changes

its sign and then conclude whether it is increasing or decreasing to identify the
relative extreme values, as follows.

Intervals (−∞, −3) (−3, −2) (−2, −1) (−1, ∞)

Test x = −4 x = −2.5 x = −1.5 x=0
Values
Sign f 0 (−4) = f 0 (−2.5) ≈ f 0 (−1.5) ≈ f 0 (0) =
of f 0 (x) −6 < 0 86.6 > 0 −0.37 < 0 6>0
Conclu- Decreasing Increasing Decreasing Increasing
sion

Now, by the first derivative test, it is concludable that the function f (x) has a
relative minimum at x = −3, −1, as the derivative changed its sign from negative
to positive and relative maximum at x = −1, as the derivative changed its sign
from negative to positive.

The End of Section 4.2

Increasing and Decreasing Functions

§ 4.3 Concavity and the Second Derivative Test

When examining the graph of a function that forms a curve, we might need to
ask whether the curve bends upwards or downwards. It might seem like the
question is simply about whether the curve is increasing or decreasing, but it
goes beyond that. For example, consider the figures provided below, where both
functions are increasing over the interval (a, b), but their shape of increase is
different.

Increasing

Concave Down
Increasing

Concave Up

a b a b

Figure 4.15 Figure 4.16

The graph of the function on the left is increasing with an increasing rate,
whereas the graph on the right is increasing with a decreasing rate. This is the
core idea behind what is referred to as the concavity of a function.

The concavity of a function is the direction in which the function curves. It can
curve upwards or downwards. For example, the graph in the left figure shown
above curves upwards, suggesting that the tangent lines to the curve get steeper
and steeper as x increases. This implies that the first derivative of the function
is increasing over that interval. When a function’s first derivative increases on
an interval, the function is said to be concave up within that interval.

Similarly, the graph in the right figure shown above curves downwards, suggest-
ing that the tangent lines to the curve get steeper and steeper as x increases.
This implies that the first derivative of the function is decreasing over that in-
terval. When a function’s first derivative decreases on an interval, the function
is said to be concave down within that interval.

Summarizing the facts above, we have the following definition of concavity.

Definition 4.3.1. (Concavity) Let f (x) be a differentiable function over an

interval I. Then

1. f (x) is concave up over I, if f 0 (x) is increasing over I.

2. f (x) is concave down over I, if f 0 (x) is decreasing over I.

Intuitively, if a function is concave up at a specific point, the graph of the func-

tion near that point is above the tangent line drawn at that point. In contrast,
if a function is concave down at a specific point, the graph of the function near
that point is below the tangent line drawn at that point.

It is important to clearly distinguish between the concepts of increasing/decreas-

ing functions and the concept of concavity, and to do that, refer to the following
figure.

a b a b
(a) (b)

a b a b
(c) (d)

Figure 4.17

In Figure 4.17(a), the function f (x) is increasing and concave up because its
first derivative, f 0 (x), is also increasing. In contrast, in Figure 4.17(b), although
the function f (x) is still increasing, but it is concave down, as the derivative
f 0 (x) is decreasing.

Similarly, in Figure 4.17(c), the function f (x) is decreasing and concave up since
f 0 (x) is increasing. In contrast, in Figure 4.17(d), the function f (x) is decreasing
and concave down because f 0 (x) is decreasing.

Now, to determine if a function is concave up or down over an interval, simply

analyze whether its first derivative f 0 (x) is increasing or decreasing over that
interval. It is known from the previous section that if f 0 (x) > 0 over an inter-
val, then f (x) is increasing over that interval. Similarly, if f 0 (x) < 0 over an
interval, then f (x) is decreasing over that interval. To determine whether f 0 (x)
itself is increasing or decreasing, we need to compute f 00 (x) and then apply the
same theorem accordingly. That is, if f 00 (x) > 0, then f 0 (x) is increasing over
an interval, and thus f (x) is concave up. In contrast, if f 00 (x) < 0, then f 0 (x) is
decreasing over an interval, and thus f (x) is concave down.

Based on the discussion above, we have the following test for the concavity of a
function as follows.

Theorem 4.3.1. (Concavity Test) Let f (x) be a function such that its second
derivative exists over an interval I. Then,

1. If f 00 (x) > 0, for all x ∈ I, then f (x) is concave up over I.

2. If f 00 (x) < 0, for all x ∈ I, then f (x) is concave down over I.

Having an understanding of the concavity of a function and a test to determine

such, we now consider a point at which the concavity of the function changes
from upwards to downwards or vice versa. Such points are referred to as the
“Point of Inflection,” which is discussed below.

Chap. 4 / Sec. 4.3 / Subsec. 4.3.1 : The Point of Inflection

To understand the Inflection Points consider the graph of the function f (x) = x3 ,
as shown below.

Figure 4.18

It is evident that the function f (x) = x3 increases in an upward shape over

the interval (−∞, 0) and a downward shape over the interval (0, ∞). Thus, the
function changes from being concave down to concave up as x passes through
the point x = 0. It is the point where a change in the concavity of the graph of
the function occurred. Such a point is called an inflection point.

Inflection points are such points where a curve changes its concavity. That is, at
those points, the second derivative changes f 00 (x) of the function f (x) changes
its sign. As a result, we have the following definition.

Definition 4.3.2. (Inflection Points) If a function f (x) is continuous at point

x = a and shows a change in concavity at that point, then the point
(a, f (a)) is a point of inflection of f (x).

Recall that to find the relative extreme values of a function, the first step was to
identify its Critical Points. Critical Points are points where the derivative f 0 (x)
equals zero or is undefined. These points act as potential candidates for relative
extreme values.

Similarly, to identify potential points of inflection, determine the values of x

where the second derivative f 00 (x) equals zero or does not exist. The following
theorem summarizes this process.

Theorem 4.3.2. (Inflection Points Test)

If c is a point of inflection on the graph of f (x), then either f 00 (c) = 0 or
f 00 (c) = does not exist.

The following example demonstrates the theorem above.

I Example 4.3.1. Identify the point of inflection of the function f (x) = sin−1 (x).
Solution. To identify the point of inflection of the function f (x) = sin−1 (x),
the first step is to compute the second derivative f 00 (x) of the function f (x).
Therefore,
d
f 0 (x) = sin−1 (x)
dx
1
=√
1 − x2
d 1
f 00 (x) = √
dx 1 − x2
d 1
=
dx (1 − x2 ) 12
d − 1
= 1 − x2 2
dx
1 − 3 d
= − 1 − x2 2 · 1 − x2

2 dx
1 − 3
= − 1 − x2 2 · (−2x)

2
1 − 2x
=−
2 (1 − x2 ) 32

x
= 3
(1 − x2 ) 2
Now, to find the inflection points, let f 00 (x) = 0. That is,
f 00 (x) = 0
x
3 = 0
(1 − x2 ) 2
32
x = 0 · 1 − x2
x=0
This suggests that f 00 (x) changes its sign from negative to positive at x = 0.
x
This is because denominator of f 00 (x) = 3 is positive on the entire inter-
(1 − x2 ) 2
val (−1, 1), however, the numerator is negative on (−1, 0) and positive on (0, 1),

x
making f 00 (x) = 3 negative on (−1, 0) and positive on (0, 1). Therefore,
(1 − x2 ) 2
f (x) is concave down on (−1, 0) and concave up on (0, 1), making x = 0 is a
inflection point of f (x).

Chap. 4 / Sec. 4.3 / Subsec. 4.3.2 : The Second Derivative Test

Previously, it was discussed how the second derivative can be used to determine
whether a function is curving upwards or downwards. Additionally, the second
derivative can also be used as a test to identify relative maximum and minimum
value of a function.

Theorem 4.3.3. (The Second Derivative Test)

Suppose a function f (x) such that f 00 (x) exists over an interval containing
c. Then,

1. If f 0 (c) = 0 and f 00 (c) < 0, then f (x) has a relative maximum at c.

2. If f 0 (c) = 0 and f 00 (c) > 0, then f (x) has a relative minimum at c.

3. If f 0 (c) = 0 and f 00 (c) = 0, then f (x) may have a relative maximum,

minimum, or neither at c.

The End of Section 4.3

Concavity and the Second Derivative Test

§ 4.4 Rolle’s Theorem and The Mean Value Theorem

The Mean Value is serves as a crucial theorem of calculus. Numerous theorem
and application of calculus rely on this theorem. In this section, we discuss the
Mean Value Theorem by first discussing a special case of it: Rolle’s Theorem.

Chap. 4 / Sec. 4.4 / Subsec. 4.4.1 : Rolle's Theorem

Rolle’s Theorem is a fundamental result in differential calculus and is really
common sense when seen visually. For example, consider a function f (x) such
that it is continuous over a closed interval [a, b] and differentiable over an open

interval (a, b) with f (a) = f (b). Then there exists a point c between a and b
such that the derivative at c is 0. That is, the slope of the tangent line at that
point is 0, as shown below.

f(x)

f(c)

x
a c b

Figure 4.19

The basic idea is that if a smooth curve starts and ends at the same height (that
is, if f (a) = f (b)), it must have at least one point where the slope of the curve
flattens out (i.e., the derivative becomes zero). This occurs because, intuitively,
the function has to “turn around” somewhere between a and b to return to the
same value at both endpoints. Therefore, we have the following theorem.

Theorem 4.4.1. (Rolle’s Theorem) Let f (x) be a function meeting the fol-
lowing conditions,

1. f (x) is continuous over [a, b]

2. f (x) is differentiable over (a, b)

3. f (a) = f (b)

then there must exist at least one point c ∈ (a, b) such that f 0 (c) = 0.

Proof. Let f (a) = f (b) = k and consider the following cases,

1. Case 1: f (x) = k, for all x ∈ (a, b)

2. Case 2: f (x) > k, for some x ∈ (a, b)

3. Case 3: f (x) < k, for some x ∈ (a, b)

Case 1: Suppose, f (x) = k, for all x ∈ (a, b), then f 0 (x) = 0 for all x ∈ (a, b), as
f (x) is a constant function.

Case 2: Suppose, f (x) > k, for some x ∈ (a, b). Since f (x) is continuous over
[a, b], therefore, by the Extreme Value Theorem, it has an absolute maximum
over [a, b]. Given that f (a) = f (b), the absolute maximum cannot occur at the
endpoints a or b. Thus, the absolute maximum must occur at some interior
point c in (a, b).
Since all absolute maximums are also relative maximums, therefore f (x) must
have a relative maximum at some point c ∈ (a, b). Now, according to the Rela-
tive/Local Extreme Value Theorem, if f (x) has a relative maximum at c and is
differentiable at c, then f 0 (c) = 0.

Case 3: Suppose, f (x) < k, for some x ∈ (a, b). Since f (x) is continuous over
[a, b], therefore, by the Extreme Value Theorem, it has an absolute minimum
over [a, b]. Given that f (a) = f (b), the absolute minimum cannot occur at the
endpoints a or b. Thus, the absolute maximum must occur at some interior
point c in (a, b).
Since all absolute minimums are also relative minimums, therefore f (x) must
have a relative minimum at some point c ∈ (a, b). Now, according to the Rela-
tive/Local Extreme Value Theorem, if f (x) has a relative minimum at c and is
differentiable at c, then f 0 (c) = 0.
The theorem above can be applied in certain situations very cleverly. For in-
stance, consider the following example.

I Example 4.4.1. Show that f (x) = x3 + x + 1 has exactly one real solution.
Solution. Notice that the value of the function f (x) = x3 + x + 1 is −5 and 1
for x = −2 and x = 0, respectively. Therefore, it follows from the Intermediate
Value Theorem (IVT) that there exists a point k between x = −2 and x = 0
such that f (k) = 0.

Now, besides of that single point k, if there were two points x = a and x = b
for which f (a) = f (b) = 0, then the Rolle’s theorem would guarantee that there
exists a point x = c between the points x = a and x = b for which f 0 (c) = 0.
However, this can never happen as the derivative of f (x) = x3 + x + 1 is,
d d
x3 + x + 1

f (x) =
dx dx

= 3x2 + 1 > 0

Which is always greater than 0 for all x. Therefore, Rolle’s theorem cannot be
met and there doesn’t exist any point other than one single x = k, for which
f (k) = 0.

Chap. 4 / Sec. 4.4 / Subsec. 4.4.2 : The Mean Value Theorem (MVT)
The Mean Value Theorem can best be understood visually. For example, con-
sider the figure below.

f(x)

f(c)
f(b)

f(a)

x
a c b

Figure 4.20

In the figure above, the slope of the secant line that passes through the point
(a, f (a)) and (b, f (b)) is
f (b) − f (a)
b−a
Now, the Mean Value Theorem essentially states that there exists at least one
point c ∈ (a, b) such that the slope of the tangent line at c is equal to the slope
of the secant line that passes through the points (a, f (a)) and (b, f (b)). That is,

f (b) − f (a)
f 0 (c) =
b−a
This implies that there is a point c where the tangent line at that point c
is parallel to the secant line that passes through the (a, f (a)) and (b, f (b)).
Therefore, we have the following theorem.

Theorem 4.4.2. (The Mean Value Theorem)

Let f (x) be a function meeting the following conditions,

1. f (x) is continuous over [a, b]

2. f (x) is differentiable over (a, b)

then there exists a point c ∈ (a, b) such that

f (b) − f (a)
f 0 (c) =
b−a
or, equivalently,
f (b) − f (b) = f 0 (c)(b − a)

Proof. Consider a straight line that goes through the points (a, f (a)) and (b, f (b)).
The slope of this line is,
f (b) − f (a)
m=
b−a
f (b) − f (a)
Since, the line goes through the point (a, f (a)) and has a slope m = ,
b−a
therefore the point-slope equation of the line yields,

y − f (a) = m(x − a)

or,
f (b) − f (a)
(x − a) + f (a)
y=
b−a
Now, let a new function g(x) be the difference between the function f (x) and
the line y as follows,

g(x) = f (x) − y

f (b) − f (a)
= f (x) − (x − a) + f (a)
b−a
f (b) − f (a)
= f (x) − (x − a) − f (a)
b−a
Now, if x = a, then g(x) yields,

f (b) − f (a)
g(x) = f (x) − (x − a) − f (a)
b−a
f (b) − f (a) :0
= f (a) − (a−a) − f (a)

b−a

= f (a) − f (a)

If x = b, then g(x) yields,

f (b) − f (a)
g(x) = f (x) − (x − a) − f (a)
b−a
f (b) − f (a)
= f (b) − (b−a)

− f (a)
b− a

= f (b) − (f (b) − f (a)) − f (a)
= f (b) − f (b) + f (a) − f (a)
=0

This implies that g(a) = 0 = g(b). Now, since f (x) and the straight line y
are a differentiable function over (a, b), therefore g(x) is also a differentiable
function over (a, b). Furthermore, since f (x) and y are a continuous function
over [a, b], therefore g(x) is also a continuous function over [a, b]. Thus, g(x)
satisfies Rolle’s Theorem implying that there exists a point c ∈ (a, b) for which
g 0 (c) = 0. Therefore, let us differentiate the function g(x),
d
g 0 (x) = g (x)
dx
d f (b) − f (a)
= f (x) − (x − a) − f (a)
dx b−a

d d f (b) − f (a) d
= f (x) − (x − a) − f (a)
dx dx b−a dx
f (b) − f (a) d
= f 0 (x) − · (x − a)
b−a dx
f (b) − f (a)
= f 0 (x) − ·1
b−a
f (b) − f (a)
∴ g 0 (x) = f 0 (x) −
b−a
Since, g(x) satisfies Rolle’s Theorem suggesting that there exists a point c ∈
(a, b) for which g 0 (c) = 0, therefore we let,

g 0 (c) = 0
f (b) − f (a)
f 0 (c) − =0
b−a
f (b) − f (a)
∴ f 0 (c) =
b−a
This completes the proof.
A few examples to demonstrate the theorem above are provided below.

I Example 4.4.2. Test the Mean Value Theorem for f (x) = x2 on the interval
[1, 4].

Solution. The function f (x) = x2 is a polynomial, therefore it is continuous and

differentiable everywhere.

Now, the slope of the secant line that passes through the points x = 1 and x = 4
is,
f (4) − f (1) 16 − 1
=
4−1 4−1
15
=
3
=5

Now, to test the Mean Value Theorem, we need to find a point c such that
f 0 (c) = 5. Therefore, we differentiate f (x) = x2 as follows.
d 2
x = 2x
dx
Setting f 0 (c) = 5, we get,

2c = 5
5
∴c=
2
= 2.5

Thus, at x = 2.5, the function f (x) = x2 has a tangent line where its slope
matches the secant line that passes through the point x = 1 and x = 4.

I Example 4.4.3. Test the Mean Value Theorem for f (x) = sin(x) on the interval
[0, π].

Solution. The function f (x) = sin(x) is both continuous and differentiable on

the interval [0, π].

The slope of the secant line passing through the points x = 0 and x = π is,
sin(π) − sin(0) 0−0
=
π−0 π

Now, to test the Mean Value Theorem, we need to find a point c such that
f 0 (c) = 0. Therefore,
d
sin(x) = cos(x)
dx
Settings f 0 (c) = 0, we get,

cos(c) = 0
∴ c = cos−1 (0)
π
=
2
π
Thus, at x = , the function f (x) = sin(x) has a tangent line where its slope
2
matches the secant line that passes through the points x = 0 and x = π.

The End of Section 4.4

Rolle's Theorem and The Mean Value Theorem

§ 4.5 Related Rates

In Chapter 3, Section 3.3, the Chain Rule was introduced. In this section, we
learn to apply the Chain Rule in a way that shows how the rate of change of
multiple variables relates to each other as they change over time. This is rooted
in the understanding that if two or more related variables are changing over time,
then the rate at which they are changing is also related. For example, suppose
the variables x and y are both changing over time. The relation between x and
y is given by the equation y = 2x3 . Now, given that the rate of change of x with
respect to time t is n, when x = a, is it possible to find the rate of change of y
with respect to t, when x = a? The answer is positive, because we know that

y = 2x3
d d
2x3

(y) =
dt dt
dy dx
= 6x2 ·
dt dt
Now, since the rate of change of x with respect to t, when x = a, is n; therefore
dx
= n. Thus, the rate of change of y with respect to t, when x = a, is,
dt
dy
= 6a2 n
dt

So, the rate of change of y with respect to t, when x = a, is 6a2 n. It is interesting

to note how we have figured out the rate of change of one variable in terms of
the other related variable, which was known to us.

In the real world, there are numerous examples of related variables that changes
over time. For example, consider inflating a balloon. As the balloon gets inflated
with respect to time, it becomes evident that its size expands, indicating an
increase in its volume V . Now, assuming the shape of the inflated balloon is
4
spherical and knowing the formula for the volume of a sphere, that is, V = πr3 ,
3
it can be said that as the volume increases over time, so does the radius r. This
dV
implies that the rate of change in the volume, that is is related to the rate
dt
dr 4
of change in the radius since V is related to r by the equation V = πr3 .
dt 3
We shall use these relations to solve the following problem.

I Example 4.5.1. A circle’s radius is increasing at a constant rate of 5 cm per

second. How fast is the area of the circle increasing when the radius is 100 cm?

Solution. Let r be the radius of the circle and A be the area of the circle at a
dr cm dA
time t, then we’re given =2 . We need to find , when r = 100 m.
dt s dt

In order to do that notice first that the area of the circle A and the radius r of
the circle is related by the following equation.

A = πr2

Now, differentiating both sides of the equation above yields,

dA d
πr2

=
dt dt
d
= π r2
dt
dr
= π · 2r
dt
dr
= 2rπ
dt
dr cm
Now, substituting the given =2 and r = 100m into the equation above
dt s
yields,
dr
2rπ = 2π (100) (2)
dt

= 2π (100) (2)
cm2
≈ 1256.64
s
cm2
Therefore, the area of the circle is increasing at a rate of approximately 1256.64 ,
s
when the radius is 100 cm.

I Example 4.5.2. A cylindrical tank with a radius of 2 meters is draining water

at a rate of 3 cubic meters per minute. How fast is the water level dropping
when the height of the water is 5 meters?

Solution. Let h be the height of the water, and V the volume of water in the
dV m3
tank. We are given = −3 (“−3” is due to the reason that the volume is
dt s
dh
decreasing) and asked to find when h = 5m.
dt

In order to do that notice that the volume V and the height h are related by
the following equation.
V = πr2 h
Since the radius r = 2m is constant, therefore, the equation above is simplified
as follows.

V = πr2 h
= π22 h
= 4πh

Now, differentiating both sides of the equation with respect to time t above
yields,
dV d
= (4πh)
dt dt
dh
= 4π
dt
dV m3
Since we are given = −3 , therefore,
dt s
dV dh
= 4π
dt dt
dh
−3 = 4π
dt

dh −3
=
dt 4π
m
≈ −0.24
s
m
Therefore, the water level is dropping at the rate of approximately 0.24 .
s

I Example 4.5.3. Sand falls from a conveyor belt at a rate of 1.5 cubic meters
per minute. This forms a conical pile of sand. The radius of the base of this
conical pile is always half the height. How fast is the height of the pile increasing
when the height is 3 meters?

Solution. Let V be the volume of the cone and h be the height of the cone. We
dV m3 h dh
are given = 1.5 and r = , and asked to find when h = 3m.
dt min 2 dt

In order to do that notice that the volume V and the height h are related by
the volume of a right circular cone formula. That is,
1
V = πr2 h
3
h
Since the radius of the pile is always half the height, that is, r = , therefore,
2
1
V = πr2 h
3
2
1 h
= π h
3 2
1
= πh3
12
Now, differentiating both sides of the equation above with respect to time t
yields,

dV d 1 3
= πh
dt dt 12
1 d
= π h3
12 dt
1 dh
= π · 3h2
12
dt
1 2 dh
= πh
4 dt

dV m3
Since = 1.5 and h = 3m,
dt min
dV π dh
= h2
dt 4 dt
π 2 dh
1.5 = (3)
4 dt
dh 6
=
dt 9π
2 m
=
3π min
2 m
Therefore, The height of the sand pile is increasing at a rate of or ap-
3π min
m
proximately 0.212 .
min

I Example 4.5.4. A 10-foot ladder is leaning against a wall. The bottom of the
ladder is being moved away from the wall at a speed of 2 feet per second. At
the moment when the bottom of the ladder is 5 feet away from the wall, how
fast is the top of the ladder sliding down the wall?

Solution. Let x be the distance from the bottom of the ladder to the wall and y
dx ft
be the height of the top of the ladder on the wall. We are given = 2 and
dt s
dy
asked to find when x = 5ft.
dt

In order to do that notice that the ladder forms a right triangle and the variables
are related by the Pythagorean theorem. That is,

x2 + y 2 = 102

Now, differentiating both sides of the equation above with respect to time t
yields,
d 2 d d
x + y 2 = 100
dt dt dt
dx dy
2x + 2y =0
dt dt
dx dy
x +y =0
dt dt
Since x = 5ft, therefore, the value of y can be found using the Pythagorean
theorem as follows.

x2 + y 2 = 100

y 2 = 100 − x2
p
y = 100 − 52
= 75ft

dx ft
Now, substituting x = 5ft, y = 75ft, and = 2 yields,
dt s
dx dy
x +y =0
dt dt
dy
5 · 2 + 75 = 0
dt
dy 10 ft
=−
dt 75 s
10 ft
Therefore, the top of the ladder is sliding down at a rate of , or approxi-
75 s
ft
mately 0.13 .
s

Having discussed the related rates problems through examples, the following can
be thought of as a general guideline for solving such problems.

1. Identify all given variables and the variables that are to be deter-
mined. Draw a figure if applicable

2. An equation that includes the variables for which the rates of change
are either given or need to be found.

3. Use the Chain Rule and differentiate implicitly both sides of the equa-
tion with respect to time.

4. Substitute all known values into the equation, then solve for the un-
known rate of change. respect to time

The End of Section 4.5

Related Rates

§ 4.6 Linearization and Differentials

In this section, we discuss a method by which a complicated function can be
approximated by a relatively simper function that is easier to work with, up
to our desired level of accuracy. This method is commonly referred to as the

“linearization/linear approximation/tangent line approximation,” and is based

on the observation that when we “zoom” into the graph of a smooth curve at
a point P that has a tangent line drawn at that point, the curve looks more
and more like the tangent line drawn at that point. Therefore, let us start our
discussion of the linearization as follows.

Chap. 4 / Sec. 4.6 / Subsec. 4.6.1 : Linearization

We are all familiar with how digital calculators have significantly enhanced our
lives, making tasks easier and more efficient than ever. Nevertheless, this in-
valuable tool was nonexistent just a few centuries ago. Is it thoughtful enough
for us to think about how individuals in earlier centuries dealt with the calcu-
√
lation of values such as 4.01? Think about it! What really is the value of
√
4.01 or at least a really good approximation of it? The answer isn’t readily
apparent. However, the derivatives can be used to deal with such problems at
√
hand. For that, let us recall that 4 = 2. Since 4.01 > 4 therefore the value of
√ √ √
4.01 > 4. So, we write 4.01 = 2 + r, where r ∈ R, r > 0. Now, the problem
√
at hand is to determine the sufficient value of r such that 2 + r ≈ 4.01. For
√
that, let us draw a tangent line to the graph of f (x) = x, at the point x = 4.
The drawn tangent line is reasonably close to the actual function near the point
x = 4.

Figure 4.21

Since the tangent line is a straight line, therefore it has a slope of

√ √
4.01 − 4
m=
√4.01 − 4√
4.01 − 4
=
0.01
√
Now, multiplying both sides by 0.01 and then solving for 4.01 yields,
√ √
4.01 − 4
0.01 × m = × 0.01

√ 0.01 √

0.01 × m = 4.01 − 4
√ √
0.01m + 4 = 4.01
√
∴ 4.01 = 2 + 0.01m

Now, a value of r is found to be r = 0.01m, where m is the slope of the tangent

line at the point x = 4. It is known that for a straight line, the slope is equal to
√
its derivative, that is, slope = m = f 0 (x). Therefore, the derivative of f (x) = x
at the point x = 4 yields,
d√
f 0 (x) = x
dx
1
= √
2 x
1
f 0 (4) = √
2 4
1
=
2·2
1
=
4
=m

This implies that

√
4.01 = 2 + r
= 2 + 0.01m

1
= 2 + 0.01 ·
4
= 2 + 0.0025
= 2.0025

This is reasonably a good approximation. We now generalize this idea further

below.

Consider a function y = f (x). Draw a tangent line to y = f (x) at the point

x = a, as shown below.

f(x)

x
a

Figure 4.22

y − f (a)
The slope of the tangent line is m = . Multiplying both sides by (x − a)
x−a
and solving for y yields,

y − f (a)
m=
x−a
y − f (a)
(x − a) · m = ·
(x−a)

x
− a
(x − a) · m = y − f (a)
(x − a) · m + f (a) = y
∴ y = f (a) + m (x − a)

It is known that for a linear function, the slope remains constant across all
points, therefore, the slope is equivalent to the derivative at any given point.
Thus,
y = f (a) + f 0 (a)(x − a)

This implies that the tangent line to the graph of the function at the point x = a
represents the graph of the following linear function.

L(x) = f (a) + f 0 (a)(x − a)

Thus, we have the following definition.

Definition 4.6.1. (Linearization) If f (x) is a differentiable function at x = a,

then the linearization of f (x) at a is defined as

L(x) = f (a) + f 0 (a)(x − a)

Where, L(x) ≈ f (x) for values of f (x) near a.

A few examples are provided below.

I Example 4.6.1. What is the linearization of f (x) = sin(x) at x = 0?

Solution. The derivative of f (x) = sin(x) is,

d
f 0 (x) = sin (x)
dx
= cos (x)

Now, since f (0) = sin(0) = 0 and f 0 (0) = cos(0) = 1, therefore the linearization
at x = 0 yields,

L(x) = f (a) + f 0 (a) (x − a)

= f (0) + f 0 (0) (x − 0)
= 0 + 1 (x)
=x

This implies that sin(x) ≈ x, near x = 0. Therefore, the linearization of

f (x) = sin(x), at the point x = 0 is L(x) = x. In other words, if the graphs
of the functions y = sin(x) and y = x are zoomed in at the point x = 0, they
become indistinguishable.

√
I Example 4.6.2. What is the linearization of f (x) = 2 1 + x at x = 3?
√
Solution. The derivative of f (x) = 2 1 + x is,
d √
f 0 (x) = 2 1+x
dx
d 1
= 2 · (1 + x) 2
dx
1 1
= 2 · (1 + x)− 2
2

1
= 1
(1 + x) 2
1
=√
1+x
√ √ 1 1 1
Since, f (3) = 2 1 + 3 = 2 4 = 4, and f 0 (3) = √ = √ = , therefore
1+3 4 2
the linearization at x = 3 yields,

L(x) = f (a) + f 0 (a) (x − a)

= f (3) + f 0 (3) (x − 3)
1
= 4 + (x − 3)
2
√ 1
This implies that 2 1 + x ≈ 4 + (x − 3), near x = 3. Therefore, the lineariza-
2
√ 1
tion of f (x) = 2 1 + x, at the point x = 3 is L(x) = 4 + (x − 3). In other
2
√ 1
words, if the graphs of the functions y = 2 1 + x and y = 4 + (x − 3) are
2
zoomed in at the point x = 3, they become indistinguishable.

Having discussed the linearization, we now shift our focus to discuss the differ-
entials that approximates the amount a function changes.

Chap. 4 / Sec. 4.6 / Subsec. 4.6.2 : Differentials

As of yet, we have seen that linearization can be used to approximate a function’s
values near a point x = a. However, they can also be used to approximate the
amount a function changes as a result of a small change in the input x. To see
how, recall that,

f (x) ≈ L(x)
= f (a) + f 0 (a)(x − a)
∴ f (x) ≈ f (a) + f 0 (a)(x − a)

where, a is a fixed point and x is its nearby points. This implies that

f (x) ≈ f (a) + f 0 (a) (x − a)

f (x) − f (a) ≈ f 0 (a) (x − a)
∆y ≈ f 0 (a) ∆x (1)

where, ∆x = x − a represents a change in the x-coordinate between a and x

and ∆y = f (x) − f (a) represents a corresponding change in the y-coordinate
between f (a) and f (x), respectively. This suggests that a change in y can be
approximated by the change in x multiplied by the derivative of the function at
that point.

Now, suppose a function y = f (x) has changed from x = a to x = a + ∆x. Then

the exact change in y, that is, ∆y, yields the following.

∆y = f (a + ∆x) − f (a)

Now, sometimes it is more difficult to calculate the exact change in y, that is,
∆y, compared to just approximating it by the use of linearization as discussed
in the previous subsection. That is, for all x near a, f (x) can more easily be
approximated by,
L(x) = f (a) + f 0 (a)(x − a)

Now, the exact change in L as x changes from x = a to x = a + ∆x, is

∆L = L (a + ∆x) − L (x)

= f (a) + f 0 (a) (a + ∆x − a) − f (a) + f 0 (a) (a − a)
| {z } | {z }
L(a+∆x) L(x)

0
= f (a) + f (a) ∆x − f (a)
= f 0 (a) ∆x
∴ ∆L = f 0 (a) ∆x (1)

Note that ∆L is an approximation of the actual change ∆y, as x changes from

x = a to x = a + ∆x. This approximation gets better and better as ∆x → 0,
that is, as the point x = a + ∆x gets closer and closer to x = a.

Now, based on this approximation ∆L of the actual change ∆y, the concept of
differentials can be defined precisely. Historically, the differential of a function
(denoted as dy) was thought of as an infinitely small (or infinitesimal) change in
y, corresponding to an infinitely small (or infinitesimal) change in x (denoted as
dx), if we were to put it in an oversimplified manner. When these infinitesimally
dy
small changes are compared together by a ratio of , the instantaneous rate of
dx
change (i.e., the derivative) of a function is found. Nevertheless, defining such
infinitesimal quantities dy and dx seemed to be all based on intuition, which

lacked the mathematical rigor.

Now, centuries later, these differentials can indeed be defined rigorously with
the concept of linearization. For that, consider the following figure, where the
change in the function y = f (x) as x changed from x = a to x = a + dx is
approximated by the tangent line drawn at x = a, using linearization.

P
S

Figure 4.23

In the figure above, as x changes from x = a to x = a + dx, where dx ∈ R is a

small increment, there occur two changes. One change occurred in the function
y = f (x) (denoted as ∆y), and the other one is in the tangent line L(x) (de-
noted as ∆L). Here, ∆y is the actual change and ∆L serves as its approximation.

Now, the differential dy can be defined as the change in this linearization, that
is, ∆L. This implies that

dy = ∆L
= f 0 (a) ∆x from equation (1)

= f 0 (a) (x + dx) − x
| {z }
∆x
0
= f (a) dx since ∆x = dx = real number

This implies that at x = a, the differential dy is dy = f 0 (a)dx. Therefore, more

generally, at any variable point x, the differential dy yields,

dy = f 0 (x) dx

Thus, we finally have the following definition of the differential of y.

Definition 4.6.2. (Differentials) If y = f (x) is a differentiable function, and

the differential of x (denoted as dx) is not 0, then the differential of y
(denoted as dy) is,
dy = f 0 (x) dx

This definition aligns with the old intuition that when dy ÷ dx, the derivative is
found. That is,

dy = f 0 (x) dx
dy f 0 (x)
dx

=
dx dx

dy
= f 0 (x)
dx
dy
This is perhaps the point where confusions come into play! Previously, was
dx
understood as the derivative of a function, defined as the limit of the quotient
∆y
, as ∆x → 0. In other words,
∆x
dy ∆y
= lim
dx ∆x→0 ∆x
dy
where was merely a notation and not an actual quotient of dy over dx.
dx
However, now that we have defined meanings for dy and dx, the derivative of a
function f 0 (x) can indeed be expressed as a quotient of dy and dx. That is,
dy
dy ÷ dx =
dx
f 0 (x)
dx

=
dx

0
= f (x)
= the derivative of f (x)

This suggests that,

dy dy
=
dx dx
dy
dy = dx
dx
Notice how dx’s in the equation above cannot simply be canceled out. This is
dy
because on the left side of the equation represents the derivative of a function,
dx

and thus it is merely a notation. In other words,

∆y
dy = lim · dx
∆x→0 ∆x

Note that every differential rule summarized before can be written in differential
form. That is,

1. d(c) = 0 13. d (tan(x)) = sec2 (x)dx

2. d(x) = dx 14. d (cot(x)) = − csc2 (x)dx

3. d(xn ) = nxn−1 dx 15. d (sec(x)) = sec(x) tan(x)dx
n+1
x
4. d = xn dx 16. d (csc(x)) = − csc(x) cot(x)dx
n+1
√ 1 1
5. d( x) = √ dx 17. d sin−1 (x) = √

dx
2 x 1 − x2
6. d (ex ) = ex dx 1
18. d cos−1 (x) = − √

dx
1 − x2
7. d (e−x ) = −e−x dx
1
19. d tan−1 (x) =

8. d (a ) = a ln(a)dx
x x
1 + x2
dx

1
9. d (ln(x)) = d (loge (x)) = dx 20. d cot−1 (x) = −
1
dx
x 1 + x2
1
10. d (loga (x)) = dx 1
x ln (a) 21. d sec−1 (x) =

√ dx
|x| x2 − 1
11. d (sin(x)) = cos(x)dx
1
22. d csc−1 (x) = −

√ dx
12. d (cos(x)) = − sin(x)dx |x| x2 − 1

The End of Section 4.6

Linearization and Differentials

§ 4.7 Newton’s Method

Newton’s Method is a numerical method to approximate the roots of a function.
A root of a function is a value of x that satisfies the equation f (x) = 0. Graph-
ically, the root represents the x-intercept of the graph of the function y = f (x).
It is known from algebra that the root of a first-degree polynomial, ax + b = 0,
is determined by the formula:
b
x=−
a

Similarly, the roots of a second-degree polynomial, ax2 + bx + c = 0, are deter-

mined by the formula: √
−b ± b2 − 4ac
x=
2a
Formulas also exist to find the roots of third-degree and fourth-degree polyno-
mials, but they aren’t as straightforward as the ones above. In fact, it is not
possible to derive a general formula for fifth-degree polynomials or polynomials
of any higher degree. As a result, if we ever need to find the roots of an equation
such as the following,
x5 + x4 + x3 + x2 + x

It may best be approximated numerically using the method discussed in this

section.

Newton’s Method is one of many methods by which the roots of a function can
be approximated. The method can best be described in a graphically. For that,
consider the figure below.

f(x)

r
x

Figure 4.24

In the figure above, the root of the function y = f (x) is r. In order to approxi-
mate the root r, we start with a value x1 such that x1 ≈ r. Since x1 6= r and
f (x1 ) 6= 0, therefore we draw a tangent line to the graph of y = f (x) at the
point (x1 , f (x1 )). Now, observe that the tangent line is close to the graph of
f (x) near the point of tangency, therefore the graph of f (x) and the tangent
line at (x1 , f (x1 )) intercepts the x-axis at approximately the same point. This

implies that the root r of f (x) is approximately equal to the root of the tangent
line, that is, x2 . Now, since the tangent line is a line and its root x2 can easily
be determined, therefore x2 can be used as a second approximation to r (where,
the first approximation was x1 ) to repeat the process of approximation with a
better precision.

Since the slope of the tangent line at the point x1 is f 0 (x1 ), therefore the equation
of the tangent line yields,
y − f (x1 ) = f 0 (x1 ) (x − x1 )
Now, since y = 0 when x = x2 (as x2 lies on the x-axis) in the equation above,
therefore,
y − f (x1 ) = f 0 (x1 ) (x − x1 )
0 − f (x1 ) = f 0 (x1 ) (x2 − x1 )
f (x1 )
− = (x2 − x1 )
f 0 (x1 )
f (x1 )
x2 = x1 − 0
f (x1 )
where, f 0 (x) 6= 0. Now, since f (x2 ) 6= 0, therefore x2 6= r, and thus x2 ≈ r. A
better approximation x3 can now be derived by repeating the process disussed
above with an initial approximation being x2 to the root r. That is, draw a
tangent line to the graph of f (x) at the point (x2 , f (x2 )), that intercepts the
x-axis at the point x3 . Therefore, the equation of the tangent line is as follows.
y − f (x2 ) = f 0 (x2 ) (x − x2 )
Since the tangent line intercepts the x-axis at the point x = x3 , therefore y = 0
when x = x3 . This implies,
y − f (x2 ) = f 0 (x2 ) (x − x2 )
0 − f (x2 ) = f 0 (x2 ) (x3 − x2 )
f (x2 )
− = x3 − x2
f 0 (x2 )
f (x2 )
x3 = x2 − 0
f (x2 )
where, f 0 (x2 ) 6= 0. If f (x3 ) 6= 0, then x3 6= r, and thus x3 ≈ r. This implies
that the process can be repeated again by considering x3 as the initial approxi-
mation. This yields a sequence of approximation x1 , x2 , x3 , x4 , · · · , as shown in
the following figure.

f(x)

r
x

Figure 4.25

In general, if the nth approximation is f (xn ) and f 0 (xn ) 6= 0, then the next
approximation yields,
f (xn )
xn+1 = xn − 0
f (xn )
This implies that if xn → r, as n → ∞, then the sequence of successive approx-
imations converges to the actual root r of the function f (x).

I Example 4.7.1. Approximate a root of f (x) = x4 − π using Newton’s Method

up to 7 iterations, where the initial approximation x1 = 1.

Solution. Since f (x) = x4 − π, therefore, f 0 (x) = 4x3 , and thus, by the Newton’s
Method,

f (xn )
xn+1 = xn −
f 0 (xn )
x4 − π
= xn − n 3
4xn

This implies that if xn is the initial approximation of a root r of f (x) = x4 − π,

x4 − π
then xn+1 = xn − n 3 is its next better approximation. Below the calculations
4xn
of these iterations are presented for n = 1, 2, · · · , 7.

f (xn )
n xn xn+1 = xn −
f 0 (xn )
1 1 1.5353981634
2 1.5353981634 1.36853218292
3 1.36853218292 1.33282459009
4 1.33282459009 1.33133785791
5 1.33133785791 1.33133536381
6 1.33133536381 1.3313353638
7 1.3313353638 1.3313353638

It is evident that as the number of iterations n increases, the sequence of ap-

proximation converges to a specific value, and that specific value is a root of the
function f (x) = x4 − π.

I Example 4.7.2. Approximate a root of f (x) = eex − e, using Newton’s Method

up to 7 iterations, where the initial approximation is x1 = 0.5.

Solution. Since f (x) = eex − e, therefore, f 0 (x) = eex+1 , and thus, by the New-
ton’s Method,
f (xn )
xn+1 = xn −
f 0 (xn )
eexn − e
= xn − exn +1
e
Calculations of the iterations are presented below for n = 1, 2, · · · , 7.

eexn − e
n xn xn+1 = xn −
eexn +1
1 0.5 0.389001924142
2 0.389001924142 0.368474393475
3 0.368474393475 0.367879922005
4 0.367879922005 0.367879441172
5 0.367879441172 0.367879441171
6 0.367879441171 0.367879441171
7 0.367879441171 0.367879441171

Although Newton’s Method is a powerful technique to find the root for finding
the roots of a function, it sometimes fails to do so. For example, if the derivative
f (xn )
f 0 (xn ) in the formula xn+1 = xn − 0 turns out to be 0 for any n, then the
f (xn )
formula involves division by zero, making it undefined altogether.

Similarly, Newton’s Method can also fail for other reasons. It might miss the
aimed root and instead converge to a different one, or not converge at all. For
example, consider the following example.

√
I Example 4.7.3. Approximate a root of f (x) = 3 x, using Newton’s Method up
to 7 iterations, where the initial approximation is x1 = 1.
√ 1
Solution. Since f (x) = 3
x, therefore f 0 (x) = 2 . Now, by the Newton’s
3x 3
Method,

f (xn )
xn+1 = xn −
f 0 (xn )
√3 x
n
= xn −
1
2
3xn3
1
xn3
= xn − 2
1 −3
x
3 n
1 1
+ 23
= xn − 1 · xn3
3
3
= xn − 3 · xn3
= xn − 3xn
= −2xn

Calculations of the iterations are presented below for n = 1, 2, · · · , 7.

n xn xn+1 = −2xn
1 1 −2
2 −2 4
3 4 −8
4 −8 16
5 16 −32
6 −32 64
7 64 −128

It is evident that the sequence of approximation does not converge to a specific

value, therefore Newton’s Method fails in this case.

The End of Section 4.7

Newton's Method

§ 4.8 L’Hôpital’s Rule

The limit of a function and its evaluation were studied in Chapter 1. However,
the techniques presented there sometimes either become cumbersome or fall
short of certain limits known as indeterminate forms. For example, consider the
following limit of a function,

ax
lim
x→∞ x

As x → ∞, the numerator ax, as well as the denominator x increases with-

∞
out a bound, resulting in an indeterminate form of . Nevertheless, a simple
∞
simplification suggests that,
ax
lim = lim a
x→∞ x x→∞

Similarly, consider the limit below,

ax
lim
x→0 x

This time, as x → 0, the numerator ax and the denominator x approaches 0, re-

0
sulting again in an indeterminate form of . Despite that, a simple simplification
0
again suggests the following,
ax
lim = lim a
x→0 x x→0

=a
∞ 0
In general, limits that take the form of or are indeterminate, and can have
∞ 0
any value. Therefore, these types of limits require a different approach when
determining their values.

The method often used in such situation is known as L’Hôpital’s Rule, and the
idea behind it is quite simple: when both the numerator and denominator of a
fraction approach either 0 or ∞ as x approaches a certain value, directly eval-
uating the limit leads to an indeterminate form. In such cases, examining the
rates at which the numerator and denominator change (i.e., their derivatives)
can provide insight into the function’s behavior. For example, if both the numer-
ator and denominator approaches infinity, we can’t immediately determine the
limit of the fraction. The outcome is ambiguous. However, if we somehow know
that the numerator of the fraction increases at a rapid pace while the denomi-
nator increases slowly, then this information becomes critical, as it shows that
the behavior of the numerator largely dictates the fraction’s limit. Therefore,
instead of directly evaluating the numerator and denominator, we analyze their
respective rates of change (i.e., derivatives) to determine the overall limit of the
fraction. In the following subsections, we go through these cases one by one.

0
Chap. 4 / Sec. 4.8 / Subsec. 4.8.1 : Indeterminate Form
0
f (x)
Consider a function h(x) = . If lim f (x) = 0 as well as lim g(x) = 0,
g(x) x→a x→a
f (x) 0 0
then the limit lim = , which is the indeterminate form of . In order to
x→a g(x) 0 0
determine the actual value of such a limit of indeterminate form, the L’Hôpital’s
Rule is used stating the following.

0
Theorem 4.8.1. L’Hôpital’s Rule for
0

Let f (x) and g(x) be differentiable functions over an open interval I con-
taining a, except possibly at a and suppose g 0 (x) 6= 0. If lim f (x) =
x→a
lim g(x) = 0, then L’Hôpital’s Rule states that
x→a

f (x) f 0 (x)
lim = lim 0
x→a g (x) x→a g (x)

provided that the limit on the right side exists (or is ±∞). The rule is also
applicable for the cases where x → ±∞, x → a+ , or x → a− .

Proof. We prove a special case of this theorem where we assume that f (a) =
g(a) = 0 and f 0 (x) and g 0 (x) are continuous at a, and g 0 (a) 6= 0. Thus, we have,

f 0 (x) f 0 (a)
lim =
x→a g 0 (x) g 0 (a)

By the definition of a derivative, we get,

f (x) − f (a)
0
f (a) lim
=
x→a x−a
0
g (a) g (x) − g (a)
lim
x→a x−a
f (x) − f (a)
= lim x−a
x→a g (x) − g (a)

x−a
f (x) − f (a) x−a
= lim

·
x→a x
−a g (x) − g (a)
f (x) − f (a)
= lim
x→a g (x) − g (a)

Since, f (a) = g(a) = 0, therefore,

f (x) − f (a) f (x) − 0

lim = lim
x→a g (x) − g (a) x→a g (x) − 0

f (x)
= lim
x→a g (x)

This completes the proof.

A few examples to demonstrate the theorem above are provided below.

ex − 1
I Example 4.8.1. Evaluate the limit lim
x→0 x

Solution. Substituting x = 0 directly into the expression yields an indeterminate

0
form of ,
0
ex − 1 e0 − 1
=
x 0
0
=
0
However, by the L’Hôpital’s Rule,
d x
ex − 1 (e − 1)
lim = lim dx
x→0 x x→0 dx
dx
d x d
e − (1)
= lim dx dx
x→0 dx
dx
ex
= lim
x→0 1

= e0
=1
ex − 1
Therefore, lim = 1.
x→0 x

Remark 4.8.1.
ex − 1
lim =1
x→0 x

ax − 1
I Example 4.8.2. Evaluate the limit lim
x→0 x
Solution. Substituting x = 0 directly into the expression yields an indeterminate
0
form of ,
0
ax − 1 a0 − 1
=
x 0
0
=
0
However, by the L’Hôpital’s Rule,
d x
ax − 1 (a − 1)
lim = lim dx
x→0 x x→0 dx
dx

d x d
a − (1)
= lim dx dx
x→0 dx
dx
ax ln(a)
= lim
x→0 1
= a ln(a)
0

= ln(a)
ax − 1
Therefore, lim = ln(a)
x→0 x

Remark 4.8.2.
ax − 1
lim = ln(a)
x→0 x

ln(1 + x)
I Example 4.8.3. Evaluate the limit lim
x→0 x
Solution. Substituting x = 0 directly into the expression yields an indeterminate
0
form of ,
0
ln(1 + x) ln(1 + 0)
=
x 0
0
=
0
However, by the L’Hôpital’s Rule,
d
ln(1 + x) ln(1 + x)
lim = lim dx
x→0 x x→0 dx
dx
1 d
· (1 + x)
= lim 1 + x dx
x→0 dx
dx
1
= lim 1 + x
x→0 1
1
=
1+0
=1
ln(1 + x)
Therefore, lim =1
x→0 x

Remark 4.8.3.
ln(1 + x)
lim =1
x→0 x

∞
Chap. 4 / Sec. 4.8 / Subsec. 4.8.2 : Indeterminate Form
∞
∞
L’Hôpital’s Rule is also applicable to the limits of the indeterminant form .
∞
f (x)
For example, consider a function h(x) = . If lim f (x) = ±∞ as well as
g(x) x→a
f (x) ∞
lim g(x) = ±∞, then the limit lim = , which is the indeterminate form
x→a x→a g(x) ∞
∞
of . Now to solve such a limit, L’Hôpital’s Rule states the following.
∞

±∞
Theorem 4.8.2. L’Hôpital’s Rule for
±∞

Let f (x) and g(x) be differentiable functions over an open interval I con-
taining a, except possibly at a and suppose g 0 (x) 6= 0. If lim f (x) = ±∞
x→a
and lim g(x) = ±∞, then L’Hôpital’s Rule states that
x→a

f (x) f 0 (x)
lim = lim 0
x→a g (x) x→a g (x)

provided that the limit on the right side exists (or is ±∞). The rule is also
applicable for the cases where x → ±∞, x → a+ , or x → a− .

A few examples to demonstrate the theorem above are presented below.

2x2 + 3x + 5
I Example 4.8.4. Evaluate the limit lim
x→∞ 2x2 − 3x − 5

∞
Solution. As x → ∞, the expression yields an indeterminate form of , there-
∞
fore, by L’Hôpital’s Rule,
d 2

2x + 3x + 52 2x + 3x + 5
lim = lim dx
x→∞ 2x2 − 3x − 5 x→∞ d
(2x2 − 3x − 5)
dx
4x + 3
= lim
x→∞ 4x − 3

∞
Now, as x → ∞, the expression again yields an indeterminate form of ,
∞
therefore, we reapply L’Hôpital’s Rule once more,
d
4x + 3 (4x + 3)
lim = lim dx
x→∞ 4x − 3 x→∞ d
(4x − 3)
dx
4
= lim
x→∞ 4

2x2 + 3x + 5
Therefore, lim =1
x→∞ 2x2 − 3x − 5

ln(1 + x)
I Example 4.8.5. Evaluate lim
x→∞ x
∞
Solution. As x → ∞, the expression yields an indeterminate form of , there-
∞
fore, by L’Hôpital’s Rule,
d
ln(1 + x) ln(1 + x)
lim = lim dx
x→∞ x x→∞ dx
dx
1 d
· (1 + x)
= lim 1 + x dx
x→∞ dx
dx
1
= lim 1 + x
x→∞ 1
1
= lim
x→∞ 1 + x

=0
ln(1 + x)
Therefore, lim =0
x→∞ x

Remark 4.8.4.
ln(1 + x)
lim =0
x→∞ x

ln (x + 1)
I Example 4.8.6. Evaluate lim
x→∞ ln (x − 1)

∞
Solution. As x → ∞, the expression yields an indeterminate form of , there-
∞
fore, by L’Hôpital’s Rule,
d
ln (x + 1) ln (x + 1)
lim = lim dx
x→∞ ln (x − 1) x→∞ d
ln (x − 1)
dx
1 d
· (x + 1)
= lim x + 1 dx
x→∞ 1 d
· (x − 1)
x − 1 dx
1
= lim x + 1
x→∞ 1
x−1
x−1
= lim
x→∞ x + 1

∞
Now, as x → ∞, the expression again yields an indeterminate form of ,
∞
therefore, we reapply L’Hôpital’s Rule once more,

d
x−1 (x − 1)
lim = lim dx
x→∞ x + 1 x→∞ d
(x + 1)
dx
1
= lim
x→∞ 1

ln (x + 1)
Therefore, lim =1
x→∞ ln (x − 1)

Chap. 4 / Sec. 4.8 / Subsec. 4.8.3 : Indeterminate Forms 0 · ∞ and ∞ − ∞

0 ∞
In addition to the limits of the indeterminate forms and , there are other
0 ∞
types of indeterminate forms. For example, if lim f (x) = 0 and lim g (x) =
x→a x→a
±∞, then the limit of a product of two functions lim f (x) g (x) results in the
x→a
indeterminate form 0 · ∞. Unlike the previously mentioned indeterminate forms,
L’Hôpital’s Rule cannot be directly applied to these limits. To address them,
0 ∞
they must first be converted into limits of the form or , as shown in the
0 ∞
examples below.

1
I Example 4.8.7. Evaluate lim x sin 2
x→∞ x

1
Solution. Since the limit lim x sin
2
is of the indeterminate form 0 · ∞,
x→∞ x
therefore L’Hôpital’s Rule cannot be directly applied to it. As a result, this
0 ∞
limit needs some modification to convert the form into either or . This can
0 ∞
1
be done by the use of reciprocals. For example, since the reciprocal of x2 is 2 ,
x
1
therefore x2 = . As a result, the limit can be rewritten as follows,
1
x2

1 1 1
lim x sin
2
= lim sin
x→∞ x x→∞ 1 x
x2
1
sin
x
= lim
x→∞ 1
x2
Now, as x → ∞, both the numerator and denominator approach 0, thus the
0
limit is converted into the indeterminate form such that L’Hôpital’s Rule now
0
can be directly applied to evaluate the limit. That is,

1 d 1
sin sin
x dx x
lim = lim
1

x→∞ x→∞ d 1
x2 dx x2

1 d 1
cos ·
x dx x
= lim
x→∞ 2
− 3
x
1 1
cos · − 2
x x
= lim
x→∞ 2
− 3
x
1
cos
x
−
= lim x2
x→∞ 2
− 3
x

1
cos
x x3
= lim ·
x→∞ x2 2
1
x cos
x
= lim
x→∞ 2
1
z }| {
1
∞ · cos
∞
=
2
=∞

1
This implies that lim x2 sin = ∞.
x→∞ x

Another type of indeterminate form is ∞ − ∞. For example, if lim f (x) = ∞

x→a
and if lim g(x) = ∞, then the limit of the function lim f (x) − g(x) yields
x→a x→a
the indeterminate form ∞ − ∞. Contrary to the previous indeterminate forms,
L’Hôpital’s Rule cannot be directly applied to these limits and to address them,
0 ∞
they must first need to be converted into limits of the form or .
0 ∞

1
I Example 4.8.8. Evaluate lim cot(x) −
x→0 x
1
Solution. As x → 0, both terms cot(x) and approach ∞, yielding an indeter-
x
minate form ∞ − ∞. As a result, we do the following simplification to convert
∞
into limits of the indeterminate form .
∞
cos(x) 1

1
lim cot(x) − = lim −
x→0 x x→0 sin(x) x
x cos(x) − sin(x)

= lim
x→0 x sin(x)
Now, we apply L’Hôpital’s Rule,
d d d
(x cos(x) − sin(x)) (x cos(x)) − sin(x)
lim dx = lim dx dx
x→0 d x→0 d
(x sin(x)) (x sin(x))
dx dx
d dx
x· cos(x) + cos(x) · − cos(x)
= lim dx dx
x→0 d dx
x· sin(x) + sin(x) ·
dx dx

−x sin(x) + cos(x)
− cos(x)

= lim

x→0 x cos(x) + sin(x)
x sin(x)
= − lim
x→0 x cos(x) + sin(x)

Reapplying L’Hôpital’s Rule again yields,

d
x sin(x) (x sin(x))
− lim = − lim dx
x→0 x cos(x) + sin(x) x→0 d
(x cos(x) + sin(x))
dx
d
(x sin(x))
= − lim dx
x→0 d d
(x cos(x)) + sin(x)
dx dx
d dx
x· sin(x) + sin(x) ·
= − lim dx dx
x→0 d dx
x· cos(x) + cos(x) · + cos(x)
dx dx
x cos(x) + sin(x)
= − lim
x→0 −x sin(x) + cos(x) + cos(x)
0 · cos(0) + sin(0)
=−
−x sin(0) + cos(0) + cos(0)
0
=−
1+1
=0

1
Therefore, lim cot(x) − =0
x→0 x

It is sometimes helpful to change the variable. That is, to evaluate lim f (x),
x→∞
1
it is sometimes helpful to let t = . Now, notice that if x → ∞, then t → 0+ .
x
As a result, we can rewrite the limit in terms of t, as follows,

1
lim f (x) = lim+ f
x→∞ t→0 t
1 1
Note that since we let t = , therefore x = . Thus, if x → +∞, then t → 0
x t
from the positive side.

√
I Example 4.8.9. Evaluate lim 2
x − x + 6x
x→∞

√
Solution. As x → ∞, both the terms x as well as x2 + 6x approach ∞, yielding
an indeterminate form ∞ − ∞. Nevertheless, to evaluate this limit, we do the
following,
s !
6x
lim x − x2 + 6x = lim x − x2 1 + 2
p
x→∞ x→∞ x
!
√
r
6
= lim x − x2 · 1 +
x→∞ x
r !
6
= lim x 1 − 1 +
x→∞ x
1 1
Now, let a new variable t = , then x = . Substituting these variables yields,
x t
r !
6 1 √
lim x 1 − 1 + = lim+ 1 − 1 + 6t
x→∞ x t→0 t
√
1 − 1 + 6t
= lim+
t→0 t
Now, as t → 0+ , both the numerator and denominator yield an indeterminate
0
form , and L’Hôpital’s Rule can now be directly applied. L’Hôpital’s Rule
0
states that,
√ d √
1 − 1 + 6t 1 − 1 + 6t
lim+ = lim+ dt
t→0 t t→0 d
t
dt
1 d
− √ · (1 + 6t)
2 1 + 6t dt
= lim+
t→0 1
1
− √ ·6
2 1 + 6t
= lim+
t→0 1
6
= lim+ − √
t→0 2 1 + 6t
6
=− √
2 1+0
6
=−
2
= −3
√
This implies that lim x − x2 + 6x = −3
x→∞

Chap. 4 / Sec. 4.8 / Subsec. 4.8.4 : Indeterminate Forms 1∞ , 00 , and ∞0

There are other indeterminate forms 1∞ , 00 , and ∞0 that arise from the limit
of the following form,
lim f (x)g(x)
x→a
The indeterminate form 1 ∞
arises if lim f (x) = 1 and lim g (x) = ±∞. The
x→a x→a
indeterminate form 0 arises if lim f (x) = 0 and lim g (x) = 0. Likewise, the
0
x→a x→a
indeterminate form ∞0 arises if lim f (x) = ∞ and lim g (x) = 0.
x→a x→a

Contrary to the previous forms, L’Hôpital’s Rule cannot directly be applied to

them. Therefore, in order to apply L’Hôpital’s Rule, they must first need to be
0 ∞
expressed in the form or by manipulation.
0 ∞
0 ∞
In order to convert them into the indeterminate forms or , the inverse
0 ∞
relationship between ln(x) and ex is used. That is, if y = f (x)g(x) , then

ln (y) = ln f (x)g(x)
= g (x) ln f (x)
eln(y) = eg(x) ln f (x)
y = eg(x) ln f (x)
f (x)g(x) = eg(x) ln f (x)

This implies that we can rewrite the limit lim f (x)g(x) as,
x→a

lim f (x)g(x) = lim eg(x) ln f (x)

x→a x→a
lim g(x) ln f (x)
= ex→a

provided that the limit lim g (x) ln f (x) exists.

x→a

I Example 4.8.10. Evaluate lim+ (sin(x))tan(x)

x→0

Solution. As x → 0 from the positive side, sin(x)tan(x) → 00 . This implies that

the limit has the indeterminate form 00 . However, notice that,

ln (sin(x))tan(x) = tan(x) ln (sin(x))

This implies that

lim+ (sin(x))tan(x) = lim+ tan(x) ln (sin(x))

x→0 x→0

0
Now, we convert this to a limit of the indeterminate form and apply L’Hôpital’s
0
Rule,

ln (sin(x))
lim+ tan(x) ln (sin(x)) = lim+
x→0 x→0 cot(x)
d
ln (sin(x))
= lim+ dx
d
cot(x)
x→0
dx
1 d
sin(x)
sin(x) dx
= lim+
x→0 −csc2 (x)
cos(x)
sin(x)
= lim+
x→0 1
− 2
sin (x)
cos(x)
= − lim+ · sin2 (x)
x→0 sin(x)

= − lim+ cos(x) sin(x)

x→0

= − cos(0) sin(0)
=0

Therefore, lim+ (sin(x))tan(x) = 0

x→0

I Example 4.8.11. Evaluate lim+ xx .

x→0

Solution. As x → 0 from the positive side, xx → 00 . This implies that the limit
has the indeterminate form 00 . To evaluate the limit we let, xx = ex ln(x) = eL
and solve the limit L = lim+ x ln (x). The limit lim+ x ln (x) has an indetermi-
x→0 x→0
nate form 0 · ∞, therefore to use L’Hôpital’s Rule, the indeterminate form 0 · ∞
0 ∞
is modified into the indeterminate form or , as shown below.
0 ∞
1
lim+ x ln (x) = lim+ · ln (x)
x→0 x→0 1
x
ln (x)
= lim+
x→0 1
x

ln (x) 0
Since the limit lim+ has the indeterminate form , therefore L’Hôpital’s
x→0 1 0
x
Rule is used stating that,
d
ln (x) ln (x)
lim = lim+ dx
1

x→0+ x→0 d 1
x dx x
1
= lim+ x
x→0 1
− 2
x
1
= lim+ · −x2

x→0 x

= lim+ −x
x→0

This implies that L = lim+ x ln (x) = 0. Since L = lim+ x ln (x) = 0, therefore

x→0 x→0

lim xx = lim+ ex ln(x)

x→0+ x→0

= lim+ e0
x→0

= lim+ 1
x→0

As a result, lim+ xx = 1.
x→0

The End of Section 4.8

L'Hôpital's Rule

Copyright © 2024 Md. Shouvik Iqbal. This book is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License (CC BY-NC-SA 4.0)
Download for free at https://md-shouvik-iqbal.github.io/calculus
Copyright © 2024 Md. Shouvik Iqbal. This book is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License (CC BY-NC-SA 4.0)
Download for free at https://md-shouvik-iqbal.github.io/calculus
Part III

Integral Calculus

293

Copyright © 2024 Md. Shouvik Iqbal. This book is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License (CC BY-NC-SA 4.0)
Download for free at https://md-shouvik-iqbal.github.io/calculus
Copyright © 2024 Md. Shouvik Iqbal. This book is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License (CC BY-NC-SA 4.0)
Download for free at https://md-shouvik-iqbal.github.io/calculus
Chapter 5

An Introduction to Integral
Calculus

Integral Calculus is a subfield of calculus that studies the concept of integra-

tion—a process of finding the integral of a given function. An integral can most
easily be understood as a way to accumulate quantities, such as the area of a
region. The area of such a region represents the measure of space that a two-
dimensional shape occupies.

Humans have long been familiar with the concept of area, even in ancient times,
and have developed various methods to calculate them. For example, the area
of a rectangle is determined by multiplying its base by its height. Similarly, the
area of a triangle is calculated as half the area of a rectangle since a triangle can
be thought of as one-half of a rectangle when divided by a diagonal. Likewise,
the area of a regular polygon can be found by dividing the polygon into smaller
triangles and summing their areas.

Contrary to the shapes above, the shape of a circle is complex for the fact
that it has no corners or edges, resulting in a smooth, continuous, and round
curve. Finding the area of a circle is thus generally more difficult than the
ones mentioned above. Documented history suggests that it was the Greeks,
particularly Archimedes, who were among the first to solve the problem of finding
the area of a circle through approximation. The method used in that process
is referred to as the “method of exhaustion” for the reason that it involves

295

inscribing1 (or circumscribing) a regular polygon into a circle and increasing its
sides in a way that the area of the polygon approaches the exact area of the
circle. For example, consider the circle below.

r
h
b

Figure 5.1

In the figure above, a regular polygon is inscribed into the circle. Since each of
1
the isosceles triangles has an area A = hb, therefore the sum of these areas
2
equals the area of the polygon, which is a close approximation of the area of the
circle. That is, the area of the polygon Ap yields,

1 1 1 1
Ap = hb + hb + hb + · · · + hb
2 2 2 2
1
= h (b + b + b + · · · + b)
2
1
= hp
2
where, p is the perimeter of the polygon. Now, if C = 2πr is the circumference
of the circle, then as the number of sides in the polygon increases, the height h in
each polygon approaches the circle’s radius r, that is, h → r; and the perimeter
of the polygon approaches the circumference C, that is, p → C. This implies
that,
1 1
Ap = hp → rC
2 2
1
to fit a smaller shape inside a larger shape in such a way that the smaller shape touches
but does not intersect the larger shape.

1
= r (2πr)
2
∴ Ap → πr2

As a result, the area A of a circle with radius r is found to be,

A = πr2

as the sides of the polygon increase indefinitely.

The method of exhaustion works effectively for circles due to their regular ge-
ometric properties. However, when applied to irregular shapes lacking such
properties, the method loses its value—it becomes impractical. Therefore, we
need to consider a different approach to calculate the area of irregular shapes.
In modern terms, the problem can be generalized to finding the area of a region
with a curved boundary.

To address this, we begin with a specific case of the general problem: find the
area under the graph of a function y = f (x) and above the x-axis, between two
vertical lines at x = a and x = b. This enclosed region has a curved boundary
only along its upper edge, making it easier to handle. This generalized problem
is commonly referred to as finding the “area under a curve.”

Solving this specific case provides a foundation for addressing more complex
regions, such as those with curved boundaries on both edges. For such regions,
the total area can be determined by subtracting the area under the lower curve
from the area under the upper curve, where both areas belong to the simpler
“area under a curve” category. These principles are discussed further in the
following.

§ 5.1 The Area Under a Curve

The term “area under a curve” refers to the total area enclosed by the graph of
a function f (x) and the x-axis, between two vertical lines from x = a to x = b
(i.e., on the interval [a, b]). For example, consider the figure of the function
x2
f (x) = below.
4

1 A

1 2 3 4

Figure 5.2

x2
Suppose, we want to determine the exact area A under the curve of f (x) =
4
and above the x-axis, from x = 1 to x = 4. In order to do that, we begin by
breaking down the area into simple, familiar shapes—rectangles.

The idea is straightforward: the area under the curve can be approximated by
adding up the areas of infinitely many narrow rectangles, where each rectangle’s
height is determined by the function’s value at different points. As we increase
the number of rectangles indefinitely, the approximation becomes more accurate,
eventually giving us the exact area.

However, since we’re dealing with infinitely many rectangles, we need a sys-
tematic way to represent their sum. That is, summing these infinitely many
rectangles will be impossible for us to do in the traditional way, so we need
some sort of mathematical “tool” to handle this situation more compactly. And
this is where the summation notation, , comes into play to help us neatly
P

organize and compute the total area. We discuss it in the following subsection
very shortly and then move to calculate the area under the curve.

P
Chap. 5 / Sec. 5.1 / Subsec. 5.1.1 : The Summation ( ) Notation
The summation notation, also known as the sigma notation, is a concise way to
represent a large sum of numbers that follow a specific pattern. For example,
if we ever need to calculate the sum of integers from 1 to 100, how would we
express it? The common approach is to place the ‘+’ sign in between each

integer and to use ‘· · · ’ to indicate skipping some of the terms as follows,

1 + 2 + 3 + 4 + 5 + · · · + 96 + 97 + 98 + 99 + 100

However, that too isn’t systematically concise enough compared to just explicitly
expressing,

“The sum of integers i, from i = 1 to i = 100”

As a result, the summation notation was developed with the purpose to address
situations like the one above. In that notation, we write,
100
X
i
i=1

to mean the sum of all i such that i ranges from 1 to 100. In other words, it is
the sum of all integers i from i = 1 to i = 100.

Here, i is called the index of summation. It may be obvious that the choice of
the variable for the index does not matter at all as long as it doesn’t confuse
with the other variables that are in use. There can be any variable in the place
100 100 100
of i to keep track of the terms to be added. For example,
X X X
i= k= j
i=1 k=1 j=1
conveys the same meaning. In fact, the index variable is called a dummy variable
as it has generally no effect upon its choice. It is, however, customary to use the
letters i, j, k for indices.

So, in general, if we want to take the sum of a set of real numbers a1 , a2 , a3 , · · · , an ,

we write it as follows,
a1 + a2 + a3 + · · · + an

. Which can be written more compactly using the summation notation as follows,
n
X
ai
i=1

where i is the index of summation, ai is the ith term of the summation, and the
start and end limits of the summation are 1 and n, respectively. Therefore,

Remark 5.1.1. (The Summation Notation)

The sum
a1 + a2 + a3 + · · · + an

can be concisely expressed as

n
X
ai
i=1

A few examples are presented below to demonstrate the use of the summation
notation discussed above.

I Example 5.1.1. Express 2 + 5 + 8 + 11 + · · · + 29 in the summation notation.

Solution. This is an arithmetic sequence where the first term is 2, and the com-
mon difference is 3. Therefore,
10
X
(2 + 3(i − 1))
i=1

10
Therefore,
X
(2 + 3(i − 1)) = 2 + 5 + 8 + 11 + · · · + 29.
i=1

I Example 5.1.2. Express 3 + 6 + 12 + 24 + 48 in the summation notation.

Solution. This is a geometric sequence with the first term a = 3 and common
ratio r = 2. Therefore,
X4
3 · 2i
i=0
4
Therefore, 3 · 2i = 3 + 6 + 12 + 24 + 48
P
i=0

1 1 1
I Example 5.1.3. Express 1 + + + in terms of the summation notation.
4 9 16
Solution. In order to use the summation notation, notice the denominator of
each term is a square of numbers from 1 to 4. That is,
1 1 1 1
2
+ 2+ 2+ 2
1 2 3 4

and there are a total of 4 terms. Thus, the sum can be expressed as,
4
X 1
k2
k=1
.
4 1 1 1 1
Therefore, .
P
2
= 1 + + +
k=1 k 4 9 16

A few summation rules are provided below to simplify the process for large and
complex summation.

Rule 5.1.1 (Sigma Notation Rules). The following rules demonstrate the lin-
earity of Sigma notation.
n n
1. Constant Multiple Rule:
X X
cai = c × ai
i=1 i=1
n n n
2. Summation Rule:
X X X
(ai + bi ) = ai + bi
i=1 i=1 i=1
n n n
3. Difference Rule:
X X X
(ai − bi ) = ai − bi
i=1 i=1 i=1

A few examples are shown below to demonstrate the rules above.

4
I Example 5.1.4. Evaluate
X
(i + 2i)
i=1

Solution. Using the summation rule, we can split the sum into two separate sums,
4
X 4
X 4
X
(i + 2i) = i+ 2i
i=1 i=1 i=1
X4 X 4
= i+2 i
i=1 i=1

4
Now, we calculate the sum
X
i,
i=1

4
X
i=1+2+3+4
i=1

= 10

This implies that,

4
X 4
X 4
X
(i + 2i) = i+2 i
i=1 i=1 i=1

= 10 + 2 · 10
= 30

3
I Example 5.1.5. Evaluate
X
2i − i2

i=1

Solution. Using the summation rule, we can split the sum into two separate sums,
3
X 3
X 3
X
2
i2

2i − i = 2i −
i=1 i=1 i=1

Now, we calculate each sum individually,

3
X 3
X
2i = 2 i
i=1 i=1

= 2 (1 + 2 + 3)
= 12

Similarly,
3
X
i2 = 1 2 + 2 2 + 3 2
i=1

=1+4+9
= 14

This implies that,

3
X 3 3
X X
2i − i2 = 2i − i2
i=1 i=1 i=1

= 12 − 14
= −2

So far, we have calculated the sum manually. For example, to solve the sum
n
i, where n = 3, we would manually calculate it as 1 + 2 + 3 = 6. However,
X

i=1
what happens when the sum is to be calculated for n = 1000, 10000, 100000,
etc.? Clearly, manual calculation falls short in such situations. Therefore, we
have the following formulas.

Theorem 5.1.1. Summation formulas are provided below,

n
1.
X
c=n×c
i=1
n
n (n + 1)
2.
X
i = 1 + 2 + 3 + ··· + n =
i=1
2
n
n (n + 1) (2n + 1)
3.
X
i2 = 12 + 22 + 32 + · · · + n2 =
i=1
6
n
n2 (n + 1)2
4.
X
3 3 3 3 3
i = 1 + 2 + 3 + ··· + n =
i=1
4

A few examples are provided below to demonstrate the formulas above.

I Example 5.1.6. Calculate 1 + 2 + 3 + · · · + 10000.

Solution. From the formula above,
10000
X 10000(10000 + 1)
i=
i=1
2
100010000
=
2
= 50005000

Therefore, 1 + 2 + 3 + · · · + 10000 = 50005000

I Example 5.1.7. Calculate 12 + 22 + 32 + · · · + 100002 .

Solution. From the formula above,
100
X 10000(10000 + 1)(2 · 10000 + 1)
i2 =
i=1
6

2000300010000
=
6
= 333383335000
Therefore, 12 + 22 + 32 + · · · + 100002 = 333383335000

Having discussed the summation notation as a sort of prerequisite to start the

discussion of the area under a curve, it is now the time that we may finally
discuss that topic. We start off by discussing the area approximation in the
following subsection.

Chap. 5 / Sec. 5.1 / Subsec. 5.1.2 : Area Approximation

With an understanding of the summation notation, we now calculate the exact
x2
area A under the curve of f (x) = and above the x-axis from x = 1 to x = 4,
4
as shown below.

1 A

1 2 3 4

Figure 5.3

There are no simple formulas to calculate the area of this particular shape. How-
ever, its area can be approximated to any desired degree of accuracy by using
the areas of familiar shapes, such as rectangles. The idea here is that the area
A can be bounded between a lower area limit L and an upper area limit U , that
is, L ≤ A ≤ U (see Figure 5.4), in a way that gives a relatively good approxi-
mation of the exact area A in terms of the lower and upper area limit L and U ,
respectively for the reason that L and U are easy to calculate.

Now, to calculate the values of these lower and upper area limits L and U , we
take a visual approach by considering the figure below.

4 4 4

3 3 3

2 ≤ 2 ≤ 2

1 1 A 1

1 2 3 4 1 2 3 4 1 2 3 4
(a) (b) (c)
Figure 5.4

The lower area limit L of the exact area A is shown in Figure 5.4(a), and the
upper area limit U is in Figure 5.4(c).

The area of the rectangle shown in Figure 5.4(a) is 0.75. This is because the
width of the rectangle is 4 − 1 = 3 and its height is the value of the function
x2 12 1
f (x) = at the point x = 1, that is, f (1) = = , thus, its area yields
4 4 4
1
× 3 = 0.75.
4

Similarly, the area of the rectangle shown in Figure 5.4(c) yields 12. This is be-
42 16
cause the width of the rectangle is 4−1 = 3 and its height is f (4) = = = 4,
4 4
resulting the area to be 4 × 3 = 12.

Notice that the shaded region shown in Figure 5.4(a) is completely contained
within Figure 5.4(b), but not all of the shaded region in Figure 5.4(b) is within
Figure 5.4(a). Therefore, the area of the shaded region in Figure 5.4(a) is strictly
smaller than that of Figure 5.4(b), thus 0.75 < A .

Similarly, notice that the shaded region shown in Figure 5.4(b) is completely
contained within Figure 5.4(c), but not all of the shaded region in Figure 5.4(c)
is within Figure 5.4(b). Therefore, the area of the shaded region in Figure 5.4(b)
is strictly smaller than that of Figure 5.4(c), thus A < 12 .

This implies that the area of the shaded region in Figure 5.4(b) is strictly less
than that of Figure 5.4(c) and greater than that of Figure 5.4(a), therefore, the
exact area A is bounded by the inequality 0.75 < A < 12 .

As a result, we’ve derived an approximation of the exact area A in the form of an

inequality. That being so, it is important to highlight a few considerations. For

example, notice that the interval on which we’re approximating the exact area
is [1, 4]. This means each rectangle’s width is 3 units as 4 − 1 = 3. The height
of which is determined by the function’s value at the left and right endpoints
of the interval [1, 4], respectively. In this way, we’ve constructed rectangles that
are either exactly under the curve or exactly above the curve. That is, we’ve
constructed rectangles such that they’re either underestimating or overestimat-
ing the exact area A. This is important to consider as choosing endpoints other
than the left or right of the interval [1, 4] would lead to rectangles that are nei-
ther exactly under the curve nor exactly over the curve, making the inequality
quite ambiguous, as shown below.

4 4

3 3

2 2

1 1

1 2 3 4 1 2 3 4

(d) (e)

Figure 5.5

Now, turning back to our approximation, we see that although the range 0.75 <
A < 12 provides a decent estimation, we can achieve a more accurate approxi-
mation of the exact area A by using a greater number of rectangles. Previously,
we used only one rectangle, but by increasing the number of rectangles, we can
get a better approximation for the reason described below using the following
figure.

4 4 4

3 3 3

2 2 2

1 1 1

1 2 3 4 1 2 3 4 1 2 3 4

Figure 5.6

Intuitively, it is clear that when we consider more rectangles, each rectangle

becomes thinner to fit within the given interval. As a result, these rectangles
x2
cover more and more of the exact area A under the graph of f (x) = . This
4
intuition can be shown to be mathematically valid by comparing the results
obtained from constructing three rectangles—each under and over the curve—
versus those obtained from a single rectangle.

For that, we need to first divide the interval [1, 4] into sub-intervals for determin-
ing the left and right endpoints of each sub-intervals, as previously explained.
Let n be the number of rectangles constructed, then if the width of the entire
interval [1, 4] is 4 − 1, then constructing n = 3 rectangles would make the width
4−1 3
of each rectangle equal to = = 1. The height of each rectangle would
3 3
be the function’s value at either endpoint of each sub-interval, as before.

With that, let us first underestimate the area A using the left-endpoint approx-
imation with 3 rectangles as follows.

1st 2nd 3rd

1 2 3 4

Figure 5.7

For the 1st rectangle:

12
1. Height: f (1) = = 0.25
4
2. Width: 1

3. Area1 : height × width = 0.25 × 1 = 0.25

For the 2nd rectangle:

22 4
1. Height: f (2) = = =1
4 4
2. Width: 1

3. Area2 : height × width = 1 × 1 = 1

For the 3rd rectangle:

32 9
1. Height: f (3) = = = 2.25
4 4
2. Width: 1

3. Area3 : height × width = 2.25 × 1 = 2.25

Total underestimation:

Total Area = Area1 + Area2 + Area3

= 0.25 + 1 + 2.25
= 3.5

Now, we overestimate the exact area A using the left-endpoint approximation

with 3 rectangles.

1st 2nd 3rd

1 2 3 4

Figure 5.8

For the 1st rectangle:

22 4
1. Height: f (2) = = =1
4 4

2. Width: 1

3. Area1 : height × width = 1 × 1 = 1

For the 2nd rectangle:

32 9
1. Height: f (3) = = = 2.25
4 4
2. Width: 1

3. Area2 : height × width = 2.25 × 1 = 2.25

For the 3rd rectangle:

42 16
1. Height: f (4) = = =4
4 4
2. Width: 1

3. Area3 : height × width = 4 × 1 = 4

Total overestimation:

Total Area = Area1 + Area2 + Area3

= 1 + 2.25 + 4
= 7.25

This implies that using three rectangles to approximate the exact area A gives
3.5 < A < 7.25 , which is clearly a better approximation compared to the pre-
vious one, where the bound for the exact area A was 0.75 < A < 12. Thus,
it has been mathematically shown that using more rectangles provides a closer
approximation of the exact area A.

Since more rectangles cover more of the area A, this means an even better
approximation is always possible as there is no restriction to the number of rect-
angles that can be fitted within a given interval. However, manually calculating
the height, width, and area of each rectangle, and then summing them up for
the total estimation, is going to be quite tedious for large values. For example,
if we fit a billion rectangles within a given interval, is it possible to manually
calculate the area of all those rectangles a billion times? Therefore, let us now
generalize the entire process for any arbitrary function f (x), in the following
subsection.

Chap. 5 / Sec. 5.1 / Subsec. 5.1.3 : Generalized Area Approximation

Let y = f (x) be any continuous, non-negative arbitrary function defined over
the interval [a, b] as shown below.

a b
Figure 5.9

To determine the exact area A bounded under the curve of f (x) and above the
x-axis, from x = a to x = b, we divide the interval [a, b] into n sub-intervals
of equal width. To do that, we use a set of increasing points a = x0 < x1 <
x2 < x3 < · · · < xn−1 < xn = b that are equally spaced between x = a and x = b.

Since the total width of the interval [a, b] is b − a, therefore dividing it into
b−a
n parts gives each sub-interval a width of . We denote this as ∆x, thus
n
b−a
∆x = . Therefore, ∆x represents the space between each of the points
n
a = x0 < x1 < x2 < x3 < · · · < xn−1 < xn = b, as shown in the figure below.

Δx Δx Δx Δx

a=x0 x1 x2 xi-1 xi xn-1 xn=b

Figure 5.10

This implies that the interval [a, b] is divided into the following n sub-intervals:

[x0 , x1 ], [x1 , x2 ], [x2 , x3 ], · · · , [xi−1 , xi ], · · · , [xn−1 , xn ]

| {z } | {z } | {z } | {z } | {z }
1st 2nd 3rd ith nth

where,

x1 = x0 + ∆x
x2 = x0 + 2∆x
x3 = x0 + 3∆x
.. ..
. .
xi−1 = x0 + (i − 1)∆x (1)
xi = x0 + i∆x (2)
.. ..
. .
xn = x0 + n∆x

The set of points P = {x0 , x1 , x2 , · · · , xn }, where a = x0 < x1 < x2 < · · · <

xn−1 < xn = b, dividing the interval [a, b] into n sub-intervals of the form
[x0 , x1 ], [x1 , x2 ], [x2 , x3 ], · · · , [xn−1 , xn ] is called a partition of [a, b]. The partition
of [a, b] becomes a regular partition of [a, b] if each sub-interval shares the same
width.

Author's Interruption. Spoiler alert: As we move through the text, we will

be required to utter “the ith sub-interval” over and over again. Therefore,
it is always a good idea to acquaint ourselves with the ith sub-interval,
that is,
[xi−1 , xi ]

from the beginning, where

xi−1 = x0 + (i − 1)∆x =⇒ the left-endpoint of the ith sub-interval

xi = x0 + i∆x =⇒ the right-endpoint of the ith sub-interval

Make sure to take a minute or two to comprehend that.

Now that we have formed the sub-intervals for an arbitrary function f (x), we
now use them to form rectangles that let us approximate the area under the
curve. This can result in either an overestimation or an underestimation of the
exact area, depending on how the rectangles are constructed.

We will first consider an underestimation. As discussed earlier, underestimation

occurs when the rectangles are constructed so that they lie entirely beneath the

curve. For that, as we saw before, the left-endpoint of each sub-interval is used
to determine the height of the rectangles. Therefore, we start our discussion
with the left-endpoint approximation and following that, we will discuss the
right-endpoint approximation.

But before that, it is important to note the ith sub-interval, that is,

[xi−1 , xi ] = [x0 , x1 ] , [x1 , x2 ] , [x2 , x3 ] , · · · , [xn−1 , xn ]

for i = 1, 2, 3, · · · , n, respectively. With that in mind, we start discussing the

left-endpoint approximation, as follows.

· · · / Subsec. 5.1.3 / Segment 5.1.3.1 : The Left-Endpoint Approximation

Since [xi−1 , xi ] = [x0 , x1 ] , [x1 , x2 ] , [x2 , x3 ] , · · · , [xn−1 , xn ] for i = 1, 2, 3, · · · , n,
respectively. Therefore, we construct a rectangle on the ith interval [xi−1 , xi ],
for i = 1, 2, 3, · · · , n, which have a width of ∆x and a height equal to the function
value at the left endpoint of [xi−1 , xi ], that is, f (xi−1 ). Therefore, the area yields,

Area = height × width = f (xi−1 ) · ∆x

Now, by summing all these areas for i = 1, 2, 3, · · · , n, we get an underesti-

mation of the exact area A. We denote this underestimation as Ln (where L
represents that we’re using the left endpoint approximation and the subscript
n represents the number of rectangles we’re using). Thus, for i = 1, 2, 3, · · · , n,
the underestimation Ln yields,

Ln = (f (x0 ) · ∆x) + (f (x1 ) · ∆x) + · · · + (f (xn−2 ) · ∆x) + (f (xn−1 ) · ∆x)

| {z } | {z } | {z } | {z }
1st rectangle 2nd rectangle (n−1)th rectangle nth rectangle

= ∆x · f (x0 ) + f (x1 ) + · · · + f (xn−2 ) + f (xn−1 )
n
for i = 1, 2, 3, · · · , n
X
= ∆x · f (xi−1 )
i=1
n
X
= f (xi−1 ) · ∆x
i=1

Therefore, for any arbitrary function f (x), we have the following definition,

Definition 5.1.1. (The Left-Endpoint Approximation)

Let f (x) be a continuous, non-negative function defined over an interval
[a, b]. Then, the left-endpoint approximation Ln to underestimate the exact
area A bounded under the curve of f (x) and over the x-axis, from x = a
to x = b, is defined as,
n
X
Ln = f (xi−1 )∆x
i=1

where, n is the number of rectangles, xi−1 is the left-endpoint of the ith

sub-interval, and ∆x is the width of each sub-interval.

Similar to the left-endpoint, we discuss the right-endpoint approximation below.

· · · / Subsec. 5.1.3 / Segment 5.1.3.2 : The Right-Endpoint Approximation

Since [xi−1 , xi ] = [x0 , x1 ] , [x1 , x2 ] , [x2 , x3 ] , · · · , [xn−1 , xn ] for i = 1, 2, 3, · · · , n,
respectively. Therefore, we construct a rectangle on the ith interval [xi−1 , xi ],
for i = 1, 2, 3, · · · , n, which have a width of ∆x and a height equal to the function
value at the right endpoint of [xi−1 , xi ], that is, f (xi ). Therefore, the area yields,

Area = height × width = f (xi ) · ∆x

Now, by summing all these areas for i = 1, 2, 3, · · · , n, we get an overestimation

of the exact area A. We denote this overestimation as Rn (where R represents
that we’re using the right endpoint approximation and the subscript n repre-
sents the number of rectangles we’re using). Thus, for i = 1, 2, 3, · · · , n, the
overestimation Rn yields,

Rn = (f (x1 ) · ∆x) + (f (x2 ) · ∆x) + · · · + (f (xn−1 ) · ∆x) + (f (xn ) · ∆x)

| {z } | {z } | {z } | {z }
1st rectangle 2nd rectangle (n−1)th rectangle nth rectangle

= ∆x · f (x1 ) + f (x2 ) + · · · + f (xn−1 ) + f (xn )
n
for i = 1, 2, 3, · · · , n
X
= ∆x · f (xi )
i=1
n
X
= f (xi ) · ∆x
i=1

Therefore, for any arbitrary function f (x), we have the following definition,

Definition 5.1.2. (The Right-Endpoint Approximation)

Let f (x) be a continuous, non-negative function defined over an interval
[a, b]. Then, the right-endpoint approximation Rn to overestimate the exact
area A bounded under the curve of f (x) and over the x-axis, from x = a
to x = b, is defined as,
Xn
Rn = f (xi )∆x
i=1

where, n is the number of rectangles, xi is the right-endpoint of the ith

sub-interval, and ∆x is the width of each sub-interval.

Having generalized the left and right endpoint approximations, we can now pro-
ceed from where we left off.

To quickly recap, recall that we aimed to mathematically show that increasing

the number of rectangles under the curve of an arbitrary function f (x) results
in a closer approximation of the exact area A. This was possible by considering
the exact area A as being bounded between two estimates—one representing un-
derestimation and the other overestimation—and showing that these bounds get
closer to the exact area as the number of rectangles increases. However, since the
manual calculation of these approximations was tedious, we generalized the pro-
cess for both underestimation and overestimation. Through this generalization,
we found that for any arbitrary function f (x), the underestimation (denoted as
Ln ) of the bounded area A is given by,
n
X
Ln = f (xi−1 )∆x
i=1

Similarly, the overestimation (denoted as Rn ) of the bounded area A is given by,

n
X
Rn = f (xi )∆x
i=1

This suggests that for any arbitrary function f (x), the exact bounded area A is
as follows,
n
X n
X
f (xi−1 )∆x < A < f (xi )∆x
i=1 i=1

where, n represents the number of rectangles.

Now, if we go back to our previous problem of finding the exact area under the
x2
curve of f (x) = and above the x-axis, from x = 1 to x = 4; we see that the
4
exact area A can be determined by the following inequality,
n
X n
X
f (xi−1 )∆x < A < f (xi )∆x
i=1 i=1

However, in order to use the inequality above, we have to first determine the
value of xi−1 , xi , and ∆x.

Since the entire width of the interval [1, 4] is 3, therefore, dividing this interval
into n sub-interval gives each sub-interval a width of

3
∆x =
n
Now, to determine xi−1 and xi , recall that,
3(i − 1) + n 3
from equation (1), xi−1 = 1 + (i − 1) · =
n n
3 3i + n
from equation (2), xi =1+i· =
n n
n
With that in hand, we now began to calculate the underestimation Ln =
X
f (xi−1 )∆x
i=1
2
x
to the exact area under the curve of the function f (x) = as follows,
4
n
X
Ln = f (xi−1 )∆x
i=1
n
X 3 (i − 1) + n
= f · ∆x
i=1
n
3 (i − 1) + n 2

n
n 3 x2
since, f (x) =
X
= ·
i=1
4 n 4
(3 (i − 1) + n)2
n
X
n2 3
= ·
i=1
4 n
(3 (i − 1))2 + 2 · 3 (i − 1) · n + n2
n
X
n2 3
= ·
i=1
4 n

n
X 9(i − 1)2 + 6n (i − 1) + n2 3
= ·
i=1
4n2 n

n 3 9(i − 1)2 + 6n (i − 1) + n2
X
=
i=1
4n3
n
3 X 2 2

= · 9(i − 1) + 6n (i − 1) + n
4n3 i=1
n n n
!
3 X 2
X X
= 3 9(i − 1) + 6n (i − 1) + n2
4n i=1 i=1 i=1
n n n
!
3 X X X
= 3 9· (i − 1)2 + 6n · (i − 1) + n2
4n i=1 i=1 i=1
n n n
!
3 X X X
i2 − 2i + 1 + 6n · n2

= 9· (i − 1) +
4n3 i=1
n n n
!i=1 n
i=1
n
! n
!
3 X X X X X X
= 3 9 i2 − 2i + 1 + 6n i− 1 + n2
4n i=1 i=1 i=1 i=1 i=1 i=1
n n n
! n n
! n
!
3 X X X X X X
= 3 9 i2 − 2 i+ 1 + 6n i− 1 + n2
4n
i=1 i=1 i=1 i=1
i=1 i=1

3 n (n + 1) (2n + 1) n (n + 1) n (n + 1) 3
= 3 9 −2 + n + 6n −n +n
4n 6 2 2

3 n (n + 1) (2n + 1) n (n + 1) n (n + 1)
Now, simplifying 3 9 −2 + n + 6n −n +n 3
4n 6 2 2
yields,
3(14n2 − 15n + 3)
Ln =
8n2
n
Similarly, we calculate the overestimation Rn = f (xi )∆x to the exact area
X

i=1
x2
A under the curve of the function f (x) = as follows,
4
n
X
Rn = f (xi )∆x
i=1
n
X 3i + n
= f · ∆x
i=1
n
3i + n 2

n
n 3 x2
since, f (x) =
X
= ·
i=1
4 n 4

(3i + n)2
n
X
n2 3
= ·
i=1
4 n
(3i)2 + 2 · 3i · n + n2
n
X
n2 3
= ·
i=1
4 n
n
X 9i2 + 6ni + n2 3
= ·
i=1
4n2 n
n
X 3 9i2 + 6ni + n 2
=
i=1
4n3
n
3 X 2
= 3· 9i + 6ni + n2
4n i=1
n n n
!
3 X X X
= 9i2 + 6ni + n2
4n3 i=1 i=1 i=1
n n n
!
3 X X X
= 3 9· i2 + 6n · i+ n2
4n i=1 i=1 i=1

3 n (n + 1) (2n + 1) n (n + 1) 3
= 3 9· + 6n · +n
4n 6 2

3 n (n + 1) (2n + 1) n (n + 1)
Now, simplifying 3 9 · + 6n · + n yields,
3
4n 6 2

3(14n2 + 15n + 3)
Rn =
8n2

This suggests that we have derived two formulas for underestimation Ln and
overestimation Rn to the exact area A, where n represents the number of rect-
x2
angles. That is, the exact area A bounded under the curve of f (x) = and
4
above the x-axis, from x = 1 to x = 4, is given by the following inequality,
3(14n2 − 15n + 3) 3(14n2 + 15n + 3)
< A <
8n2 8n2
To verify these formulas, set n = 1 and n = 3 to see if the resulting inequalities
match with the previously manually calculated inequalities: 0.75 < A < 12 for
n = 1 and 3.5 < A < 7.25 for n = 3.

If n = 1, then

3 14n2 − 15n + 3 3 14n2 + 15n + 3
<A<
8n2 8n2

3 14 · 12 − 15 · 1 + 3 3 14 · 12 + 15 · 1 + 3
<A<
8 · 12 8 · 12
0.75 < A < 12

Which matches with the previously manually done inequality. Similarly, if n = 3,

then,

3 14n2 − 15n + 3 3 14n2 + 15n + 3
<A<
8n2 8n2
2
3 14 · 3 − 15 · 3 + 3 3 14 · 32 + 15 · 3 + 3
<A<
8 · 32 8 · 32
3.5 < A < 7.25

Which also matches with the previously manually done inequality of the exact
area A.

Notice that as the number of rectangles n increases, both the underestimation

and overestimation seem to converge to a particular value around 5. That is,

For n = 1 rectangles:

3(14n2 + 15n + 3) 3(14 · 12 + 15 · 1 + 3)

1. Overestimation: = = 12
8n2 8 · 12
3(14n2 − 15n + 3) 3(14 · 12 − 15 · 1 + 3)
2. Underestimation: = = 0.75
8n2 8 · 12
3. Inequality: 0.75 < A < 12

For n = 3 rectangles:

3(14n2 + 15n + 3) 3(14 · 32 + 15 · 3 + 3)

1. Overestimation: = = 7.25
8n2 8 · 32
3(14n2 − 15n + 3) 3(14 · 32 − 15 · 3 + 3)
2. Underestimation: = = 3.5
8n2 8 · 32
3. Inequality: 3.5 < A < 7.25

For n = 9 rectangles:

3(14n2 + 15n + 3) 3(14 · 92 + 15 · 9 + 3)

1. Overestimation: = = 5.89
8n2 8 · 92
3(14n2 − 15n + 3) 3(14 · 92 − 15 · 9 + 3)
2. Underestimation: = = 4.6389
8n2 8 · 92
3. Inequality: 4.639 < A < 5.889

For n = 81 rectangles:

3(14n2 + 15n + 3) 3(14 · 812 + 15 · 81 + 3)

1. Overestimation: = = 5.3196
8n2 8 · 812
3(14n2 − 15n + 3) 3(14 · 812 − 15 · 81 + 3)
2. Underestimation: = = 5.1807
8n2 8 · 812
3. Inequality: 5.1807 < A < 5.3196

For n = 6561 rectangles:

3(14n2 + 15n + 3) 3(14 · 65612 + 15 · 6561 + 3)

1. Overestimation: = = 5.2509
8n2 8 · 65612
3(14n2 − 15n + 3) 3(14 · 65612 − 15 · 6561 + 3)
2. Underestimation: = =
8n2 8 · 65612
5.249

3. Inequality: 5.249 < A < 5.2509

In fact, as n → ∞, both the underestimation and overestimation converge to

the exact area A. This can be proven by taking the limit of the underestimation
and overestimation as n → ∞.

For that, let us first take the limit of the underestimation as n → ∞ as follows.

3(14n2 − 15n + 3) 3 14n2 − 15n + 3

lim = lim ·
n→∞ 8n2 n→∞ 8 n2
3 14n2 − 15n + 3
= · lim
8 n→∞ n2
2
14n 15n 3
3 2
− 2 + 2
= · lim n n n
8 n→∞ n2
n2
15 3
3 14 − +
= · lim n n2
8 n→∞ 1
3 15 3
= · lim 14 − + 2
8 n→∞ n n
3
= · (14 − 0 + 0)
8
42
=
8
= 5.25

Similarly, taking the limit of the overestimation as n → ∞ yields,

3(14n2 + 15n + 3) 3 14n2 + 15n + 3
lim = lim ·
n→∞ 8n2 n→∞ 8 n2
2
3 14n + 15n + 3
= · lim
8 n→∞ n2
2
14n 15n 3
3 2
+ 2 + 2
= · lim n n
2
n
8 n→∞ n
n2
15 3
3 14 + + 2
= · lim n n
8 n→∞
1
3 15 3
= · lim 14 + + 2
8 n→∞ n n
3
= · (14 + 0 + 0)
8
42
=
8
= 5.25
Now, since both the underestimation and overestimation approach a value of
5.25 as the number of rectangles n increases indefinitely, that is,
3(14n2 − 15n + 3) 3(14n2 + 15n + 3)
lim = lim = 5.25 = A
n→∞ 8n2 n→∞ 8n2
x2
therefore the exact area A bounded under the curve of f (x) = and above the
4
x-axis, from x = 1 to x = 4 is 5.25.

Having determined the exact area A in the previously mentioned problem, it’s
important to notice something significant. In finding the exact area, we con-
structed an infinite number of rectangles under the curve and observed that
both the underestimation and overestimation approach the same value. This
implies that to determine the exact area, the use of just one approximation
is sufficient. Moreover, the approximation doesn’t need to be specifically an
overestimation or an underestimation—it can be any approximation when the
number of rectangles is unbounded.

Previously, to perform an underestimation, we used the left endpoint of the

ith sub-interval, and for an overestimation, we used the right endpoint. This
was necessary because using endpoints other than the right or left would nei-
ther completely result in an underestimation nor an overestimation. However,

we later realized that as the number of rectangles n increases indefinitely, both

approximations converge to the same value. Therefore, it is now unnecessary
to use only the right or left endpoints as long as the number of rectangles n
increases indefinitely.

It is the best moment to introduce the Riemann Sum and define the area under
a curve in terms of that in the following subsection.

Chap. 5 / Sec. 5.1 / Subsec. 5.1.4 : Riemann Sum and the Definition of the
Area Under a Curve
The Riemann sum resolves a few restrictions. For example, we can now choose
any point x∗i ∈ [xi−1 , xi ] to determine the height of the rectangle over the ith
sub-interval, for i = 1, 2, 3, · · · , n, as shown below,

1 2 3 4

Figure 5.11

Doing so would not change the limit of the approximation, if the number of
rectangle n → ∞ is considered. Therefore, let f (x∗i ) represent the height of
the rectangle over the ith sub-interval [xi−1 , xi ], for i = 1, 2, 3, · · · , n, where,
x∗i ∈ [xi−1 , xi ] is any arbitrary point. Then, the area of the rectangle over that
sub-interval yields,

height × width = f (x∗i ) · ∆x for i = 1, 2, 3, · · · , n

Now, summing these quantities for i = 1, 2, 3, · · · , n, we get an approximation

to the exact area A, which is called the Riemann Sum:

A ≈ (f (x∗1 ) · ∆x) + (f (x∗2 ) · ∆x) + · · · + f (x∗n−1 ) · ∆x + (f (x∗n ) · ∆x)

| {z } | {z } | {z } | {z }
1st rectangle 2nd rectangle (n−1)th rectangle nth rectangle

∗ ∗ ∗ ∗
= ∆x · f (x1 ) + f (x2 ) + · · · + f (xn−1 ) + f (xn )
| {z }
n term summations
n
for i = 1, 2, 3, · · · , n
X
= ∆x · f (x∗i )
i=1
n
X
= f (x∗i ) · ∆x
i=1
Xn
∴A≈ f (x∗i )∆x
i=1

n
The approximation f (x∗i )∆x is called the Riemann sum, named after Ger-
X

i=1
man mathematician Georg Friedrich Bernhard Riemann. Therefore, we have
the following definition of the approximation above.

Definition 5.1.3. (Riemann Sum)

Let f (x) be a continuous, non-negative function defined over an interval
[a, b]. Then, the Riemann sum S is defined as
n
X
S= f (x∗i )∆x
i=1

where, n is the number of rectangles, x∗i is any point in the ith sub-interval
[xi−1 , xi ], and ∆x is the width of each sub-interval.

The point x∗i ∈ [xi−1 , xi ], for i = 1, 2, 3, · · · , n, used in the approximation above

is called a “sample point”. Now, based on the definition of the Riemann sum,
we define the area under a curve. That is, if the number of rectangles n → ∞,
then the limit n
lim
X
f (x∗i )∆x
n→∞
i=1

represents the exact area A in terms of the Riemann sum. As a result, we can
define the exact area A bounded under the curve of f (x) and above the x-axis,
from x = a to x = b, in terms of the Riemann sum as follows.

Definition 5.1.4. (Area Under a Curve)

Let f (x) be a continuous, non-negative function defined over an interval
n
[a, b], and let S = f (x∗i )∆x be the Riemann sum for f (x). Then, the
X

i=1
exact area A under the curve of f (x) and over the x-axis, from x = a to
x = b, is defined as,
n
A = lim
X
f (x∗i )∆x
n→∞
i=1

Now, notice that the current definition of the “area under a curve” comes with
a great limitation: it only applies when the function f (x) is non-negative. In
other words, for this definition to apply, the curve has to lie entirely above the
x-axis. If any part of the curve lies below the x-axis, the definition above can’t
be applied there.

To resolve the issue, the concept of “Definite Integral” comes into play. Definite
integral expands the definition of the area under a curve to include regions
bounded by curves that can lie below the x-axis by considering both positive
and negative regions bounded by the curve of f (x). To make this distinction, we
introduce the idea of “net signed area,” where the curve lying above the x-axis
contributes positively, and below the x-axis contributes negatively, as discussed
in the following section.

The End of Section 5.1

The Area Under a Curve

§ 5.2 The Definite Integral

In the preceding section we defined the area under a curve in terms of Riemann
sums as follows: n
A = lim
X
f (x∗i )∆x
n→∞
i=1

However, that definition required the function f (x) to be non-negative. Since not
all functions satisfy these conditions, we may need to remove these requirements.
As a result, f (x) can now sometimes be negative. But this raises the question:
how do we interpret Riemann sums when f (x) is negative at some or all of its

points over the interval [a, b]? The answer is given through the concept of net
signed area, as discussed below.

Chap. 5 / Sec. 5.2 / Subsec. 5.2.1 : Net Signed Area and the Definition of the
Definite Integral
To understand the net signed area, approximate the area bounded by the curve
of f (x) = −x2 from x = 0 to x = 2, using n = 4 rectangles.

Since the interval is [0, 2], therefore after dividing the interval into n sub-intervals
2−0
results in the width of each sub-interval being ∆x = = 0.5. Therefore,
4
the sub-intervals yields,

[0, 0.5], [0.5, 1], [1, 1.5], [1.5, 2]

Now, to use the Riemann sum approximation, we choose the sample points
(which can be any points within each respective sub-interval),

x∗1 = 0.3 ∈ [0, 0.5]

x∗2 = 0.7 ∈ [0.5, 1]
x∗3 = 1.2 ∈ [1, 1.5]
x∗4 = 1.9 ∈ [1.5, 2]

As a result, the resulting Riemann sums yield the following,

n
X 4
X
f (x∗i )∆x = f (x∗i ) · (0.5)
i=1 i=1

= f (0.3) · (0.5) + f (0.7) · (0.5) + f (1.2) · (0.5) + f (1.9) · (0.5)

= (f (0.3) + f (0.7) + f (1.2) + f (1.9)) · (0.5)
= (−0.09 − 0.49 − 1.44 − 3.61) · (0.5)
= (−5.63) · (0.5)
= −2.815

Here, all the values of the height f (x∗i ) are negative. As a result, the Riemann
sum—which approximates the area under the curve—turns out to be also neg-
ative. However, if the Riemann sum is used to approximate the “area” under
the curve, then how can the area—being a physical measurement—be a negative

quantity?

The key point to realize here is that the Riemann sum, in this case, does not
approximate the area in the physical sense. Instead, it approximates the “net
signed area,” that is, a difference of areas that accounts for both the positive
and the negative contributions of areas.

Although the positive contribution of area implies the actual positive area, the
negative contribution of area does not literally imply a “negative area,” as if
there were an absence or void of space between the curve and the x-axis. In-
stead, it represents the negative of the actual area bounded above the curve
and below the x-axis (see the red shaded region in Figure 5.12). This happens
because, in the Riemann sum, the height of the ith rectangle is determined by
the the values of the function f (x∗i ), where x∗i ∈ [xi−1 , xi ]. Now, if f (x∗i ) falls
below the x-axis, then the height of the ith rectangle, that is, −f (x∗i ) represents
the negative of the actual height |f (x∗i )|, and thus the product, −f (x∗i )∆x, rep-
resents the negative of the actual product |f (x∗i )∆x|, which is the area of the
ith rectangle when f (x) falls below the x-axis.

Now, times occur when f (x) is positive over some parts of [a, b] and negative
over other parts of [a, b]. In such cases, the Riemann sum approximates the pos-
itive area when f (x) is positive and negative of the area when f (x) is negative.
Therefore, the Riemann sum over the interval [a, b] approximates the sum of the
positive areas that lie above the x-axis and the negative of the areas that lie
beneath the x-axis, which is precisely what we refer to as the net signed area,
as shown below.

f(x)

Figure 5.12

In the figure above, the area bounded under the curve of f (x) and above the
x-axis is A1 . Similarly, the area bounded above the curve of f (x) and below the
x-axis is −A2 . Now, the Riemann sum, as the number of rectangles n → ∞,
represents the sum of these areas A1 and −A2 . Therefore, we write,
n
lim
X
f (x∗i )∆x = A1 + (−A2 )
n→∞
i=1

= A1 − A2
The quantity A1 − A2 is what we call the net signed area, which can be positive,
negative, or even zero.

If there’s multiple positive and negative of the areas; for example, suppose,
A1 , A2 , A3 , · · · , An are the areas bounded under the curve of f (x) and above the
x-axis, and −B1 , −B2 , −B3 , · · · , −Bk are the areas bounded above the curve of
f (x) and below the x-axis. Then the net signed area yields,
n
lim
X
f (x∗i )∆x
n→∞
i=1

= (A1 + A2 + A3 + · · · + An ) + ((−B1 ) + (−B2 ) + (−B3 ) + · · · + (−Bk ))

= (A1 + A2 + A3 + · · · + An ) + (−B1 − B2 − B3 − · · · − Bk )
= (A1 + A2 + A3 + · · · + An ) + (− (B1 + B2 + B3 + · · · + Bk ))
= (A1 + A2 + A3 + · · · + An ) − (B1 + B2 + B3 + · · · + Bk )
= sum of areas above the x-axis − sum of areas below the x-axis
As a result, we have the following definition of the net signed area.

Definition 5.2.1. (Net Signed Area)

Let R be the region bounded by the graph of f (x) and the x-axis over the
interval [a, b]. Then the net signed area of R is defined as,
n
lim
X
f (x∗i )∆x = R1 − R2
n→∞
i=1

where, R1 is the sum of areas of R that lie above the x-axis and R2 is the
sum of areas of R that lie below the x-axis.

This suggests that the Riemann sum approximates the net signed area bounded
by the curve of f (x) and the x-axis over the interval [a, b]. This approximation
gets closer to the exact net signed area as the number of rectangles n → ∞.

Notice that as the number of rectangles n → ∞, the width of each rectangle

∆x → 0. This idea is intuitive but important to emphasize because the Riemann
sum we have used so far considers regular partitions where all sub-intervals have
the same width ∆x. However, in reality, the interval [a, b] can be divided into
sub-intervals of unequal widths. In such cases, each sub-interval has a different
width, denoted as ∆xi for i = 1, 2, 3, . . . , n.
n
When n → ∞ in the limit lim f (x∗i )∆xi , all ∆xi → 0. We denote the largest
X
n→∞
i=1
width among the sub-intervals as max ∆xi . This implies that if max ∆xi → 0,
then all ∆xi also approach 0, for i = 1, 2, 3, · · · , n. Therefore, we write the net
signed area bounded by the curve of f (x) and the x-axis over a given interval
in terms of the following Riemann sum,
n
lim
X
f (x∗i )∆xi
max ∆xi →0
i=1

where, max ∆xi is the largest width among all sub-intervals. This idea works
because as max ∆xi → 0, all ∆xi → 0, for xi = 1, 2, 3, · · · , n. As all ∆xi → 0,
n must also increase indefinitely.

With that in mind, it is finally the time that we define the “definite integral” to
expand the previously discussed definition of the area under a curve.

Definition 5.2.2. (The Definite Integral)

Let f (x) be a function defined on an interval [a, b]. If the Riemann sum,
n
lim
X
f (x∗i )∆xi
max ∆xi →0
i=1

exists, then the definite integral of f (x), from x = a to x = b, symbolically,

Zb
f (x) dx
a

is defined as the following limit:

Zb n
lim
X
f (x) dx = f (x∗i )∆xi
max ∆xi →0
a i=1

and f (x) is said to be integrable on [a, b].

The definition above suggests that for a definite integral to exist, the limit used
to define that integral must first exist. The precise interpretation of this limit
is that,

For all ε > 0, there exists an integer N such that,

Zb n
lim
X
f (x) dx − f (x∗i )∆xi < ε
max ∆xi →0
a i=1

for all n > N and for all x∗i ∈ [xi−1 , xi ].

The notation for the definite integral has historical origins

Z and is read as “the
definite integral from a to b of f (x) dee x.” The symbol is called the integral
sign and was introduced by Gottfried Wilhelm Leibniz. This symbol represents
an elongated S, as in the word ‘sum’ and it was selected because an integral
represents the limit of a sum. The numbers a and b attached to the integral sign
are called the lower and upper limits of integration, respectively. The function
f (x) is called the integrand, and dx indicates that the integration is performed
with respect to x. The figure below illustrates these components.

Upper limit

of integration

x is the variable

The integral sign of integration

Integrand

Lower limit

of integration

The variable of integration is a dummy variable, it has no effect on the compu-

tation of the integral, as long as it does not conflict with other variables that
are in use. That is, the following notation all means the same thing.

Zb Zb Zb Zb
f (x) dx = f (t) dt = f (u) du = f (s) ds
a a a a

It is important to note that we have defined the definite integral for integrable
functions but not all functions fall into this criteria. A function f (x) is integrable
if f (x) is continuous on the interval [a, b]. Additionally, if f (x) is bounded on
[a, b], meaning that there exists a number M such that |f (x)| < M for all
x ∈ [a, b] and it has only a finite number of discontinuities, then f (x) is also
integrable on [a, b]. This implies that even if a function is not continuous, it can
still be integrable on [a, b]. As a result, we have the following theorem.

Theorem 5.2.1. (Integrability of Continuous Functions)

If f (x) is continuous on [a, b] or bounded on [a, b] with a finite number of
discontinuities, then f (x) is integrable on [a, b].

The theorem above suggests that if f (x) is continuous on [a, b], then the definite
Zb
integral f (x) dx represents the net signed area bounded by the curve f (x)
a
and the x-axis over that given interval. This means that the integral takes into
account both the areas above the x-axis (which are considered positive) and
those below the x-axis (which are considered negative). As a result, the definite
Zb
integral f (x) dx gives a single numerical value, which corresponds to the total
a
quantity accumulated, considering the signs of the areas.

A few examples to evaluate the definite integral using the definition above are
provided below.

Z1
I Example 5.2.1. Evaluate 2xdx
0

Solution. In order to evaluate the definite integral, we divide the interval [0, 1]
into n sub-intervals of equal width. The width of each sub-interval yields,
1−0 1
=
n n
Now, we let the ith sample point xi be the right endpoint of each sub-interval
∗

for calculation simplicity. Then, we get,

x∗i = x0 + i∆x
1
=0+i·
n
∗ i
∴ xi =
n
Therefore, the Riemann sum for the function f (x) = 2x yields,
n n
X
∗
X i 1
f (xi )∆xi = 2 ·
i=1 i=1
n n
n
2 X
= 2· i
n i=1
2 n(n + 1)
· =
n2 2
n+1
=
n
Now, as the number of rectangles n → ∞, then,
n
n+1
lim f (x∗i )∆xi = lim
X
n→∞
i=1
n→∞ n
1
1+
= lim n
n→∞ 1
1
= lim 1 + lim
n→∞ n→∞ n

=1
Z1
Therefore, 2xdx = 1
0

Z2
I Example 5.2.2. Evaluate x2 dx
0

Solution. In order to evaluate the definite integral, we divide the interval [0, 2]
into n sub-intervals of equal width. The width of each sub-interval yields,
2−0 2
=
n n
Now, we let the ith sample point x∗i be the right endpoint of each sub-interval
for calculation simplicity. Then, we get,

x∗i = x0 + i∆x
2
=0+i·
n
2i
=
n
Therefore, the Riemann sum for the function f (x) = x2 yields,
n n 2
X X 2i 2
f (x∗i )∆xi = ·
i=1 i=1
n n
n
X 4i2 2
= ·
i=1
n2 n
n
X 8i2
=
i=1
n3
n
8 X 2
= 3 i
n i=1
8 n(n + 1)(2n + 1)
= ·
n3 6
4(n + 1)(2n + 1)
=
3n2
Now, as the number of rectangles n → ∞, then,
n
4(n + 1)(2n + 1)
lim f (x∗i )∆xi = lim
X
n→∞
i=1
n→∞ 3n2
4 (n + 1)(2n + 1)
= · lim
3 n→∞ n2
4 2n2 + 3n + 1
= · lim
3 n→∞ n2

2n2 3n 1
4 2
+ 2+ 2
= · lim n n
2
n
3 n→∞ n
n2
3 1
4 2+ + 2
= · lim n n
3 n→∞
1
4 3 1
= · lim 2 + lim + lim 2
3 n→∞ n→∞ n n→∞ n
8
=
3
Z2
8
Therefore, x2 dx = .
3
0

Having discussed the definite integral and its evaluation, we now discuss its
properties in the next subsection.

Chap. 5 / Sec. 5.2 / Subsec. 5.2.2 : The Properties of the Definite Integral
The properties of the definite integral acts as a set of rules that govern how
definite integrals can be manipulated and evaluated. We start our discussion by
introducing the the reversal of limits of integration and the identical limits of
integration, as follows.

· · · / Subsec. 5.2.2 / Segment 5.2.2.1 : The Reversal of Limits of Integration

and Identical Limits of Integration
Recall that we have defined the definite integral as:

Zb n
lim
X
f (x) dx = f (x∗i )∆xi
max ∆xi →0
a i=1

where we indirectly assumed that the lower limit of integration a is less than the
upper limit of integration b, that is, a < b. However, it is certainly possible that
a > b. In such cases, the width of each sub-interval changes from ∆xi to −∆xi ,
for i = 1, 2, 3, · · · , n, in the Riemann sum on the right side of the equation. This
is due to the fact that a − b < 0 is divided into n sub-intervals, rather than

b − a > 0. As a result, we have,

Za n
lim
X
f (x) dx = f (x∗i ) (−∆xi )
max ∆xi →0
b i=1
n
lim
X
=− f (x∗i )∆xi
max ∆xi →0
i=1
Zb
=− f (x) dx
a
Za Zb
∴ f (x) dx = − f (x) dx
b a

Therefore, we have the following theorem.

Theorem 5.2.2. (The Reversal of Limits of Integration)

Let f (x) be an integrable function defined over an interval [a, b], then

Za Zb
f (x) dx = − f (x) dx
b a

Now, if the lower limit of integration a is equal to the upper limit b (i.e., a = b),
then ∆x = 0. This happens because the interval a − a = 0 is divided into n
0
sub-intervals, which is simply = 0. Therefore,
n
Za n
lim
X
f (x) dx = f (x∗i ) · 0
max ∆xi →0
a i=1

=0
Za
∴ f (x) dx = 0
a

Therefore, we have the following theorem.

Theorem 5.2.3. (Identical Limits of Integration)

Let f (x) be an integrable function defined over an interval [a, b], then
Za
f (x) dx = 0
a

· · · / Subsec. 5.2.2 / Segment 5.2.2.2 : Integrals Involving a Sum or Difference

of Functions
If f (x) and g(x) are integrable functions, then the definite integral of the function
f (x) + g(x) is given by the following theorem.

Theorem 5.2.4. (Integrals Involving a Sum of Functions)

Let f (x) and g(x) be integrable functions defined over an interval [a, b],
then
Zb Zb Zb
f (x) + g (x) dx = f (x)dx + g (x)dx
a a a

Proof. We will proof this directly.

Zb n
lim
X
(f (x) + g(x)) dx = (f (x∗i ) + g (x∗i )) ∆xi
max ∆xi →0
a i=1
n n
!
lim
X X
= f (x∗i ) ∆xi + g (x∗i ) ∆xi
max ∆xi →0
i=1 i=1
n n
lim lim
X X
= f (x∗i ) ∆xi + g (x∗i ) ∆xi
max ∆xi →0 max ∆xi →0
i=1 i=1
Zb Zb Zb
∴ (f (x) + g(x)) dx = f (x)dx + g(x)dx
a a a

This completes the proof.

Similar to the theorem above, we also have the following theorem for integrals
involving a difference of functions.

Theorem 5.2.5. (Integrals Involving a Difference of Functions)

Let f (x) and g(x) be integrable functions defined over an interval [a, b],
then
Zb Zb Zb
f (x) − g (x) dx = f (x)dx − g (x)dx
a a a

Proof. We will proof this directly.

Zb n
lim
X
(f (x) − g(x)) dx = (f (x∗i ) − g (x∗i )) ∆xi
max ∆xi →0
a i=1

n n
!
lim
X X
= f (x∗i ) ∆xi − g (x∗i ) ∆xi
max ∆xi →0
i=1 i=1
n n
lim lim
X X
= f (x∗i ) ∆xi − g (x∗i ) ∆xi
max ∆xi →0 max ∆xi →0
i=1 i=1
Zb Zb Zb
∴ (f (x) − g(x)) dx = f (x)dx − g(x)dx
a a a

This completes the proof.

· · · / Subsec. 5.2.2 / Segment 5.2.2.3 : Integrals Involving a Constant

If f (x) is an integrable function defined over an interval [a, b], then the definite
integral of the function cf (x), where c is a constant, is given by the following
theorem.

Theorem 5.2.6. (Integrals Involving a Constant)

Let f (x) be an integrable function defined over an interval [a, b], then

Zb Zb
cf (x)dx = c · f (x)dx
a a

Proof. We will proof this directly.

Zb n
lim
X
cf (x)dx = cf (x)∆xi
max ∆xi →0
a i=1
n
!
lim
X
= c· f (x)∆xi
max ∆xi →0
i=1
n
lim lim
X
= c· f (x)∆xi
max ∆xi →0 max ∆xi →0
i=1
n
lim
X
=c· f (x)∆xi
max ∆xi →0
i=1
Zb
=c· f (x)dx
a

This completes the proof.

· · · / Subsec. 5.2.2 / Segment 5.2.2.4 : Integrals over Multiple Sub-intervals

If f (x) is an integrable function defined over an interval [a, b] and there exists
a point c in between a and b, then the integral on [a, b] may be split into the
following two integrals over [a, c] and [c, b], respectively.

Theorem 5.2.7. (Integrals over Multiple Sub-intervals)

Let f (x) and g(x) be integrable functions defined over an interval [a, b] and
c be a constant such that a < c < b, then

Zb Zc Zb
f (x)dx = f (x)dx + f (x)dx
a a c

The theorem above is pretty clear when seen visually as follows.

f(x) f(x)

x x
a b a c b

Figure 5.13

· · · / Subsec. 5.2.2 / Segment 5.2.2.5 : Integrals Involving Absolute Values

In addition to the properties mentioned above, it is important to consider the
property of the definite integral involving an absolute value. For example, con-
Zb
sider the definite integral of the form |f (x)| dx. The absolute value function,
a
|f (x)|, represents the distance of the function f (x) from the x-axis, ensuring that
the result is always non-negative. This means that |f (x)| returns the magnitude

of f (x) without regard to whether f (x) is positive or negative.

Zb
As a result, the integral |f (x)| , dx calculates the “total area” between the
a
curve f (x) and the x-axis over the interval [a, b]. This includes the area above
the x-axis where f (x) is positive, as well as the area below the x-axis where f (x)
is negative, since the absolute value converts negative values of f (x) to positive.
Therefore, this integral provides the sum of the areas of both regions, treating
both as positive contributions.

Zb
That is, the definite integral f (x)dx represents the following net signed area.
a

f(x)

x
a b

Figure 5.14

Zb
While the definite integral |f (x)| dx represents the following area under the
a
curve of |f (x)|.

f(x)

x
a b

Figure 5.15

Therefore, we have the following theorem.

Theorem 5.2.8. (Integrals Involving Absolute Values)

Let f (x) be a function such that |f (x)| is integrable on the interval [a, b],
then the definite integral,
Zb
|f (x)| dx
a

represents the sum of areas of the region bounded the function f (x) and
the x-axis, from x = a to x = b.

Having discussed the properties above, they are now summarized below for a
quick reference.

Property 5.2.1. Let f (x) and g(x) be integrable functions on an interval

[a, b] and k be a constant, then
Za
1. f (x) dx = 0
a

Za Zb
2. f (x) dx = − f (x) dx
b a

Zb Zb
3. k · f (x) dx = k · f (x) dx
a a

Zb Zb Zb
4. f (x) + g (x) dx = f (x)dx + g (x)dx
a a a

Zb Zb Zb
5. f (x) − g (x) dx = f (x)dx − g (x)dx
a a a

Zb Zc Zb
6. f (x)dx = f (x)dx + f (x)dx
a a c

A few examples of the properties above are provided below.

Z 1 Z 2
7
I Example 5.2.3. If x dx = − , then what is
2
x2 dx?
2 3 1
Z 1
7
Solution. If x2 dx = − , then by the property above,
2 3
Z 2 Z 1
2
x dx = − x2 dx
1
2
7
=− −
3
7
=
3
Z 2
7
Therefore, x2 dx =
1 3

Z1 Z1 Z1
I Example 5.2.4. If f (x)dx = e and g(x)dx = π, then what is (πf (x) + eg(x)) dx
0 0 0

Z1 Z1
Solution. If f (x)dx = e and g(x)dx = π, then by the properties above,
0 0

Z1 Z1 Z1
(πf (x) + eg(x)) dx = π f (x)dx + e g(x)dx
0 0 0

= πe + eπ

= 2eπ

Z1
Therefore, (πf (x) + eg(x)) dx = 2eπ.
0

The End of Section 5.2

The Definite Integral

§ 5.3 Numerical Integration

In the previous sections, we used rectangles to approximate the area of a region
bounded by a curve and the x-axis. In those approximations, each rectangle
generally had an equal width, but their heights varied depending on the type
of approximation used. In the right endpoint approximation, the height of each
rectangle was determined by the function’s value at the right endpoint of each
sub-interval. Similarly, in the left endpoint approximation, the height was de-
termined by the function’s value at the left endpoint of each sub-interval.

We then introduced the Riemann sum, where the height of each rectangle could
be based on any point within each sub-interval. This method is valid based on
the fact that all these approximations converge to the same limit as the number
of rectangles approaches infinity.

Following these methods, we now introduce a few new approximations.

Chap. 5 / Sec. 5.3 / Subsec. 5.3.1 : The Midpoint Rule

In the Midpoint Rule, we approximate the area bounded by the curve and the
x-axis using rectangles as done before. However, this time, each rectangle’s
height is determined by the value of the function at the midpoint mi of each
sub-interval. For instance, for the ith sub-interval [xi−1 , xi ], the height of the
xi + xi−1
rectangle is the value of the function at the midpoint mi = , as shown
2
below.

a b

Figure 5.16

That being said, we now generalize this approximation. Suppose we want to

approximate the area bounded by the curve of f (x) and the x-axis from x = a
to x = b, by the Midpoint rule.

In order to do that we partition the interval [a, b] into n sub-interval of equal

b−a
width ∆x = , yielding the following sub-intervals:
n
[x0 , x1 ], [x1 , x2 ], [x2 , x3 ], · · · , [xi−1 , xi ], · · · , [xn−1 , xn ]

This implies that the ith sub-interval is [xi−1 , xi ], where i = 1, 2, 3, · · · , n. Recall

that the midpoint rule approximates the area by rectangles whose height is
determined by the function’s value at the midpoint of each sub-interval. We
denote such height as f (mi ), where mi represents the midpoint of the ith sub-
xi−1 + xi
interval, that is, . Therefore, the area of each rectangle yields the
2
following:

xi−1 + xi
height × width = f (mi )∆x = f ∆x
2

Now, summing these quantities for i = 1, 2, 3, · · · , n, we get the following sum

denoted by Mn .

Mn = f (m1 ) ∆x + f (m2 ) ∆x + · · · + f (mn ) ∆x

| {z } | {z } | {z }
1st rectangle 2nd rectangle nth rectangle

= f (m1 ) + f (m2 ) + · · · + f (mn ) ∆x

n
for i = 1, 2, · · · , n
X
= ∆x · f (mi )
i=1
n
X
= f (mi ) ∆x
i=1
n
X xi−1 + xi
= f ∆x
i=1
2

The Midpoint rule approximation gets better and better as the number of rect-
angles n increases. This forms the following definition.

Definition 5.3.1. (Midpoint Rule)

Let Mn represent the Midpoint rule approximation to the definite integral
Zb
f (x) dx on the interval [a, b]. Then,
a

n
X
Mn = f (mi ) ∆x
i=1

xi−1 + xi
where, mi = is the midpoint of the ith interval, [xi−1 , xi ], and
2
∆x is the width of each sub-interval.

A few examples to demonstrate the definition above are provided below.

Zπ
I Example 5.3.1. Use the Midpoint Rule with 6 sub-intervals to estimate sin(x)dx
0

Solution. In order to approximate, we divide the interval [0, π] into n = 6 sub0in-

tervals of equal width. That is,
π−0 π
=
6 6
Thus, we have the following sub-intervals,
h π i h π π i h π π i π 2π 2π 5π 5π
0, , , , , , , , , , ,π
6 6 3 3 2 2 3 3 6 6

Thus, we have the following midpoints of each sub-intervals,

π
0+
m1 = 6 = π
2 12

π π
+ π
m2 = 6 3 =
2 4
π π
+ 5π
m3 = 3 2 =
2 12
π 2π
+
m4 = 2 3 = 7π
2 12
2π 5π
+
m5 = 3 6 = 3π
2 4
5π
+π 11π
m6 = 6 =
2 12
Now, we approximate the integral as follows,
Zπ 6
sin(x)dx ≈
X
f (mi )∆x
0 i=1

= f (m1 )∆x + f (m2 )∆x + f (m3 )∆x + f (m4 )∆x + f (m5 )∆x + f (m6 )∆x

π π 5π 7π 3π 11π
= ∆x f +f +f +f +f +f
12 4 12 12 4 12

π π π 5π 7π 3π 11π
= · sin + sin + sin + sin + sin + sin
6 12 4 12 12 4 12
≈ 2.02303
Rπ
Therefore, sin(x)dx ≈ 2.02303.
0

Notice that in all of our approximations discussed so far, we used rectangles to

estimate areas under curves. However, it is certainly not the case that we are
limited solely to the rectangles to approximate the area—other geometric shapes
can also be used instead. A common alternative of using rectangles is to the
trapezoids. We discuss this in the following subsection.

Chap. 5 / Sec. 5.3 / Subsec. 5.3.2 : The Trapezoidal Rule

In the Trapezoidal rule approximation, we approximate the net area bounded
by the curve of f (x) and the x-axis on an interval [a, b] with trapezoids instead
of rectangles. For that, recall that the area of a trapezoid with a height of h and
the lengths of the two parallel sides a and b, that is,

a b

Figure 5.17

is given by the following formula,

a+b
area = h
2

Now, in order to use the Trapezoidal rule approximation to approximate the net
area bounded by the curve of f (x) and the x-axis on the interval [a, b], we divide
b−a
the interval [a, b] into n sub-intervals of equal width ∆x = . This forms
n
the following sub-intervals:

[x0 , x1 ], [x1 , x2 ], [x2 , x3 ], · · · , [xi−1 , xi ], · · · , [xn−1 , xn ]

This implies that the ith sub-intervals is [xi−1 , xi ], for i = 1, 2, 3, · · · , n. We

now construct trapezoids on the ith sub-interval for i = 1, 2, 3, · · · , n, as shown
below.

a b

Figure 5.18

The height of each trapezoid is equal to the width of each sub-interval, that is,
b−a
h = ∆x = and the lengths of each trapezoid’s sides a and b are f (xi−1 ) and
n
f (xi ), respectively, for i = 1, 2, 3, · · · . Therefore, the area of the ith trapezoid
yields,
f (xi−1 ) + f (xi )
area = ∆x
2
Now, summing these quantities for i = 1, 2, 3, · · · , n, we get the following sum
denoted by Tn .

f (x0 ) + f (x1 ) f (x1 ) + f (x2 ) f (xn−1 ) + f (xn )
Tn = ∆x + ∆x + · · · + ∆x
2 2 2
| {z } | {z } | {z }
1st trapezoid 2nd trapezoid nth trapezoid

f (x0 ) + f (x1 ) f (x1 ) + f (x2 ) f (xn−1 ) + f (xn )
= + + ··· + ∆x
2 2 2

f (x0 ) f (x1 ) f (x1 ) f (x2 ) f (xn−1 ) f (xn )
= + + + + ··· + + ∆x
2 2 2 2 2 2
 
 f (x0 ) f (x1 ) f (x1 ) f (xn−1 ) f (xn−1 ) f (xn ) 
=
 2 + + + · · · + + +  ∆x
| 2 {z 2 } | 2 {z 2 } 2 
f (x1 ) f (xn−1 )
 
 
 f (x0 ) f (x n ) 
=
 2 + f (x 1 ) + f (x 2 ) + · · · + f (xn−1 ) +  ∆x
 | {z } 2 

n−1 P
f (xi )
i=1

n−1
!
f (x0 ) X f (xn )
= + f (xi ) + ∆x
2 i=1
2
The Trapezoidal rule approximation gets better and better as the number of
rectangles n increases. This leads to the following definition.

Definition 5.3.2. (Trapezoidal Rule) Let f (x) be continuous over an interval

Zb
[a, b]. The Trapezoidal Rule to approximate f (x) dx is given by,
a

∆x
Tn = f (x0 ) + 2f (x1 ) + 2f (x2 ) + · · · + 2f (xn−1 ) + f (xn )
2
Zb
Where, lim Tn = f (x) dx
n→∞
a

A few examples of the definition above are presented below.

I Example 5.3.2. Use the trapezoidal rule with 4 sub-intervals to estimate

Z1
x2 dx.
0

Solution. The width of each sub-interval is,

1−0
∆x = = 0.25
4
This implies that,

x0 = 0 + 0 · 0.25 = 0
x1 = 0 + 1 · 0.25 = 0.25
x2 = 0 + 2 · 0.25 = 0.5
x3 = 0 + 3 · 0.25 = 0.75
x4 = 0 + 4 · 0.25 = 1

Now, compute the function values,

f (0) = 02 = 0
f (0.25) = (0.25)2 = 0.0625
f (0.5) = (0.5)2 = 0.25
f (0.75) = (0.75)2 = 0.5625
f (1) = (1)2 = 1

Using the composite trapezoidal rule,

Z1
0.25
x2 dx ≈ (f (0) + 2(f (0.25) + f (0.5) + f (0.75)) + f (1))
2
0
0.25
= (0 + 2(0.0625 + 0.25 + 0.5625) + 1)
2
0.25
= (0 + 2(0.875) + 1)
2
0.25
= (2.75)
2
= 0.34375
Z1
Therefore, x2 dx ≈ 0.34375
0

The End of Section 5.3

Numerical Integration

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Download for free at https://md-shouvik-iqbal.github.io/calculus
Chapter 6

The Connection Between

Differential and Integral Calculus

As of now, we have been introduced to two main branches of calculus: differential

calculus and integral calculus. As discussed in the previous chapters, differential
calculus concerns itself with the rate of change of a quantity as in the slopes of
curves, while integral calculus focuses on the accumulation of quantities as in
the areas under curves.

These two concepts may seem unrelated—how could the slope of a curve be
possibly connected to the area under it? However, the relation between these
counter-intuitive ideas can ultimately be shown by the Fundamental Theorem of
Calculus, which we aim to study shortly. But, before we dive into that, though, it
is essential to understand the concepts of antiderivatives and indefinite integrals.
Therefore, we begin our discussion with these foundational concepts.

§ 6.1 Antidifferentiation
The concept of an inverse operation is a key concept in mathematics. For ex-
ample, in arithmetic, addition and subtraction are inverse of one another as one
operation can reverse the effect of the other. Similarly, multiplication and di-
vision are inverses of one another as applying one operation can be undone by
applying the other.

349

Similarly, in calculus, since differentiation is an operation, and like other opera-

tions, it has an inverse. This inverse of differentiation is known as antidifferenti-
ation. While differentiation is the process of finding the derivative of a function,
antidifferentiation is the process of finding an antiderivative of a function. This
prompts us to study the concept of antiderivative.

We discuss the concept of antiderivative in the following subsection.

Chap. 6 / Sec. 6.1 / Subsec. 6.1.1 : Antiderivative

The concept of an antiderivative of a function is unique in that it is a differen-
tiable function whose derivative is equal to the originally given function. That
is, an antiderivative of a given function f (x) is a differentiable function F (x)
such that when F (x) is differentiated, it equals the original function f (x).

This concept may seem complicated and, quite frankly, somewhat confusing.
Nevertheless, to properly understand the concept, consider the following.

Consider a function f (x). If f (x) is differentiable then its derivative is denoted

as f 0 (x). That is,

derivative

Now, suppose another function F (x). If F (x) is differentiable, then its derivative
is denoted as F 0 (x). That is,

derivative

Now, here is the connection: F (x) is said to be an antiderivative of f (x) if there’s

this specific relationship between F 0 (x) and f (x) such that F 0 (x) = f (x). That
is, if

equ
al

is true, then the following holds,

antiderivative

That is, F (x) is an antiderivative of f (x). Based on that, we have the following
definition of an antiderivative of a function.

Definition 6.1.1. (Antiderivative)

A function F (x) is an antiderivative of f (x) over an interval I, if

F 0 (x) = f (x)

for all x ∈ I

The definition above may be better understood by the following examples.

I Example 6.1.1. What is an antiderivative of the function f (x) = 2x?

Solution. If F (x) = x2 , then the derivative is F 0 (x) = 2x. Now, if f (x) = 2x,
then an antiderivative of f (x) is F (x) = x2 , for the reason that F 0 (x) = f (x).

I Example 6.1.2. What is an antiderivative of the function f (x) = x2 ?

x3
Solution. If F (x) = , then the derivative is
3
d
F 0 (x) = F (x)
dx
d x3
=
dx 3
1
= · 3x2
3
= x2

x3
Now, if f (x) = x , then an antiderivative of f (x) is F (x) = , for the reason
2
3
that F 0 (x) = f (x).

x3
Now, notice that although is an antiderivative of x2 , it is not—however—the
3
x3
antiderivative of x2 . This is because the function F (x) = is not the only
3
function whose derivative is x2 . Actually, every function of the form F (x) =
x3
+ C, where C ∈ R a constant, has a derivative of x2 . This is because,
3
d x3

d
F (x) = +C
dx dx 3
0
d x3

d >
= + (C)
dx 3 dx

= x2 + 0
= x2

x3
This implies that the most general antiderivative of x is + C, where C ∈ R, 2
3
and thus we have the following theorem.

Theorem 6.1.1. (The Family of Antiderivatives)

If F (x) is an antiderivative of f (x) over an interval I, then all the an-
tiderivative of f (x) over I is of the form,

F (x) + C

where, C ∈ R is an arbitrary constant.

This implies that all the antiderivatives of a function differ only by a constant C
and therefore, they belong to one general family of functions, that is, functions
of the form F (x) + C.

It’s fascinating how each differentiation formula, when reversed, transforms into
an antidifferentiation formula. For example, as we’ve just seen, an antiderivative
x3 xn+1
of x2 is , suggesting an antiderivative of xn to be , where n 6= −1,
3 n+1
because
n+1
d x 1 d
= · xn+1
dx n + 1 n + 1 dx
1
· (n + 1) · xn+1−1

=
n+1

1
= ·
(n+1)

· xn
n+1

= xn where n 6= −1

d
Similarly, an antiderivative of sin(x) is − cos(x), because (− cos(x)) = −(− sin(x)) =
dx
d x
sin(x). An antiderivative of e is e , because
x x
(e ) = ex , and so on. In the
dx
table below, a list of functions along with their particular antiderivatives are
presented for a quick reference, where each of the particular antiderivatives can
be verified by differentiating and matching with the original function.
Table 6.1

Particular
Function
antiderivative

0 C∈R

1 x

n nx

xn+1
xn , n 6= 1
n+1
nxn−1 xn
1 √
√ x
2 x

ex ex

ax · ln(a) ax
1
ln(x) = loge (x)
x
1
loga (x)
x ln (a)

cos(x) sin(x)

sin(x) − cos(x)

sec2 (x) tan(x)

csc2 (x) − cot(x)

tan(x) sec(x) sec(x)

cot(x) csc(x) − csc(x)

1
√ sin−1 (x)
1 − x2
1
−√ cos−1 (x)
1 − x2
1
tan−1 (x)
1 + x2
1
− cot−1 (x)
1 + x2
1
√ sec−1 (x)
|x| x2 − 1
1
− √ csc−1 (x)
|x| x2 − 1

A few examples are provided below.

1
I Example 6.1.3. What is the general antiderivative of f (x) = √
2 x
1
Solution. The antiderivative of f (x) = √ is,
2 x
√
F (x) = x+C

Because,
d√ d
F 0 (x) = x+ C
dx dx
1
= √
2 x
∴ F 0 (x) = f (x)

1 √
Therefore, the antiderivative of f (x) = √ is F (x) = x+C, where C ∈ R.
2 x

I Example 6.1.4. What is the general antiderivative of f (x) = sin(x)

Solution. The general antiderivative of f (x) = sin(x) is,

F (x) = − cos(x) + C

Because,
d d
F 0 (x) = (− cos(x)) + C
dx dx
= sin(x)
∴ F 0 (x) = f (x)

Therefore, the antiderivative of f (x) = sin(x) is F (x) = − cos(x) + C, where

C ∈ R.

1
I Example 6.1.5. What is the general antiderivative of f (x) = √
1 − x2
1
Solution. The general antiderivative of f (x) = √ is,
1 − x2

F (x) = sin−1 (x) + C

Because,
d d
F 0 (x) = sin−1 (x) + C
dx dx
1
=√
1 − x2
∴ F 0 (x) = f (x)

1
Therefore, the antiderivative of f (x) = √ is F (x) = sin−1 (x) + C, where
1−x2
C ∈ R.

So far, we have used F (x) to denote a particular antiderivative of the function

f (x). However, since there are infinitely many antiderivatives of a single function
f (x) that can be written in the form F (x) + C, therefore, it becomes necessary
to establish a notation that represents all these antiderivatives of a function
f (x) collectively. The notation that denotes this concept is referred to as the
“Indefinite Integral,” we discuss it further below.

Chap. 6 / Sec. 6.1 / Subsec. 6.1.2 : Indefinite integral

An indefinite integral of a function f (x) represents the family of all antideriva-
tives, that is, the set of all functions that are an antiderivative of the function
f (x).
Z
For historical reasons, the elongated S symbol —referred to as the integral
sign—is used to denote the collection of all antiderivatives of a function f (x) in
the following manner, Z
f (x)dx

The notation above is called the indefinite integral and read as “the integral of
f (x) dee x.”

Now, observe carefully that the notation for a collection of antiderivatives is the
Zb
same as the notation for a definite integral f (x)dx, except for the core differ-
a
ence that a definite integral is a numerical value that represents the net signed
area of the region bounded by the curve of a function f (x) and the x-axis, from
x = a to x = b. On the other hand, an indefinite integral is a family of functions
that represents the collection of all antiderivatives of the function f (x). Clearly,
a numerical value and a family of functions cannot be the same thing and there-
fore the definite and indefinite integral belongs to two distinct concepts. This
distinction is also reflected in their notation for the limits of integration. That
Zb
is, a definite integral includes limits of integration in its integral sign , while
Z a

an indefinite integral does not .

The reason for using such similar notation for two distinct concepts and causing
temporary confusion will become clear once we understand the Fundamental
Theorem of Calculus. Without giving too much away, the Fundamental The-
orem of Calculus essentially shows that a definite integral can be calculated
without using any limit of approximation by the use of an indefinite integral.
And, therefore such similar notation for an indefinite integral is strategically
chosen for later convenience.
Z
Similar to the notation of a definite integral, the symbol is also followed by

the function f (x) called the integrad and the differential dx called the variable
of integration in the notation of an indefinite integral. The differential dx is
simply there to remind the fact that x is the independent variable, similar to
d
the role in the Leibniz notation for derivative, f (x).
dx

Based on that, we have the following definition of an indefinite integral.

Definition 6.1.2. (Indefinite Integral)

The collection of all antiderivatives of a function f (x) is referred to as the
indefinite integral of f (x) with respect to x, and is denote as,
Z
f (x) dx
Z
where, the symbol represents the integral sign. The function f (x) is
called the integrand of the integral and x is the variable of integration.

This implies that given a function f (x), the indefinite integral of f (x), that is,
Z
f (x)dx

represents all the antiderivative of the function f (x) that are of the form F (x) +
C. This suggests that

Remark 6.1.1. Z
f (x)dx = F (x) + C

Having discussed the concept of antiderivatives and its notation, we now aim to
show how Differentiation and Antidifferentiation are indeed inverse operations
of each other. For that, recall that for a given function f (x), the family of all
antiderivatives of f (x) is represented as,
Z
f (x)dx = F (x) + C

Since, f (x) = F 0 (x) by the definition of an antiderivative (as F (x) is an an-

tiderivative of f (x)), therefore, we can write,
Z
F 0 (x)dx = F (x) + C

Now, differentiating both sides of the equation above yields,

Z
d d
F 0 (x)dx = (F (x) + C)
dx dx
= F 0 (x)

Since, f (x) = F 0 (x), therefore, we have the following,

Z
d
f (x)dx = f (x)
dx

The result above suggests that when a function f (x) is first antidifferentiated
and then differentiated, it results in the same function f (x) or the function f (x)
is unchanged.

Similarly, we now show that when a function is first differentiated and then
antidifferentiated, it also results in the same function or does not change at all.
In order to do that, recall that for a given function F (x), if its antiderivative is
f (x) and is differentiated, then we get,

d
f (x) = f 0 (x)
dx
Now, antidifferentiating both sides of the equation above (that is, finding the
family of all antiderivatives of the function f 0 (x)) yields,
Z Z
d
f (x)dx = f 0 (x)dx
dx

Since, F (x) = f 0 (x) by the definition of an antiderivative (as f (x) is an an-

tiderivative of F (x)), therefore,
Z Z
d
f (x)dx = f 0 (x)dx
dx Z
= F (x)dx (1)
Z
Now, since the notation F (x)dx represents the family of all antiderivative of
the function F (x) that are of the form f (x) + C, therefore, by the definition of
an indefinite integral, we write,

Z
F (x)dx = f (x) + C

And this implies, from equation (1), that

Z
d
f (x)dx = f (x) + C
dx

As a result, we have the following theorem,

Theorem 6.1.2.
Z Z
d d
f (x)dx = f (x) and f (x)dx = f (x) + C
dx dx

The theorem above suggests that differentiation and antidifferentiation are in-
deed inverse operations of each others. Since they are inverses of each other,
therefore a few indefinite integrals (i.e., the collection/family of all antideriva-
tives) can easily be obtained by reversing the derivative of various functions, as
listed below.

Note that each indefinite integral listed below can be verified by differentiat-
ing the function on the right-hand side of the equation and confirming that it
matches the integrand of the corresponding indefinite integral.

Therefore, we have the following informal intuition,

Z
derivative dx = antiderivative + C

This intuition suggests that if the “antiderivative” on the right side of the equa-
tion is differentiated, it equals the “derivative” in the integrand on the left side.
Z Z
1. dx = x + C 6. ax ln(a)dx = ax + C

xn+1
Z
dx
Z
2. n
x dx = +C 7. = ln(x) + C
n+1 x

√
Z Z
1 dx
3. √ dx = x + C 8. = loga (x) + C
2 x x ln(a)
Z
1 x
Z
4. x
a dx = a +C 9. sin(x)dx = − cos(x) + C
ln(a)
Z Z
5. x
e dx = e + C x
10. cos(x)dx = sin(x) + C

Z Z
dx
11. sec (x)dx = tan(x) + C
2
15. √ = sin−1 (x) + C
1−x 2
Z
12. csc2 (x)dx = − cot(x) + C Z
dx
16. = tan−1 (x) + C
1 + x2
Z
13. sec(x) tan(x)dx = sec(x) + C
Z
Z dx
14. csc(x) cot(x)dx = − csc(x)+C 17. √ = sec−1 (x) + C
2
|x| x − 1

Similar to the constant multiple, addition, and subtraction rules in differenti-

ation, we have the following elementary rules for indefinite integrals (i.e., the
collection/family of all antiderivatives).

Theorem 6.1.3. (Elementary Rules for Indefinite Integrals)

Z Z
1. cf (x)dx = c f (x)dx
Z Z Z
2. f (x) + g(x) dx = f (x)dx + g(x)dx
Z Z Z
3. f (x) − g(x) dx = f (x)dx − g(x)dx

A few examples of the indefinite integral are provided below.

Z
I Example 6.1.6. Evaluate 7xdx

Solution. Using the indefinite integrals, we write,

Z Z
7xdx = 7 xdx
7x2
= +C
2
7x2
Z
Therefore, 7xdx = +C
2

Z
0
I Example 6.1.7. Evaluate dx
x
Solution. Using the indefinite integrals, we write,
Z Z
0
dx = 0dx
x

Z
=0· dx

=0·x+C
=0+C
Z
0
Therefore, dx = 0 + C
x

Z
I Example 6.1.8. Evaluate x2 + x3 + x5 dx

Solution. Using the indefinite integrals, we write,

Z Z Z Z
x + x + x dx = x dx + x dx + x5 dx
2 3 5 2 3

x3 x4 x6
= + + +C
3 4 6
x3 x4 x6
Z
Therefore, 2 3 5

x +x +x dx = + + +C
3 4 6

Z
I Example 6.1.9. Evaluate sin(5x)dx

Solution. To solve the integral, we apply the Chain rule as follows.

d d
cos(5x) = − sin(5x) · (5x)
dx dx
= −5 sin(5x)

Therefore,
Z Z
d
−5 sin(5x)dx = cos(5x)dx
Z dx
−5 · sin(5x)dx = cos(5x) + C
Z
1
sin(5x)dx = − cos(5x) + C
5
Z
1
Therefore, sin(5x)dx = − cos(5x) + C
5

For the cases illustrated in the example above, let us now generalize the indefinite
integral for any constant a 6= 0, as follows.

Remark 6.1.2. (Generalized Indefinite Integrals)

Z
1
1. eax dx = eax + C
a
Z
dx 1
2. = ln |x| + C
ax a
Z
1
3. sin(ax)dx = − cos(ax) + C
a
Z
1
4. cos(ax)dx = sin(ax) + C
a
Z
1
5. sec2 (ax)dx = tan(ax) + C
a
Z
1
6. csc2 (ax)dx = − cot(ax) + C
a
Z
1
7. sec(ax) tan(ax)dx = sec(ax) + C
a
Z
1
8. csc(ax) cot(ax)dx = − csc(ax) + C
a
Z
dx −1 x

9. √ = sin +C
a2 − x2 a
Z
dx 1 −1 x

10. = tan +C
a2 + x 2 a a
Z
dx 1 x
11. √ = sec−1 +C
x x 2 − a2 a a

Z
dx
I Example 6.1.10. Evaluate
2x
Z
dx 1
Solution. If we let a = 2, then the integral matches the form = ln |x|+C.
ax a
Therefore,
Z
1 1
dx = ln |x| + C
2x 2
Z
1 1
Therefore, dx = ln |x| + C.
2x 2

Z
3
I Example 6.1.11. Evaluate √ dx
25 − x2
Z
dx
Solution. If we let a = 5, then the integral matches the form √ =
x a2 − x2
sin−1 + C. Therefore,
a
Z Z
3 1
√ dx = 3 · √ dx
25 − x2 52 − x2

−1 x

= 3sin +C
5
Z
3 −1 x

Therefore, √ dx = 3sin +C
25 − x2 5

√
x2 + x x2 − 25 + 9
Z
I Example 6.1.12. Evaluate √ √ dx
9x x2 − 25 + x3 x2 − 25
√
x2 + x x2 − 25 + 9
Z
Solution. In order to evaluate the integral √ √ dx, we
9x x2 − 25 + x3 x2 − 25
simplify the integrand as follows,
√ √
x2 + x x2 − 25 + 9 x2 + x x2 − 25 + 9
√ √ = √
9x x2 − 25 + x3 x2 − 25 x x2 − 25 (9 + x2 )
√
x x2 − 25 + x2 + 9
= √
x x2 − 25 (x2 + 9)
√
x x2− 25 x2+9

= √
+ √ 2

x x2 − 25 (x2 + 9) x x2 − 25x
+9
1 1
= 2 + √
x + 9 x x2 − 25
This implies that,
√
x2 + x x2 − 25 + 9
Z Z
1 1
√ √ dx = + √ dx
9x x2 − 25 + x3 x2 − 25 x2 + 9 x x2 − 25
Z Z
1 1
= dx + √ dx
x2 + 9 x x2 − 25
Z Z
1 1
= 2 2
dx + √ dx
x +3 x x − 52
2
Z Z
dx 1 −1 x
dx
The integrals above matches the form = tan +C and √ =
a2 + x2 a a x x2 − a2
1 −1 x
sec + C, for a = 3 and a = 5, respectively. Therefore, we write,
a a
√
x2 + x x2 − 25 + 9
Z Z Z
1 1
√ √ dx = 2 2
dx + √ dx
9x x2 − 25 + x3 x2 − 25 3 +x x x2 − 52

1 x 1 x
= tan−1 + sec−1 +C
3 3 5 5
√
x2 + x x2 − 25 + 9
Z
1 x 1 x
Therefore, √ √ dx = tan−1 + sec−1 +C
9x x2 − 25 + x3 x2 − 25 3 3 5 5

The End of Section 6.1

Antidifferentiation

§ 6.2 The Fundamental Theorem of Calculus (FTC)

The Fundamental Theorem of Calculus (FTC) is, in a way, the most important
of all theorems in calculus. In simple terms, it states that differentiation and
integration are complete opposites of each other, much like how addition and
subtraction are opposites of one another.

To recap, “differentiation” refers to the process of finding the derivative of a

function f (x), which represents the instantaneous rate of change of that func-
tion with respect to its independent variable. On the other hand, “integration”
is the process of calculating the definite integral of a function f (x), which repre-
sents the net signed area between the graph of f (x) and the x-axis over a given
interval. This is done by taking the limit of Riemann sum approximations as
the number of rectangles n to approach infinity.

Now, up to this point, these two concepts may have seemed unrelated. How-
ever, the Fundamental Theorem of Calculus now states that differentiation and
integration are actually opposite processes. This might seem contradictory, as
we have discussed, in the previous section that, it is antidifferentiation—not
integration—that serves as the inverse of differentiation.

However, the key point to realize here is that the Fundamental Theorem of Cal-
culus now allows us to perform integration in terms of antidifferentiation. That
is, we can now evaluate definite integrals using antiderivatives, rather than re-
lying on the limit of a Riemann Sum.

It is for this reason that we can say—by the Fundamental Theorem of Calculus—
that “integration” and “antidifferentiation” are essentially the same thing when
evaluating a definite integral.

Having introduced the Fundamental Theorem of Calculus and its core idea that
integration can be performed through antidifferentiation, we will now explore
how this is possible. Specifically, we will derive the first part of the theorem,
which then leads to the derivation of the second part. This, in turn, will enable
us to evaluate definite integrals using antiderivatives.

In order to do that, consider a function y = f (t) defined over an interval [a, b]. If
the function f (t) is integrable over [a, b], then the exact area A under the curve
f (t) and above the t-axis, over the interval [a, b], that is, from t = a to t = b, is
given by the definite integral A provided below,
Zb
A= f (t) dt
a

Now, say we want to compute the area from t = a to an unspecified bound

t = x. That is, we want to compute the area A over the interval [a, x], where x
is a variable, varying over a set of values greater than a. If that’s the case, then
the definite integral would now be,
Zx
f (t) dt
a

This is shown below,

moveable
fixed bound bound

a x

Figure 6.1

Notice here that the definite integral above is not able to produce any value
as an area unless the upper limit of integration is specified. That is, since the
upper limit of integration is variable x, therefore for each input in the upper limit
x, there would be a corresponding output as a numerical value of the definite

integral. Therefore, the definite integral is now a function of x. We denote this

function by A(x). That is,
Zx
A(x) = f (t) dt
a

This is the exact area from a to x. Since x is a variable that can take on values
greater than a, therefore, an increment h > 0 in x would yield,

x+h
Z
A(x + h) = f (t) dt
a
This implies that A(x + h) is the exact area from a to x + h as shown as the
shaded region in the figure below,

a x x+h

Figure 6.2

Recall that A(x) denotes the exact area from a to x. Therefore, subtracting
A(x + h) from A(x) would yield the exact area from x to x + h as shown below,

h
subtracted area

a x x+h

Figure 6.3

Now, if the increment h > 0 is very small then the exact area of A(x + h) − A(x)
is,

A(x + h) − A(x) ≈ area of a thin rectangle at t = x

= height × width
= f (x) × h
∴ A(x + h) − A(x) ≈ f (x) · h (1)

as shown below,

a x+h
x

Figure 6.4

Now, dividing both sides by h and then taking the limit as h → 0 yields,

A(x + h) − A(x)
≈ f (x)
h
A(x + h) − A(x)
lim = lim f (x)
h→0 h h→0
0
A (x) = f (x) (2)

A(x + h) − A(x)
Here, lim = A0 (x) by the definition of a derivative. Since
h→0 h
A0 (x) = f (x) in the equation above, therefore A(x) is an antiderivative of f (x),
by the definition of antiderivatives. As a result, we have the following theorem.

Theorem 6.2.1. (The Fundamental Theorem of Calculus, Part I)

Let f be a function continuous on [a, b], then the area function,
Zx
A(x) = f (t) dt
a

where, a ≤ x ≤ b, is continuous on [a, b] and differentiable on (a, b), and

A0 (x) = f (x)

That is, the area function A(x) is an antiderivative of f (x), by the definition
of antiderivatives.

The theorem above suggests that, since

d
A0 (x) = A(x)
dx
Zx
d
= f (t)dt
dx
|a {z }
A(x)

= f (x) since A0 (x) = f (x)

Zx
d
∴ A0 (x) = f (t)dt = f (x) (3)
dx
a

therefore, the area function A(x) is an antiderivative of f (x) on the interval

[a, b]. Based on this part of the Fundamental theorem of Calculus, we derive the
second part of the theorem by which a definite integral can be evaluated using
antiderivatives only.

For that, notice that the area function A(x) above is an antiderivative of f (x).
It is not, however, the only antiderivative of f (x). From previous discussions, we
know that all antiderivatives of a function differ by a constant. So, we let F (x)
be another antiderivative of f (x), such that F (x) = A(x)−C or A(x) = F (x)+C.

Now, to derive the second part of the theorem, we need to determine the value
of C in the equation above. In order to do that, notice that when x = a, the
area function yields,
Za
A(x) = f (t) dt
a

due to the zero-width property of definite integrals. Therefore,

A(a) = 0

F (a) + C = 0
C = −F (a)
Now, substituting the value of C in A(x) = F (x) + C yields,
A(x) = F (x) + (−F (a))
= F (x) − F (a)
This implies that the definite integral A(x) from t = a to t = x (where x is a
variable point) can be calculated by evaluating and subtracting the antideriva-
tive F (a) from F (x), without using the limit of an approximation.

This further implies that to calculate the definite integral A(x) from t = a to
t = x, that is, over the interval [a, b]; we need to evaluate the antiderivative F (x)
at both endpoints of the interval [a, b] and subtract F (a) from F (b). Therefore,
we have the following theorem.

Theorem 6.2.2. (The Fundamental Theorem of Calculus, Part II)

If f (x) is a continuous function over an interval [a, b] and F (x) is any
antiderivative of f (x), then,

Zb
f (x) dx = F (b) − F (a)
a

The part two of the Fundamental Theorem of Calculus above suggests that
Zb
the definite integral f (x) dx can be evaluated in terms of antiderivatives by
a
evaluating F (x) at both endpoints of [a, b], that is, F (a) and F (b), and then
subtracting F (a) from F (b).

Notice that if it’s shown that the integral in the theorem above is being evaluated,
then the following notation is conveniently used.
Zb ib
f (x) dx = F (x)
a
a

= F (b) − F (a)
That is, the difference F (b) − F (a) is denoted as follows,
ib h ib
F (x) = F (b) − F (a) or F (x) = F (b) − F (a)
a a

The Fundamental Theorem of Calculus allows us to evaluate definite integrals

using antiderivatives, which is based on the concept of derivatives. This creates
a similar relationship between integration and antidifferentiation, while creating
an inverse relationship between integration and differentiation, as antidifferen-
tiation has always been the inverse operation of differentiation. We discuss this
matter further below.

Chap. 6 / Sec. 6.2 / Subsec. 6.2.1 : The Inverse Relationship between Differ-
entiation and Integration
Previously, we have seen, in antidifferentiation, that when a function f (x) is first
antidifferentiated and then differentiated, it results in the same function f (x).
That is, Z
d
f (x)dx = f (x)
dx
Similarly, we see in the first part of the Fundamental Theorem of Calculus that
when a function f (t) is first integrated and then differentiated, it results in the
same. That is, from the equation (3), we get,
Zx
d
f (t)dx = f (x)
dx
a

Likewise, we have seen, in antidifferentiation, that when a function is first dif-

ferentiated and then antidifferentiated, it results in the same function. That
is, Z
d
f (x)dx = f (x) + C
dx
d
where, f (x) is an antiderivative of f (x) = f 0 (x) (i.e., simply, F (x)).
dx

Similarly, we see in the second part of the Fundamental Theorem of Calculus that
when a function f (x) is first differentiated and then integrated over the interval
[a, b], it results in the same function f (x), evaluated at both the endpoints a and
b. That is,
Zb
d
f (x)dx = f (b) − f (a)
dx
a
d
where, f (x) is an antiderivative of f (x) = f 0 (x) (i.e., simply, F (x))
dx

This is the reason why the terms “antidifferentiation” and “integration” are of-
ten used interchangeably, and both “integration” and “antidifferentiation” are
considered the inverse operations of “differentiation” when evaluating a definite
integral using the Fundamental Theorem of Calculus.

To demonstrate the Fundamental Theorem of Calculus, a few examples are

presented below.

Z1
I Example 6.2.1. Evaluate x2 dx
0

Z1
Solution. In order to evaluate the definite integral x2 dx, we only need to find
0
the antiderivative F (x) of the function f (x) = x , so that F 0 (x) yields f (x). We 2

know, from the previous discussion, that an antiderivative F (x) of functions of

xn+1
the form f (x) = x is
n
. Therefore,
n+1
xn+1
Z
xn dx = +C
n+1
Letting n = 2 and using the Fundamental Theorem of Calculus yields the fol-
lowing,
Z1 1
2 x3
x dx =
3 0
0
(1)3 (0)3
= −
3 3
1
=
3
Z1
1
Therefore, x2 dx =
3
0

Now, notice carefully that since definite integrals yield a specific numerical value,
the constant of integration (C) is not relevant as it cancels out during the eval-
uation process. That is,
Zb h ib
f (x)dx = F (x) + C
a
a

= (F (b) + C) − (F (a) + C)
= F (b) + C − F (a) − C
= F (b) − F (a)

Notice also that, since theZ difference F (b) − F (a) can be written shortly as
ib
F (x) , and since F (x) = f (x)dx, therefore, we can write,
a

Zb
f (x)dx = F (b) − F (a)
a
ib
= F (x)
a
Z b
= f (x)dx
a

This implies that

Remark 6.2.1.
Zb Z b
f (x)dx = f (x)dx
a
a

Zπ
I Example 6.2.2. Evaluate sin(x)dx
0

Solution. We know that,

Z
sin(x)dx = − cos(x) + C

Therefore, by the Fundamental Theorem of Calculus, we write the following,

Zπ h iπ
sin(x)dx = − cos(x)
0
0

= − (cos(π) − cos(0))
= − (−1 − 1)
=2
Zπ
Therefore, sin(x)dx = 2
0

Zπ
I Example 6.2.3. Evaluate e2x dx
e

Solution. We know that,

e2x
Z
e2x dx = +C
2
Therefore, by the Fundamental Theorem of Calculus, we write the following,
Zπ π
2x e2x
e dx =
2 e
e
e2π e2e
= −
2 2
e2π − e2e
=
2
Zπ
e2π − e2e
Therefore, e2x dx =
2
e

Z+e
edx
I Example 6.2.4. Evaluate
x
−e

Solution. We know that, Z

dx
= ln |x| + C
x
Therefore, by the Fundamental Theorem of Calculus, we write the following,

Z+e Z+e
edx dx
=e
x x
−e −e

= e · [ln |x|]+e
−e
= e · (ln |e| − ln |−e|)
= e · (1 − 1)
=0

Z+e
edx
Therefore, =0
x
−e

Z1
dx
I Example 6.2.5. Evaluate
1 − sin(x)
0

Solution. In order to evaluate the integral, we simplify it as follows,

Z1 Z1
dx 1 (1 + sin(x))
= · dx
1 − sin(x) (1 − sin(x)) (1 + sin(x))
0 0
Z1
1 + sin(x)
= dx
1 − sin2 (x)
0
Z1
1 + sin(x)
= dx
cos2 (x)
0
Z1
sin(x)

1
= + dx
cos2 (x) cos2 (x)
0
Z1
sin(x)

1 1
= + · dx
cos2 (x) cos(x) cos(x)
0
Z1
sec2 (x) + tan(x) · sec(x) dx

=
0
Z1 Z1
= sec (x)dx +
2
tan(x) sec(x)dx
0 0

Now, we know, from the list of indefinite integral, that

Z Z
sec (x)dx = tan(x) + C
2
and sec(x) tan(x)dx = sec(x) + C

Therefore, we write the following,

Z1 Z1 i1 h i1 h
sec (x)dx +
2
tan(x) sec(x)dx = tan(x) + sec(x)
0 0
0 0
h i h i
= tan(1) − tan(0) + sec(1) − sec(0)
≈ 2.40
Z1
dx
Therefore, ≈ 2.40
1 − sin(x)
0

The End of Section 6.2

The Fundamental Theorem of Calculus (FTC)

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Chapter 7

Methods of Integration

The methods of “integration” discussed in this chapter are more accurately be

said as the methods of “antidifferentiation.” This is because, as outlined in the
previous chapter, the Fundamental Theorem of Calculus shows that any definite
integral can typically be evaluated using antiderivatives, eliminating the need
for the limit of approximations. From our experience, we can tell that using the
limit of an approximation to evaluate a definite integral can be as impractical as
to be totally impossible due to its complexity for complex functions. Fortunately,
though, the Fundamental Theorem provides a simpler path—antidifferentiation.

However, this approach has its own limitations. So far, we only know the an-
tiderivatives for a limited set of functions, specifically those whose derivatives
we’ve encountered before. This is far from sufficient. For example, what is the
antiderivative of f (x) = sec(x)? In other words, what is that function F (x)
such that when it is differentiated, it equals f (x)? As of now, we don’t have an
answer, which means we are unable to evaluate any definite integral involving
f (x) = sec(x). This presents a fundamental challenge.

Nevertheless, to address such challenges, several techniques have been developed

to integrate—or more accurately, to antidifferentiate—functions, and this chap-
ter will explore those methods, starting with the integration by substitution.

377

§ 7.1 Integration by Substitution

Integration by substitution, or commonly known as “u-substitution,” is funda-
mentally about simplifying an indefinite integral (i.e., the collection/family of
antiderivatives) by changing its variable and then solving it. This method allows
us to transform a complicated integral into a simpler one by introducing a new
variable that makes it easier to solve.

Conceptually, this method can be thought of as a “backward rule” for the Chain
Rule in differentiation. In differentiation, the Chain Rule allows us to differen-
tiate a composite function by breaking it down into its constituent parts. For
example, if y = f (g(x)), then the Chain rule states that the derivative of y
with respect to x is f 0 (g(x)) · g 0 (x). Now, if we were to integrate the function
f 0 (g(x)) · g 0 (x), the substitution rule allows us to simplify the integral by chang-
ing its variables.

In order to understand this method, consider a composite function F (u), where

u = g(x) and that F (u) is an antiderivative of some function f (u), that is,
F 0 (u) = f (u). Now, differentiating F (u) = F (g(x)), we get,
d
F (g (x)) = F 0 (g (x)) · g 0 (x)
dx
Since F 0 (u) = f (u), where u = g(x), therefore,

F 0 (g (x)) · g 0 (x) = f (g (x)) · g 0 (x)

d
∴ F (g (x)) = f (g (x)) · g 0 (x)
dx
This implies that F (g (x)) is an antiderivative of f (g (x)) g 0 (x). In notation, we
write the following to denote the collection of all such antiderivatives.
Z
f (g (x)) g 0 (x) dx = F (g (x)) + C

du
Now, since u = g(x), therefore = g 0 (x) or in the differential form:
dx
du = g 0 (x)dx
Z
This suggests that f (g (x)) g 0 (x) dx = F (g (x))+C can be expressed in terms
of the variable u by substituting u = g(x) and du = g(x)dx, as follows:
Z
f (g (x)) g 0 (x) dx = F (g (x)) +C
| {z } | {z } | {z }
u du u

Z
f (u) du = F (u) + C
Z
By doing so, we can transform a complex integral f (g(x))g 0 (x) dx = F (g(x))+
Z
C with respect to x, into a simpler integral f (u) du = F (u) + C with respect
to a new variable u. This technique is useful because the new integral with
respect to u is often easier to solve than the original integral with respect to x.
Therefore, we have the following theorem.

Theorem 7.1.1. (Integration by Substitution)

Let u = g(x) be a function such that g 0 (x) is continuous over an interval,
and let f (u) be a continuous function over the corresponding range of g(x).
Then, if F (u) is an antiderivative of f (u), then
Z Z
0
f (g (x)) g (x) dx = f (u) du

= F (u) + C
= F (g (x)) + C

The theorem above provides a method to simplify complex integrals involving

composite functions. By introducing a new variable, we can transform these
integrals into a simpler form that is easier to evaluate with respect to that new
variable.

The challenge in using the Substitution Rule is selecting a suitable substitution

for u. The key idea behind substitution is to simplify the integral, therefore,
ideally, u should be a function whose differential (or a part of its differential) is
present in the integrand. If this isn’t possible, consider choosing u to be a more
complex part of the integrand, such as the inner function of a composite function.

The bottom line is that finding the correct substitution often requires some trial
and error. It’s common to make an incorrect guess, so if the first choice doesn’t
work, simply try a different one.

To demonstrate the method, a few examples are provided below.

Z
I Example 7.1.1. Evaluate (3x + 4)4 dx

Solution. In order to evaluate this integral, let the variable u = 3x + 4, so that,

du
=3
dx
du = 3dx in differential form
1
∴ dx = du
3
This implies that,
Z Z
1
4
(3x + 4) dx = · u4 du
3
1 u5
= · +C
3 5
u5
= +C
15
Since, u = 3x + 4, therefore,

u5
Z
(3x + 4)4 dx = +C
15
(3x + 4)5
= +C
15
(3x + 4)5
Z
Therefore, (3x + 4)4 dx = +C
15

Z
I Example 7.1.2. Evaluate
p
x x2 + 2dx
√
Solution. In order to solve this integral, let the variable u = x2 + 2, so that,
du d
x2 + 2

=
dx dx
= 2x
du = 2xdx
1
∴ dx = du
2x
This implies that,
√ 1
Z p Z
x x2 + 2dx = x u du
2x

Z
1 1
= x · u 2 du
2x
3
1 u2
= · +C
2 3
2
3
2u 2
= +C
6
3
u2
= +C
3
3
x2 + 2 2
= +C
3
3
x2 + 2
Z 2

Therefore,
p
x x2 + 2dx = +C
3

Z
1
I Example 7.1.3. Evaluate dx
x ln(x)
Solution. In order to solve this integral, let the variable u = ln(x), so that,

du d
= ln(x)
dx dx
1
=
x
1
∴ du = dx
x
This implies that,
Z Z
1 1 1
dx = · dx
x ln(x) ln(x) x
| {z } |{z}
u du
Z
1
= du
u
= ln |u| + C
= ln |ln(x)| + C
Z
1
Therefore, dx = ln |ln(x)| + C
x ln(x)

Sometimes the choice of u works for more than one substitution. For example,
consider the following example.

Z
x
I Example 7.1.4. Evaluate √ dx
x+1
Solution. The integral above can be evaluated in multiple ways using the substi-
tution rule. This is because the choice of u is not apparently visible. Whether
√
we choose u = x + 1 or u = x + 1, both yield the same answer.

Choosing u = x + 1: if u = x + 1, then x = u − 1, and

du
=1
dx
∴ du = dx

This implies that,

u−1
Z Z
x
√ dx = √ du
x+1 u
Z
u 1
= √ −√ du
u u
Z Z
u 1
= √ du − √ du
u u
√
Z Z
1
= udu − √ du
u
Z Z
1 1
= u 2 du − u− 2 du
Z Z
1 1
= u du − u− 2 du
2

3 3 1
= u 2 − 2u 2 + C
2
3 3 1
= (x + 1) 2 − 2(x + 1) 2 + C
2
√ √
Choosing u = x + 1: if u = x + 1, then u2 = x + 1 or x = u2 − 1, and

d d 2 d
x= u − 1
du du du
dx
= 2u
du
∴ dx = 2udu

This implies that,

u2 − 1
Z Z
x
√ dx = 2udu
x+1 u

u2 − 1
Z
2udu =
Z u
u2 − 1 du

=2·
Z Z
2
=2· u du − du
3
u
=2· −u +C
3
2u3
= − 2u + C
3
√ 3
2 x+1 √
= −2 x+1+C
3
2 3 √
= (x + 1) 2 − 2 x + 1 + C
3
√
Z
x 2
Which yields the same result. Therefore,
3
√ dx = (x + 1) 2 −2 x + 1+
x+1 3
C

Using the integration by substitution, a few formulas for integration can be

derived as follows.

Z
I Example 7.1.5. Evaluate (ax + b)n dx

Solution. In order to evaluate the integral, we let,

u = ax + b =⇒ du = adx
du
This implies that dx = . Thus, we write,
a
Z Z
n n du
(ax + b) dx = u
a
Z
1
= · un du
a
1 un+1
= · +C
a n+1
un+1
= +C
a (n + 1)
Since u = ax + b, therefore
(ax + b)n+1
Z
(ax + b)n dx = +C
a (n + 1)

(ax + b)n+1
Z
Therefore, (ax + b)n dx = +C
a (n + 1)

Remark 7.1.1.

(ax + b)n+1
Z
(ax + b)n dx = +C n 6= 1
a (n + 1)

Z
I Example 7.1.6. Evaluate sin (ax + b)dx

Solution. In order to evaluate the integral, we let,

u = ax + b =⇒ du = adx
du
This implies that dx = . Thus, we write,
a
Z Z
du
sin (ax + b) dx = sin (u)
a
Z
1
= · sin (u) du
a
1
= · (− cos (u))
a
1
= − cos (u)
a
Since u = ax + b, therefore
Z
1
sin (ax + b)dx = − cos (ax + b)
a
Z
1
Therefore, sin (ax + b)dx = − cos (ax + b) + C
a

Remark 7.1.2.
Z
1
sin (ax + b)dx = − cos (ax + b) + C
a

Z
I Example 7.1.7. Evaluate cos (ax + b) dx

Solution. In order to evaluate the integral, we let,

u = ax + b =⇒ du = adx

du
This implies that dx = . Thus, we write,
a
Z Z
du
cos (ax + b) dx = cos (u)
a
Z
1
= · cos (u) du
a
1
= · sin (u) + C
a
Since u = ax + b, therefore
Z
1
cos (ax + b) dx = sin (ax + b) + C
a
Z
1
Therefore, cos (ax + b) dx = sin (ax + b) + C
a

Remark 7.1.3.
Z
1
cos (ax + b) dx = sin (ax + b) + C
a

Z
I Example 7.1.8. Evaluate tan (ax + b) dx

Solution. In order to evaluate the integral, we let,

u = ax + b =⇒ du = adx

du
This implies that dx = . Thus, we write,
a
Z Z
du
tan (ax + b) dx = tan (u)
a
Z
1
= · tan (u) du
a
sin (u)
Z
1
= · du
a cos (u)

sin (u)
Z
Now, to solve the integral du, we apply the substitution rule once
cos (u)
again. That is, we let

v = cos(u) =⇒ dv = − sin(u)du

Thus, we write,
sin (u)
Z Z
1
du = (−dv)
cos (u) v
Z
1
=− dv
v
= − ln |v| + C

Since v = cos(u) and u = ax + b, therefore, v = cos(ax + b). As a result, we

Remark 7.1.4.
Z
1 1
tan (ax + b) dx = − ln |cos (ax + b)| + C = ln |sec (ax + b)| + C
a a

Z
I Example 7.1.9. Evaluate cot (ax + b) dx

Solution. In order to evaluate the integral, we let,

u = ax + b =⇒ du = adx

du
This implies that dx = . Thus, we write,
a
Z Z
du
cot (ax + b) dx = cot (u)
a

Z
1
= · cot (u) du
a
cos (u)
Z
1
= · du
a sin (u)

cos (u)
Z
Now, to solve the integral du, we apply the substitution rule once
sin (u)
more. That is, we let,

v = sin(u) =⇒ dv = cos(u)du

Thus, we write,
cos (u)
Z Z
1
du = dv
sin (u) v
= ln |v| + C

Since v = sin(u) and u = ax + b, therefore, v = sin(ax + b). As a result, we

Remark 7.1.5.
Z
1 1
cot (ax + b) dx = ln |sin (ax + b)| + C = − ln |csc (ax + b)| + C
a a

Z
I Example 7.1.10. Evaluate sec (ax + b) dx

Solution. In order to evaluate the integral, we let,

u = ax + b =⇒ du = adx

du
This implies that dx = . Thus, we write,
a
Z Z
du
sec (ax + b) dx = sec (u)
a
Z
1
= · sec (u) du
a
tan (u) + sec (u)
Z
1
= · sec (u) · du
a tan (u) + sec (u)
sec (u) tan (u) + sec2 (u)
Z
1
= · du
a tan (u) + sec (u)

sec (u) tan (u) + sec2 (u)

Z
Now, to solve the integral du, we apply the substitu-
tan (u) + sec (u)
tion rule once more. That is, we let,

v = tan(u) + sec(u) =⇒ dv = sec(u) tan(u) + sec2 (u) du

Thus, we write,
sec (u) tan (u) + sec2 (u)
Z Z
1
du = dv
tan (u) + sec (u) v
= ln |v| + C

Since v = tan(u)+sec(u) and u = ax+b, therefore, v = tan(ax+b)+sec(ax+b).

Remark 7.1.6.
Z
1
sec (ax + b) dx = ln |tan (ax + b) + sec (ax + b)| + C
a

Z
I Example 7.1.11. Evaluate csc (ax + b) dx

Solution. In order to evaluate the integral, we let,

u = ax + b =⇒ du = adx
du
This implies that dx = . Thus, we write,
a
Z Z
du
csc (ax + b) dx = csc (u)
a
Z
1
= · csc (u) du
a
csc (u) + cot (u)
Z
1
= · csc (u) · du
a csc (u) + cot (u)
csc2 (u) + cot (u) csc (u)
Z
1
= · du
a csc (u) + cot (u)
csc2 (u) + cot (u) csc (u)
Z
Now, to solve the integral du, we apply the substitu-
csc (u) + cot (u)
tion rule once more. That is, we let,

v = csc(u) + cot(u) =⇒ dv = − csc2 (u) − cot(u) csc(u) du

Thus, we write,
csc2 (u) + cot (u) csc (u)
Z Z
1
du = (−dv)
csc (u) + cot (u) v
Z
1
=− dv
v
= − ln |v| + C

Since v = csc(u) + cot(u) and u = ax + b, therefore, v = csc(ax + b) + cot(ax + b).

Remark 7.1.7.
Z
1
csc (ax + b) dx = − ln |csc(ax + b) + cot(ax + b)| + C
a

f 0 (x)
Z
I Example 7.1.12. Evaluate dx, where f (x) 6= 0 is any function.
f (x)
Solution. In order to evaluate the integral, we let,

u = f (x) =⇒ du = f 0 (x)dx

Thus, we write,
f 0 (x)
Z Z
1
dx = du
f (x) u
= ln |f (x)| + C

f 0 (x)
Z
Therefore, dx = ln |f (x)| + C
f (x)

Remark 7.1.8.
f 0 (x)
Z
dx = ln |f (x)| + C
f (x)

A few common integrals above are summarized below for a quick reference.

Z
1
7. csc (ax + b) dx = − ln |csc (ax + b) + cot (ax + b)| + C
a
f 0 (x)
Z
8. dx = ln |f (x)| + C
f (x)

Z
I Example 7.1.13. Evaluate (5x + 5)5 dx

Solution. We know that,

(ax + b)n+1
Z
n
(ax + b) dx = +C
a (n + 1)
where, n 6= −1. Therefore,
(5x + 5)5+1
Z
(5x + 5)5 dx = +C
5(5 + 1)
(5x + 5)6
= +C
30
(5x + 5)6
Z
Therefore, (5x + 5) dx =
5
+C
30

Chap. 7 / Sec. 7.1 / Subsec. 7.1.1 : Substitution Rule for Definite Integrals
The substitution rule can also be applied to definite integrals. For that, when
changing variables from x to u, the limits of integration are also changed accord-
ingly. That is, if u = g(x), the original lower limit x = a becomes u = g(a), and

the upper limit x = b becomes u = g(b). After substituting both the variable
and the limits, we can complete the integration with respect to u. As a result,
we have the following theorem.

Theorem 7.1.2. Let u = g(x) be a differentiable function over an interval I,

and let f (u) be a continuous function over the rage of g(x). If F (u) is an
antiderivative of f (u) on I, then

Zb Zg(b)
f (g(x))g 0 (x)dx = f (u)du
a g(a)

Z2
2x
I Example 7.1.15. Evaluate dx
2 + x2
0

Solution. In order to solve the integral, let the variable u = 2 + x2 , then

du d
= (2 + x2 )
dx dx
= 2x
∴ du = 2xdx

Now, since u = 2 + x2 , therefore,

lower limit x = 0 becomes u = 2 + 02 = 2

upper limit x = 2 becomes u = 2 + 22 = 6

This implies that,

Z2 Z2
2x 1
dx = 2 2xdx
2 + x2 2| + x | {z }
0 0 {z } du
u
Z6
1
= du
u
2
i6
= ln |u|
2
= ln |6| − ln |2|
≈ 1.0986

Z2
2x
dx ≈ 1.0986
2 + x2
0

Ze2
ex − e−x
I Example 7.1.16. Evaluate dx
ex + e−x
e

Solution. In order to solve the integral, let the variable u = ex + e−x , then
du d x
e + e−x

=
dx dx
du d x d
= e + e−x
dx dx dx
du
= ex − e−x
dx
du = ex − e−x dx differential form

Now, since u = ex + e−x , therefore,

lower limit x = e becomes u = ee + e−e

upper limit x = e2 becomes
2 2
u = ee + e−e

This implies that,

Ze2 x −x Ze2
e −e 1 x −x

dx = −x e − e dx
ex + e−x e|x +
{ze } | {z }
e e du
u
2 2
+e−e
ee Z
1
= du
u
ee +e−e
h iee2 +e−e2
= ln |u|
ee +e−e

= ln e + e−e − ln ee + e−e
e2 2

≈ 4.666

Ze2
ex − e−x
Therefore, dx ≈ 4.666
ex + e−x
e

The End of Section 7.1

Integration by Substitution

§ 7.2 Integration by Parts

Integration by parts is a technique used to solve integrals involving products of
functions that are otherwise difficult to integrate directly. It can be thought of
as a “backward rule” for the Product rule in differentiation. The core idea of this
technique is to transform a difficult integral into the difference between a simple
product and an integral that is easier to solve. The method is derived as follows.

Suppose, we’re given h(x) = f (x) · g(x), then the product rule in differentiation
states that,
d d
h(x) = f (x) · g(x)
dx dx
d d
= f (x) · g(x) + g(x) · f (x)
dx dx
0 0 0
∴ h (x) = f (x) g (x) + g(x)f (x) (1)

This implies—by the definition of an antiderivative—that h(x) is an antideriva-

tive of f (x) g 0 (x) + g(x)f 0 (x), therefore, writing it in the integral form yields
the following,
Z
f (x) g 0 (x) + g(x)f 0 (x) dx = h (x)
Z Z
f (x) g (x) dx + g(x)f 0 (x) dx = f (x) g (x)
0

Z Z
f (x) g (x) dx = f (x) g (x) − g(x)f 0 (x) dx
0

Now, if we let u = f (x), then the differential of u yields du = f 0 (x)dx. Similarly,

if v = g(x), then the differential of v yields dv = g 0 (x)dx. This implies the
following,
Z Z
f (x) g (x) dx = f (x) g(x) − g(x) f 0 (x) dx
0
| {z } | {z } |{z} |{z} |{z} | {z }
u dv u v v du

Therefore, we have the following,

Z Z
u dv = uv − v du
Z
The method is based on the idea that if the integral udv is challenging to
Z Z
evaluate, it can be simplified by rewriting it as uv − vdu, where vdu is
easier to solve. Thus, we have the following theorem.

Theorem 7.2.1. (Integration by Parts)

If u and v are differentiable functions of x, then
Z Z
u dv = uv − v du

A few examples of the theorem above are provided below.

Z
I Example 7.2.1. Evaluate x sin(x)dx
Z
Solution. Notice that the integral sin(x)dx is easier to solve, however, the
Z
integral x sin(x)dx is not. In order to solve such an integral, we use integration
by parts, stating the following.
Z Z
u dv = uv − v du
Z Z
This necessitates us to write the integral x sin(x)dx in the form udv. That
Z
is to identify which part of the integral x sin(x)dx is u and which part is dv.

Now to choose u and dv, notice that if we choose u = x, then dv = sin(x)dx fits
a basic integration formula. Therefore, we let the following,

u=x =⇒ du = dx
Z Z
dv = sin(x)dx =⇒ v = dv = sin(x)dx = − cos(x)

As a result, by integration by parts, we write the following.

Z Z
x sin (x) dx = |{z}
|{z} x · (− cos (x)) − (− cos (x)) |{z}
dx
| {z } | {z } | {z }
u dv u v v du
Z
= −x cos (x) + cos (x) dx

= −x cos (x) + sin (x) + C

Z
Therefore, x sin(x)dx = −x cos (x) + sin (x) + C

Z
I Example 7.2.2. Evaluate x cos(x)dx

Solution. In order to solve the integral above, we will use integration by parts,
stating the following. Z Z
u dv = uv − v du
Z Z
This necessitates us to write the integral x sin(x)dx in the form udv. That
Z
is, to identify which part of the integral is x cos(x)dx is u and which part is dv.

If we let u = x, then dv = sin(x)dx fits a basic integration formula. Therefore,

we let the following,

u=x =⇒ du = dx
Z Z
dv = cos(x)dx =⇒ v = dv = cos(x)dx = sin(x)

As a result, by integration by parts, we write the following.

Z Z
x cos (x) dx = |{z}
|{z} x sin (x) − sin (x) |{z}dx
| {z } | {z } | {z }
u dv u v v du

= x sin (x) − (− cos (x)) + C

= x sin (x) + cos (x) + C
Z
Therefore, x cos(x)dx = x sin (x) + cos (x) + C

Z
I Example 7.2.3. Evaluate ln(x)dx

Solution. In order to solve this, we use integration by parts. Therefore, we let,

1
u = ln(x) =⇒ du = dx
Z x Z
dv = dx =⇒ v = dv = dx = x

Therefore, by integration by parts, we write the following,

Z Z
1
ln(x) |{z}
dx = ln(x) · |{z}
x − |{z} x · dx
| {z } | {z } x
u dv
|{z} u v v
du
Z
= x ln(x) − dx

= x ln(x) − x + C

Z
Therefore, ln(x)dx = x ln(x) − x + C

Remark 7.2.1. Z
ln(x)dx = x ln(x) − x + C

Integration by parts is a handy method, especially when dv is easy to integrate.

However, picking the right u and dv isn’t straightforward and often comes down
to experience and practice. A useful technique is to select u such that it becomes
simpler when differentiated, and dv such that it remains easy to integrate, as
done before. However, this technique is somewhat vague. A more structured
method is described below.

Chap. 7 / Sec. 7.2 / Subsec. 7.2.1 : The LIATE Rule

The LIATE rule helps determine which part of an integrand to choose as u when
using integration by parts. The acronym LIATE stands for the following set of
functions.
1. Logarithmic Functions (e.g. ln(x), loga (b), etc.)

2. Inverse Trigonometric Functions (e.g. sin−1 (x), cos−1 (x), etc.)

√
3. Algebraic Functions (e.g. x2 , x, etc.)

4. Trigonometric Functions (e.g. sin(x), cos(x), etc.)

5. Exponential Functions (e.g. ex , 2x , etc.)

According to this rule, u is chosen as the first function that appears in the LIATE
order (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponen-
tial). The remaining part of the integrand becomes dv.
Z
For example, to integrate the integral xex dx using integration by parts, we
use the LIATE rule. Since x (an Algebraic function) appears before ex (an Ex-
ponential function) in the LIATE order, therefore we let u = x and dv = ex dx.

It’s important to note that while this method isn’t foolproof, it works well enough
to be considered useful. Below are a few examples where this rule applies per-
fectly.

Z
I Example 7.2.4. Evaluate xex dx

Solution. Notice that the integrand is a product of an algebraic function x and

an exponential function ex . According to the LIATE rule, we choose,

u=x =⇒ du = dx
Z Z
dv = ex dx =⇒ v = dv = ex dx = ex

Thus, we write the following,

Z Z
x
xe dx = u dv
Z
= uv − v du
Z
= xe − ex dx
x

= xex − ex + C
= (x − 1) ex + C
Z
Therefore, xex dx = (x − 1) ex + C

Sometimes, evaluating an integral requires using integration by parts multiple

times. This repeated process is necessary when the resulting integral after the
first application can still be integrated using integration by parts. By repeatedly
applying this method, we can gradually simplify the integral until we find a
solution. For example, consider the following integral.

Z
I Example 7.2.5. Evaluate x2 ex dx

Solution. By the LIATE rule, we choose,

u = x2 =⇒ du = 2xdx
Z Z
dv = ex dx =⇒ v = dv = ex dx = ex

Thus, we write the following,

Z Z
x
e dx = u dv

Z
= uv − v du
Z
= x2 ex − ex 2x dx
Z Z
Now, notice that the integral 2xe dx is of the form
x
f (x)g(x)dx such that
it can be further integrated by integration by parts. In order to do that, we
choose, by the LIATE rule, the following.

u = 2x =⇒ du = 2dx
Z Z
dv = ex dx =⇒ v = dv = ex dx = ex

Thus, we write the following,

Z Z
x
e 2x dx = u dv
Z
= uv − v du
Z
= 2xe − ex 2dx
x

Z
= 2xe − 2 ex dx
x

= 2xex − 2ex

This implies that

Z
x2 ex dx = x2 ex − 2xex − 2ex + C

= (x2 − 2x − 2)ex + C
Z
Therefore, x2 ex dx = (x2 − 2x − 2)ex + C

Z
I Example 7.2.6. Evaluate x2 sin(x)dx

Solution. By the LIATE rule, we choose,

u = x2 =⇒ du = 2x dx
Z Z
dv = sin(x)dx =⇒ v = dv = sin(x) dx = − cos(x)

Thus, we write the following,

Z Z
x sin(x)dx = u dv
2

Z
= uv − v du
Z
= x (− cos(x)) −
2
(− cos(x))2x dx
Z
= −x cos(x) + 2
2
x cos(x) dx
Z Z
Now, notice that the integral x cos(x) dx is of the form f (x)g(x)dx such
that it can be further integrated by integration by parts. In order to do that,
we choose, by the LIATE rule, the following.

u=x =⇒ du = dx
Z Z
dv = cos(x)dx =⇒ v = dv = cos(x)dx = sin(x)

Thus, we write the following,

Z Z
x cos(x)dx = u dv
Z
= uv − v du
Z
= x sin(x) − sin(x)dx

= x sin(x) − (− cos(x))
= x sin(x) + cos(x)

This implies the following,

Z
x2 sin(x)dx = −x2 cos(x) + 2 (x sin(x) + cos(x))

= −x2 cos(x) + 2x sin(x) + 2 cos(x) + C

Z
Therefore, x2 sin(x)dx = −x2 cos(x) + 2x sin(x) + 2 cos(x) + C

Chap. 7 / Sec. 7.2 / Subsec. 7.2.2 : Integration by Parts for Definite Integrals
Integration by parts can also be applied to definite integrals as stated by the
following theorem.

Theorem 7.2.2. Let u and v be differentiable functions of x, then

Zb ib Zb
u(x)v 0 (x)dx = u(x)v(x) − v(x)u0 (x)dx
a
a a

A few examples are provided below.

Z2
I Example 7.2.7. Evaluate ln(x)dx
1

Solution. In order to integrate, we choose the following,

1
u = ln(x) =⇒ du = dx
Zx Z
dv = dx =⇒ v = dv = dx = x

Therefore, we write the following,

Z2 Z2
ln(x)dx = u(x) v 0 (x)dx
|{z} | {z }
1 1 u dv
2
Z2
= u(x) v(x) − v(x) u0 (x)dx
|{z} |{z} |{z} | {z }
u v 1 1 v du
Z2
i2 1
= ln(x) · x − x dx
1 x
1

i2 Z2
= x · ln(x) − dx
1
1

= 2 · ln(2) − 1 · ln(1) − (2 − 1)
= 2 ln(2) − 1
≈ 0.386

Z2
Therefore, ln(x)dx ≈ 0.386
1

Chap. 7 / Sec. 7.2 / Subsec. 7.2.3 : Integration by Reduction Formulas

Reduction formulas are formulas that express an integral involving a power of a
function in terms of an integral involving a lower power of that same function.
By repeatedly applying these formulas, we can reduce a complex integral into
simpler forms, eventually arriving at a solvable
Z integral. For example, we can
derive a reduction formula for the integral sinn (x)dx, by noting that sinn (x)
can be written as sinn−1 (x) · sin(x). This implies
Z Z
sin (x) dx = sinn−1 (x) sin (x) dx
n

Now, if we let

u = sinn−1 (x) =⇒ du = (n − 1) sinn−2 (x) cos(x)dx

Z Z
dv = sin(x)dx =⇒ v = dv = sin(x)dx = − cos(x)

then we have the following,

Z Z
udv =uv − vdu
Z
sinn−1 (x) sin (x) dx =sinn−1 (x) · (− cos (x))
Z
− (− cos (x)) · (n − 1) sinn−2 (x) cos (x) dx
Z
(sin (x))n−1 (sin (x))1 dx = − sinn−1 (x) cos (x)
Z
+ (n − 1) cos2 (x) · sinn−2 (x) dx
Z
(sin (x))n−1+1 dx = − sinn−1 (x) cos (x)
Z
1 − sin2 (x) · sinn−2 (x) dx

+ (n − 1)
Z
sinn (x) dx = − sinn−1 (x) cos (x)
Z
+ (n − 1) sinn−2 (x) − sin2 (x) sinn−2 (x) dx
Z
sinn (x) dx = − sinn−1 (x) cos (x)
Z
+ (n − 1) sinn−2 (x) − sinn (x) dx
Z
sinn (x) dx = − sinn−1 (x) cos (x)

Z Z
+ (n − 1) sin n−2
(x) dx − (n − 1) sinn (x) dx
Z Z
sin (x) dx + (n − 1)
n
sinn (x) dx = − sinn−1 (x) cos (x)
Z
+ (n − 1) sinn−2 (x) dx
Z Z
n sin (x) dx = − sin
n
(x) cos (x) + (n − 1) sinn−2 (x) dx
n−1

−sinn−1 (x) cos (x) + (n − 1) sinn−2 (x) dx

Z R
sin (x) dx =
n
n
n−1
Z
1 n−1
= − sin (x) cos (x) + sinn−2 (x) dx
n n
This implies that
n−1
Z Z
1 n−1
sin dx = − sin
n
(x) cos (x) + sinn−2 (x) dx
n n
Similar formulas can be obtained for other trigonometric functions as well.

n−1
Z Z
1 n−1
1. sin (x) dx = − sin
n
(x) cos (x) + sinn−2 (x) dx
n n
n−1
Z Z
1 n−1
2. cos (x) dx = cos
n
(x) sin (x) + cosn−2 (x) dx
n n
Z Z
1
3. tann (x) dx = tann−1 (x) − tann−2 (x) dx, n 6= 1
n−1
Z Z
1
4. cot (x) dx = −
n
cotn−1
(x) − cotn−2 (x) dx, n 6= 1
n−1
n−2
Z Z
1
5. sec (x) dx =
n
secn−2
(x) tan (x) − secn−2 (x) dx,
n−1 n−1
n 6= 1
n−2
Z Z
1
6. csc (x) dx =
n
csc n−2
(x) cot (x) − cscn−2 (x) dx,
n−1 n−1
n 6= 1

The End of Section 7.2

Integration by Parts

§ 7.3 Trigonometric Integrals

So far, we have discussed various methods to integrate integrals that are rela-
tively easy to solve. These methods have so far been instrumental, but when

faced with integrals involving complicated functions such as trigonometric func-

tions raised to any powers or in products, we may fall short of techniques to
integrate them. For example, from our current knowledge, we can easily inte-
grate the following types of integral,
Z
ex sin(x)dx

The same, however, cannot be said for integrating the following types of inte-
grals, Z Z
sin (x)dx
2
or sin(x) cos(x)dx

This is because neither the substitution rule nor integration by parts is immedi-
ately useful in this case, as the trigonometric expressions don’t lend themselves
to obvious substitutions or algebraic simplifications. This limitation becomes
even more apparent when considering even complex examples, such as:
Z
sin4 (x) cos3 (x)dx

These types of integrals fall under the category of trigonometric integrals, which
require different strategies than the ones we have discussed so far. In this section,
we aim to develop techniques for integrating such integrals. We will do so by
manipulating trigonometric expressions using identities, reducing powers, and
transforming products into more manageable forms.

Chap. 7 / Sec. 7.3 / Subsec. 7.3.1 : Integrals Involving Powers of sin(x) and
cos(x)
In order to integrate integrals of the form,
Z Z
sinm (x)dx and cosn (x)dx

two strategies are primarily used based on whether the power is odd or even. For
example, with even powers of sin(x) or cos(x), we use the half-angle formulas,
1 − cos(2x) 1 + cos(2x)
sin2 (x) = and cos2 (x) =
2 2
to reduce the powers of sin(x) or cos(x) in the integrand.

Similarly, with odd powers of sin(x) or cos(x), the best idea is to split off a single
factor of sin(x) or cos(x) and to use the Pythagorean identities in a way that

makes the substitution rule useful.

These techniques can best be explained through the following examples.

Z
I Example 7.3.1. Evaluate cos3 (x)dx

Solution. For odd powers, we use substitution by expressing the cosine function
in terms of sine. For that, we rewrite cos3 (x) as the following,

cos3 (x) = cos2 (x) · cos(x)

Now, since cos2 (x) = 1 − sin2 (x), therefore, the integral becomes,
Z Z
cos (x)dx = cos2 (x) cos(x)dx
3

Z
1 − sin2 (x) cos(x)dx

=

Now, let u = sin(x), so, du = cos(x)dx. Then,

Z Z
1 − sin (x) cos(x)dx =
2
1 − u2 du

Z Z
= du − u2 du
u3
=u− +C
3
sin3 (x)
= sin(x) − +C
3
sin3 (x)
Z
Therefore, cos3 (x)dx = sin(x) − +C
3

Z
I Example 7.3.2. Evaluate sin4 (x)dx

Solution. For even powers, we express sin4 (x) as (sin2 (x))2 , and then use the
half-angle formula for sin2 (x). That is, since,
1 − cos(2x)
sin2 (x) =
2
therefore, we write the following,
Z Z
sin (x)dx = sin2 (x) dx
4
2

1 − cos(2x)
Z 2
= dx
2
(1 − cos(2x))2
Z
dx =
4
1 − 2 cos(2x) + cos2 (2x)
Z 2
= dx
Z 4
1
1 − 2 cos(2x) + cos2 (2x) dx

= ·
4
Z Z Z
1
= · dx − 2 cos(2x)dx + cos2 (2x)dx
4
Z
Now, to evaluate the integral cos2 (2x)dx, we use the following half-angle
formula for cos2 (2x), that is,

1 + cos(4x)
cos2 (2x) =
2
This implies that
Z Z Z Z
1
sin (x)dx = ·
4
dx − 2 cos(2x)dx + cos (2x)dx 2
4
1 + cos(4x)
Z Z Z
1
= · dx − 2 cos(2x)dx + dx
4 2
Z Z Z
1 1
= · dx − 2 cos(2x)dx + · (1 + cos(4x)) dx
4 2
Z Z Z Z
1 1
= · dx − 2 cos(2x)dx + · dx + cos(4x)dx
4 2
sin(2x) 1 sin(4x)

1
= · x−2· + · x+ +C
4 2 2 4
1 sin(4x)

1 1
= · x − sin(2x) + x + +C
4 2 2 4
x sin(2x) x sin(4x)
= − + + +C
4 4 8 32
x sin(2x) x sin(4x)
Z
Therefore, sin4 (x)dx = − + + +C
4 4 8 32

Chap. 7 / Sec. 7.3 / Subsec. 7.3.2 : Integrals Involving Product of Powers of

sin(x) and cos(x)
We now consider integrals involving products of sin(x) and cos(x) of the following
form,
Z
sinm (x) cosn (x) dx

where, m and n are positive integers.

In order to solve such integrals, we refer to the following steps.

Z
1. If m in the integral sinm (x) cosn (x) dx is odd, then

(a) Split off a factor of sin(x)

(b) Use the identity sin2 (x) = 1 − cos2 (x)
(c) Make the substitution u = cos(x) and du = − sin(x)dx
Z
2. If n in the integral sinm (x) cosn (x) dx is odd, then

(a) Split off a factor of cos(x)

(b) Use the identity cos2 (x) = 1 − sin2 (x)
(c) Make the substitution u = sin(x) and du = cos(x)dx
Z
3. If both m, n in the integral sinm (x) cosn (x) dx is even, then use
the following identities to reduce the powers on sin(x) and cos(x).

1 − cos (2x) 1 + cos (2x)

sin2 (x) = and cos2 (x) =
2 2

Z
I Example 7.3.3. Evaluate sin4 (x) cos3 (x) dx

Solution. To solve the integral, we follow the procedure above.

Z Z
sin (x) cos (x) dx = sin4 (x) cos2 (x) cos (x) dx
4 3

Z
= sin4 (x) 1 − sin2 (x) cos (x) dx

Now, let u = sin(x), then du = cos(x)dx. Thus, we write the following,

Z Z
sin (x) 1 − sin (x) cos (x) dx = u4 1 − u2 du
4 2

Z
u4 − u6 du

=
Z Z
= u du − u6 du
4

u5 u7
= − +C
5 7
sin6 (x) sin8 (x)
= − +C
6 8
sin6 (x) sin8 (x)
Z
Therefore, sin4 (x) cos3 (x) dx = − +C
6 8

Z
I Example 7.3.4. Evaluate sin3 (x) cos4 (x) dx

Solution. In order to solve the integral, we follow the procedure above.

Z Z
sin (x) cos (x) dx = cos4 (x) sin2 (x) sin (x) dx
3 4

Z
= cos4 (x) 1 − cos2 (x) sin (x) dx

Now, let u = cos(x), then du = − sin(x)dx. Thus, we write the following,

Z Z
cos (x) 1 − cos (x) sin (x) dx = u4 1 − u2 (−du)
4 2

Z
= − u4 1 − u2 du

Z
u4 − u6 du

=−
Z Z
= − u du + u6 du
4

u5 u7
=− + +C
5 7
cos7 (x) cos5 (x)
= − +C
7 5
cos7 (x) cos5 (x)
Z
Therefore, sin3 (x) cos4 (x) dx = − +C
7 5

Chap. 7 / Sec. 7.3 / Subsec. 7.3.3 : Integrals Involving Powers of tan(x) and
sec(x)
The idea behind integrating integrals involving powers of tan(x) and sec(x) is
similar to that of integrals involving powers of sin(x) and cos(x) as discussed
before. For example, consider integrating the following integral.

Z
I Example 7.3.5. Evaluate sec3 (x)dx

Solution. For odd powers, we use integration by parts by rewriting sec3 (x) as
sec(x) sec2 (x) and letting,

u = sec(x) =⇒ du = sec(x) tan(x)dx

Z Z
dv = sec2 (x)dx =⇒ v = dv = sec2 (x)dx = tan(x)

Therefore,
Z Z
sec (x)dx =
3
sec(x)sec2 (x)dx
Z
= udv
Z
= uv − vdu
Z
= sec(x) tan(x) − tan(x) · sec(x) tan(x)dx
Z
= sec(x) tan(x) − tan2 (x) sec(x)dx

Now, since tan2 (x) = sec2 (x) − 1, therefore,

Z Z
sec (x)dx = sec(x) tan(x) − tan2 (x) sec(x)dx
3

Z Z
sec (x)dx = sec(x) tan(x) −
3
sec2 (x) − 1 sec(x)dx

Z Z Z
sec (x)dx = sec(x) tan(x) − sec (x)dx + sec(x)dx
3 3

Z Z Z
sec (x)dx + sec (x)dx = sec(x) tan(x) + sec(x)dx
3 3

Z
2 · sec3 (x)dx = sec(x) tan(x) + ln |sec(x) + tan(x)| + C
sec(x) tan(x) + ln |sec(x) + tan(x)| + C
Z
sec3 (x)dx =
2

Z
1 1
sec3 (x)dx = sec(x) tan(x) + ln |sec(x) + tan(x)| + C
2 2
Z
1 1
Therefore, sec3 (x)dx = sec(x) tan(x) + ln |sec(x) + tan(x)| + C
2 2

Z
I Example 7.3.6. Evaluate tan4 (x)dx

Solution. For even powers, we rewrite tan4 (x) as tan2 (x) · tan2 (x) and use the
following identity,
tan2 (x) + sec2 (x) = 1

This implies that,

Z Z
tan (x)dx =
4
tan2 (x) · tan2 (x)dx
Z
tan2 (x) · sec2 (x) − 1 dx

=
Z
tan2 (x)sec2 (x) − tan2 (x) dx

=
Z Z
= tan (x)sec (x)dx − tan2 (x)dx
2 2

Z Z
tan (x)sec (x)dx −
2 2
sec2 (x) − 1 dx

=
Z Z Z
= tan (x)sec (x)dx − sec (x)dx + dx
2 2 2

Now, let u = tan(x), then du = sec2 (x)dx in the first integral. Therefore,
Z Z Z Z
tan (x)dx = tan (x)sec (x)dx − sec (x)dx + dx
4 2 2 2

Z Z Z
= u du − sec (x)dx + dx
2 2

u3
Z Z
= − sec (x)dx + dx
2
3
tan3 (x)
= − tan(x) + x + C
3
tan3 (x)
Z
Therefore, tan (x)dx =
4
− tan(x) + x + C
3

Chap. 7 / Sec. 7.3 / Subsec. 7.3.4 : Integrals Involving Product of Powers of

tan(x) and sec(x)
We now consider integrals involving tan(x) and sec(x) of the following form,
Z
tanm (x)secn (x)dx

where, m and n are positive integers.

In order to solve such integrals, we refer to the following steps.

Z
1. If m in the integral tanm (x) secn (x) dx is odd, then

(a) Split off a factor of sec(x) tan(x)

(b) Use the identity tan2 (x) = sec2 (x) − 1
(c) Make the substitution u = sec(x) and du = sec(x) tan(x)dx
Z
2. If n in the integral tanm (x) secn (x) dx is even, then

(a) Split off a factor of sec2 (x)

(b) Use the identity sec2 (x) = tan2 (x) + 1
(c) Make the substitution u = tan(x) and du = sec2 (x)dx
Z
3. If m is even and n is odd in the integral tanm (x) secn (x) dx, then
use the following identities to reduce the powers on sec(x).

(a) tan2 (x) = sec2 (x) − 1

(b) Use the reduction formula for powers of sec(x).

Z
I Example 7.3.7. Evaluate tan3 (x) sec4 (x) dx

Solution. In order to solve the integral, we follow the procedure above.

Z Z
tan (x) sec (x) dx = tan3 (x) sec2 (x) sec2 (x) dx
3 4

Z
= tan3 (x) tan2 (x) + 1 sec2 (x) dx

Now, let u = tan(x), then du = sec2 (x)dx. Thus, we write the following,
Z Z
tan (x) sec (x) dx = tan3 (x) sec2 (x) sec2 (x) dx
3 4

Z
tan3 (x) tan2 (x) + 1 sec2 (x) dx

=

Z Z
tan (x) tan (x) + 1 sec (x) dx =
3 2 2
u3 u2 + 1 du

Z
u5 + u3 du

=
Z Z
= u du + u3 du
5

u6 u4
= + +C
6 4
tan6 (x) tan4 (x)
= + +C
6 4
tan6 (x) tan4 (x)
Z
Therefore, tan3 (x) sec4 (x) dx = + +C
6 4

The End of Section 7.3

Trigonometric Integrals

§ 7.4 Trigonometric Substitution

Trigonometric substitution is a technique to simplify integrals that are otherwise
difficult to solve. It is used in cases when the square root of a sum or difference
√ √
of two squares is not easy to simplify (that is, of the form: a2 + x2 , a2 − x2 ,
√
and x2 − a2 ), but the square root of a product of two squares is (that is, of
√
the form: A2 B 2 = |AB|). For that reason, integrals such as the following
Z p
1 + x2 dx

√
is quite difficult to solve. Nevertheless, if we can replace 1 + x2 with a product
of squares, then it becomes easier to solve.

The
p goal of this√section is to transform the sum or difference of squares,
p that is,
a ± x , and x − a , into products of squares. It turns out that a2 ± x2 ,
2 2 2 2
√
and x2 − a2 can be transformed into products of squares by seemingly unex-
pected substitution: trigonometric substitution. Really, what do trigonometric
functions have to do with the sum or difference of squares? This is what we see
in the following subsections.

√
Chap. 7 / Sec. 7.4 / Subsec. 7.4.1 : Integrals Involving a2 − x2
√
Suppose, we’re given an integral whose integrand involves the term a2 − x2 ,
where a is a positive constant. Now, notice what happens when we substitute
√
x = a sin(θ) in the expression a2 − x2 , as shown below.
q
a − x = a2 − (a sin (θ))2
p
2 2
q
= a2 − a2 sin2 (θ)
q
= a2 1 − sin2 (θ)

It is known from the Pythagorean identity that sin2 (θ) + cos2 (θ) = 1, or,
cos2 (θ) = 1 − sin2 (θ). Therefore,

= a2 1 − sin2 (θ)
s
| {z }
cos2 (θ)

= a2 · cos2 (θ)
p

= |a · cos (θ)|
= a cos (θ) (1)
√
This implies that the substitution x = a sin(θ) transform the difficult a2 − x 2
into an easier product a cos(θ). That is,

a2 − x2 = a cos(θ)
p

if we let x = a sin(θ).

√
The substitution x = a sin(θ) for a2 − x2 works because, if we consider a right-
angled triangle whose one side is x and the hypotenuse is a, the length of the
√
other side becomes a2 − x2 from the Pythagorean theorem, as shown below.

a
x

Figure 7.1

This implies that,

x
sin(θ) =
a
∴ x = a sin(θ)

A few examples of the technique above are provided below.

Z
dx
I Example 7.4.1. Evaluate √
4 − x2
Solution. Since a = 2 in the integral above, therefore to solve the integral, let
x = 2 sin(θ), then dx = 2 cos(θ)dθ. This implies that,
2 cos(θ)dθ
Z Z
dx
√ = q
4 − x2 4 − (2 sin(θ))2
2 cos(θ)
Z
= q dθ
4 − (2 sin(θ)) 2

2 cos(θ)
Z
= p dθ
4 − 4sin2 (θ)
2 cos(θ)
Z
= q dθ
4 1 − sin2 (θ)
2 cos(θ)
Z
= √ p dθ
4 · cos2 (θ)
2 cos(θ)
Z

= dθ

2 cos(θ)

Z
= dθ

=θ+C

−1 x

Since x = 2 sin(θ), therefore, θ = sin . As a result,
2
Z
dx
√ =θ+C
4 − x2 x
= sin−1 +C
2
Z
dx x
Therefore, √ = sin−1 +C
4 − x2 2

Z p
I Example 7.4.2. Evaluate 9 − x2 dx

Solution. Since a = 3 in the integral above, therefore, let x = 3 sin(θ), then

dx = 3 cos(x)dθ. This implies the following,
Z p Z q
2
9 − x dx = 9 − (3 sin(θ))2 3 cos(θ)dθ
Z q
= 9 − 9sin2 (θ)3 cos(θ)dθ
Z q
9 1 − sin2 (θ) 3 cos(θ)dθ

=
Z p
= 9 (cos2 (θ))3 cos(θ)dθ
Z p
= 9 (cos2 (θ))3 cos(θ)dθ
Z √ p
= 9 · cos2 (θ)3 cos(θ)dθ
Z
= 3 cos(θ) · 3 cos(θ)dθ
Z
= 9 cos2 (θ)dθ

1 + cos(2θ)
Since it is known, from the double-angle identity, that cos2 (θ) = ,
2
therefore,
1 + cos(2θ)
Z Z
9 cos (θ)dθ = 9
2
dθ
Z 2
9
= (1 + cos(2θ)) dθ
2
Z Z
9
= dθ + cos(2θ)dθ
2
sin(2θ)

9
= θ+ +C
2 2
Since it is known, from the double-angle identity, that sin(2θ) = 2 sin(θ) cos(θ),
therefore,
sin(2θ)
Z
9
9 cos (θ)dθ =
2
θ+ +C
2 2
2 sin(θ) cos(θ)

9
= θ+ +C
2 2
−1 x

Since x = 3 sin(θ), therefore θ = sin . Thus,
3
2 sin(θ) cos(θ)
Z p
2
9
9 − x dx = θ+ +C
2 2

−1 x −1 x

sin cos sin 2 sin
 
9 x
= sin−1 + 3 3 + C
2 3 2

9 −1 x x x
= sin + sin sin−1 cos sin−1 +C
2 3 3 3
√
Since sin(sin−1 (x)) = x and cos(sin−1 (x)) = 1 − x2 , therefore,
Z p
9 −1 x
−1 x

−1 x

2
9 − x dx = sin + sin sin cos sin +C
2 3 3 3
!
9 x x r x 2
= sin−1 + · 1− +C
2 3 3 3
r
9 −1 x 3x x 2
= sin + 1− +C
2 3 2 3
Z p
9 x 3x r x 2
Therefore, 9 − x dx = sin
2 −1
+ 1− +C
2 3 2 3

We now generalize the above integrals into formulas.

Z p
I Example 7.4.3. Evaluate a2 − x2 dx

Solution. To evaluate the integral, we let x = a sin(θ). The the differential of x

becomes dx = a cos(θ)dθ. Therefore,
Z p Z q
2 2
a − x dx = a2 − (a sin (θ))2 a cos (θ) dθ
Z q
= a2 − a2 sin2 (θ)a cos (θ) dθ
Z q
a2 1 − sin2 (θ) a cos (θ) dθ

=
Z √ q
1 − sin2 (θ) a cos (θ) dθ

= 2
a ·
Z
= a · cos2 (θ)a cos (θ) dθ
p

Z
= a cos (θ) cos (θ) dθ
Z
=a 2
cos2 (θ) dθ (E1)

To deal with the integrand of the above integral, we make use of the half-angle
identity, stating that,
1 + cos(2θ)
cos2 (θ) =
2

Therefore, continuing on with equation (E1), we get,

1 + cos (2θ)
Z Z
a 2
cos (θ) dθ = a
2 2
dθ
2
1 cos (2θ)
Z
2
=a + dθ
2 2
cos (2θ)
Z Z
2 1 2
=a dθ + a dθ
2 2
a2 a2
Z Z
= dθ + cos (2θ) dθ
2 2
a2 sin (2θ)

= θ+ +C
2 2
a2 2 sin (θ) cos (θ)

= θ+ +C
2 2
a2
= (θ + sin (θ) cos (θ)) + C (E2)
2
Now, from the Pythagorean identity, it is known that,

sin2 (θ) + cos2 (θ) = 1

cos2 (θ) = 1 − sin2 (θ)
q
cos(θ) = 1 − sin2 (θ) (E3)

Substituting equation (E3) into (E2) yields the following.

a2 a2
q
(θ + sin (θ) cos (θ)) + C = θ + sin (θ) 1 − sin (θ) + C
2
(E4)
2 2
Now, since x = a sin(θ), therefore,

x = a sin(θ)
x
= sin(θ)
a x
∴ θ = sin−1
a
x
Substituting θ = sin −1
into equation (E4) yields the following,
a
a2
q
= θ + sin (θ) 1 − sin (θ) + C
2
2
a2 x r
x x
= sin−1 + sin sin−1 1 − sin2 sin−1 +C
2 a a a
!
a2 x x r x 2
= sin−1 + · 1− +C
2 a a a

r !
a2 x x x 2
= sin−1 + · 1− 2 +C
2 a a a
r
a2 −1 x a2 x a2 − x2
= sin + +C
2 a 2 ar a2
a2 −1 x x 1 p 2
= sin + a · a − x2 + C
2 a 2 a2
a2 −1 x x a p 2
= sin + √ · a − x2 + C
2 a 2 a2
a2 −1 x x p 2
= sin + a − x2 + C
2 a 2
This suggests the following,

a2
Z p x xp
a2 − x2 dx = sin−1 + a2 − x 2 + C
2 a 2

Remark 7.4.1.
a2
Z p x xp
a2 − x2 dx = sin−1 + a2 − x 2 + C
2 a 2

Z
dx
I Example 7.4.4. Evaluate √
a2 − x 2
Solution. To evaluate the integral, we let x = a sin(θ). Then, the differential
dx = a cos(θ)dθ. Therefore,

a cos (θ) dθ
Z Z
dx
√ = q
a2 − x 2 a2 − (a sin (θ))2
a cos (θ) dθ
Z
=
a2 − a2 sin2 (θ)
p

a cos (θ) dθ
Z
= q
a2 1 − sin2 (θ)

a cos (θ) dθ
Z
= √ q
1 − sin2 (θ)

a2 ·
a cos (θ) dθ
Z
= q
a 1 − sin2 (θ)

a cos (θ) dθ
Z
=
a cos2 (θ)
p

a cos (θ) dθ
Z
=
a cos (θ)
Z
= dθ

=θ+C (E1)

Now, since x = a sin(θ), therefore,

x = a sin(θ)
x
= sin(θ)
a x
∴ θ = sin−1
a
x
Now, substituting θ = sin−1 into equation (E1), we have the following,
a
−1 x

θ + C = sin +C
a
This suggests the following,
Z
dx x
√ = sin−1 +C
a2 − x2 a

Remark 7.4.2. Z
dx −1 x

√ = sin +C
a2 − x2 a

√
Chap. 7 / Sec. 7.4 / Subsec. 7.4.2 : Integrals Involving a2 + x2
√
Suppose, we’re given an integral whose integrand involves the term a2 + x2 ,
where a is a positive constant. If we substitute x = a tan(θ), then
q
a + x = a2 + (a tan(θ))2
p
2 2
q
= a2 + a2 tan2 (θ)
q
= a2 (1 + tan2 (θ))

It is known from the Pythagorean identity that sec2 (x) = 1+tan2 (x). Therefore,

= a2 1 + tan2 (θ)
s
| {z }
sec2 (θ)

a2 · sec2 (θ)
p
=
= |a · sec (θ)|
= a sec (θ) (2)
√
This implies that the substitution x = a tan(θ) transform the difficult a2 + x 2
into an easier product a sec(θ). That is,

a2 + x2 = a sec (θ)
p

if we let x = a tan(θ).

√
The substitution x = a tan(θ) for a2 + x2 works because, if we consider a
right-angled triangle whose opposite side is x and adjacent side is a, then the
hypotenuse is given by the Pythagorean theorem, as follows.

opposite side2 + adjacent side2 = hypotenuse2

x2 + a2 = hypotenuse2
∴ hypotenuse = x2 + a2
p

This is shown below.

Figure 7.2

This implies that

x
tan(θ) =
a

∴ x = a tan(θ)

A few examples are provided below.

Z
dx
I Example 7.4.5. Evaluate √
x2 + 9
√
Solution. Notice that the integral above involves the term x2 + a2 , where a = 3.
Therefore, to solve the integral, we let x = 3 tan(θ), then dx = 3 sec2 (θ)dθ.
Therefore,

3sec2 (θ)dθ
Z Z
dx
√ = q
x2 + 9 (3 tan(θ))2 + 9
3sec2 (θ)dθ
Z
= p
9tan2 (θ) + 9
3sec2 (θ)dθ
Z
= p
9 (tan2 (θ) + 1)
3sec2 (θ)dθ
Z
= √ p
9 · sec2 (θ)
3sec2 (θ)dθ
Z

=
3 sec(θ)
Z
= sec(θ)dθ
Z
Since sec(θ)dθ = ln |sec(θ) + tan(θ)| + C and if x = 3 tan(θ), then θ =
x
tan−1 , therefore,
3
Z
dx
√ = ln |sec(θ) + tan(θ)| + C
x2 + 9
−1 x −1 x

= ln sec tan + tan tan +C
3 3
√
Since tan(tan−1 (x)) = x and sec(tan−1 (x)) = 1 + x2 , therefore,
Z
dx
−1 x

−1 x

√ = ln sec tan + tan tan +C
x2 + 9 3 3
r x 2 x
= ln 1+ + +C
3 3
r
x2 x
= ln 1+ + +C
9 3

r
9 + x2 x
= ln + +C
9 3
√
x2 + 9 x
= ln √ + +C
9 3
√
x x2 + 9
= ln + +C
3 3
√
x2 + 9
Z
dx x
Therefore, √ = ln + +C
x2 + 9 3 3

We now generalize the integrals above into formulas.

Z
dx
I Example 7.4.6. Evaluate
a2 + x2
Solution. To evaluate the integral, we let x = a tan(θ). Then, the differential of
x becomes dx = a sec2 (θ)dθ. Therefore,
Z Z
dx dx
=
2
a +x 2
a + (a tan (x))2
2

asec2 (θ) dθ
Z
=
a2 + a2 tan2 (θ)
asec2 (θ) dθ
Z
=
a2 sec2 (θ)
Z
1
= dθ
a
θ
= +C (E1)
a
Now, since x = a tan(θ), therefore,

x = a tan(θ)
x
= tan(θ)
a
x
∴ θ = tan−1 ( )
a
x
Substituting θ = tan−1 into equation (E1) yields the following,
a
x
θ tan−1
+C = a +C
a a

This suggests the following,

Z
dx 1 −1 x

= tan +C
a2 + x2 a a

Remark 7.4.3. Z
dx 1 −1 x

= tan +C
a2 + x2 a a

Z p
I Example 7.4.7. Evaluate a2 + x2 dx

Solution. To evaluate the integral, we let x = a tan(θ). Then, the differential dx

yields dx = a sec2 (θ)dθ. Therefore,
Z p Z q
2 2
a + x dx = a2 + (a tan (θ))2 asec2 (θ) dθ
Z q
= a2 + a2 tan2 (θ)asec2 (θ) dθ
Z q
= a2 (1 + tan2 (θ))asec2 (θ) dθ
Z p
= a2 sec2 (θ)asec2 (θ) dθ
Z
= (a sec (θ)) asec2 (θ) dθ

Z
= a2 sec3 (θ) dθ
Z
=a 2
sec3 (θ) dθ (E1)
Z
Now, to evaluate sec3 (θ)dθ in the integral above, we apply the reduction
formula, stating that,

secn−2 (x) tan (x) n − 2

Z Z
sec (x) dx =
n
+ secn−2 (x) dx
n−1 n−1
for n = 3. That is,
sec (x) tan (x) 1
Z Z
sec (x) dx =
3
+ sec (x) dx
2 2
sec (x) tan (x) ln |tan (x) + sec (x)|
= + +C
2 2

sec (θ) tan (θ) ln |tan (θ) + sec (θ)|

Z
Now, substituting + + C for sec3 (θ) dx in
2 2
the equation (E1), we get,

2 sec (θ) tan (θ) ln |tan (θ) + sec (θ)|

Z
a 2
sec (θ) dθ = a
3
+ +C
2 2
a2
= (sec (θ) tan (θ) + ln |tan (θ) + sec (θ)|) + C
2
This implies,
a2
Z p
a2 + x2 dx = (sec (θ) tan (θ) + ln |tan (θ) + sec (θ)|) + C (E2)
2
Now, from the Pythagorean theorem, it is known that,

sec2 (x) − tan2 (x) = 1

sec2 (x) = 1 + tan2 (x)
q
∴ sec(x) = 1 + tan2 (x) (E3)

Substituting equation (E3) into (E2) yields the following.

a2
= (sec (θ) tan (θ) + ln |tan (θ) + sec (θ)|) + C
2
a2
q q
= 1 + tan (θ) tan (θ) + ln tan (θ) + 1 + tan (θ) + C
2 2 (E4)
2
Now, to express the integral in terms of x, notice that x = a tan(θ), therefore,

x = a tan(θ)
x
= tan(θ)
a
−1 x

∴ θ = tan
a
x
Substituting θ = tan −1
into equation (E4), we get,
a
a2
q q
= 1 + tan (θ) tan (θ) + ln tan (θ) + 1 + tan (θ) + C
2 2
2
a2
r x x
= 1 + tan2 tan−1 tan tan−1
2 a a
x r x
+ ln tan tan−1 + 1 + tan2 tan−1 +C
a a
r r !
a2 x 2 x x x 2
= 1+ · + ln + 1 + +C
2 a a a a

r r !
a2 x2 x x x2
= 1 + 2 · + ln + 1 + 2 +C
2 a a a a
r r !
2 2 2 2
1 a + x x x a + x
= a2 · + a2 ln + +C
2 a2 a a a2
r r !
1 1 x x 1
· a2 + x2 · + a2 ln +
p p
= a2 2
· a2 + x2 + C
2 a a a a2
√ !
1 x a 2 + x2
x a2 + x2 + a2 ln +
p
= +C
2 a a
√ !
1 x + a 2 + x2
x a2 + x2 + a2 ln
p
= +C
2 a

This suggests the following,

√ !
a2 + x2
Z p
1 x+
x a2 + x2 + a2 ln
p
a2 + x2 dx = +C
2 a

Remark 7.4.4.
√ !
a2 x2
Z p
1 x+ +
x a2 + x2 + a2 ln
p
a2 + x2 dx = +C
2 a

√
Chap. 7 / Sec. 7.4 / Subsec. 7.4.3 : Integrals Involving x2 − a2
√
Suppose, we’re given an integral whose integrand involves the term x2 − a2 ,
where a is a positive constant. If we substitute x = a sec(θ), then
q
x − a = (a sec (θ))2 − a2
p
2 2

= a2 sec2 (θ) − a2
p
p
= a2 (sec2 (θ) − 1)

It is known from the Pythagorean identity that sec2 (x) = 1+tan2 (x) or, sec2 (x)−
1 = tan2 (x). Therefore,

= a2 sec2 (θ) − 1
s
| {z }
tan2 (θ)

q
= a2 · tan2 (θ)
= |a · tan (θ)|
= a tan (θ)
√
This implies that the substitution x = a sec(θ) transforms the difficult x 2 − a2
into an easier product a tan(θ). That is,

x2 − a2 = a tan (θ)
p

if we let x = a tan(θ).

The reason why the substitution x = a sec(θ) works is that if we consider a

right-angled triangle whose hypotenuse is x and the adjacent side is a, then the
opposite side is given by the Pythagorean theorem, stating that,

opposite side2 + adjacent side2 = hypotenuse2

opposite side2 + a2 = x2
opposite side = x2 − a2
p

This is shown below.

Figure 7.3

This implies that

x
sec(θ) =
a
∴ x = a sec(θ)

A few examples are provided below.

Z
xdx
I Example 7.4.8. Evaluate √
x2 − 16

√
Solution. Notice that the integral above involves the term x2 − a2 , where a = 4.
Therefore, to solve the integral, we let x = 4 sec(θ), then dx = 4 sec(θ) tan(θ)dθ.
Thus,

4 sec(θ) · 4 sec(θ) tan(θ)dθ

Z Z
xdx
√ = q
2
x − 16 (4 sec(θ))2 − 16
4 sec(θ) · 4 sec(θ) tan(θ)dθ
Z
= p
16sec2 (θ) − 16
4 sec(θ) · 4 sec(θ) tan(θ)dθ
Z
= p
16 (sec2 (θ) − 1)
4 sec(θ) · 4 sec(θ) tan(θ)dθ
Z
= √
16 · sec2 (θ) − 1
p

4 sec(θ) · 4 sec(θ) tan(θ)dθ

Z
=
4 · tan2 (θ)
p

4 sec(θ) · 4 sec(θ) tan(θ)dθ

Z
=
4 tan(θ)
4 sec(θ) · 4 sec(θ)
tan(θ)dθ
Z

=
tan(θ)
4
Z
= 4sec2 (θ)dθ
Z
= 4 sec2 (θ)dθ

= 4tan2 (θ) + C
x
Since x = 4 sec(θ), therefore θ = sec −1
. Thus,
4
Z
xdx
√ = 4tan2 (θ) + C
2
x − 16 x
= 4tan2 sec−1 +C
4
√
Since tan(sec−1 ) = x1 − 1, therefore,
Z
xdx
−1 x

√ = 4tan sec
2
+C
x2 − 16 4
r
x 2
=4· −1+C
4
r
x2 − 16
=4· +C
p 16
= x2 − 16 + C

Z
xdx
Therefore,
p
√ = x2 − 16 + C
x2 − 16

We now generalize the integrals above into formulas.

Z p
I Example 7.4.9. Evaluate x2 − a2 dx

Solution. To evaluate the integral, we let x = a sec(θ). Then, the differential dx

yields dx = a sec(θ) tan(θ)dθ. Therefore,
Z p Z q
2 2
x − a dx = (a sec (θ))2 − a2 a sec (θ) tan (θ) dθ
Z q
= (a sec (θ))2 − a2 a sec (θ) tan (θ) dθ
Z p
= a2 sec2 (θ) − a2 a sec (θ) tan (θ) dθ
Z p
= a2 (sec2 (θ) − 1)a sec (θ) tan (θ) dθ
Z √ q
= a2 tan2 (θ)a sec (θ) tan (θ) dθ
Z
=a 2
sec (θ) tan2 (θ) dθ
Z
2
sec (θ) sec2 (θ) − 1 dθ

=a
Z
2
sec3 (θ) − sec (θ) dθ

=a
Z Z
=a 2
sec (θ) dθ − a
3 2
sec (θ) dθ (E1)
Z
Now, to evaluate sec3 (θ)dθ in the integral above, we apply the reduction
formula:
secn−2 (x) tan (x) n − 2
Z Z
sec (x) dx =
n
+ secn−2 (x) dx
n−1 n−1
for n = 3. That is,

sec (x) tan (x) 1

Z Z
sec (x) dx =
3
+ sec (x) dx
2 2
sec (x) tan (x) ln |tan (x) + sec (x)|
= + +C
2 2

sec (θ) tan (θ) ln |tan (θ) + sec (θ)|

a2
Z p
x2 − a2 dx = (sec (θ) tan (θ) − ln | sec(θ) + tan(θ)|) + C (E2)
2
Now, from the Pythagorean theorem, it is known that,

sec2 (x) − tan2 (x) = 1

tan2 (x) = sec2 (x) − 1
tan(x) = sec2 (x) − 1 (E3)
p

Substituting equation (E3) into (E2) yields the following.

a2
= (sec (θ) tan (θ) − ln | sec(θ) + tan(θ)|) + C
2
a2
sec (θ) sec (θ) − 1 − ln | sec(θ) + sec (θ) − 1| + C (E4)
p p
= 2 2
2
Now, to express the integral in terms of x, notice that x = a sec(θ), therefore,

x = a sec(θ)
x
= sec(θ)
a x
∴ θ = sec−1
a

x
Substituting θ = sec−1 into equation (E4), we obtain the following,
a
a2
sec (θ) sec2 (θ) − 1 − ln sec(θ) + sec2 (θ) − 1 + C
p p
=
2
a2 x r x
= sec sec−1 sec2 sec−1 −1
2 a a
x r x
− ln sec sec −1
+ sec sec 2 −1 −1 +C
a a
r r !
a2 x x 2 x x 2
= − 1 − ln + −1 +C
2 a a a a
r r !
2 2 2
a x x x x
= 2
− 1 − ln + −1 +C
2 a a a a2
r r !
a2 x x 2 − a2 x x 2 − a2
= − ln + +C
2 a a2 a a2
r r !
2 2 2 2
1 x x − a x x − a
= a2 · − a2 · ln + +C
2 a a2 a a2
√ √ !
1 x 2
x −a 2 x 2
x −a 2
= a2 · · − a2 · ln + +C
2 a a a a
√ !
1 x + x 2 − a2
x x2 − a2 − a2 ln
p
= +C
2 a

This suggests the following.

√ !
x2 − a2
Z p
1 x+
x x2 − a2 − a2 ln
p
x2 − a2 dx = +C
2 a

Remark 7.4.5.
√ !
x2 − a2
Z p
1 x+
x x2 − a2 − a2 ln
p
x2 − a2 dx = +C
2 a

The End of Section 7.4

Trigonometric Substitution

§ 7.5 Partial Fractions

Partial fraction decomposition is a powerful technique used in calculus to inte-
grate rational functions. For example, consider the following function,

P (x)
f (x) =
Q(x)
where, P (x) and Q(x) 6= 0 are both polynomials. In that case, f (x) is said to
be a rational function and is expressed as the ratio of polynomial functions. A
polynomial is a function of the form,

P (x) = an xn + an−1 xn−1 + an−2 xn−2 + · · · + a2 x2 + a1 x + a0

where, n represents a non-negative integer and a0 , a1 , a2 , · · · , an−1 , an are con-

stants, where an 6= 0.

Integrating functions of this nature isn’t always straightforward. It is beneficial

to decompose them into simpler functions, which can be easily integrated. For
example, suppose we have to integrate the following integral,
Z
6x
2
dx
x +x−2
Other methods to integrate the integral above might exist, but no method is as
6x 2 4
simple as if we realize that 2 can be written as + for the
x +x−2 x−1 x+2
reason that,
2 4 2 (x + 2) + 4 (x − 1)
+ =
x−1 x+2 (x − 1) (x + 2)
2x + 4 + 4x − 4
=
(x − 1) (x + 2)
2x + 4x
=
(x − 1) (x + 2)
6x
=
(x − 1) (x + 2)
6x
= 2
x + 2x − x − 2
6x
= 2
x +x−2
This suggests that,
Z Z
2x 2 4
2
dx = + dx
x +x−2 x−1 x+2

Now, notice that the integral on the right side of the equation is straightforward
to solve, whereas the integral on the left side is not. That is,
Z Z Z
2 4 2 4
+ dx = dx + dx
x−1 x+2 x−1 x+2
Z Z
1 1
=2 dx + 4 dx
x−1 x+2
= 2 ln |x − 1| + 4 |x + 2| + C
Z
2x
∴ dx = 2 ln |x − 1| + 4 |x + 2| + C
x2 +x−2
While the method is efficient, its practicality relies on the ability to factor
6x 2 4
2
into fractions and , referred to as partial fractions. For
x +x−2 x−1 x+2
2 4
example, if and are given, then it is of no effort that we may deter-
x−1 x+2
6x
mine 2 , however, the converse is not true. Therefore, in this section,
x +x−2
various techniques are discussed regarding that issue.

However, that comes with a few considerations. First of all, we can only decom-
P (x)
pose a rational function if the degree of P (x) is less than that of Q(x),
Q(x)
symbolically, deg P (x) < deg Q(x) . Such a rational function is called a
proper rational function. In contrast, if we ever face an improper rational func-
tion, where
the degree
of P (x) is greater than Q(x) or equal to that, that is,
deg P (x) ≥ deg Q(x) , then we’ll first need to perform long polynomial
P (x) R(x)
division to express as S(x) + , where R(x) is the remainder and
Q(x) Q(x)
deg R(x) < deg Q(x) .

So, in general,

suppose
(x) and Q(x) are both polynomial functions, where
P
deg P (x) < deg Q(x) , also suppose that Q(x) can be factored as a product
of linear factors (of the form ax + b) or/and irreducible quadratic factors (of the
P (x)
form ax2 + bx + c, where b2 − 4ac < 0); then, can be expressed as a sum
Q(x)
of partial fractions.

Chap. 7 / Sec. 7.5 / Subsec. 7.5.1 : The denominator Q(x) is a product of lin-
ear factors

P (x)
Remark 7.5.1. If Q(x) in the rational function can be factored into
Q(x)
a product of linear factors, where each linear factor is distinct, that is,
factors of the form,

(a1 x + b1 ) (a2 x + b2 ) · · · (an x + bn )

then, there exist constant A1 , A2 , · · · , An such that the rational function

P (x)
has a partial fraction decomposition of the form,
Q(x)

P (x) P (x)
=
Q(x) (a1 x + b1 ) (a2 x + b2 ) · · · (an x + bn )
A1 A2 An
= + + ··· +
a1 x + b1 a2 x + b2 an x + bn

x2 + 2x + 1
Z
I Example 7.5.1. Evaluate dx
x3 + 2x2 + x2 + 2x
Solution. In order to integrate the integral above, notice first that x3 + 2x2 + x2 + 2x
can be factored as x(x + 1)(x + 2). As a result,
x2 + 2x + 1 x2 + 2x + 1
=
x3 + 2x2 + x2 + 2x x(x + 1)(x + 2)
A B C
= + +
x (x + 1) (x + 2)
Now, the goal is to determine the values of A, B, and C. For that multiply
x(x + 1)(x + 2) in the both sides of the equation below,
x2 + 2x + 1 A B C
= + +
x(x + 1)(x + 2) x (x + 1) (x + 2)
A B C
x2 + 2x + 1 = ·
x(x + 1)(x + 2) + · x
(x+1)(x

+ 2) + · x(x + 1)
(x+2)

x
(x+ 1)

(x+ 2)

x2 + 2x + 1 = A(x + 1)(x + 2) + Bx(x + 2) + Cx(x + 1)

= A(x2 + 2x + x + 2) + B(x2 + 2x) + C(x2 + x)
= Ax2 + 2Ax + Ax + 2A + Bx2 + 2Bx + Cx2 + Cx
= (Ax2 + Bx2 + Cx2 ) + (2Ax + Ax + 2Bx + Cx) + 2A
x2 + 2x + 1 = (A + B + C)x2 + (3A + 2B + C)x + 2A

Since (A + B + C)x2 + (3A + 2B + C)x + 2A is identical to x2 + 2x + 1, therefore,

their coefficients must also be equal. Thus, we write the following,

A+B+C =1 (1)
3A + 2B + C = 2 (2)
2A = 1 (3)

1
Since 2A = 1, therefore, A = . Substituting the value of A into equation (1)
2
yields the following,
1
+B+C =1
2
1
B =1−C − (4)
2
Substituting the value of A and B into equation (2) yields the following,

1 1
3 +2 1−C − +C =2
2 2

1 1
C =2−2 1−C − −3
2 2
3
= 2 − 2 + 2C + 1 −
2
3
= 2C + 1 −
2
1
= 2C −
2
1
C − 2C = −
2
1
−C = −
2
1
∴C=
2
Substituting the value of A and C into equation (4) yields the following,
1
B =1−C −
2
=0

1 1
Therefore, we have found the value of A, B, and C to be , 0, and , respectively.
2 2
Thus, we write,

x2 + 2x + 1 A B C
= + +
x(x + 1)(x + 2) x (x + 1) (x + 2)

1 1
0
= 2 + + 2
x (x + 1) (x + 2)
1 1
= +
2x 2(x + 2)

As a result,

x2 + 2x + 1
Z Z
1 1
dx = + dx
x3 + 2x2 + x2 + 2x 2x 2(x + 2)
Z Z
1 1
= dx + dx
2x 2(x + 2)
Z Z
1 1 1 1
= · dx + · dx
2 x 2 x+2
1 1
= ln |x| + ln |x + 2| + C
2 2
x2 + 2x + 1
Z
1 1
Therefore, 3 2 2
dx = ln |x| + ln |x + 2| + C
x + 2x + x + 2x 2 2

Chap. 7 / Sec. 7.5 / Subsec. 7.5.2 : The denominator Q(x) contains repeated
product of linear factors

P (x)
Remark 7.5.2. If Q(x) in the rational function has at least one repeated
Q(x)
liner factor, that is, factor of the form,

(ax + b)n where n ≥ 2

then, there exist constants A1 , A2 , · · · , An such that the rational function

P (x)
has a partial fraction decomposition of the form,
Q(x)

P (x) P (x)
=
Q(x) (ax + b)n
A1 A2 An
= + + ··· +
ax + b (ax + b)2 (ax + b)n

x2 + x + 1
Z
I Example 7.5.2. Evaluate dx
(x + 1)2 (x − 1)

Solution. In order to integrate the integral above, we express the given rational
function as a sum of partial fractions as follows.
x2 + x + 1 A B C
2 = + +
(x + 1) (x − 1) (x − 1) (x + 1) (x + 1)2

Now, multiply both sides by the denominator (x + 1)2 (x − 1) to eliminate the

denominators.

x2 + x + 1 = A(x + 1)2 + B(x + 1)(x − 1) + C(x − 1)

= A(x2 + 2x + 1) + B(x2 − 1) + C(x − 1)
= Ax2 + 2Ax + A + Bx2 − B + Cx − C
= Ax2 + Bx2 + 2Ax + Cx + A − B − C
∴ x2 + x + 1 = (A + B)x2 + (2A + C)x + (A − B − C)

Now, we compare the coefficients of x2 , x, and the constant terms on both sides
of the equations, as follows.

A+B =1 (1)
2A + C = 1 (2)
A−B−C =1 (3)

Now, to determine the values of A, B, and C, notice that since A + B = 1,

therefore, B = 1 − A. Substituting the value of B into equation (3) yields the
following.

A−B−C =1
A − (1 − A) − C = 1
A−1+A−C =1
2A − C = 2 (4)

Adding equations (2) and (4) together, we get,

4A = 3
3
A=
4
3
Substituting A = into equation (2), we get,
4

3
2 +C =1
4

3
+C =1
2
3
C =1−
2
1
=−
2
1
This implies that C = − . Substituting the value of A into equation (1) yields
2
the following,

A+B =1
3
B =1−
4
1
=
4
1
This implies that B = . Thus, we have the partial fraction decomposition as
4
follows,
3 1 1
x2 + x + 1 4 4 2
2 = + −
(x + 1) (x − 1) (x − 1) (x + 1) (x + 1)2
This implies that
3 1 1
2
x +x+1 4 4 2
= + −
(x + 1)2 (x − 1) (x − 1) (x + 1) (x + 1)2

3 1 1
 
x2 + x + 1
Z Z
 4 4 2
dx = + −  dx

(x + 1)2 (x − 1) (x − 1) (x + 1) (x + 1) 2


3 Z 1Z Z 1
= 4 dx + 4 dx − 2 dx
(x − 1) (x + 1) (x + 1)2
Z Z Z
3 1 1 1 1 1
= · dx + · dx − · dx
4 (x − 1) 4 (x + 1) 2 (x + 1)2
Z
3 1 1 1
= ln |x − 1| + C1 + ln |x + 1| + C2 − · dx
4 4 2 (x + 1)2
(5)
Z
1
Now, to solve the integral dx, we use the u-substitution rule. There-
(x + 1)2

fore, let u = x + 1, then du = dx, thus, we write the following,

Z Z
1 1
dx = du
(x + 1)2 u2
Z
= u−2 du
u−2+1
+ C3 =
−2 + 1
u−1
= + C3
−1
1
= − + C3
u
1
=− + C3 (6)
x+1
Now, substituting the result in equation 6 into equation 5, we get,
x2 + x + 1
Z Z
3 1 1 1
2 dx = ln |x − 1| + C1 + ln |x + 1| + C2 − · dx
(x + 1) (x − 1) 4 4 2 (x + 1)2

3 1 1 1
= ln |x − 1| + C1 + ln |x + 1| + C2 − · − + C3
4 4 2 x+1
3 1 1
= ln |x − 1| + C1 + ln |x + 1| + C2 + + C3
4 4 2 (x + 1)
3 1 1
= ln |x − 1| + ln |x + 1| + +C
4 4 2 (x + 1)
x2 + x + 1
Z
3
where, C = C1 + C2 + C3 . Therefore, 2 dx = ln |x − 1| +
(x + 1) (x − 1) 4
1 1
ln |x + 1| + + C.
4 2 (x + 1)

Chap. 7 / Sec. 7.5 / Subsec. 7.5.3 : The denominator Q(x) contains irreducible
quadratic factors

P (x)
Remark 7.5.3. If Q(x) in the rational function has at least one irre-
Q(x)
ducible quadratic factor, that s, factor of the form,
ax2 + bx + c where, b2 − 4ac < 0
P (x)
then, there exist constants A and B such that the rational function
Q(x)

has a partial fraction decomposition of the form,

P (x) P (x)
= 2
Q(x) ax + bx + c
Ax + B
= 2
ax + bx + c

x2 + x + 5
Z
I Example 7.5.3. Evaluate dx
x3 + 5x
Solution. Notice that x3 + 5x can be factored as x(x2 + 5). Since x2 + 5 cannot
be factored any further as b2 − 4ac < 0, therefore we write the integrand as,
x2 + x + 5 x2 + x + 5
=
x3 + 5x x(x2 + 5)
A Bx + C
= + 2
x x +5
Now, we multiply both sides by the denominator x3 + 5x to eliminate the de-
nominators. That is,

x2 + x + 5 = A(x2 + 5) + (Bx + C)x

= Ax2 + 5A + Bx2 + Cx
= Ax2 + Bx2 + Cx + 5A
= (A + B)x2 + Cx + 5A

Now, we compare the coefficients of x2 , x, and the constant term both sides of
the equation,

A+B =1 (1)
C=1 (2)
5A = 5 (3)

It is given that C = 1 . Since 5A = 5, therefore A = 1 . Substituting the value

of A into equation (1), we get the value of B,

A+B =1
B=0

Therefore, we have found the value of A, B, and C to be 1, 0, and 1, respectively.

Thus, we write,
x2 + x + 5 A Bx + C
3
= + 2
x + 5x x x +5

1 0·x+1
= + 2
x x +5
1 1
= + 2
x x +5
As a result,

x2 + x + 5
Z Z
1 1
dx = + dx
x3 + 5x x x2 + 5
Z Z
1 1
= dx + 2
dx
x x +5
√
Z
1 x
Now, to solve the integral dx, we let u = √ , then x = 5u and
x2 + 5 5
1
du = √ dx. Thus, we get,
5
Z Z √
1 5
dx = 2√ du
x2 + 5 5u + 5
Z √
5
= du
5 (u2 + 1)
√ Z
5 1
= · du
5 u2 + 1
Z
1
Now, to solve the integral du, we let u = tan(z), then du = sec2 (z)dz.
u2 +1
Thus,

sec2 (z)
Z Z
1
du = dz
u2 + 1 tan2 (z) + 1
sec2 (z)
Z
= dz
sec2 (z)
Z
= dz

=z+C

Since u = tan(z), therefore, z = tan−1 (u). Thus,

Z √ Z
1 5 1
dx = · du
x2 + 5 5 u2 + 1
√
5
= ·z+C
5
1
= √ · tan−1 (u) + C
5

1 x
= √ tan−1 √ +C
5 5
As a result,
x2 + x + 5
Z Z Z
1 1
dx + dx = dx
x x3 + 5x 2
x +5

1 x
= ln |x| + √ tan −1
√ +C
5 5
Z 2
x +x+5 1 x
Therefore, dx = ln |x| + √ tan −1
√ +C
x3 + 5x 5 5

Chap. 7 / Sec. 7.5 / Subsec. 7.5.4 : The denominator Q(x) contains repeated
irreducible quadratic factors

P (x)
Remark 7.5.4. If Q(x) in the rational function has at least one irre-
Q(x)
ducible quadratic factor, that is, factor of the form,

(ax2 + bx + c)n where b2 − 4ac < 0 and n ≥ 2

them, there exist constants A1 , A2 , · · · , An and B1 , B2 , · · · , Bn such that

P (x)
the rational function has a partial fraction decomposition of the form,
Q(x)

P (x) P (x)
=
Q(x) (ax + bx + c)n
2

A1 x + B1 A2 x + B2 An x + Bn
= 2 + + ··· +
ax + bx + c (ax2 + bx + c)2 (ax2 + bx + c)n

x3 + x2 + x + 1
Z
I Example 7.5.4. Evaluate dx
x(x2 + 1)2
Solution. Since x2 + 1 cannot be factored any further as b2 − 4ac < 0, therefore
we write the integral as,
x3 + x2 + x + 1 A Bx + C Dx + E
2 = + 2 +
x(x2 + 1) x x +1 (x2 + 1)2
Now, we multiply both sides by the denominator x(x2 + 1) to eliminate the
2

denominators on the right side of the equation. That is,

x3 + x2 + x + 1 = A(x2 + 1)2 + (Bx + C)x(x2 + 1) + (Dx + E)x

= A(x4 + 2x2 + 1) + (Bx + C)(x3 + x) + Dx2 + Ex

= (Ax4 + 2Ax2 + A) + (Bx4 + Bx2 + Cx3 + Cx) + Dx2 + Ex
= (Ax4 + Bx4 ) + Cx3 + (2Ax2 + Bx2 + Dx2 ) + (Cx + Ex) + A
= (A + B)x4 + Cx3 + (2A + B + D)x2 + (C + E)x + A

Now, we compare the coefficients of x4 , x3 , x2 , x, and the constant term both

sides of the equation,

A+B =0
C=1
2A + B + D = 1
C +E =1
A=1

Solving these equations, we get A = 1, B = −1, C = 1, D = 0, and E = 0.

Thus, we write,

x3 + x2 + x + 1 A Bx + C Dx + E
= + +
x(x2 + 1)2 x x2 + 1 (x2 + 1)2
1 −x + 1 0·x+0
= + 2 +
x x +1 (x2 + 1)2
1 1−x
= + 2
x x +1
1 1 x
= + 2 − 2
x x +1 x +1
As a result,

x3 + x2 + x + 1
Z Z
1 1 x
dx =
+ − dx
x(x2 + 1)2 x x2 + 1 x2 + 1
Z Z Z
1 1 x
= dx + 2
dx − 2
dx
x x +1 x +1
Z Z
1 1
Here, dx = ln |x| + C1 and 2
dx = tan−1 (x) + C2 . However, to
xZ x +1
x
integrate 2
dx, we use the substitution rule and let u = x2 + 1, then
x +1
du
du = 2xdx or = xdx. Thus, we write,
2
Z Z
x 1 du
dx =
x2 + 1 u 2

Z
1 1
= · du
2 u
1
= · ln |u| + C3
2
1
= ln x2 + 1 + C3
2
Therefore, we write,
Z 3
x + x2 + x + 1
Z Z Z
1 1 x
2 dx = dx + 2
dx − 2
dx
2
x(x + 1) x x +1 x +1
1
= ln |x| + tan−1 (x) − ln x2 + 1 + C
2
Z 3
x + x2 + x + 1 1
Therefore, 2 dx = ln |x| − ln x2 + 1 + tan−1 (x) + C
x(x2 + 1) 2

The End of Section 7.5

Partial Fractions

§ 7.6 Improper Integrals

So far, we have discussed the definite integrals where the function being inte-
grated was continuous and the intervals of integration were finite. However,
there are situations where the function becomes infinite within the interval of
integration or the interval itself extends to infinity. These are called improper
integrals.

Chap. 7 / Sec. 7.6 / Subsec. 7.6.1 : Infinite Intervals

1
Consider the area A under the curve y = 2 and above the x-axis, over the
x
interval [1, ∞), as shown below.

x
1

Figure 7.4

Since the interval of integration is infinite, therefore it may be intuitive to say

that the area A would also be infinite. However, to our surprise, this is not the
case. This can be tested. For example, if we let x = b be a line such that b > 1,
then the area between x = 1 and x = b yields,
Zb
1
A= dx
x2
1
Zb
= x−2 dx
1
b
x−2+1
=
−2 + 1 1
1 b

=−
x 1
1
∴A=1−
b
Now, notice that the area A over the finite interval [1, b] will always be less than
1
1 for all b > 1 because of the term . If we let b → ∞ to find the area A over
b
the interval [1, ∞), then

1
lim 1 − =1−0=1
b→∞ b
1
That is, the area A → 1 as b → ∞. Thus, the area A under the curve y = 2 ,
x
above the x-axis and right to the vertical line x = 1 is 1, and we write,

Z∞ Zb
1 1
dx = lim dx
x2 b→∞ x2
1 1

Based on that observation, we now define the following.

Definition 7.6.1. (Infinite Intervals)

1. If f (x) is continuous over [a, ∞), then

Z∞ Zb
f (x) dx = lim f (x) dx
b→∞
a a

given that the limit exists.

2. If f (x) is continuous over (−∞, b], then

Zb Zb
f (x) dx = lim f (x) dx
a→−∞
−∞ a

given that the limit exists.

3. If f (x) is continuous over (−∞, ∞), then

Z∞ Zc Zb
f (x) dx = lim f (x) dx + lim f (x) dx
a→−∞ b→∞
−∞ a c

given that the both limit exists.

In each case, if the limit exists, then the improper integral is called
convergent, if not, it is called divergent.

Z∞
1
I Example 7.6.1. Is the integral dx convergent or divergent?
x
1

Solution. From the definition above, we write,

Z∞ Zb
1 1
dx = lim dx
x b→∞ x
1 1
ib
= lim ln(x)
b→∞ 1

= lim (ln(b) − ln(1))

b→∞

= lim ln(b)
b→∞

=∞

Since the limit does not yield a finite value, therefore the limit does not exist.
Z∞
1
As a result, the integral dx is divergent.
x
1

Z∞
1
I Example 7.6.2. Evaluate dx
1 + x2
0

Solution. According to the definition of improper integrals, we write,

Z∞ Zb
1 1
dx = lim dx
1 + x2 b→∞ 1 + x2
0 0
ib
= lim tan (x)
−1
b→∞ 0
= lim tan (b) − tan−1 (0)
−1

b→∞

= lim tan−1 (b)

b→∞
π
=
2
Z∞
1 π
Therefore, dx =
1 + x2 2
0

Chap. 7 / Sec. 7.6 / Subsec. 7.6.2 : Discontinuous Integrands

In the previous subsection, we observed that the area of the region extended
infinitely horizontally because the interval of integration was infinite. In this

subsection, we will examine how the area of the region extends infinitely in the
1
vertical direction. To illustrate, consider the area A under the curve y = √
x
and above the x-axis, from x = 0 to x = 1, as shown below.

f(x)

x
1

Figure 7.5

1
Notice that y = √ increases without a bound as x → 0+ , and is not defined at
x
Z1
1
the exact point x = 0. However, we can still determine the area A = √ dx,
x
0
Z1
1
if we first determine the integral √ dx and take the limit as a → 0+ . That
x
a
is,
Z1
1 √ i1
√ dx = 2 x
x a
a
√ √
=2 1−2 a
√
=2−2 a
√
=2 1− a
√
Now, take the limit of 2 (1 − a) as a → 0+ .
√
lim+ 2 1 − a = 2 (1 − 0)
a→0

Observe how the area of the bounded region is finite, even though the length of
the boundary curve is infinite!

Based on the observation above, we define the following.

Definition 7.6.2. (Discontinuous Integrands)

1. If f (x) is continuous over (a, b], where lim+ f (x) = ±∞. Then
x→a

Zb Zb
f (x) dx = lim+ f (x) dx
c→a
a c

given that the limit exists.

2. If f (x) is continuous over [a, b), where lim− f (x) = ±∞. Then
x→b

Zb Zc
f (x) dx = lim− f (x) dx
c→b
a a

given that the limit exists.

3. If f (x) is continuous over [a, b], except at a point c such that a < c <
b. Then
Zb Zc Zb
f (x) dx = f (x) dx + f (x) dx
a a c

Zc Zb
given that improper integrals f (x) dx and f (x) dx exist.
a c

In every case, if the limits exist, the improper integral is called con-
vergent, if not, it is called divergent.

Z1
I Example 7.6.3. Evaluate ln(x)dx
0

Solution. Since ln(x) has a vertical asymptote at the point x = 0, therefore we

write the following,

Z1 Z1
ln(x)dx = lim+ ln(x)dx
a→0
0 a

Since we know that,

Z
ln(x)dx = x ln(x) − x + C

Therefore, we write the following,

Z1 h i1
lim ln(x)dx = lim+ x ln(x) − x
a→0+ a→0 a
a

= lim+ ((1 · ln(1) − 1) − (a ln(a) − a))

a→0

= lim+ (ln(1) − 1 − a ln(a) + a)

a→0

= lim+ (a − a ln(a) − 1)
a→0

= lim+ a − lim+ a ln(a) − lim+ 1

a→0 a→0 a→0

= 0 − lim+ a ln(a) − 1
a→0

Now, to find the limit lim+ a ln(a), we apply the L’Hôpital’s Rule as follows,
a→0

ln(a)
lim+ a ln(a) = lim+
a→0 a→0 1
a
d
ln(a)
= lim+ da
a→0 d 1
da a
1
= lim+ a
a→0 1
− 2
a
= lim+ (−a)
a→0

As a result, we get,

Z1 Z1
ln(x)dx = lim+ ln(x)dx
a→0
0 a

= 0 − lim+ a ln(a) − 1
a→0

=0−0−1
= −1

Z1
Therefore, ln(x)dx = −1
0

The End of Section 7.6

Improper Integrals

§ 7.7 Table of Integrals

Differentiation has some straightforward rules for finding the derivative of a
complicated function. Integration, however, lacks such straightforwardness. As
a result, tables of known integrals often prove helpful to aid in finding antideriva-
tives.

In integration, finding the antiderivative does not follow a set of simple, uni-
versal rules. Integrating complex functions directly is often challenging and less
intuitive. For this reason, mathematicians have compiled tables of integrals.
These tables list the integrals of many common functions. By referencing these
tables, we can quickly find the antiderivative of a function that matches one of
the known forms, without having to derive each one from scratch.

Author's Interruption. While most textbook authors place the table of inte-
grals in the appendices to serve as a reference and avoid distracting the
reader, I have chosen to include them in the main text. This is due to the
reason I felt that the integrals listed in the table are often overlooked by
readers. Although memorizing every single integral listed in the table is not
necessary at a beginner to intermediate level, I believe that a thoughtful
review of them can be valuable for those interested in advanced calculus
or mathematical research. For beginners, the elementary integrals should
be of top priority.

Chap. 7 / Sec. 7.7 / Subsec. 7.7.1 : Elementary Integrals

xn+1
Z
1. xn dx = +C n 6= −1
n+1
Z
2. dx = x + c
Z Z Z
dx 1
3. = dx = x−1 dx = ln |x| + C
x x
eax
Z
4. eax dx = +C a 6= 0
a
bax
Z
5. bax dx = +C b > 0, b 6= 1, a 6= 0
a ln (b)
Z
6. ln (x) dx = x ln (x) − x + C
Z
1
7. logb (x) dx = x ln (x) − x + C b > 0, b 6= 1
ln (b)
Z
8. sin (x) dx = − cos (x) + C
Z
9. cos (x) dx = sin (x) + C
Z
10. tan (x) dx = − ln cos (x) + C
Z
11. cot (x) dx = ln sin (x) + C
Z
12. sec (x) dx = ln tan (x) + sec (x) + C
Z
13. csc (x) dx = − ln cot (x) + csc (x) + C
Z
dx x
14. √= sin−1 +C a > 0, |x| < 0
a2 − x2 a
Z
dx 1 −1 x

15. = tan +C a>0
a2 + x2 a a
Z
dx 1 x
16. √ = sec−1 +C a > 0, |x| > 0
x x2 − a2 a a

Chap. 7 / Sec. 7.7 / Subsec. 7.7.2 : Trigonometric Integrals

Z
1
1. sin (ax) dx = − cos (ax) + C a 6= 0
a
Z
x 1
2. sin2 (ax)dx = − sin(2ax) + C
2 4a
Z
1
3. cos (ax) dx = sin (ax) + C a 6= 0
a
Z
x 1
4. cos2 (ax)dx = + sin(2ax) + C
2 4a
Z
1
5. tan (ax) dx = − ln cos (ax) + C a 6= 0
a
Z
1
6. tan2 (ax)dx = tan(ax) − x + C
a
Z
1
7. cot (ax) dx = ln sin (ax) + C a 6= 0
a
Z
1
8. cot2 (ax)dx = − cot(ax) − x + C
a
Z
1
9. sec (ax) dx = ln tan (ax) + sec (ax) + C a 6= 0
a
Z
1
10. sec2 (ax)dx = tan(ax) + C
a
Z
1
11. csc (ax) dx = − ln cot (ax) + csc (ax) + C a 6= 0
a
Z
1
12. csc2 (ax)dx = − cot(ax) + C
a
sin(ax) x cos(ax)
Z
13. x sin(ax)dx = − +C
a2 a
cos(ax) x sin(ax)
Z
14. x cos(ax)dx = + +C
a2 a
x3
Z 2
x 1 x
15. x sin (ax)dx =
2 2
− − 3 sin(2ax) − 2 cos(2ax) + C
6 4a 8a 4a
x3
Z 2
x 1 x
16. x cos (ax)dx =
2 2
+ − 3 sin (2ax) + 2 cos (2ax) + C
6 4a 8a 4a

Chap. 7 / Sec. 7.7 / Subsec. 7.7.3 : Inverse Trigonometric Integrals

Z
1. sin−1 (x) dx = xsin−1 (x) +
p
1 − x2 + C
√
1 − a2 x2
Z
2. sin (ax)dx = xsin (ax) +
−1 −1
+C
a
Z
3. cos−1 (x) dx = xcos−1 (x) −
p
1 − x2 + C
√
1 − a2 x2
Z
4. cos (ax)dx = xcos (ax) −
−1 −1
+C
a
ln 1 + x2
Z
5. tan (x) dx = xtan (x) −
−1 −1
+C
2
ln 2 2

1 + a x
Z
6. tan−1 (ax) dx = xtan−1 (ax) − +C
2a
ln 1 + x2
Z
7. cot (x) dx = xcot (x) +
−1 −1
+C
2
ln 1 + a2 x2
Z
8. cot (ax) dx = xcot (ax) +
−1 −1
+C
2a
Z
9. sec−1 (x) dx = xsec−1 (x) − ln x + x2 − 1 + C
p

Z
1
10. sec−1 (ax) dx = xsec−1 (ax) −
ln ax + a2 x2 − 1 + C
p
a
Z
11. csc (x) dx = xcsc (x) + ln x + x − 1 + C
p
−1 −1 2

Z
1
12. csc−1 (ax) dx = xcsc−1 (ax) + ln ax + a2 x2 − 1 + C
p
a

Chap. 7 / Sec. 7.7 / Subsec. 7.7.4 : Reduction Formula for Trigonometric Func-
tions
sinn−1 (x) cos (x) n − 1
Z Z
1. sin (x) dx = −
n
+ sinn−2 (x) dx
n n
cosn−1 (x) sin (x) n − 1
Z Z
2. cos (x) dx =
n
+ cosn−2 (x) dx
n n

tann−1 (x)
Z Z
3. tan (x) dx =
n
− tann−2 (x) dx, n 6= 1
n−1
cotn−1 (x)
Z Z
4. cot (x) dx = −
n
− cotn−2 (x) dx, n 6= 1
n−1
secn−2 (x) tan (x) n − 2
Z Z
5. sec (x) dx =
n
− secn−2 (x) dx, n 6= 1
n−1 n−1
cscn−2 (x) cot (x) n − 2
Z Z
6. csc (x) dx =
n
− cscn−2 (x) dx, n 6= 1
n−1 n−1
sinm+1 (x) cosn−1 (x) n − 1
Z Z
7. sin (x) cos (x) dx =
m n
+ sinm (x) cosn−2 (x) dx,
m+n m+n
m 6= −n
xn cos (ax) n
Z Z
8. x sin (ax) dx = −
n
+ xn−1 cos (ax) dx, a 6= 0
a a
xn sin (ax) n
Z Z
9. x cos (ax) dx =
n
− xn−1 sin (ax) dx, a 6= 0
a a

Chap. 7 / Sec. 7.7 / Subsec. 7.7.5 : Integrals of the Form a2 − x2 , a > 0

a2 −1 x
Z p
xp 2
1. 2 2
a − x dx = a − x + sin
2 +C
2 2 a
a4 −1 x
Z
x
2. a − x + sin
p p
2 2 2 2 2 2 2
x a − x dx = 2x − a +C
8 8 a
√
a + a2 − x 2
Z
dx 1
3. √ = − ln +C
x a2 − x2 a x
√
a2 − x2
Z
dx
4. √ =− +C
x2 a2 − x2 a2 x
x2 a2 −1 x
Z
xp 2
5. √ dx = − a − x + sin
2 +C
a2 − x2 2 2 a
Z √ 2
a − x2 1p 2 x
6. dx = − a − x2 − sin−1 +C
x2 x a
Z
dx 1 x+a
7. 2 2
= ln +C
a −x 2a x−a

Chap. 7 / Sec. 7.7 / Subsec. 7.7.6 : Integrals of the Form x2 − a2 , a > 0

a2
Z p
xp 2
1. ln x + x2 − a2 + C
p
x2 − a2 dx = x − a2 −
2 2
a4
Z
x
2. ln x + x2 − a2 + C
p p p
x2 x2 − a2 dx = 2x2 − a2 x2 − a2 −
8 8
Z
dx
3. = ln x + x2 − a2 + C
p
√
x2 − a2
√
x2 − a2
Z
dx
4. √ = +C
x2 x2 − a2 a2 x
x2 a2
Z
xp 2
5. ln x + x2 − a2 +
p
√ dx = x − a2 + C
x2 − a2 2 2
Z √ 2 √
x − a2 x2 − a2
6. ln
p
dx = x + x 2 − a2 − +C
x2 x
x−a
Z
dx 1
7. = ln +C
x2 − a2 2a x+a
x2 − a2
Z
dx 1
8. = ln +C
x (x2 − a2 ) 2a2 x2

Chap. 7 / Sec. 7.7 / Subsec. 7.7.7 : Integrals of the Form a2 + x2 , a > 0

a2
Z p
xp 2
1. ln x + a + x + C
p
2 2
a + x dx = a +x + 2 2 2
2 2
a4
Z
x 2
2. ln x + a + x + C
p p p
2 2 2 2 2 2 2 2
x a + x dx = a + 2x a +x −
8 8
Z
dx
3. = ln x + a + x + C
p
√ 2 2
a2 + x2
Z
dx 1
4. = ln a − a2 + x2 + C
p
√
x a2 + x2 a
√
a2 + x2
Z
dx
5. √ =− +C
x2 a2 + x2 a2 x
x2 a2
Z xp
6. dx = − ln x + a2 + x2 +
p
√ a2 + x2 + C
2
a +x 2 2 2

Z √ √
a2 + x2 a+ a2 + x 2
7. dx = a2 + x2 − a ln
p
+C
x x
Z √ √
a2 + x2 a2 + x2
8. ln
p
dx = x + a2 + x2 − +C
x2 x
Z
dx x
9. q = √ +C
3 a 2 a2 + x2
2 2
(a + x )

x2
Z
dx 1
10. = 2 ln +C
x (a2 + x2 ) 2a a2 + x2

Chap. 7 / Sec. 7.7 / Subsec. 7.7.8 : Integrals of the Form ax ± b, a 6= 0, b >

0
(ax + b)n+1
Z
1. (ax + b)n dx = + C, n 6= −1
a (n + 1)
Z √ √ n+2
n 2 ax + b
2. ax + b dx = + C, n 6= −2
a (n + 2)
√ √
ax + b − b
Z
dx 1
3. √ = √ ln √ √ +C
x ax + b b ax + b + b
r !
−
Z
dx 2 ax b
4. √ = √ tan−1 +C
x ax − b b b

x b ln |ax + b|
Z
x
5. dx = − +C
ax + b a a2
x2 (ax + b)2 − 4b (ax + b) + 2b2 ln |ax + b|
Z
6. dx = +C
ax + b 2a3
Z
dx 1 x
7. = ln +C
x (ax + b) b ax + b
Z
dx 1 a ax + b
8. = − + 2 ln +C
x2 (ax
+ b) bx b x
√
2 (ax − 2b) ax + b
Z
x
9. √ dx = +C
ax + b 3a2
Z √
2
q
10. x ax + bdx = 2
(3ax − 2b) (ax + b)3 + C
15a

(ax + b)n+1
Z
ax + b b
11. x(ax + b) dx = n
− + C, n 6= −1, −2
a2 n+2 n+1

Chap. 7 / Sec. 7.7 / Subsec. 7.7.9 : Integrals with Exponential and Trigono-
metric Functions
eax (a sin (bx) − b cos (bx))
Z
1. eax sin (bx) dx = +C
a2 + b2
eax (a cos (bx) − b sin (bx))
Z
2. eax cos (bx) dx = +C
a2 + b2

Chap. 7 / Sec. 7.7 / Subsec. 7.7.10 : Integrals with Exponential and Logarith-
mic Functions
Z
dx
1. = ln |ln (x)| + C
x ln (x)

1
Z x n+1
ln (x) −
n+1
2. x ln (x) dx =
n
+ C, n 6= −1
n+1
Z
3. xex dx = xex − ex + C

xn eax n
Z Z
4. n ax
x e dx = − xn−1 eax dx, a 6= 0
a a
Z Z
5. ln (x) dx = xln (x) − n
n n
lnn−1 (x) dx

Chap. 7 / Sec. 7.7 / Subsec. 7.7.11 : Miscellaneous Integral Forms

xn+1
Z
Z xn+1
sin −1
(x) − √ dx
1 − x2
1. xn sin−1 (x) dx = , n 6= −1
n+1
xn+1
Z
Z x cos (x) + √
n+1 −1
dx
1 − x2
2. x cos (x) dx =
n −1
, n 6= −1
n+1

xn+1
Z
Z x n+1
tan −1
(x) − dx
x2 + 1
3. xn tan−1 (x) dx = , n 6= −1
n+1

x−a
√
2
a2 sin−1
− −
Z p
(x a) 2ax x a
4. 2ax − x2 dx = + + C, a > 0
2 2

−1 x − a
Z
dx
5. √ = sin + C, a > 0
2ax − x2 a

The End of Section 7.7

Table of Integrals

Applications of Integration

In the previous chapters, definite integrals were discussed in terms of the Rie-
mann Sum, and it was shown how they can be evaluated using antiderivatives
through the Fundamental Theorem of Calculus. This chapter will now focus on
the definite integral as a computational tool and demonstrate its importance in
solving and understanding a wide range of mathematical problems.

§ 8.1 Average Value of a Function

In mathematics, a fundamental challenge when dealing with a continuous func-
tion defined over an interval is determining the average of its values throughout
that interval. Unlike discrete data, where we can calculate the average by sum-
ming a finite number of values and dividing it by the count, continuous functions
pose a different challenge. A function on a closed interval, say [a, b], has infinitely
many values between a and b, so the traditional notion of averaging cannot di-
rectly apply.

To handle this, we define the average value of a function in terms of an integral.

For this, we let f (x) be a continuous function defined over an interval [a, b].
b−a
We divide the interval [a, b] into n sub-intervals of equal width ∆x = .
n
Let xi be a sample point in each sub-interval [xi−1 , xi ] and calculate f (xi ) for
∗ ∗

i = 1, 2, 3, · · · , n. Then,

f (x∗1 ) + f (x∗2 ) + f (x∗3 ) + · · · + f (x∗n )

459

is the total value of the function. Now, to take its average, we divide it by the
number of terms, that is, divide it by n,

f (x∗1 ) + f (x∗2 ) + f (x∗3 ) + · · · + f (x∗n )

n
b−a b−a
Now, since ∆x = , therefore n = , and thus,
n ∆x
f (x∗1 ) + f (x∗2 ) + f (x∗3 ) + · · · + f (x∗n )
b−a
∆x
n
Since, can be concisely written as
X
f (x∗1 ) + f (x∗2 ) + f (x∗3 ) + · · · + f (x∗n ) f (x∗i ),
i=1
therefore,
n
f (x∗i )
P
f (x∗1 ) + f (x∗2 ) + f (x∗3 ) + · · · + f (x∗n ) i=1
=
b−a b−a
∆x ∆x
n
1 X
= · f (x∗i )
b−a
i=1
∆x
n
∆x X
= · f (x∗i )
b − a i=1
n
!
1 X
= ∆x · f (x∗i )
b−a i=1
n
1 X
= · f (x∗i )∆x
b−a
|i=1 {z }
Riemann sum

Now, taking n → ∞ to get the exact value, we get,

n
!
1
lim
X
= · f (x∗i )∆x
b−a n→∞
i=1
 
Zb
1
= · f (x) dx
b−a
a

This is the average value of a function. We denote this as fave

Definition 8.1.1. (The Average Value of a Function)

Let f (x) be a continuous function over an interval [a, b], then the average
value of f (x), denoted as fave , is defined as,

Zb
1
fave = · f (x) dx
b−a
a

A few examples are shown below to demonstrate the definition above.

I Example 8.1.1. Find the average value of the function f (x) = x2 over the
interval [0, 2].
x3
Z
Solution. Since f (x) = x , therefore x2 dx =
2
+ C. Thus, we write,
3
Z2
1
fave = · x2 dx
2−0
0
2
1 x3

= ·
2 3 0
1 23
= ·
2 3
4
=
3
Therefore, the average value of the function f (x) = x2 over the interval [0, 2] is
4
.
3

I Example 8.1.2. Find the average value of the function f (x) = ex over the
interval [0, e].
Z
Solution. Since f (x) = e , therefore ex dx = ex + C. Thus, we write,
x

Ze
1
fave = · ex dx
e−0
0
1 i e
= · ex
e 0
1
= · ee − e0

e

ee − 1
=
e
≈ 5.207

Therefore, the average value of the function f (x) = ex over the interval [0, e] is
approximately 5.207.

Chap. 8 / Sec. 8.1 / Subsec. 8.1.1 : The Mean Value Theorem for Integrals
When we were discussing the average value of a function over a specific interval,
we might have wondered if there was a particular point within that interval
where the function’s value is exactly equal to its average value over the entire
interval, as the function can take on infinitely many values. That is, if there
exists a specific point, say c, within the interval such that f (c) is the same as
the average values of the function fave within that range. This observation brings
us to the mean value theorem for integrals.

Theorem 8.1.1. (The Mean Value Theorem For Integrals) If f (x) is a con-
tinuous function over an interval [a, b], then there exists a point c ∈ [a, b],
such that

f (c) = fave
Zb
1
= · f (x) dx
b−a
a

Equivalently,
Zb
f (x) dx = f (c) · (b − a)
a

A few examples to demonstrate the theorem above are provided below.

I Example 8.1.3. Find a number c in the interval [0, 2] such that the average
value of f (x) = x + 2 equals f (c).

Solution. First, we will find the average value of f (x) = x + 2 on [0, 2] as follows,
Z2
1
fave = · (x + 2)dx
2−0
0
2 2
1 x
= · + 2x
2 2 0
2
1 2
= · +2·2
2 2
=3

Now, according to the mean value theorem for integrals, there must be a point
c ∈ [0, 2] such that f (c) = 3. That is,

f (c) = c + 2 = 3
c=3−2
=1

Therefore, at c = 1, the function reaches its average value of 3.

I Example 8.1.4. Find a number c in the interval [0, 1] such that the average
value of f (x) = ex equals f (c).

Solution. First, we will find the average value of f (x) = ex on [0, 1] as follows,
Z1
1
fave = · ex dx
1−0
0
i1
x
=e
0
= e − e01

=e−1

Now, according to the mean value theorem for integrals, there must be a point
c ∈ [0, 1] such that f (c) = e − 1. That is,

f (c) = ec = e − 1
ln(ec ) = ln(e − 1)
c · ln(e) = ln(e − 1)
| {z }
1

c = ln(e − 1)

Therefore, at c = ln(e − 1), the function reaches its average value of e − 1.

The End of Section 8.1

Average Value of a Function

§ 8.2 Area of a Region Between Two Curves

The idea of finding the area of a region enclosed between two curves is essentially
an extension of the basic idea of calculating the area under a curve. For example,
consider two functions f (x) and g(x) continuous over an interval [a, b], where
f (x) ≥ g(x).
y

upper curve

x
a b
lower curve

Figure 8.1

Now, in order to find the area of the region that is enclosed by the curves over
the interval [a, b], we partition the interval [a, b] into n sub-interval of equal
b−a
width ∆x = . On each sub-interval, we construct a rectangle spanning
n
from the lower curve to the upper curve. Thus, the height of the ith rectangle
yields f (x∗i )−g(x∗i ), where, x∗i is a sample point chosen from the ith sub-interval.
Therefore, the area of the ith rectangle yields (f (x∗i ) − g(x∗i )) · |{z}
∆x . Summing
width
| {z }
height
the areas of the n rectangles provides an estimate for the exact area enclosed by
the curves, that is,
Xn
A≈ (f (x∗i ) − g(x∗i )) · ∆x
i=1

upper curve

x
a b
lower curve

Figure 8.2

This forms the Riemann sum. Now, letting n → ∞, we get the exact area A
bounded by the curves and the vertical lines x = a and x = b. That is,
n
A = lim
X
(f (x∗i ) − g(x∗i )) · ∆x
x→∞
i=1

As a result, we have the following definition of the area of a region between two
curves.

Definition 8.2.1. (Area of a Region Between Two Curves) Let f (x) and g(x)
be continuous functions such that f (x) ≥ g(x) over the interval [a, b]. The
area of the region bounded by the graphs of f (x) and g(x) over [a, b] is

Zb
A= (f (x) − g(x))dx
a

A few examples to demonstrate the definition above are shown below.

I Example 8.2.1. Find the area of the region bounded by f (x) = sin(x) and
g(x) = cos(x), over the interval [1, 3].

Solution. The graph of f (x) = sin(x) is the upper curve and the graph of g(x) =
cos(x) is the lower curve on the interval [1, 3]. Therefore, the area of the region
between these curves is,

Z3 Z3 Z3
(sin(x) − cos(x)) dx = sin(x)dx − cos(x)dx
1 1 1

Z3 Z3
= sin(x)dx − cos(x)dx
1 1
i3 i3
= − cos(x) − sin(x)
1 1
= − cos(3) + cos(1) − sin(3) + sin(1)
≈ 2.23

Therefore, the area of the region bounded by f (x) = sin(x) and g(x) = cos(x),
over the interval [1, 3] is approximately 2.23.

I Example 8.2.2. Find the area of the region bounded by f (x) = x and g(x) = x2 ,
over the interval [0, 1].

Solution. The graph of f (x) = x is the upper curve and the graph of g(x) = x2 is
the lower curve on the interval [0, 1]. Therefore, the area of the region between
these curves is,
Z1 Z1 Z1
(x − x2 )dx = xdx − x2 dx
0 0 0
2 1 3 1

x x
= −
2 0 3 0
1 1
= −
2 3
≈ 0.167

Therefore, the area of the region bounded by f (x) = x and g(x) = x2 , over the
interval [0, 1] is approximately 0.167.

Chap. 8 / Sec. 8.2 / Subsec. 8.2.1 : Area of a Region between Two Curves with
respect to y
There are times when it is easier to integrate the area with respect to y instead
of x. For example, consider integrating the area of a region enclosed by the
graph of x = f (y) and x = g(y) over the interval [c, d] lying on the y-axis, where
f (y) ≥ g(y), implying that the graph of f (y) lies to the right of the graph of g(y)

and the lower and upper bound of the region are y = c and y = d, respectively,
as shown below.

f(x)

Figure 8.3

Here, in order to integrate, we treat the dependent variable y as the independent

variable and divide the interval [c, d] into n sub-interval of equal height ∆y =
d−c
. On each sub-interval, we construct a rectangle spanning from the left
n
curve to the right curve. Thus, the width of the ith rectangle yields f (yi∗ )−g(yi∗ ),
where yi∗ is a sample point chosen from the ith sub-interval. Therefore, the area
of the ith rectangle yields f (yi∗ ) − g(yi∗ ) · ∆y . Summing the areas of the n
| {z } |{z}
width height
rectangles and taking the limit as n → ∞ provides the exact area of a region
enclosed between two curves with respect to y, that is,
n
A = lim
X
(f (yi∗ ) − g (yi∗ )) ∆y
n→∞
i=1

As a result, we have the following definition of the area of a region enclosed

between two curves with respect to the y-axis.

Definition 8.2.2 (Area of a Region Between Two Curves with respect to y). Let f (y)
and g(y) be continuous functions such that f (y) ≥ g(y) over the interval
[c, d]. The area of the region bounded by the graphs of f (y) and g(y) over

[c, d] is
Zd
A= (f (y) − g(y))dy
c

A few examples to demonstrate the definition above are provided below.

I Example 8.2.3. Find the area of the region bounded by x = y 2 and x = 4y.

Solution. In order to integrate, we must first identify the integration intervals by

finding the points of intersection of the two functions, which is done by equating
the functions as shown below.

y 2 = 4y
y 2 − 4y = 0
y(y − 4) = 0

Therefore, the points of intersection are y = 0 and y = 4.

Since x = 4y is to the right of x = y 2 in that region, therefore, we have the

following,
Z4 Z4 Z4
2
y 2 dy

4y − y dy = 4ydy −
0 0 0
Z4 Z4
=4 ydy − y 2 dy
0
!0
2 4
4
y3

y
=4 −
2 0 3 0
2 3

4 4
=4 −
2 3
32
=
3
32
Therefore, the area of the region bounded by x = y 2 and x = 4y is .
3

I Example 8.2.4. Find the area of the region bounded by the graph of y = x2
and y = x + 2.

Solution. We will integrate the area of the bounded region with respect to y.
Therefore, we first need to solve for x in terms of y, as shown below.
√
y = x2 =⇒ x= y
y =x+2 =⇒ x=y−2

Now, to find the limits of integration, we set x2 = x + 2 as follows,

x2 = x + 2
x2 − x − 2 = 0
(x + 1)(x − 2) = 0

Therefore, the points of intersection are x = 2 and x = −1.

Now, we substitute these x-values into the original functions to find the corre-
sponding y-values, as shown below.

x=2 =⇒ y = 22 = 4
x = −1 =⇒ y = (−1) + 2 = 1
√
Now, notice that since x = y − 2 is to the right of x = y, therefore, we have
the following,
Z4 Z4 Z4
√ √
y − (y − 2) dy = ydy − (y − 2)dy
1 1 1
 4 
Z4 Z4
√
Z
= ydy −  ydy − 2dy 
1 1 1
Z4 Z4 Z4
1
= y 2 dy − ydy + 2 dy
1 1 1
4
3 4
y2  y2
= − + 2 y]41
3 2 1
2 1 
3 3 2
12

42 12  4
= − − − + 2 (4 − 1)
3 3 2 2
2 2

8 1 1
= − − 8− +6
3 3 2
2 2

19
=
6
Therefore, the area of the region bounded by the graph of y = x2 and y = x + 2
19
is .
6

The End of Section 8.2

Area of a Region Between Two Curves

§ 8.3 Volume: The Slicing Method

Area, as we intuitively understand it, is the amount of space that a two-dimensional
shape occupies. For example, the area of a rectangle is calculated as the product
of its length and width, quantifying the amount of space it covers. Extending
this concept to the three dimensions introduces the idea of “volume.” That is,
volume is the amount of space that a three-dimensional object occupies. For
example, the volume of a rectangular box is found by multiplying its length,
width, and height, or equivalently, multiplying the area of its base by its height.

Although intuitive, these ideas are too vague to serve as a formal definition,
as they lack the necessary level of mathematical rigor. Nevertheless, having
previously defined the concept of area rigorously in terms of the definite inte-
gral, we can now use that definition to define the concept of volume and discuss
methods for calculating a specific type of volume: the volume of a solid of rev-
olution. In order to do that, three primary techniques will be discussed in this
chapter—slicing, disks, and washers—each appropriate for solids with particular
characteristics.

Having this on mind, we now begin our discussion of defining the volume of a
solid.

Chap. 8 / Sec. 8.3 / Subsec. 8.3.1 : The Definition of the Volume of a Solid
To precisely define the volume of a solid, we make use of the slicing method.
This is a technique that determines the volume by integrating the cross-sectional
areas of the solid. Put simply: this method works by “slicing” the solid into thin
cross-sectional pieces, calculating the area of each slice, and then summing up

these areas across the entire volume.

A natural question is now to follow—what is a cross-section? Informally speak-

ing, a cross-section is essentially a slice of the solid, which has a certain shape
depending on the solid itself. For example, consider the following solid S of a
right circular cylinder.

Figure 8.4

When this solid S is intersected with a two-dimensional plane P , the resulting

region (colored in blue) is called the cross-section of the solid S, as shown below.

P
S

Figure 8.5

Here, the cross-section turns out to be a circle, because the solid S is a right
circular cylinder. The area of this cross-section is what we refer to as the “cross-
sectional area” (the region highlighted in blue, in this case).

We can use this cross-sectional area to define the volume of the solid S. The
idea is that since the area of the cross-section can exactly be determined using
definite integrals, the volume of S can also be determined by summing up all
the area slices of the cross-section.

With that idea in mind, now consider calculating the volume of the solid shown
below, from x = a to x = b.

a b

Figure 8.6

To calculate the volume of the solid within the interval [a, b], the solid is in-
tersected with a plane to form a cross-section. The area of this cross-section
is represented by a known, integrable function A(x). Next, the interval [a, b]
b−a
is divided into n sub-intervals, each with an equal thickness of ∆x = .
n ∗
The ith sub-interval is denoted by [xi−1 , xi ]. Within each sub-interval, we let xi
represent an arbitrary point, as shown below.

a b

Figure 8.7

Now, let the area of the cross-section (i.e., the cross-sectional area) at the point
x∗i is A(x∗i ) and thus, the volume Vi at the point x∗i can be approximated as

Vi ≈ A(x∗i ) · ∆x

Adding these approximations together we can derive an approximation for the

total volume V over the interval [a, b]. That is,

n
X
V ≈ A(x∗i )∆x
i=1

This forms the Riemann sum. Now, as n increases indefinitely, the approxima-
tion gets better and better to the actual volume V over the interval [a, b]. Thus,
n
we take the limit of the sum A(x∗i )∆x as n → ∞ as follows,
X

i=1
n
V = lim
X
A(x∗i )∆x
n→∞
i=1
Zb
= A(x∗i )dx
a

As a result, we have derived a definition of the volume of a solid in terms of a

definite integral.

Definition 8.3.1. (The Definition of the Volume of a Solid)

Let S be a solid enclosed between x = a and x = b. If the cross-sectional
area of S perpendicular to the x-axis at each point x ∈ [a, b] is given by
A(x), then the volume V of the solid is

Zb
V = A(x)dx
a

provided that A(x) is integrable.

Having defined the definition of the volume of a solid in terms of the definite
integral, we now turn our attention to discuss a specific type of solids known as
solids of revolution. These are three-dimensional shapes formed by rotating a
two-dimensional region around a specified axis. Our goal in the following sub-
section will be to discuss the methods used to calculate the volumes of such
solids systematically.

Chap. 8 / Sec. 8.3 / Subsec. 8.3.2 : The Disk Method

The Disk Method is used to calculate the volume of a solid of revolution. For
that, suppose f (x) ≥ 0 is a continuous function on the interval [a, b]. Let

R represent the region enclosed by the graph of f (x) and the x-axis over the
interval [a, b], as shown below.

x
a b

Figure 8.8

If we rotate this region R around the x-axis, it forms a three-dimensional solid

known as a solid of revolution. Our goal is to find the volume of this solid, which
can be done using the slicing method aforementioned.

Figure 8.9

Now, in order to determine the volume of this solid of revolution over the interval
b−a
[a, b], we divide the interval into n sub-interval of equal thickness ∆x = .
n

Then [xi−1 , xi ] is the ith sub-interval. Let x∗i be a point in the ith sub-interval
[xi−1 , xi ], as shown below.

Figure 8.10

The area of the cross-section at the point x∗i is A(x∗i ). Notice that A(x∗i ) has
a specific form. That is, all the cross-sections perpendicular to the x-axis form
a circular disk whose radius is the function’s value f (x). For this reason, the
cross-section area yields,

A(x∗i ) = π · (radius)2
= πf (x∗i )2

Therefore, by the previously discussed slicing method, the volume of the solid
yields,

Zb
V = A (x) dx
a
Zb
= πf (x)2 dx
a

Thus, we have the following definition.

Theorem 8.3.1. (The Disk Method)

Let f (x) be a non-negative continuous function on the interval [a, b]. If
R is the region bounded by the graph of f (x) and the x-axis, over the

interval [a, b]; then the volume V of the solid of revolution that is obtained
by rotating R about the x-axis is the following,
Z b
V = πf (x)2 dx
a

A few examples are provided below.

I Example 8.3.1. Let R be the region bounded by the graph of f (x) = x2 and
the x-axis, over the interval [0, 5]. What is the volume of the solid of revolution
obtained by rotating R about the x-axis?
Solution. We know that the volume of a solid of revolution is,
Z b
V = πf (x)2 dx
a

Where, f (x) = x and a = 0 and b = 5. Therefore, the volume yields,

Z 5
V = π(x2 )2 dx
0
Z 5
=π x4 dx
0
5 5
x
=π
5 0
= 625π

Therefore, the volume of the solid of revolution obtained by rotating R about

the x-axis is 625π.

√
I Example 8.3.2. Let R be the region bounded by the graph of f (x) = x
and the x-axis, over the interval [0, 100]. What is the volume of the solid of
revolution obtained by rotating R about the x-axis?
Solution. We know that the volume of a solid of revolution is,
Z b
V = πf (x)2 dx
a

Where, f (x) = x and a = 0 and b = 100. Therefore, the volume yields,

Z 100
√
V = π xdx
0

Z 100
1
=π x 2 dx
0
" 3
#100
2x 2
=π
3
0
2000π
=
3
Therefore, the volume of the solid of revolution obtained by rotating R about
2000π
the x-axis is .
3

Chap. 8 / Sec. 8.3 / Subsec. 8.3.3 : The Washer Method

A slight variation on the disk method enables us to compute the volume of more
complicated solids of revolution. For example, consider R is a region bounded
by the graph of f (x) and g(x) on the interval [a, b], where f (x) ≥ g(x) ≥ 0, as
shown below.
y

x
a
b

Figure 8.11

If we revolve R about the x-axis, the solid that forms has a hole through it. In
other words, we get a hollowed-out solid of revolution, as shown below.

x
a
b

Figure 8.12

Now, in order to determine the volume of this solid of revolution over the interval
b−a
[a, b], we divide the interval into n sub-interval of equal width ∆x = . Then
n
[xi−1 , xi ] is the ith sub-interval. Let xi be a point in the ith sub-interval [xi−1 , xi ],
∗

as shown below.
y

x
a
b

Figure 8.13

Notice that, this time, the cross-section through the solid perpendicular to the
x-axis forms a circular washer (a ring-shaped solid). This circular washer has
an outer radius R = f (x) and an inner radius r = g(x), where a ≤ x ≤ b. This
is shown in more detail in the following,

a
b

Figure 8.14

For this reason, the cross-sectional area of the circular washer is the area of the
entire disk (that is, πf (x)2 ) minus the area of the hole (that is, πg(x)2 ). This
implies that the area of the circular washer is as follows,

A (x) = πR2 − πr2

= πf (x)2 − πg(x)2

2 2
= π f (x) − g(x)

Since we’ve found the area of the cross-section, we can apply the slicing method
to determine the volume of the solid as follows.

Theorem 8.3.2. (The Washer Method)

Let f (x) and g(x) be a continuous function on the interval [a, b] with f (x) ≥
g(x) ≥ 0. If R is the region bounded by the graph of y = f (x) and y = g(x)
over the interval [a, b]; then the volume V of the solid of revolution that is
obtained by rotating R about the x-axis is the following,

Zb
π f (x)2 − g(x)2 dx

V =
a

√
I Example 8.3.3. Suppose R is a region bounded by the graph of f (x) = x and
g(x) = x2 on the interval [0, 1]. What is the volume of the solid that is obtained
by rotating the region R about the x-axis?

Solution. Since f (x) ≥ g(x) on the interval [0, 1], we’ll use the washer method.
The area of the cross-section at the point x is,

A (x) = π f (x)2 − g(x)2
√ 2
2
=π x − x2
= π x − x4

By the washer method, the volume yields,

Z1
π x − x4 dx

V =
0
Z1
x − x4 dx

=π
0
1
x2 x5

=π −
2 5 0
3π
=
10

The End of Section 8.3

Volume: The Slicing Method

§ 8.4 Volume: The Method of Cylindrical Shells

The Method of Cylindrical Shells is another method to calculate the volume of a
solid of revolution. This method is particularly useful when the solid is difficult
to calculate in terms of the previously explained slicing method (whether, it is
disks or washers). To demonstrate the method, let us consider R be the region
bounded by the graph of f (x) and the x-axis over the interval [a, b], where
0 ≤ a < b and f (x) ≥ 0, as shown below.

x
a b

Figure 8.15

It is evident that when this region R revolves around the y-axis, a solid of
revolution is generated. Our objective is to determine the volume of this solid.
y

Figure 8.16

To do this, we divide the interval [a, b] into n sub-interval of equal width ∆x =

b−a
. Then [xi−1 , xi ] is the ith sub-interval for i = 1, 2, 3, · · · , n. Let x∗i be any
n
arbitrary point on the ith sub-interval. Now, we build a rectangle on the ith
sub-interval with a height of f (x∗i ) and a width of ∆x, as shown below.

x
a b

Figure 8.17

Notice that as the region R revolves around the y-axis, the rectangle built on
the ith sub-interval generates a thin cylindrical shell, as shown below.
y

Figure 8.18

When this cylindrical shell is unwrapped, it approximately forms a thin rect-

angular slab. The height of this rectangular slab is the height of the original
rectangle f (x∗i ) and the width is also the original width of the rectangle x∗i . The
length of the rectangular slab is the circumference of the cylindrical shell’s circle
that has a radius of (x∗i − 0) = x∗i . Therefore, the length yields 2πx∗i . As a
result, the volume of the ithe shell is approximately,

2πx∗i · f (x∗i ) · |{z}

∆x = 2πx∗i f (x∗i )∆x
width
| {z } | {z }
length height

Summing the volume of the n cylindrical shells gives an approximation to the

actual volume of the entire solid of revolution. That is,

n
X
V ≈ 2πx∗i f (x∗i )∆x
i=1

Now, as n → ∞, we obtain the exact volume of the solid of revolution as a

definite integral.
n
V = lim
X
2πx∗i f (x∗i )∆x
n→∞
i=1
Zb
= 2πxf (x)dx
a

The formulation above can be generalized for a revolving region bounded by two
curves f (x) and g(x), where f (x) ≥ g(x), instead of a region bounded by the
only function f (x) and the x-axis on the interval [a, b].

In order to do this, only the height of the rectangle built on an arbitrary point
x∗i needs to be adjusted in the previous formulation. That is, instead of height
being f (x∗i ) on the ith interval, the height now is the difference between two
functions, that is, (f (x∗i ) − g(x∗i )). So, the generalized formulation yields,
n
V = lim
X
2πx∗i (f (x∗i ) − g(x∗i )) ∆x
n→∞
i=1
Zb
= 2πx (f (x) − g(x)) dx
a

As a result, we have the following,

Theorem 8.4.1. (The Method of Cylindrical Shells)

Let f (x) and g(x) be continuous functions where f (x) ≥ g(x) on the in-
terval [a, b]. If R is the region bounded by the graph of f (x) and g(x),
over the interval [a, b]; then the volume V of the solid of revolution that is
obtained by rotating R about the y-axis is the following,

Zb
V = 2πx f (x) − g(x) dx
a

√
I Example 8.4.1. Let R be the region bounded by the graph of f (x) = x and
the x-axis on the interval [0, 4]. What is the volume of the solid of revolution
generated when the region R is rotated about the y-axis?

Solution. The volume of the solid of revolution can be determined by the shell
√
method, where x is the radius and f (x) = x is the height of the cylindrical
shell. Therefore, the volume of the solid by the shell method on the interval
[0, 4] yields,
Zb
V = 2πxf (x)dx
a
Z4
√
= 2πx xdx
0
Z4
1
= 2π x1 x 2 dx
0
Z4
3
= 2π x 2 dx
0
" 5
#4
x2
= 2π 5
2 0
" 5
#4
2x 2
= 2π
5
0
" 5
#4
2x 2
= 2π
5
0
64π
= 2π ·
5
128π
=
5
128π
Therefore, the volume of the solid of revolution is .
5

I Example 8.4.2. Let R be hthe region bounded by the graph of f (x) = sin(x) and
πi
the x-axis on the interval 0, . What is the volume of the solid of revolution
2
generated when the region R is rotated about the y-axis?

Solution. The volume of the solid of revolution can be determined by the shell
method, where x is the radius and f (x) = sin(x) is the height of the cylindrical
shell. Therefore, the volume of the solid by the shell method on the interval
π
[0, ] is,
2
Zb
V = 2πxf (x)dx
a
π
Z2
= 2πx sin(x)dx
0
π
Z2
= 2π x sin(x)dx
0
π
Z2
Now, to solve the integral x sin(x)dx, integration by parts is used as follows.
0

u=x =⇒ du = dx
Z Z
dv = sin(x)dx =⇒ v = dv = sin(x)dx = − cos(x)

As a result, by integration by parts, we write the following.

π π
Z2 h i π Z2
x sin(x)dx = −x cos(x) + cos(x)dx
2

0
0 0
h iπ h iπ
= −x cos(x) + sin(x)
2 2

0 0
=0+1
=1

This implies the following,

π
Z2
V = 2π x sin(x)dx
0

= 2π · 1
= 2π

Therefore, the volume of the solid of revolution is 2π.

The End of Section 8.4

Volume: The Method of Cylindrical Shells

§ 8.5 Arc Length

Curves, unlike straight lines, change their direction throughout their path. There-
fore, calculating the precise length of a curve, such as y = f (x), requires more
than simply measuring the overall span from one point to another. That is, it
requires us to account for every subtle bend its path.

Nevertheless, we can calculate the precise length of a curve in terms of an integral

using the formula for arc length. For that, let f (x) be a curve such that f 0 (x)
exists and is continuous over an interval [a, b]. Such a curve f (x) is said to be
smooth on [a, b]. Now, our objective is to determine the length of the curve from
the point (a, f (a)) to (b, f (b)), as shown below.

Figure 8.19

b−a
We partition the interval [a, b] into n sub-intervals of equal width ∆x = us-
n
ing points a = x0 , x1 , x2 , · · · , xn−1 , xn = b, and connect the corresponding points
(x0 , f (x0 )), (x1 , f (x1 )), (x2 , f (x2 )), · · · , (xn−1 , f (xn−1 )), (xn , f (xn )) on the graph
of f (x) with line segments, resulting in a polygonal line with n line segments.
When n is sufficiently large (and hence ∆x is small), the length of this polygonal
line is a good approximation to the actual curve.

Since we’ve used a regular partition, the change in the horizontal distance on
every interval is fixed, that is, ∆x. However, the change in the vertical distance
varies from interval to interval. In general, the change in the vertical distance
on the ith interval is ∆yi = f (xi ) − f (xi−1 ), for i = 1, 2, 3, · · · , n. This implies
that the line segment over the ith sub-interval represents the hypotenuse of a
right triangle with sides of length ∆x and |∆yi | = |f (xi ) − f (xi−1 )|, as shown
below,

Figure 8.20

Thus, by the Pythagorean theorem,

hypontenuse2 = (∆x)2 + |∆yi |2

q
hypontenuse = (∆x)2 + |∆yi |2

Since, the hypotenuse is the ith line segment for i = 1, 2, 3, · · · , n, therefore the
length of each line segment is,

for, i = 1, 2, 3, · · · , n
p
(∆x)2 + |∆yi |2

This can be written as,

s
(∆yi )2
q
(∆x)2 + |∆yi |2 = (∆x)2 + (∆x)2 ·
(∆x)2
v
u 2 !
u ∆yi
= t(∆x)2 1+
∆x

s
∆yi 2
q
2
= (∆x) · 1 +
∆x
s
∆yi 2

= ∆x · 1 +
∆x
Now, summing these lengths together for i = 1, 2, 3, · · · , n, we get the length of
the polygonal lines that approximates the actual length L of the curve. that is,
s
n 2
X ∆yi
L≈ ∆x · 1+
i=1
∆x
∆yi
Notice that represents the slope of the line segment over the ith interval.
∆x
Now, by the Mean Value Theorem, this slope equals the derivative f 0 (x∗i ) for
some point x∗i ∈ [xi−1 , xi ]. Therefore,
s
n
∆yi 2
X
L≈ ∆x · 1 +
i=1
∆x
s
n
∆yi 2
X
= 1+ · ∆x
i=1
∆x
X n q
= 1 + f 0 (x∗i )2 ∆x
i=1

This forms the Riemann sum. Now, letting n → ∞, the sum approaches a
definite integral which is the exact length L of the curve. That is,
n q
L = lim
X
1 + f 0 (x∗i )2 ∆x
n→∞
i=1
Zb q
= 1 + f 0 (x)2 dx
a

As a result, we have the following definition for the arc length of y = f (x).

Definition 8.5.1. (Arc Length of a Function) Let f (x) be a function such

that f 0 (x) exists and is continuous over an interval [a, b], then the length
L of the curve f (x) from the point (a, f (a)) to (b, f (b)) is

Zb q
L= 1 + f 0 (x)2 dx
a

A few examples are provided below to demonstrate the definition above.

I Example 8.5.1. Find the length of the graph of f (x) = 5x + 3 from x = 0 to

x = 5.

Solution. Since f (x) = 5x+3, therefore, f 0 (x) = 5. Thus, we write the following,

Z5 p
L= 1 + 52 dx
0
5
√ Z
= 26 · dx
0
√ i5
= 26 · x
0
√
= 26 · 5
≈ 25.495

Therefore, the length of the graph of f (x) = 5x + 3 from x = 0 to x = 5 is

approximately 25.495.

I Example 8.5.2. Find the length of the graph of f (x) = ln(cos(x)) from x = 0
to x = 1.
sin(x)
Solution. Since f (x) = ln(cos(x)), therefore, f 0 (x) = − . Thus, we write
cos(x)
the following,

Z1
s
sin(x)
2
L= 1+ − dx
cos(x)
0
Z1 q
= 1 + (− tan(x))2 dx
0
Z1 q
= 1 + tan2 (x)dx
0
Z1 p
= sec2 (x)dx
0

Z1
= sec(x)dx
0
i1
= ln |sec(x) + tan(x)|
0
= ln |sec(1) + tan(1)| − ln |sec(0) + tan(0)|
≈ 1.23

Therefore, the length of the graph of f (x) = ln(cos(x)) from x = 0 to x = 1 is

approximately 1.23.

The End of Section 8.5

Arc Length

§ 8.6 The Area of a Surface of Revolution

In the previous section, we discussed solids of revolution and explored various
methods to calculate their volumes. Now, we turn our attention to calculate the
area of the surface of such solids of revolution. However, before diving into this
topic, we need to first understand the formula for the lateral surface area of a
frustum of a right circular cone, which in turn requires us to know the formula
for the lateral surface area of a right circular cone.

It is known from geometry that the lateral surface area A of the following right
circular cone

Figure 8.21

is given by the following formula:

A = πr`

where, r is the radius of the base of the cone and ` is the slant height of the
√
cone, which is equal to r2 + h2 by the Pythagorean theorem. As a result, we
have the formula for the lateral surface area A of a right circular cone as,

A = πr`
p
= πr r2 + h2

Now, using the formula for the lateral surface area of a right circular cone, we
will derive the formula for the lateral surface area of a frustum of a right circular
cone. A frustum of a right circular cone is a part of a cone that remains after
the top portion of the cone is cut off by a plane parallel to the base.

In order to derive the formula for the frustum, consider a linear function f (x) =
nx over an interval [a, b], where 0 < a < b and n > 0, as shown below.

f(x)

x
a b

Figure 8.22

Now, rotate this line segment about the x-axis resulting in a frustum of a cone
as shown below.

f(x)

x
a b

Figure 8.23

We need to determine the surface area S of this frustum. The idea is to subtract
the small surface area of the frustum over [0, a] from the larger surface area of
the frustum over [0, b]. This allows us to calculate the surface area of the frustum
over [a, b].

Now, notice carefully that the radius r of the cone over [0, b] is f (b) = nb and
the height h over [0, b] is b. Thus, the surface area of the cone over [0, b] yields,
p
S[0,b] = πr r2 + h2
q
= π (nb) (nb)2 + b2
p
= π (nb) b2 (n2 + 1)
√ p
= π (nb) b2 · (n2 + 1)
p
= π (nb) b · (n2 + 1)
p
= πb2 n n2 + 1

Similarly, the radius r of the cone over [0, a] is f (a) = na and the height h over
[0, a] is a. Thus, the surface area of the cone [0, a] yields,
p
S[0,a] = πr r2 + h2
q
= π (na) (na)2 + a2
p
= π (na) a2 (n2 + 1)
√ p
= π (na) a2 · (n2 + 1)
p
= π (na) a · (n2 + 1)

p
= πa2 n n2 + 1

Now, the surface area S of the frustum over [a, b] yields the following,

S = S[0,b] − S[0,a]
p p
= πb2 n n2 + 1 − πa2 n n2 + 1
p
= πn b2 − a2 n2 + 1

Now, we calculate the line segment ` between (a, f (a)) and (b, f (b)) for fur-
ther simplification in the equation above. For that, note that the line segment
between the point (a, f (a)) = (a, an) and (b, f (b)) = (b, bn) yields the following.
q
` = (b − a)2 + (bn − an)2
q
= b2 − 2ab + a2 + (bn)2 − 2anbn + (an)2
p
= b2 − 2ab + a2 + b2 n2 − 2abn2 + a2 n2
p
= b2 − 2ab + a2 + n2 (b2 − 2ab + a2 )
p
= (b2 − 2ab + a2 ) (1 + n2 )
p p
= (b2 − 2ab + a2 ) · (1 + n2 )
q
= (b − a)2 · (1 + n2 )
p
p
= (b − a) (n2 + 1)

As a result, we have line segment ` = (b − a)

p
(n2 + 1)

√
Now, since the surface area S of the frustum over [a, b] is πn b2 − a2 n2 + 1.
Therefore,
p
S = πn b2 − a2 n2 + 1
p
= πn (b + a) (b − a) n2 + 1
| {z }
(b2 −a2 )
p
= πn (b + a) (b − a) n2 + 1
| {z }
`

= πn (b + a) `
 

= π |{z}
bn + |{z}
an  `
f (b) f (a)

= π f (b) + f (a) `

Thus, we have derived the formula for the lateral surface area S of a frustum of
a right circular cone that was obtained by rotating the function f (x) over the
interval [a, b]. With this result, we are now equipped to derive a general formula
for the surface area of a surface of revolution.

Chap. 8 / Sec. 8.6 / Subsec. 8.6.1 : The Generalized Area Formula for a Sur-
face of Revolution
The formula for the surface area S of a frustum of a cone in the previous sub-
section was found to be the following,

S = π f (b) + f (a) `

Based on this result, we now derive the general formula for the surface area of
a surface of revolution. In order to do that, consider a continuous function f (x)
over an interval [a, b], as shown below.

f(x)

x
a b

Figure 8.24

Now, rotate this curve of f (x) about the x-axis, resulting in the following solid
of revolution,

f(x)

x
b
a

Figure 8.25

Now, we need to determine the surface area S of this solid of revolution. For
b−a
that, we divide the interval [a, b] into n sub-interval of equal width ∆x = .
n
Then [xi−1 , xi ] becomes the ith sub-interval, as shown below.

f(x)

x
b
a

Figure 8.26

Now, notice that the line segment ` over the ith sub-interval, that is, the line
segment between the points (xi , f (xi−1 )) and (xi , f (xi )) is calculated as follows,
p
` = (xi − xi−1 )2 + (f (xi ) − f (xi−1 ))2
p
= (∆x)2 + (∆yi )2

Here, since all the sub-intervals have equal width, therefore width ∆x is the
same for all i and thus ∆x doesn’t have any subscripts. Notice, however, that
this is not the case for the change in y, that is, ∆yi = f (xi ) − f (xi−1 ).

As a result, the surface area of this surface of revolution over [xi−1 , xi ] is given
by the formula of the surface area of a frustrum of a cone over [xi−1 , xi ], that is,

Si = π (f (xi ) + f (xi−1 )) `
q
= π (f (xi ) + f (xi−1 )) (∆x)2 + (∆yi )2

Now, summing these individual frustums for i = 1, 2, 3, · · · , n yields the total

surface area of the surface of revolution S,
n
X
S≈ Si
i=1
n
X q
= π (f (xi ) + f (xi−1 )) (∆x)2 + (∆yi )2
i=1
n q
(1)
X
= π (f (xi ) + f (xi−1 )) (∆x)2 + (f (xi ) − f (xi−1 ))2
i=1

It is known—from the mean value theorem—that,

f (xi ) − f (xi−1 )
= f 0 (x∗i )
∆x
f (xi ) − f (xi−1 ) = f 0 (x∗i ) ∆x (2)

Now, substituting equ. (2) into (1), yields,

n
X n
X q
Si = π (f (xi ) + f (xi−1 )) (∆x)2 + (f (xi ) − f (xi−1 ))2
i=1 i=1
Xn q
= π (f (xi ) + f (xi−1 )) (∆x)2 + (f 0 (x∗i ) ∆x)2
i=1
Xn q
= π (f (xi ) + f (xi−1 )) (∆x)2 + (∆x)2 f 0 (x∗i )2
i=1
Xn r
2
0 ∗ 2
= π (f (xi ) + f (xi−1 )) (∆x) 1 + f (xi )
i=1
n
X
r
= π (f (xi ) + f (xi−1 )) 1 + f 0 (x∗i )2 ∆x
i=1

Now, when n (the number of sub-intervals within the interval [a, b]) is sufficiently
large, then xi−1 ≈ xi ≈ x∗i . This implies that f (xi−1 ) ≈ f (xi ) ≈ f (x∗i ), for
i = 1, 2, 3, · · · , n, given that the function f (x) is continuous. Therefore,
Xn r
S≈ π (f (x∗i ) + f (x∗i )) 1 + f 0 (x∗i )2 ∆x
i=1
n
X
r
= 2πf (x∗i ) 1 + f 0 (x∗i )2 ∆x
i=1

This forms the Riemann sum. Now if n → ∞, then ∆x → 0, and we have the
following,
n r
S = lim
X
∗
2πf (xi ) 1 + f 0 (x∗i )2 ∆x
n→∞
i=1
Zb r
= 2πf (x) 1+ f 0 (x)2 dx
a

As a result, we have the following definition of the area of a surface of revolution.

Definition 8.6.1. (Area of a Surface of Revolution)

Let f (x) be a differentiable and non-negative function over an interval [a, b].
The area of the surface S formed by rotating f (x) over the interval [a, b]
about the x-axis is given by

Zb r
S= 2πf (x) 1 + f 0 (x)2 dx
a

A few examples to demonstrate the definition above are provided below.

I Example 8.6.1. Find the surface area generated by rotating the function f (x) =
2 about the x-axis over the interval [0, 3].

Solution. Since f (x) = 2, therefore f 0 (x) = 0. As a result, we write the following,

Z3 r
S= 2πf (x) 1+ f 0 (x)2 dx
0
Z3 p
= 2π · 2 · (1 + 0)dx
0

Z3
= 4πdx
0
Z3
= 4π · dx
0
i3
= 4π · x
0
= 4π · 3
= 12π

Therefore, the surface area is 12π.

√
I Example 8.6.2. If the graph of f (x) = x is rotated about the x-axis over the
interval [0, 2], then what is the area of the surface of this revolution?
√ 1
Solution. Since f (x) = x, therefore f 0 (x) = √ . As a result, we write the
2 x
following,
Z2
s 2
√

1
S= 2π x 1+ √ dx
2 x
0
Z2
s 2
√

1
= 2π x· 1+ √ dx
2 x
0
Z2 r
√ 1
= 2π x· 1+ dx
4x
0
Z2 r
√ 4x + 1
= 2π x· dx
4x
0
Z2 √
√ 4x + 1
= 2π x · √ dx
2x
0
Z2 √
4x + 1
= 2π dx
2
0
Z2
1 √
= 2π 4x + 1dx
2
0

Z2
(1)
1
=π· (4x + 1) 2 dx
0

Z2
Now, in order to solve the integral π · (4x + 1) 2 dx, we use the substitution
1

0
rule (that is, the u-substitution). For that, let u = 4x + 1, then
du d
= (4x + 1)
dx dx
=4
du
∴ dx =
4
Now, for the limit of integration, we do the following,
For x=0 =⇒ u=4·0+1=1
For x=2 =⇒ u=4·2+1=9
du
Now, substituting u = 4x + 1, dx = , and the limits of integration into the
4
equation (1), we get the following,
Z2 Z9
1 1 du
π· (4x + 1) dx = π ·2 u2
4
0 1
Z9
π 1
= · u 2 du
4
1
9
3
π u  2
= ·
4 3


2 1 
3 3
π 9 1  2 2
= · − 
4 3 3
2 2
π 52
= ·
4 3
13π
=
3
13π
Therefore, the surface area is .
3

The End of Section 8.6

The Area of a Surface of Revolution

§ 8.7 Work
In everyday language, we often use the word “work” to mean any kind of task
or activity we do, whether it involves any sort of movement. For example, when
we say, “I did a lot of work today” we typically emphasize the fact that a lot of
energy or effort has been used to get something done.

However, this common understanding of work doesn’t align with the scientific
definition of work. For example, simply holding a heavy object in place requires
energy and effort, but in the scientific sense, no work is done if there is no
movement. Mathematically, work is defined as a measurable quantity that only
occurs when a force F moves an object in the direction of that force. That is,
work is defined as follows.

W =F ·d (1)

where, W = work, F = force, and d = displacement.

In defining work, the equation above only works when the force remains constant
throughout its entire application. But in reality, the application of force often
changes. For example, imagine we’re pushing an object along a path, and the
force we exert changes from point to point as we move along. Let’s say we’re
moving the object from x = a to x = b along the x-axis by a continuously
varying force F (x), as shown in the figure below.

f(x)

x
a b

Figure 8.27

Now the question is how much work is done in moving that object based on
the definition of work we have seen so far? Equation (1) cannot be directly

applied to this problem because it assumes a constant force. However, by the

help of integration, we can adapt Equation (1) to solve the problem even when
the force is not constant. The idea lies in the realization that the area of the
region bounded by the graph of y = F (x) and the x-axis over the interval [a, b]
represents the work done by the force F (x) over that distance, as shown below.

f(x)

x
a b

Figure 8.28

Now, if we calculate this area under the graph, we can determine the work done
by the force F (x) over that interval. To calculate the area under the graph, we
divide the interval [a, b] into n sub-intervals that are sufficiently small such that
the force does not change much, that is, remains nearly constant within each
sub-divided sub-interval. This approach allows us to treat the force as constant
over each sub-interval, making it easier to approximate the work done on each
sub-interval using the formula for work. By adding together these small approx-
imations for each sub-interval, we create a Riemann sum. This Riemann sum
gives us an approximation of the total work W done over the entire interval [a, b].
Finally, we take the limit of these Riemann sums as the number of sub-intervals
increases without a bound, and thus obtain the actual work W done in moving
the object.

In order to do this, let us divide the interval [a, b] into n sub-interval of equal
b−a
width ∆x = by using points a = x0 , x1 , x2 , · · · , xn−1 , xn = b between
n
x = a and x = b. Let x∗i be any point in the ith sub-interval, for i = 1, 2, 3, · · · , n.
Since, the interval [a, b] is divided into sub-intervals in a way that the force
within each sub-interval remains approximately constant, therefore the force in
the ith sub-interval is roughly constant with a value F (x∗i ). Thus, the work done
in moving the object across the ith sub-interval is approximately F (x∗i ) · ∆x,

according to Equation (1). Similarly, summing the work done over every sub-
interval, we get an approximation of the total work W done over the interval
[a, b], that is,
Xn
W ≈ F (x∗i )∆x
i=1

n Zb
Now, if we let the number of sub-interval n → ∞, then
X
F (x∗i )∆x → F (x)dx =
i=1 a
W . Thus, taking the limit as n → ∞, we get
n
W = lim
X
F (x∗i )∆x
n→∞
i=1
Zb
= F (x)dx
a

As a result, we have the following definition of work done by a variable force.

Definition 8.7.1. (Work Done by a Variable Force) If an object is moved by

a varying force F from x = a to x = b in the direction of the x-axis, then
the work W done in moving the object is
n
W = lim
X
F (x∗i )∆x
n→∞
i=1
Zb
= F (x)dx
a

A few examples to demonstrate the definition above are shown below.

I Example 8.7.1. Find the work done by a force F (x) = ex applied over the
distance from x = 0 to x = 1.

Solution. Using the definition of work, we write the following,

Z1 i1
ex dx = ex
0
0

= e1 − e0
≈ 1.718

Therefore, the work done by a force F (x) = ex applied over the distance from
x = 0 to x = 1 is approximately 1.718 units.

N
I Example 8.7.2. A spring has a spring constant k = 200 m . How much work
is required to stretch the spring from its natural length to a displacement of
x = 0.1m?

Solution. To calculate the work done in stretching the spring, we integrate the
force F (x) over the displacement of 0.1m. That is,
Z0.1
W = F (x)dx
0

From Hooke’s Law, we know that

F (x) = kx
N
Since, k = 200 m , therefore,

F (x) = kx
= 200x

This implies that

Z0.1
W = F (x)dx
0
Z0.1
= 200xdx
0
Z0.1
= 200 · xdx
0
0.1
x2
= 200 ·
2 0
0.12
= 200 ·
2
= 1J

Therefore, the work required to stretch the spring is 1J.

The End of Section 8.7

Work

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Index

Absolute extreme values, 224 Continuous Functions, 101

Absolute maximum and minimum, Critical Points, 228
224 Cylindrical Shells, 481
Algebraic Manipulation, 44
Definite Integral, 323
Analytic Evaluation of Limit of a
Definition
Function, 34
of Arc Length of a Function, 489
Antiderivative, 350
of Concavity, 246
Antidifferentiation, 349
of the Absolute Extreme Values,
Arc Length, 487
224
Area Approximation, 304
of the Antiderivative, 351
Area of a Region Between Two
of the Area of a Region Between
Curves, 464
Two Curves, 465
Area of a Surface of Revolution, 491
of the Area of a Surface of
Area Under a Curve, 297
Revolution, 498
Average Value of a Function, 459
of the Area Under a Curve, 323
Chain Rule, 172 of the Average Value of a
Concavity and the Second Function, 461
Derivative Test, 245 of the Slope of a Tangent Line,
Constant Multiple Rule, 145 124
Constant Rule, 144 Derivative of a Function, 129
Continuity, 94, 136 of Continuity at a Point, 97
at a Point, 94 of Continuity at Endpoints, 98
of Combined Functions, 100 of Continuity over an Closed
of Composite Functions, 102 Interval, 99

507

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of Differentiability, 137 of Vertical Asymptote, 71
of Differentials, 270 of Work Done by a Variable
of Discontinuous Integrands, 448 Force, 503
of Indefinite Integral, 357 Definition of the Area Under a
of Infinite Intervals, 445 Curve, 321
of Infinite Limits, 69 Definition of the Volume of a Solid,
of Infinite Limits at Infinity, 76 470
of Linearization, 266 Derivation of Derivatives, 143
of One-sided Infinite Limits, 69 Difference Quotient, 116
of One-sided Limits, 31 Difference Rule, 152
of the Critical Point, 231 Differentiability, 136
of the Definite Integral, 328 Differential Calculus, 115
of the Definition of the Volume Differentials, 267
of a Solid, 474 Differentiation
of the Derivative of a Function, of Composite Functions, 172
128 of the Cosecant Function, 171
of the Epsilon - Delta Definition of the Cosine Function, 164
of Limit, 108 of the Cotangent Function, 167
of the Formal Definition of of the Exponential Function
Limit, 108 with Base a, 192
of the Exponential Function
of the Horizontal Asymptote, 74
with Base e, 188
of the Instantaneous Rate of
of the Natural Logarithm, 195
Change, 124, 125
of the Secant Function, 169
of the Left-Endpoint
of the Sine Function, 161
Approximation, 313
of the Tangent Function, 165
of the Limit at Infinity, 74
of the Common Logarithm, 198
of the Midpoint Rule, 342
Differentiation Rules
of the Net Signed Area, 327
for Exponential Functions, 188
of the Relative Extreme Values,
for Logarithmic Functions, 195
228
for Trigonometric Functions, 161
of the Riemann Sum, 322
Direct Substitution, 37
of the Right-Endpoint
Discontinuous Integrands, 446
Approximation, 314
Disk Method, 474
of the Slope of a Tangent Line,
Dividing Out Technique, 46
125
of the Tangent Line, 126 Elementary Rules of Differentiation,
of the Trapezoidal Rule, 345 143

508

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End Behaviors, 75 involving a Sum of Functions,
Epsilon - Delta Definition of Limit, 334
104 involving Absolute Values, 336
Existence of a Limit, 32 involving Powers of Sine and
Extreme Value Theorem, 226 Cosine Function, 404
Extreme Values of a Function, 223 involving Powers of Tangent and
Secant Function, 409
First Derivative Test, 241
involving Product of Powers of
Formal Definition of Limit, 104
Sine and Cosine Function,
Fundamental Theorem of Calculus,
407
364
involving Product of Powers of
General Properties of Limit of a Tangent and Secant
Function, 34 Function, 411
Generalized Area Approximation, over Multiple Sub-intervals, 336
310 Integration
Generalized Area Formula for a by Parts, 394
Surface of Revolution, 495 by Parts for Definite Integrals,
Higher-Order Derivatives, 139 400
by Reduction Formulas, 402
Identical Limits of Integration, 332
by Substitution, 378
Implicit Differentiation, 182
Intermediate Value Theorem, 102
Improper Integrals, 443
Inverse Relationship between
Increasing and Decreasing
Differentiation and
Functions, 236
Integration, 370
Test, 239
Increasing, Decreasing, and L’Hôpital’s Rule, 277
Constant Functions, 237 Left-Endpoint Approximation, 312
Indefinite integral, 356 Leibniz Notation, 134
Infinite LIATE Rule, 397
Intervals, 443 Limit, 24
Limits, 69 at Infinity, 73
Limits at Infinity, 75 involving Infty, 68
Inflection Points, 248 of a Composite Function, 42
Informal Definition of Limit, 29 of a Function, 29
Integrals of a Reciprocal, 39
involving a Constant, 335 of Functions Involving Radicals
involving a Difference of at Infinity, 88
Functions, 334 of Polynomial Functions, 37

509

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of Polynomial Functions at Relative (Local) Extreme Values,
Infinity, 78 227
of Power functions at Infinity, 77 Reversal of Limits of Integration,
of Rational Functions, 37 332
of Rational Functions at Riemann Sum, 321
Infinity, 80 Right-Endpoint Approximation, 313
of Transcendental Functions, 40 Rolle’s Theorem, 250
of Transcendental Functions at Rolle’s Theorem and The Mean
Infinity, 91 Value Theorem, 250
Linearization, 263
Linearization and Differentials, 262 Sandwich Theorem, 52
Location of Absolute (Global) Second Derivative Test, 250
Extreme Values, 233 Slicing Method, 470
Logarithmic Differentiation, 200 Squeeze Theorem, 52
Substitution Rule
Mean Value Theorem, 253 for Definite Integrals, 391
for Integrals, 462 Summation Notation, 298
Midpoint Rule, 340 Summation Rule, 150

Net Signed Area , 324 Table of Integrals, 450

Newton’s Method, 271 Tangent Line Problem, 117
Notation for Derivatives, 134 Theorem
Numerical Integration, 340 of Concavity Test, 247
of Cylindrical Shells, 484
One-Sided Limits, 30 of Differentiability Implies
Continuity, 137
Partial Fractions, 431
of Existence of a Limit, 32
Pinching Theorem, 52
of Family of Antiderivatives, 352
Point of Inflection, 247
of General Power Rule, 201
Power Rule, 147
of Identical Limits of
Product Rule, 153
Integration, 333
Properties
of Integrability of Continuous
of Continuity, 100
Functions, 329
of the Definite Integral, 332
of Integrals Involving a
Quotient Rule, 158 Constant, 335
of Integrals Involving a
Rationalizing Technique, 48 Difference of Functions, 334
Related Rates, 257 of Integrals Involving a Sum of

510

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Functions, 334 of the Limit of Polynomial
of Integrals Involving Absolute Functions, 38
Values, 338 of the Limit of Transcendental
of Integrals over Multiple Functions, 41
Sub-intervals, 336 of the Power Rule, 148
of Integration by Parts, 395 of the Product Rule, 156
of Integration by Substitution, of the Quotient Rule, 159
379 of the Second Derivative Test,
of Not Continuous Implies Not 250
Differentiable, 138 of the Summation Rule, 150
of Reversal of Limits of of the Washer Method, 480
Integration, 333 Extreme Value Theorem, 226
of the Constant Multiple Rule, Fundamental Theorem of
146 Calculus, Part I, 367
of the Derivative of the Fundamental Theorem of
Cosecant Function, 171 Calculus, Part II, 369
of the Derivative of the Cosine Intermediate Value Theorem,
Function, 164 103
of the Derivative of the Mean Value Theorem, 254
Cotangent Function, 168 Mean Value Theorem For
of the Derivative of the Integrals, 462
Exponential Function with of Continuity of Combined
Base a, 193 Functions, 101
of the Derivative of the of Limit Laws, 34
Exponential Function with of the Chain Rule, 175
Base e, 191 of the Constant Rule, 144
of the Derivative of the Natural of the Derivative of Identity
Logarithm, 196 Function, 147
of the Derivative of the Sine of the Derivative of the
Function, 162 Common Logarithm, 198
of the Difference Rule, 152 of the Derivative of the Secant
of the Disk Method, 476 Function, 169
of the First Derivative Test, 242 of the Derivative of the Tangent
of the Inflection Points Test, 249 Function, 166
of the Limit of a Composite of the Limit of Rational
Function, 42 Functions, 38
of the Limit of a Reciprocal, 40 Relative/Local Extreme Value

511

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Theorem, 229 Integrals, 403
Rolle’s Theorem, 251 Manipulation, 60
Squeeze Theorem, 54 Substitution, 412
Trapezoidal Rule, 343
Travel, 23 Washer Method, 478
Trigonometric Work, 501

512

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