Introduction to Integration

Two main operations in calculus are taking a derivative and taking an integral. Ironically, these
operations can be thought of as being “inverses of each other”.
• When we differentiate, we find a rate of change at a particular point on a curve.
• When we integrate, we are attempting to find the area between the curve and the x-axis
in some particular interval [𝑎𝑎, 𝑏𝑏]

𝑏𝑏
Warning: ∫𝑎𝑎 𝑓𝑓(𝑥𝑥)𝑑𝑑𝑑𝑑 measures the difference in area in the interval [𝑎𝑎, 𝑏𝑏] and NOT the area
between the curve and the x-axis.

How did all of this come about?

Let us suppose that we are given a function 𝑓𝑓(𝑥𝑥) and we want to

find the area enclosed between the curve 𝑦𝑦 = 𝑓𝑓(𝑥𝑥), the x-axis, and
the lines 𝑥𝑥 = 𝑎𝑎 and 𝑥𝑥 = 𝑏𝑏.

In order to do this, mathematicians decided to break up the area

into subdivisions.

𝑏𝑏−𝑎𝑎
𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿ℎ 𝑜𝑜𝑜𝑜 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 = # 𝑜𝑜𝑜𝑜 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑒𝑒𝑒𝑒

𝑏𝑏−𝑎𝑎
Δ𝑥𝑥 = 𝑛𝑛

The area of each section can’t be calculated due

curved top. In order to account for this, we can
approximate the area by crafting rectangles.
• If we take the smaller rectangles we get
the lower sum.
• If we take the larger rectangles we get
the upper sum. Lower Sum [17.44 𝑢𝑢2 ] Upper Sum [20.69 𝑢𝑢2 ]
Since the lower sum will always give us an area smaller than what we want and the upper sum will
give us an area always larger than what we want we get the following:

� 𝑎𝑎𝑎𝑎𝑎𝑎 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑠𝑠𝑠𝑠𝑠𝑠 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 ≤ 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝑈𝑈𝑈𝑈𝑈𝑈𝑈𝑈𝑈𝑈 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 ≤ � 𝑎𝑎𝑎𝑎𝑎𝑎 𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢 𝑠𝑠𝑠𝑠𝑠𝑠 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟

𝑛𝑛 𝑛𝑛

� 𝑓𝑓�𝑥𝑥𝑖𝑖# �Δ𝑥𝑥 ≤ 𝐴𝐴 ≤ � 𝑓𝑓(𝑥𝑥𝑖𝑖∗ )Δ𝑥𝑥 𝑤𝑤ℎ𝑒𝑒𝑒𝑒𝑒𝑒 𝑛𝑛 𝑟𝑟𝑟𝑟𝑟𝑟. # 𝑜𝑜𝑜𝑜 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑒𝑒𝑠𝑠 𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢.

𝑖𝑖=1 𝑖𝑖=1

Note: As the number of rectangles used increases, our approximations must get closer and
closer to the actual area.

Therefore, our area in the interval [𝑎𝑎, 𝑏𝑏] must be given by

𝑛𝑛

𝐴𝐴 = lim � 𝑓𝑓(𝑥𝑥𝑖𝑖 )Δ𝑥𝑥

𝑛𝑛→∞
𝑖𝑖=1

We will use a much simpler notation for this 

𝑏𝑏
𝐴𝐴 = � 𝑓𝑓(𝑥𝑥)𝑑𝑑𝑑𝑑
𝑎𝑎 18.33 𝑢𝑢2 . Actual is 19.1667 𝑢𝑢2

Note: the ∫ sign is an elongated ‘s’ and stands for ‘sum’, just as the ∑ did previously. However,
∫ means limit of a sum whereas ∑ means finite sum

A Crude Version of The Fundamental Theorem of Calculus

𝑑𝑑
�� 𝑓𝑓(𝑥𝑥)𝑑𝑑𝑑𝑑� = 𝑓𝑓(𝑥𝑥)
𝑑𝑑𝑑𝑑

This means, if I take the derivative of an integral, both operations cancel each other out and I
get back the original function.
• Note: The opposite is also true, if I integrate a derivative, I get back 𝑓𝑓(𝑥𝑥)

𝑑𝑑
Moreover, if you want to solve for ∫ 𝑓𝑓(𝑥𝑥)𝑑𝑑𝑑𝑑 then that means you need to move the 𝑑𝑑𝑑𝑑 to the
other side. Therefore, we must “anti-differentiate 𝒇𝒇(𝒙𝒙)” to solve our integral.

Suppose that 𝐹𝐹(𝑥𝑥) is the anti-derivative of 𝑓𝑓(𝑥𝑥). This means, if you take the derivative of 𝐹𝐹(𝑥𝑥)
you get 𝑓𝑓(𝑥𝑥). If we anti-differentiate both sides of the above equation we end up getting
� 𝑓𝑓(𝑥𝑥)𝑑𝑑𝑑𝑑 = 𝐹𝐹(𝑥𝑥) + 𝑐𝑐 𝑤𝑤ℎ𝑒𝑒𝑒𝑒𝑒𝑒 𝑐𝑐 𝑖𝑖𝑖𝑖 𝑎𝑎𝑎𝑎 𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
• Note: 𝑓𝑓(𝑥𝑥) in the integral is called an integrand.
Consider the following

𝑑 3
[𝑥 ] = 3𝑥 2 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 � 3𝑥 2 𝑑𝑥 = 𝑥 3
𝑑𝑥

𝑑
[sin(𝑥)] = cos(𝑥 ) 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 � cos(𝑥 ) 𝑑𝑥 = sin(𝑥 )
𝑑𝑥

𝑑 𝑑 𝑑
Note: [𝑥 3 + 5] = 3𝑥 2 [𝑥 3 − 12] = 3𝑥 2 [𝑥 3 + 1337] = 3𝑥 2 … etc.
𝑑𝑥 𝑑𝑥 𝑑𝑥

• We cannot simply say ∫ 3𝑥 2 𝑑𝑥 = 𝑥 3 since we do not know if there was a constant after the
𝑥 3 . Moreover, it is impossible to determine what it is without additional information.

Therefore, must include a constant after we integrate!

Two important rules for derivatives are

• The Constant Multiple Rule: ∫ 𝑘𝑓 (𝑥 )𝑑𝑥 = 𝑘 ∫ 𝑓(𝑥)𝑑𝑥

• Sum or Difference Rule: ∫[𝑓(𝑥 ) ± 𝑔(𝑥 )]𝑑𝑥 = ∫ 𝑓(𝑥 )𝑑𝑥 ± ∫ 𝑔(𝑥 )𝑑𝑥

Ex) Find the following indefinite integrals

a) ∫ 𝑥𝑑𝑥 b) ∫ 𝑥 3 𝑑𝑥 c) ∫ 4𝑥 3 𝑑𝑥 d) ∫ 3𝑥 2 + 6𝑥 + 5 𝑑𝑥

3 3𝑥 3 𝑑𝑥
e) ∫ 𝑎𝑥 𝑛 𝑑𝑥 f) ∫ 10 √𝑥 2 𝑑𝑥 g) ∫
4𝑥 5
 5 𝑥 5 −3𝑥+7 𝑥 9 −10
h) ∫ 𝑑𝑥 i) ∫ 𝑑𝑥 j) ∫ 𝑑𝑥
𝑥 2𝑥 2 √𝑥

2 3
k) ∫ 𝑒 𝑥 𝑑𝑥 l) ∫ 3𝑒 2𝑥+5 𝑑𝑥 m) ∫ 5 sin(𝑥 ) 𝑑𝑥 n) ∫ cos(𝑥 ) 𝑑𝑥
5 4

o) ∫(𝑥 2 + 5)2 𝑑𝑥 p) ∫(𝑥 + 3)4 𝑑𝑥

 Overview of Today’s Lesson – Introduction to Integration
Today we talked about and learned
 How the integral finds the difference in area in an interval
[a, b]
 What the integral symbolism means
 The Fundamental theorem of calculus which states that the
integral is the inverse operation of the derivative
 How to take the integral of a polynomial
 How to take the integral of basic functions that have
𝑒 𝑥 , sin(𝑥) , 𝑜𝑟 cos(𝑥) in them.
o Much harder functions require integration rules which
are not taught in the course.
 Why you must put in the constant of integration when
solving an integral problem.
o To solve for this constant, you must be given
additional information (generally an initial condition)

Sample Questions

2
(𝑥 2 +3)
1. Integrate ∫ 𝑑𝑥
3𝑥 3

2 +5
2. A) Integrate ∫ 2𝑥 ∙ 𝑒 𝑥 𝑑𝑥 B) Integrate ∫ 2 sin(5𝑥) 𝑑𝑥
Integration Using Substitution
Sometimes when we wish to integrate a function, the function can be horrific!

𝑥 1+𝑥
Ex) ∫ 4𝑥 3 cos(𝑥 4 + 2) 𝑑𝑥 Ex) ∫ 𝑑𝑥 Ex) ∫ 𝑑𝑥
√1−4𝑥 2 1+𝑥 2

Fortunately, mathematicians have developed particular strategies for solving such integrals.
Such strategies include
• Substitution • Trigonometric substitution
• Integration by parts • Brute force table of integrals
• Partial fraction decomposition

Today, we will focus on using substitution. It is important to note that the above 3 integrals
can all be solved using substitution.

The Substitution Rule

If we have an integral of the form ∫ 𝑓�𝑔(𝑥 )�𝑔′ (𝑥 )𝑑𝑥 (see 1st example) then let 𝑢 = 𝑔(𝑥).
𝑑𝑢
If 𝑢 = 𝑔(𝑥 ) then = 𝑔′(𝑥) or 𝑔′ (𝑥)𝑑𝑥 = 𝑑𝑢 and therefore
𝑑𝑥

� 𝑓�𝑔(𝑥 )�𝑔′ (𝑥 )𝑑𝑥 = � 𝑓(𝑢)𝑑𝑢

Note: In some cases, the Substitution rule will be easy to spot (1st example) but in other cases
it might not be. One of the first steps in integration is to see if substitution will work since it’s
the easiest strategy to use.

Ex) Find the following indefinite integrals

a) ∫(2𝑥 + 1)4 𝑑𝑥 b) ∫ 2𝑥(𝑥 2 + 1)3 𝑑𝑥

 𝑥2 3𝑙𝑛𝑥
c) ∫ √𝑥 3 𝑑𝑥 d) ∫ 𝑥√𝑥 − 1 𝑑𝑥 e) ∫ 𝑑𝑥
−4 𝑥

3𝑥 2 +1 2+1
f) ∫(4 + 𝑠𝑖𝑛𝑥 )9 𝑐𝑜𝑠𝑥 𝑑𝑥 g) ∫ 3 +𝑥+5 𝑑𝑥 h) ∫ 𝑥𝑒 𝑥 𝑑𝑥
𝑥

𝑑𝑥 𝑐𝑜𝑠𝑥
i) ∫ 2 j) ∫ (5+𝑠𝑖𝑛𝑥)2 𝑑𝑥
√𝑥�1+√𝑥�

Challenge Questions:

𝑥3 cos3 (𝑥)
k) ∫ √𝑥 2 𝑑𝑥 l) ∫ 𝑑𝑥 m) ∫ cos3 (𝑥 )𝑑𝑥
+4 �sin(𝑥)
 Overview of Today’s Lesson – Integration Using Substitution
Today we talked about and learned
 How a substitution may be used in order to evaluate an
integral.
o Generally to use this method, you must see a
function and its derivative in the integral.

Sample Questions

12𝑥 3
1. Integrate ∫ 𝑑𝑥
3𝑥 4 +1

𝑒 𝑥 −𝑒 −𝑥
2. Integrate ∫ 𝑥 −𝑥 𝑑𝑥
𝑒 +𝑒

𝑥
3. Challenge: Integrate ∫ √𝑥+1 𝑑𝑥
The Area between Two Curves
So far, we have only considered areas bounded by a single curve and the x-axis. Sometimes it is useful
to be able to find areas bounded by two curves.

There are a few scenarios that you will have to consider when answering a question
1. Finding the area of a region between two curves
17
Ex) Find the area of the region bounded by the graphs of 𝑦 = 𝑥 2 + 2, 𝑦 = −𝑥, 𝑥 = 0, 𝑎𝑛𝑑 𝑥 = 1. � 6 �

2. Area of a region between intersecting curves

9
Ex) Find the area of the region bounded by the graphs of 𝑓 (𝑥) = 2 − 𝑥 2 𝑎𝑛𝑑 𝑔(𝑥) = 𝑥. �2�

3. Region lying between two intersecting curves

Ex) The sine and cosine curves intersect infinitely many times, bounding regions of equal areas. Find
the area of one of these regions. �2√2�

4. Curves that intersect at more than two points

Ex) Find the area of the region between the graphs of 𝑓(𝑥) = 3𝑥 3 − 𝑥 2 − 10𝑥 and 𝑔(𝑥) = −𝑥 2 + 2𝑥. [24]

Area of a Region between Two Curves

If f and g are continuous on [𝑎, 𝑏] and 𝑓 (𝑥) ≥ 𝑔(𝑥) for all 𝑥𝜖[𝑎, 𝑏], then the area of the region bounded by
the graphs of f and g and the vertical lines 𝑥 = 𝑎 and 𝑥 = 𝑏 is

𝑏 𝑏 𝑏
𝐴 = � 𝑓(𝑥)𝑑𝑥 − � 𝑔(𝑥)𝑑𝑥 = � [𝑓(𝑥) − 𝑔(𝑥)]𝑑𝑥
𝑎 𝑎 𝑎

In order to do these problems, you must

1. Find the intersection points of 𝑓 (𝑥) 𝑎𝑛𝑑 𝑔(𝑥).
2. Determine when each curve is above the other in the specified interval [𝑎, 𝑏].
3. Write down separate integrals to calculate the area where appropriate.
4. Calculate each integral and find the total area.

For difficult polynomial problems (case 4), the best tool for completing steps 1-3 is by solving
𝑓 (𝑥) ≥ 𝑔(𝑥) using an interval table
• Ie) Create an interval table to solve [𝑓(𝑥) − 𝑔(𝑥)] ≥ 0. In doing this, you will find the x-values
where f(x) and g(x) intersect. When you get +, this means f(x)>g(x) and when you get -, this
means f(x)<g(x). This allows you to set up your integrals properly.
Sample Problems
343
1. Find the area enclosed between the graphs of 𝑦 = 2 − 𝑥 and 𝑦 = 12 − (𝑥 − 2)2 � 6
�

11
2. Find the area enclosed by the graphs of 𝑦 = 𝑒 𝑥 and 𝑦 = 𝑥 2 , the y-axis, and the line 𝑥 = 2. �𝑒 2 − 3
�

3. Find the area enclosed between the graphs of 𝑦 = 2 cos(𝑥) + 1 and 𝑦 = 1 − 2sin(𝑥) where 0 ≤ 𝑥 ≤ 𝜋.
�4√2�
• Hint: you will have to solve tan(𝑥) = −1

11
4. Find the area enclosed between the graphs of 𝑦 = 𝑥 3 + 𝑥 and 𝑦 = 2𝑥 where −2 ≤ 𝑥 ≤ 1. � 4 �

5. Find the total area enclosed between the graphs of 𝑦 = 𝑥 (𝑥 − 4)2 and 𝑦 = 𝑥 2 − 7𝑥 + 15. [8]

32
6. The area enclosed between the curve 𝑦 = 𝑥 2 and the line 𝑦 = 𝑚𝑥 is 3
. Find the value of m if 𝑚 > 0.
[𝑚 = 4]
 Overview of Today’s Lesson – Area between 2 Curves

Today we talked about and learned

 The area between two curves f(x) and g(x) is given by
∫ f(x)dx − ∫ g(x)dx = ∫[f(x) − g(x)]dx where f(x) ≥ g(x).

 To find the area enclosed, you must determine where the

curves intersect and take multiple integrals

 It is not necessary to orient f(x) and g(x) in the integral if you

use absolute value integrals in your calculation.

Sample Questions

1. Determine the area of the region bounded by 𝑦 = −𝑥 and 𝑦 = 𝑥 2 + 2, 𝑥 = 0 and 𝑥 = 1.

2. Determine the area between the curves 𝑓(𝑥) = 𝑥 − 1 and 𝑔(𝑥) = 𝑥 2 + 3𝑥 − 4 from 𝑥 = −5 to 𝑥 = 2.
Assignment – Nov 20, 2019

1) Sketch the solid obtained by rotating the given shaded region about the y-axis, and find an
integral that represents the volume of the solid. [ 32 ]

2) Find the volume of the solid obtained by rotating the region bounded by the curve y  e x and the
lines x  0 and y  2 about the y-axis. [0.592]