Integration
Lyndon Booc Decasa
August 6, 2018
Lyndon Booc Decasa PSHS
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Lyndon Booc Decasa PSHS
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Lyndon Booc Decasa PSHS
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Lyndon Booc Decasa PSHS
Integration
Outline of my discussions
Lyndon Booc Decasa PSHS
Integration
Outline of my discussions
1 Overview of the problem of finding areas.
Lyndon Booc Decasa PSHS
Integration
Outline of my discussions
1 Overview of the problem of finding areas.
2 Following this overview, we will discuss the Fundamental
Theorem of Calculus which is the theorem that relates the
problems of finding tangent lines and areas and we will discuss
techniques for calculating areas.
Lyndon Booc Decasa PSHS
Integration
Outline of my discussions
1 Overview of the problem of finding areas.
2 Following this overview, we will discuss the Fundamental
Theorem of Calculus which is the theorem that relates the
problems of finding tangent lines and areas and we will discuss
techniques for calculating areas.
3 Lastly, we will define and discuss integration and its
importance in real-life situation.
Lyndon Booc Decasa PSHS
AN OVERVIEW OF THE AREA PROBLEM
Introductory
In this section we will consider the problem of calculating areas of
the plane regions with curvilinear boundaries. Our purpose here
is simply to introduce and motivate the fundamental concepts.
Lyndon Booc Decasa PSHS
AN OVERVIEW OF THE AREA PROBLEM
Introductory
In this section we will consider the problem of calculating areas of
the plane regions with curvilinear boundaries. Our purpose here
is simply to introduce and motivate the fundamental concepts.
The Area Problem
1 Formulas for the areas of polygons: square, rectangle, triangle
and others
Lyndon Booc Decasa PSHS
AN OVERVIEW OF THE AREA PROBLEM
Introductory
In this section we will consider the problem of calculating areas of
the plane regions with curvilinear boundaries. Our purpose here
is simply to introduce and motivate the fundamental concepts.
The Area Problem
1 Formulas for the areas of polygons: square, rectangle, triangle
and others
2 However, the problem of finding formulas for regions with
curves boundaries caused difficulties for early mathematicians.
Lyndon Booc Decasa PSHS
AN OVERVIEW OF THE AREA PROBLEM
Introductory
In this section we will consider the problem of calculating areas of
the plane regions with curvilinear boundaries. Our purpose here
is simply to introduce and motivate the fundamental concepts.
The Area Problem
1 Formulas for the areas of polygons: square, rectangle, triangle
and others
2 However, the problem of finding formulas for regions with
curves boundaries caused difficulties for early mathematicians.
Lyndon Booc Decasa PSHS
Archimedes
Greek mathematician Archimedes made the first real progress in
dealing with the general area problem - using an ingenious
procedure
Lyndon Booc Decasa PSHS
Ingenious Procedure
Legend: A(n) denote the area of a regular n-sided polygon
inscribed in a circle of radius 1.
1.png
Therefore,
lim A(n) = π.
n→∞
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1 As a result, computation of area using the ingenious
procedure was a very curbersome procedure.
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It remained to Isaac Newton and Gottfried Leibniz
to obtain a general method of finding areas
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THE AREA PROBLEM
Lyndon Booc Decasa PSHS
THE AREA PROBLEM
We will discuss their method in the context of the following
problem.
Lyndon Booc Decasa PSHS
THE AREA PROBLEM
We will discuss their method in the context of the following
problem.
Problem
Given a function f that is continuous and nonnegative on an
interval [a, b], find the area between the graph of f and the
interval [a, b] on the x-axis.
Lyndon Booc Decasa PSHS
THE AREA PROBLEM
We will discuss their method in the context of the following
problem.
Problem
Given a function f that is continuous and nonnegative on an
interval [a, b], find the area between the graph of f and the
interval [a, b] on the x-axis.
Methods for Finding Areas
1 Rectangle Method
2 Antiderivative Method
Lyndon Booc Decasa PSHS
THE AREA PROBLEM
We will discuss their method in the context of the following
problem.
Problem
Given a function f that is continuous and nonnegative on an
interval [a, b], find the area between the graph of f and the
interval [a, b] on the x-axis.
Methods for Finding Areas
1 Rectangle Method
2 Antiderivative Method
Lyndon Booc Decasa PSHS
The Rectangle Method for Finding Areas
Lyndon Booc Decasa PSHS
The Rectangle Method for Finding Areas
1 If A denotes the exact area under the curve and An denotes
the approximation to A using n rectangles, then
lim An = A
n→∞
Lyndon Booc Decasa PSHS
The Rectangle Method for Finding Areas
1 If A denotes the exact area under the curve and An denotes
the approximation to A using n rectangles, then
lim An = A
n→∞
We will call this the rectangle method for computing A.
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The Antiderivative Method for Finding Areas
Lyndon Booc Decasa PSHS
The Antiderivative Method for Finding Areas
Although the rectangle method is appealing intuitively,
Lyndon Booc Decasa PSHS
The Antiderivative Method for Finding Areas
Although the rectangle method is appealing intuitively, the limits
that result can only be evaluated in certain cases.
Lyndon Booc Decasa PSHS
The Antiderivative Method for Finding Areas
Although the rectangle method is appealing intuitively, the limits
that result can only be evaluated in certain cases. For this reason,
progress on the area problem remained at a rudimentary level
until the latter part of the seventeenth century when Isaac Newton
and Gottfield Leibniz independently discovered a fundamental
relationship between areas and derivatives.
Lyndon Booc Decasa PSHS
The Antiderivative Method for Finding Areas
Although the rectangle method is appealing intuitively, the limits
that result can only be evaluated in certain cases. For this reason,
progress on the area problem remained at a rudimentary level
until the latter part of the seventeenth century when Isaac Newton
and Gottfield Leibniz independently discovered a fundamental
relationship between areas and derivatives.
Relationship Between Area and Derivatives
If f is a nonnegative continuous function on the interval [a, b], and
if A(x) denotes the area under the graph of f over the interval
[a, x], where x is any point in the interval [a, b], then
0
A (x) = f (x).
Lyndon Booc Decasa PSHS
Illustration
For each of the function f , find the area A(x) between the graph
0
of f and the interval [a, x] = [−1, x], and find the derivative A (x)
of this area function.
1 f (x) = 2
2 f (x) = x + 1
Lyndon Booc Decasa PSHS
Illustration
For each of the function f , find the area A(x) between the graph
0
of f and the interval [a, x] = [−1, x], and find the derivative A (x)
of this area function.
1 f (x) = 2
2 f (x) = x + 1
Note
Equation
0
A (x) = f (x)
is very important because it relates the area function A and the
region-bounding function f .
Lyndon Booc Decasa PSHS
Illustration
For each of the function f , find the area A(x) between the graph
0
of f and the interval [a, x] = [−1, x], and find the derivative A (x)
of this area function.
1 f (x) = 2
2 f (x) = x + 1
Note
Equation
0
A (x) = f (x)
is very important because it relates the area function A and the
region-bounding function f . Although a formula for A(x) may be
difficult to obtain directly, its derivative f (x), is given.
Lyndon Booc Decasa PSHS
Illustration
For each of the function f , find the area A(x) between the graph
0
of f and the interval [a, x] = [−1, x], and find the derivative A (x)
of this area function.
1 f (x) = 2
2 f (x) = x + 1
Note
Equation
0
A (x) = f (x)
is very important because it relates the area function A and the
region-bounding function f . Although a formula for A(x) may be
difficult to obtain directly, its derivative f (x), is given. But if a
formula for A(x) can be recovered from the given formula for
0
A (x),
Lyndon Booc Decasa PSHS
Illustration
For each of the function f , find the area A(x) between the graph
0
of f and the interval [a, x] = [−1, x], and find the derivative A (x)
of this area function.
1 f (x) = 2
2 f (x) = x + 1
Note
Equation
0
A (x) = f (x)
is very important because it relates the area function A and the
region-bounding function f . Although a formula for A(x) may be
difficult to obtain directly, its derivative f (x), is given. But if a
formula for A(x) can be recovered from the given formula for
0
A (x), then the area under the graph of f over the interval [a, b]
can be obtained by computing A(b).
Lyndon Booc Decasa PSHS
Illustration
The process of finding a function from its derivative is called
antidifferentiation,
Lyndon Booc Decasa PSHS
Illustration
The process of finding a function from its derivative is called
antidifferentiation, and a procedure for finding areas via
antidifferentiation is called the antiderivative method.
Lyndon Booc Decasa PSHS
Illustration
The process of finding a function from its derivative is called
antidifferentiation, and a procedure for finding areas via
antidifferentiation is called the antiderivative method.
Example
Use the antiderivative method to find the area under the graph of
y = x over the interval [0, x].
Lyndon Booc Decasa PSHS
Antiderivatives
In the last discussion, we saw how antidifferentiation could be
used to find exact areas. In this discussion, we will develop some
fundamental results about antidifferentiation.
Lyndon Booc Decasa PSHS
Antiderivatives
In the last discussion, we saw how antidifferentiation could be
used to find exact areas. In this discussion, we will develop some
fundamental results about antidifferentiation.
Definition
A function F is called an antiderivative of a function f on a given
open interval if
0
F (x) = f (x)
for all x in the interval.
Lyndon Booc Decasa PSHS
Antiderivatives
In the last discussion, we saw how antidifferentiation could be
used to find exact areas. In this discussion, we will develop some
fundamental results about antidifferentiation.
Definition
A function F is called an antiderivative of a function f on a given
open interval if
0
F (x) = f (x)
for all x in the interval.
Theorem
If F (x) is any antiderivative of f (x) on an open interval, then for
any constant C the function F (x) + C is also antiderivative on
that interval. Moreover, each antiderivative of f (x) on the interval
can be expressed in the form F (x) + C by choosing the constant C
appropriately.
Lyndon Booc Decasa PSHS
Antiderivatives
In the last discussion, we saw how antidifferentiation could be
used to find exact areas. In this discussion, we will develop some
fundamental results about antidifferentiation.
Definition
A function F is called an antiderivative of a function f on a given
open interval if
0
F (x) = f (x)
for all x in the interval.
Theorem
If F (x) is any antiderivative of f (x) on an open interval, then for
any constant C the function F (x) + C is also antiderivative on
that interval. Moreover, each antiderivative of f (x) on the interval
can be expressed in the form F (x) + C by choosing the constant C
appropriately.
Lyndon Booc Decasa PSHS
The Indefinite Integral
Lyndon Booc Decasa PSHS
The Indefinite Integral
What is the meaning of integration?
Lyndon Booc Decasa PSHS
The Indefinite Integral
What is the meaning of integration?
Integration or antidifferentiation is a process of finding all
antiderivatives.
Lyndon Booc Decasa PSHS
The Indefinite Integral
What is the meaning of integration?
Integration or antidifferentiation is a process of finding all
antiderivatives.
Lyndon Booc Decasa PSHS
The Indefinite Integral
What is the meaning of integration?
Integration or antidifferentiation is a process of finding all
antiderivatives.
If
d
[F (x)] = f (x)
dx
Lyndon Booc Decasa PSHS
The Indefinite Integral
What is the meaning of integration?
Integration or antidifferentiation is a process of finding all
antiderivatives.
If
d
[F (x)] = f (x)
dx
then integrating or antidifferentiating the function f (x)
produces an antiderivative of the form F (x) + C , where C is a
constant.
Lyndon Booc Decasa PSHS
The Indefinite Integral
What is the meaning of integration?
Integration or antidifferentiation is a process of finding all
antiderivatives.
If
d
[F (x)] = f (x)
dx
then integrating or antidifferentiating the function f (x)
produces an antiderivative of the form F (x) + C , where C is a
constant.
Integral notation
To emphasize this process, we have the symbol
Z
f (x)dx = F (x) + C
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Area Under a Curve
Definition
If the function f is continuous on [a, b] and if f (x) ≥ 0 for all
x ∈ [a, b], then the AREA under the curve y = f (x) over the
interval [a, b] is defined by
n
X
A = lim f (xk∗ ) M x.
n→∞
k=1
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Integrable
Definition
A function f is said to be integrable on a finite closed interval
[a, b] if the limit
X n
lim f (xk∗ ) M xk
maxMxk →0
k=1
exists and does not depend on the choice of partitions or on the
choice of the points xk∗ in the subintervals. When this is the case
we denote the limit by the symbol
Z b n
X
f (x)dx = lim f (xk∗ ) M xk
a maxMxk →0
k=1
which is called the definite integral of f from a to b. The
numbers a and b are called the lower and upper limits of
integration, and f (x) is called the integrand.
Lyndon Booc Decasa PSHS
Fundamental Theorem of Calculus
Theorem
If f is continuous on [a, b] and F is any antiderivative of f on
[a, b], then
Z b
f (x)dx = F (b) − F (a).
a
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Theorems on Antidifferentiation
Theorem
Z
1 dx = x + C
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Theorems on Antidifferentiation
Theorem
Z
1 dx = x + C
2 If a is any constant, then
Z Z
af (x)dx = a f (x)dx.
Lyndon Booc Decasa PSHS
Theorems on Antidifferentiation
Theorem
Z
1 dx = x + C
2 If a is any constant, then
Z Z
af (x)dx = a f (x)dx.
3 If f and g are defined on the same interval, then
Z Z Z
(f (x) ± g (x))dx = f (x)dx ± g (x)dx.
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Theorems on Antidifferentiation
Theorem
Z
1 dx = x + C
2 If a is any constant, then
Z Z
af (x)dx = a f (x)dx.
3 If f and g are defined on the same interval, then
Z Z Z
(f (x) ± g (x))dx = f (x)dx ± g (x)dx.
4 If n is any rational number and n 6= −1, then
x n+1
Z
x n dx = + C.
n+1
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Theorems on Antidifferentiation
Theorem
Z
1 dx = x + C
2 If a is any constant, then
Z Z
af (x)dx = a f (x)dx.
3 If f and g are defined on the same interval, then
Z Z Z
(f (x) ± g (x))dx = f (x)dx ± g (x)dx.
4 If n is any rational number and n 6= −1, then
x n+1
Z
x n dx = + C.
n+1
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Antiderivatives of Trigonometric Functions
Theorem
Z
1 sin xdx = − cos x + C
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Antiderivatives of Trigonometric Functions
Theorem
Z
1 sin xdx = − cos x + C
Z
2 cos xdx = sin x + C
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Antiderivatives of Trigonometric Functions
Theorem
Z
1 sin xdx = − cos x + C
Z
2 cos xdx = sin x + C
Z
3 sec2 xdx = tan x + C
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Antiderivatives of Trigonometric Functions
Theorem
Z
1 sin xdx = − cos x + C
Z
2 cos xdx = sin x + C
Z
3 sec2 xdx = tan x + C
Z
4 csc2 xdx = − cot x + C
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Antiderivatives of Trigonometric Functions
Theorem
Z
1 sin xdx = − cos x + C
Z
2 cos xdx = sin x + C
Z
3 sec2 xdx = tan x + C
Z
4 csc2 xdx = − cot x + C
Z
5 sec x tan xdx = sec x + C
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Antiderivatives of Trigonometric Functions
Theorem
Z
1 sin xdx = − cos x + C
Z
2 cos xdx = sin x + C
Z
3 sec2 xdx = tan x + C
Z
4 csc2 xdx = − cot x + C
Z
5 sec x tan xdx = sec x + C
Z
6 csc x cot xdx = − csc x + C
Lyndon Booc Decasa PSHS
Antiderivatives of Trigonometric Functions
Theorem
Z
1 sin xdx = − cos x + C
Z
2 cos xdx = sin x + C
Z
3 sec2 xdx = tan x + C
Z
4 csc2 xdx = − cot x + C
Z
5 sec x tan xdx = sec x + C
Z
6 csc x cot xdx = − csc x + C
Lyndon Booc Decasa PSHS
EXAMPLES
Evaluate the following
Z
1 (sin x + cos x)dx
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EXAMPLES
Evaluate the following
Z
1 (sin x + cos x)dx
Z
2 tan2 xdx
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EXAMPLES
Evaluate the following
Z
1 (sin x + cos x)dx
Z
2 tan2 xdx
Z
cos x
3 dx
sin2 x
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EXAMPLES
Evaluate the following
Z
1 (sin x + cos x)dx
Z
2 tan2 xdx
Z
cos x
3
2
dx
Z sin x
dx
4 dx
1 + sin x
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EXAMPLES
Evaluate the following
Z
1 (sin x + cos x)dx
Z
2 tan2 xdx
Z
cos x
3
2
dx
Z sin x
dx
4 dx
1 + sin x
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Integration by Substitution
Substitution Rule
If u = g (x) is a differentiable function whose range is an interval I
and f is continuous on I , then
Z Z
0
f (g (x)) · g (x)dx = f (u)du.
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EXAMPLES
Evaluate the following
√
Z
1 1 − 4xdx
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EXAMPLES
Evaluate the following
√
Z
1 1 − 4xdx
Z
10
2 x 2 x 3 − 1 dx
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EXAMPLES
Evaluate the following
√
Z
1 1 − 4xdx
Z
10
2 x 2 x 3 − 1 dx
Z
x
3 dx
(x 2 + 1)3
Lyndon Booc Decasa PSHS
EXAMPLES
Evaluate the following
√
Z
1 1 − 4xdx
Z
10
2 x 2 x 3 − 1 dx
Z
x
3 dx
(x 2 + 1)3
t3
Z
4 √ dt
t2 + 3
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EXAMPLES
Evaluate the following
√
Z
1 1 − 4xdx
Z
10
2 x 2 x 3 − 1 dx
Z
x
3 dx
(x 2 + 1)3
t3
Z
4 √ dt
t 2+3
√
Z
5 t t − 1dt
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EXAMPLES
Evaluate the following
√
Z
1 1 − 4xdx
Z
10
2 x 2 x 3 − 1 dx
Z
x
3 dx
(x 2 + 1)3
t3
Z
4 √ dt
t 2+3
√
Z
5 t t − 1dt
Z
6 sec3 x tan xdx
Lyndon Booc Decasa PSHS
EXAMPLES
Evaluate the following
√
Z
1 1 − 4xdx
Z
10
2 x 2 x 3 − 1 dx
Z
x
3 dx
(x 2 + 1)3
t3
Z
4 √ dt
t 2+3
√
Z
5 t t − 1dt
Z
6 sec3 x tan xdx
Lyndon Booc Decasa PSHS
APPLICATIONS
Particular Antiderivatives
0
1 Given that F (x) = 2x and F (2) = 6, find F (x).
2 The slope of the tangent line at any point (x, y ) on a curve is
√
given by 3 x. Find the equation of a curve if the point (9, 4)
is on the curve.
Lyndon Booc Decasa PSHS
APPLICATIONS
Particular Antiderivatives
0
1 Given that F (x) = 2x and F (2) = 6, find F (x).
2 The slope of the tangent line at any point (x, y ) on a curve is
√
given by 3 x. Find the equation of a curve if the point (9, 4)
is on the curve.
Lyndon Booc Decasa PSHS
APPLICATIONS
Rectilinear Motion Revisited
Suppose that a particle is traveling along a strait line and s(t),
v (t) and a(t) are equations of motion, velocity and acceleration,
respectively, of the particle.
Lyndon Booc Decasa PSHS
APPLICATIONS
Rectilinear Motion Revisited
Suppose that a particle is traveling along a strait line and s(t),
v (t) and a(t) are equations of motion, velocity and acceleration,
respectively, of the particle. Recall that
0 0
v (t) = s (t) and a(t) = v (t)
Lyndon Booc Decasa PSHS
APPLICATIONS
Rectilinear Motion Revisited
Suppose that a particle is traveling along a strait line and s(t),
v (t) and a(t) are equations of motion, velocity and acceleration,
respectively, of the particle. Recall that
0 0
v (t) = s (t) and a(t) = v (t)
1 Suppose a particle moves along the x-axis under the given
pieces of information. Find the position function x(t) of the
particle in each of the following:
velocity v (t) = 3t 2 − 2t with initial position x(0) = 1.
acceleration a(t) = t − 2; x(0) = 1 and v (0) = 3.
acceleration a(t) = 5 sin(5t); initial velocity v (0) = 5 and
initial position x(0) = 2
2 A heavy projectile is fired straight up from a platform 3
m
meters above the ground with an initial velocity of 160 .
s
m
Find an equation of motion of the particle. Use −10 2 for
s
acceleration due to gravity.
Lyndon Booc Decasa PSHS
APPLICATIONS
Rectilinear Motion Revisited
Suppose that a particle is traveling along a strait line and s(t),
v (t) and a(t) are equations of motion, velocity and acceleration,
respectively, of the particle. Recall that
0 0
v (t) = s (t) and a(t) = v (t)
1 Suppose a particle moves along the x-axis under the given
pieces of information. Find the position function x(t) of the
particle in each of the following:
velocity v (t) = 3t 2 − 2t with initial position x(0) = 1.
acceleration a(t) = t − 2; x(0) = 1 and v (0) = 3.
acceleration a(t) = 5 sin(5t); initial velocity v (0) = 5 and
initial position x(0) = 2
2 A heavy projectile is fired straight up from a platform 3
m
meters above the ground with an initial velocity of 160 .
s
m
Find an equation of motion of the particle. Use −10 2 for
s
acceleration due to gravity.
Lyndon Booc Decasa PSHS
Integrals Yielding Logarithmic and Exponential Functions
1
Integrals of f (x) = and of the other Circular Functions
x
1
1 Recall: Dx (ln |x|) = , where x 6= 0.
x
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Integrals Yielding Logarithmic and Exponential Functions
1
Integrals of f (x) = and of the other Circular Functions
x
1
1 Recall: Dx (ln |x|) = , where x 6= 0.
Z x
1
2 du = ln |u| + C
u
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Integrals Yielding Logarithmic and Exponential Functions
1
Integrals of f (x) = and of the other Circular Functions
x
1
1 Recall: Dx (ln |x|) = , where x 6= 0.
Z x
1
2 du = ln |u| + C
u
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Examples
Evaluate the following integrals.
Z
dx
1
3x + 2
Z 4
xdx
2
4 − x2
Z3
dx
3
xlog2 x
2x 3 + 9x 2 − 3x + 2
Z
4 dx
2x − 1
x −1
Z
5
2
dx
x + 2x + 1
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