single variable
TRANSCRIPT
SINGLE VARIABLE
CALCULUS:
An Introduction
To
Integration 2nd ed.
Delia D. Samuel, Ph.D.
This text are available and licensed under Creative Commons license CC BY-NC-SA allowing
anyone to copy and redistribute the material in any medium or format, remix, transform, and
build upon the material.
Table of Contents
PREFACE ..................................................................................................................................................... i
INTRODUCTION ....................................................................................................................................... ii
CHAPTER 1: AN INTRODUCTION TO INTEGRATION .................................................................. 1
CHAPTER 2: THE DEFINITE INTEGRAL ......................................................................................... 10
CHAPTER 3: THE FUNDAMENTAL THEOREM OF CALCULUS ................................................ 26
CHAPTER 4: THE INDEFINITE INTEGRAL .................................................................................... 35
CHAPTER 5: TECHNIQUES OF INTEGRATION I .......................................................................... 43
CHAPTER 6: TECHNIQUES OF INTEGRATION II ......................................................................... 56
CHAPTER 7: TECHNIQUES OF INTEGRATION III ....................................................................... 66
CHAPTER 8: TECHNIQUES OF INTEGRATION IV ....................................................................... 85
CHAPTER 9: REVIEW OF TECHNIQUES OF INTEGRATION ................................................... 102
CHAPTER 10: APPROXIMATING DEFINITE INTEGRALS ........................................................ 109
CHAPTER 11: IMPROPER INTEGRALS ......................................................................................... 117
CHAPTER 12: AREA ............................................................................................................................ 125
CHAPTER 13: VOLUME ...................................................................................................................... 142
CHAPTER 14: ARC LENGTH ............................................................................................................. 157
CHAPTER 15: HYDROSTATIC PRESSURE AND FORCE ............................................................ 163
CHAPTER 16: MOMENTS AND CENTER OF MASS ..................................................................... 172
CHAPTER 17: POLAR COORDINATES ........................................................................................... 185
CHAPTER 18: SEQUENCES ............................................................................................................... 209
CHAPTER 19: SERIES I β GEOMETRIC SERIES ........................................................................... 221
CHAPTER 20: SERIES II β CONVERGENCE OF SERIES ............................................................ 230
CHAPTER 21: SERIES III β CONVERGENCE OF SERIES ........................................................... 237
CHAPTER 22: SERIES IV β CONVERGENCE OF SERIES ........................................................... 245
CHAPTER 23: POWER SERIES ......................................................................................................... 258
CHAPTER 24: SERIES EXPANSIONS ............................................................................................... 268
CHAPTER 25: AN INTRODUCTION TO DIFFERENTIAL EQUATIONS .................................. 280
CHAPTER 26: SOLVING DIFFERENTIAL EQUATIONS ............................................................. 291
REFERENCES ........................................................................................................................................ 304
i
PREFACE
Welcome to Calculus II. It promises to be an exciting time for you. You will learn much, work
hard, but also have fun.
For some of you, this may be the first class that you are taking at college, having just graduated
from high school. Some of you are looking forward to this class and canβt wait to start learning
about integration, what you can do with it, and how you can apply it to the discipline you are
currently studying. Yet, others may have a fear for Calculus. You may have heard some horror
stories about how difficult Calculus is, especially Calculus II. However, I would like to put your
fears to rest. Indeed, Calculus II is more challenging than Calculus I (Differential Calculus) but
you will enjoy it, and if you apply certain principles to learning and studying this course, you
will excel, and in hindsight, you will wonder what you were afraid of.
This book explains Calculus II concepts adequately, comprehensively and concisely. It is
student-friendly and will cater to your individual needs of the students. It also makes use of
illustrations like graphs and tables which will increase your comprehension of concepts.
The concepts taught in this book gives an adequate picture of Calculus II and prepares you for
other disciplines like Engineering and Physics, as well as higher-level Mathematics courses.
This is second edition of the textbook which was published in 2017. Some edits were made and
a new chapter added (Ch. 23 β Power Series).
Enjoy this book! Have fun learning a very interesting area of Calculus.
ii
INTRODUCTION
In this preliminary chapter, I will show you how you can excel at Calculus. You will learn how
to eliminate math anxiety, what professors expect of Calculus students, and tips for becoming a
Calculus expert. Read these tips throughout this course and practice them consistently, and you
will be amazed at how much you will enjoy this course and how well you will do.
Overcoming Math Anxiety
Math anxiety is characterized by an overwhelming feeling of fear of mathematics. An individual
experiencing such anxiety feels frustrated and helpless and shudders at the thought of being
tested in mathematics. He/she feels incapable of understanding mathematics.
Students with math anxiety invariably perform poorly in mathematics. A terrified student taking
a math exam will develop a mental block and will perform below his potential on the exam.
When he/she obtains the exam results, he/she will feel justified for the fear of mathematics and
will again declare emphatically, βI just cannot do mathematics. I hate it.β This negative attitude
will lower the studentβs self-esteem and increase math anxiety, which will in turn produce low
test scores. It is a self-fulfilling prophecy.
Math anxiety
Poor performance
Lower self esteem
iii
Thus, it is very important to control such anxiety in all students. Here are some tips for
overcoming math anxiety:
1. Be aware of your math anxiety and find the reasons or sources for this anxiety. Some
reasons include:
β’ An unpleasant experience in mathematics
β’ A lack of encouragement from your role models
2. Work on having a positive attitude towards mathematics.
β’ Speak positively to yourself reminding yourself again and again that you can
succeed, in spite of possibly having a history of failure or lower-than-your-
potential scores in mathematics.
3. Develop a rapport with your instructor.
β’ Ask questions no matter how trivial you may find them to be.
β’ Visit your instructor during office hours.
4. Seek help β from instructors, peers, tutors, online.
5. Have a solid foundation in basic mathematics.
β’ Review all material that are pre-requisites for learning Calculus II. Some of this
includes algebra, differential calculus, and trigonometry.
6. Do homework every night.
β’ You must spend 2 β 3 hours on homework for every hour spent in class.
β’ Read your textbook actively. Take notes. Highlight important points.
7. Practice anxiety reduction techniques.
β’ Find your happy place and visit that place often.
β’ Meditate.
β’ Practice deep breathing when you find yourself getting anxious.
iv
β’ Relax and eliminate tension.
8. Do not compare yourself with others.
β’ You are unique with your own talents. There will always be students who are
better at mathematics than you and there will always be students who are not as
good as you. Instead of comparing yourself with others, work on your
mathematical deficiencies and focus on your strengths.
9. Practice proper examination techniques.
β’ Study regularly.
β’ Do not cram right before an exam.
β’ Get a good nightβs sleep the night before an exam.
β’ Do something fun before an exam to release any tension.
10. Never give up!
Instructorsβ expectations of Calculus II students
All students should:
β’ Have a solid background in algebra, trigonometry and differential calculus
β’ Have a positive attitude and be motivated and willing to learn
β’ Be an active reader able to analyze and dissect information
β’ Have problem-solving skills
β’ Be able to communicate mathematically, think critically and logically
β’ Write clearly, legibly and sequentially
β’ Practice effective time-management skills
β’ Complete daily homework
β’ Purchase required materials β textbook, calculator, notebook for notetaking
β’ Take notes
β’ Attend class and be on time
v
More Tips for learning Calculus II
1. Find a study group made up of peers you can work with and who can be of support.
2. Always ask questions when you do not understand concepts taught.
3. Learn the vocabulary of Calculus and use it.
4. Practice problems often.
5. Make note cards of important formulas and concepts.
6. Do not wait until test time to study. Study regularly.
7. Seek help.
8. Do not procrastinate and get behind with work.
9. Use textbook effectively. Read the section being taught before coming to class and after
class. Review material needed to understand concepts being learned.
10. Develop an effective schedule and plan for study and homework
11. Read the syllabus. It spells out exactly what is expected of you.
1
CHAPTER 1: AN INTRODUCTION TO INTEGRATION
The problem of finding the line tangent to a given curve was solved through differential
calculus. The problem of finding the area enclosed by a given curve is solved through integral
calculus.
We can easily find the area under a line graph. This bounded region would either be a triangle or
trapezoid which we can find the area of. Recall:
The area of a triangle = π
π Γ ππππ Γ ππππππ
The area of a trapezoid = π
π Γ (πππ ππ ππππππππ πππ ππ) Γ ππππππ
Let us look at some examples:
Example 1.1
1. Find the area under the line π¦ = 3 β π₯ from π₯ = 0 to π₯ = 3.
2
The shaded region under the graph is a triangle with
Height = 3
Base = 3
Area under line = Area of shaded triangle = 1
2 Γ 3 Γ 3 = 4.5
2. Find the area under the line π¦ = π₯ + 2 from π₯ = 0 to π₯ = 3.
Area of region = Area of trapezoid = 1
2 Γ (2 + 5) Γ 3 = 10.5
3
AREA UNDER A CURVE
How do we find the area under a curve? First, we shall approximate the area by dividing the
region into rectangles and finding the sum of the areas of the rectangles. We can form the
rectangles in a few ways: by taking the right end points of each interval, the left end points, or
the midpoints.
Example 1.2
1. Estimate the area under the parabola π¦ = π₯2 from 1 to 3 by dividing the area into 4
approximating rectangles and using:
a) The right end points
b) The left end points
c) The midpoints of each interval
a) We first form rectangles by using the right end points of each interval.
4
Note that the width of each rectangle is 1
2 .
The length of each rectangle is the value of the y-coordinate at the right end points.
For example, the length of the first rectangle = π (3
2) = (
3
2)
2.
The length of the second rectangle = π(2) = (2)2, and so onβ¦.
Thus the area under the curve is approximately equal to the total area of the rectangles,
which is
1
2(
3
2)
2
+1
2(2)2 +
1
2(
5
2)
2
+1
2(3)2 = 10
3
4
Note that this is an overestimation of the exact area under the curve.
b) We form rectangles by using the left end points of each interval.
5
Again, the width of each rectangle is 1
2 .
The length of each rectangle is the value of the y-coordinate at the left end points.
For example, the length of the first rectangle = π(1) = (1)2.
The length of the second rectangle = π (3
2) = (
3
2)
2, and so onβ¦.
Thus the area under the curve is approximately equal to the total area of the rectangles,
which is
1
2(1)2 +
1
2(
3
2)
2
+1
2(2)2 +
1
2(
5
2)
2
= 63
4
Note that this is an underestimation of the exact area under the curve.
c) We now form rectangles by using the mid-points of each interval.
6
The width of each rectangle is 1
2 .
The length of each rectangle is the value of the y-coordinate at the mid-points.
For example, the length of the first rectangle = π (5
4) = (
5
4)
2.
The length of the second rectangle = π (7
4) = (
7
4)
2, and so onβ¦.
Thus the area under the curve is approximately equal to the total area of the rectangles,
which is
1
2[(
5
4)
2
+ (7
4)
2
+ (9
4)
2
+ (11
4)
2
] =69
8
This estimation is closer to the exact area under the curve than the previous two
estimations.
2. Estimate the area under the parabola π¦ = π₯2 from 1 to 3 by dividing the area into 8
approximating rectangles and using:
a) The right end points
b) The left end points of each interval
a)
7
Area β
0.25[(1.25)2 + (1.5)2 + (1.75)2 + (2)2 + (2.25)2 + (2.5)2 + (2.75)2 + (3)2] = 9.6875
b)
Area β
0.25[(1)2 + (1.25)2 + (1.5)2 + (1.75)2 + (2)2 + (2.25)2 + (2.5)2 + (2.75)2] = 7.6875
8
HOMEWORK ON CHAPTER 1
1. Estimate the area under the parabola π¦ = π₯3 from 0 to 2 by dividing the area into 4
approximating rectangles and using:
a) The right end points
b) The left end points of each interval.
Show a sketch of the graph and the rectangles in each case.
Classify each case as either an overestimation or an underestimation of the actual
area.
2. Estimate the area under the curve π¦ = sin π₯ from 0 to π
2 by dividing the area into 6
approximating rectangles and using:
a) The right end points
b) The left end points
c) The midpoints of each interval.
Show a sketch of the graph and the rectangles in each case.
3. Estimate the area under the line π¦ = 4 β π₯ from 0 to 4 by dividing the area into 8
approximating rectangles and using:
a) The right end points
b) The left end points
c) The midpoints of each interval.
Show a sketch of the graph and the rectangles in each case.
Find the exact area under the line and then classify each case as either an
overestimation or an underestimation of the actual area.
9
4. Estimate the area under the graph of π¦ =1
π₯ from 1 to 6 by dividing the area into 5
approximating rectangles and using:
a) The left end points
b) The right end points
c) The mid points of each interval.
10
CHAPTER 2: THE DEFINITE INTEGRAL
Review of limits at infinity
In this chapter we will make use of your previous knowledge of finding limits of rational
functions at infinity. Let us review it here:
For rational functions, if substitution results in indeterminate form β
β, divide each term by the
highest power of x in the expression, then find the limit.
This simply means that we can first find the highest power of x in the rational function. The limit
is the quotient of the coefficients of this power of x in the numerator and denominator.
Example 2.1
1. limπ₯ββ
π₯2β2π₯+1
3π₯2+4π₯β5=
1
3
2. limπ₯ββ
4π₯3β2π₯+1
5π₯3+4π₯β5=
4
5
3. limπ₯ββ
7π₯6β2π₯+1
4π₯6+4π₯2β5=
7
4
4. limπ₯ββ
2π₯+1
3π₯2+4π₯β5=
0
3= 0
Riemann Sums
Recall that the problem of finding the line tangent to a given curve was solved through
differential calculus. The problem of finding the area enclosed by a given curve was solved
through integral calculus. The method of Riemann Sums was used to calculate such an area. It
led to the definition of the definite integral.
11
In the previous chapter, we estimated the area under the parabola π¦ = π₯2 from 1 to 3 by dividing
the area into 4 and 8 approximating rectangles and using left and right end points and midpoints
of each interval. We can increase the number of rectangles and find even more approximations.
Let us examine approximations by using the left and right end points of the intervals. Denote the
area obtained by using n rectangles and their left end points by πΏπ, and π π, the area obtained by
using n rectangles and their right end points. The table below shows the approximate areas
obtained based on a specified number of rectangles:
n π³π πΉπ
10 7.88 9.48
20 8.27 9.07
30 8.40148148148148 8.93481481481481
40 8.4675 8.8675
50 8.5072 8.8272
100 8.5868 8.7468
200 8.6267 8.7067
10000 8.66586668 8.66746668
100000 8.6665866668 8.6667466668
Notice that as the number of rectangles increase, the estimates get better, and πΏπ gets closer to
π π. This suggests that as8
π β β, πΏπ β π π β ππ₯πππ‘ ππππ
So, π π β ? as π β β
It appears that as π β β, π π β 8.667 or 82
3
We can show that
12
limπββ
π π = 82
3
Recall when we divided the area into 4 rectangles, we found that:
π 4 =1
2(
3
2)
2
+1
2(2)2 +
1
2(
5
2)
2
+1
2(3)2
=2
4(1 +
2
4)
2
+2
4(1 +
4
4)
2
+2
4(1 +
6
4)
2
+2
4(1 +
8
4)
2
Similarly, for 8 rectangles, we obtained:
π 8 = 0.25[(1.25)2 + (1.5)2 + (1.75)2 + (2)2 + (2.25)2 + (2.5)2 + (2.75)2 + (3)2]
=2
8[(1 +
2
8)
2
+ (1 +4
8)
2
+ (1 +6
8)
2
+ (1 +8
8)
2
+ (1 +10
8)
2
+ (1 +12
8)
2
+ (1 +14
8)
2
+ (1 +16
8)
2
]
Observing the pattern in both formulas, we can write a general formula when we divide the area
into n rectangles, where n is any positive integer. Thus,
π π =2
π[(1 +
2
π)
2
+ (1 +4
π)
2
+ (1 +6
π)
2
+ β― + (1 +2π
8)
2
]
It can then be shown that
π π = 82
3
We have just demonstrated the method of Riemann Sums:
1. Approximate the area under the curve by drawing n rectangles which can be formed by
choosing an arbitrary point within each rectangle.
2. Find the total area of the rectangles.
3. Find the limit as π β β of this total area. This gives the exact area of the region.
13
Riemann Sum
Let π¦ = π(π₯) be a function which is positive and continuous on the interval [a, b]. This is how
the area bounded by the curve π¦ = π(π₯), the x-axis and the lines π₯ = π and π₯ = π can be
approximated.
Divide [a, b] into n equal intervals with end points π₯0, π₯1, β¦ , π₯π. Each interval has length βπ₯,
where βπ₯ =πβπ
π.
Approximate the area of the region by drawing rectangles whose total area is close to the actual
area. We will choose an arbitrary point π₯π in each interval, which can be the left end point,
midpoint or right end point of each interval. For our purposes we will use the right end point of
each interval.
Since the arbitrary point (say, the right end point) chosen to form the rectangles is denoted by π₯π,
then the length of each rectangle is π(π₯π). As mentioned previously, the width of each rectangle
is βπ₯π . So the area of each rectangle is π(π₯π)βπ₯π. The area under the curve is approximately equal
to the sum of the areas of each of the n rectangles, which is:
β π(π₯π)
π
π=1
βπ₯
This is called the Riemann Sum.
14
We obtain the exact area of the region by taking the limit of this sum as π β β. This leads to the
definition of the definite integral.
The definite integral
The definite integral of a function f from π₯ = π to π₯ = π, denoted by
β« π(π₯) ππ₯
π
π
finds the area of the region under the graph π¦ = π(π₯) from π₯ = π to π₯ = π.
π(π₯) is called the integrand.
The limits of integration are a and b.
dx is called the differential of the variable x. It indicates that we are integrating with respect to x.
15
Example 2.2
Find the area under the line π¦ = 2 β π₯ from π₯ = 0 to π₯ = 2.
Let us first sketch the region:
So
β« (2 β π₯) ππ₯ =1
2Γ 2 Γ 2 = 2
2
0
Definition
Suppose f is continuous on [a, b]. Then the definite integral of f from a to b is defined as:
β« π(π₯) ππ₯π
π
= limπββ
(β π(π₯π) βπ₯
π
1=1
)
16
where βπ₯ =πβπ
π. π₯π is an arbitrary point within each interval.
For uniformity, we will use π₯π as the right end point of each interval. So
π₯π = π + βπ₯ . π
You may find these formulas useful:
β π = 1 + 2 + 3 + 4 + β― + π =π(π + 1)
2
π
π=1
β π2 = 12 + 22 + 32 + 42 + β― + π2 =π(π + 1)(2π + 1)
6
π
π=1
β π3 = 13 + 23 + 33 + 43 + β― + π3 =π2(π + 1)2
4
π
π=1
Examples 2.3
Use Riemann sums to find:
1. β« π₯ ππ₯4
0
Recall:
β« π(π₯) ππ₯π
π
= limπββ
(β π(π₯π) βπ₯
π
1=1
)
π(π₯) = π₯, βπ₯ =π β π
π=
4 β 0
π=
4
π, π₯π = π + βπ₯. π = 0 +
4
ππ =
4π
π
So
β« π₯ ππ₯ = limπββ
(β π (4π
π)
4
π
π
1=1
)4
0
17
= limπββ
(β4π
π.4
π
π
1=1
)
= limπββ
(β16π
π2
π
1=1
)
= limπββ
16
π2(β π
π
1=1
)
= limπββ
16
π2 .
π(π + 1)
2
=16
2
= 8
2. β« π₯2 ππ₯3
0
Recall:
β« π(π₯) ππ₯π
π
= limπββ
(β π(π₯π) βπ₯
π
1=1
)
π(π₯) = π₯2, βπ₯ =π β π
π=
3 β 0
π=
3
π, π₯π = π + βπ₯. π = 0 +
3
ππ =
3π
π
So
β« π₯2 ππ₯ = limπββ
(β π (3π
π)
3
π
π
1=1
)3
0
18
= limπββ
(β9π2
π2.3
π
π
1=1
)
= limπββ
(β27π2
π3
π
1=1
)
= limπββ
27
π3(β π2
π
1=1
)
= limπββ
27
π3 .
π(π + 1)(2π + 1)
6
=27 Γ 2
6
= 9
3. β« π₯3 ππ₯2
0
β« π(π₯) ππ₯π
π
= limπββ
(β π(π₯π) βπ₯
π
1=1
)
π(π₯) = π₯3, βπ₯ =π β π
π=
2 β 0
π=
2
π, π₯π = π + βπ₯. π = 0 +
2
ππ =
2π
π
So
β« π₯3 ππ₯ = limπββ
(β π (2π
π)
2
π
π
1=1
)2
0
= limπββ
(β8π3
π3.2
π
π
1=1
)
19
= limπββ
(β16π3
π4
π
1=1
)
= limπββ
16
π4(β π3
π
1=1
)
= limπββ
16
π4 .
π2(π + 1)2
4
=16
4
= 4
4. β« (3 β π₯) ππ₯4
1
β« π(π₯) ππ₯π
π
= limπββ
(β π(π₯π) βπ₯
π
1=1
)
π(π₯) = 3 β π₯, βπ₯ =π β π
π=
4 β 1
π=
3
π, π₯π = π + βπ₯. π = 1 +
3
ππ = 1 +
3π
π
So
β« (3 β π₯) ππ₯ = limπββ
(β π (1 +3π
π)
3
π
π
1=1
)4
1
= limπββ
(β [3 β (1 +3π
π)] .
3
π
π
1=1
)
= limπββ
(β [2 β3π
π]
π
1=1
.3
π)
20
= limπββ
(β [6
πβ
9π
π2]
π
1=1
)
= limπββ
(β6
πβ β
9π
π2
π
π=1
π
1=1
)
= limπββ
[6
πβ 1 β
9
π2β π
π
π=1
π
π=1
]
= limπββ
[6
π. π β
9
π2.π(π + 1)
2]
= 6 β9
2
=3
2
Interpreting Areas of regions
Consider the graph of π¦ = π(π₯) which lies above the x-axis. So π(π₯) is positive.
Consider the area of the region bounded by π¦ = π(π₯), the x-axis, and the lines π₯ = π and π₯ = π.
21
Since π(π₯) is positive, then β« π(π₯) ππ₯π
π is positive.
Now consider the area of this region S bounded by π¦ = π(π₯), the x-axis, and the lines π₯ = π and
π₯ = π, but this time, π(π₯) and the region lie below the π₯-axis. So π(π₯) is negative. Thus,
β« π(π₯) ππ₯π
π is negative. But area is positive. Thus, the area of this region is
|β« π(π₯) ππ₯π
π
|
Example 2.4
Consider the region below bounded by π¦ = π(π₯) and the x-axis.
Suppose β« π(π₯) ππ₯ = β81
4
3
0. Find:
a) β« π(π₯) ππ₯0
β3
Note that since the region corresponding to β« π(π₯) ππ₯3
0 lies below the x-axis, then its area is
81
4.
Since this graph is symmetrical about the origin, then
β« π(π₯) ππ₯0
β3
=81
4
22
b) β« π(π₯) ππ₯3
β3
β« π(π₯) ππ₯ = β« π(π₯) ππ₯ +0
β3
3
β3
β« ππ₯3
0
= β81
4+
81
4= 0
c) the area of the shaded region.
The area of the shaded region is:
81
4+
81
4=
81
2
Examples 2.5
Find each definite integral by interpreting it in terms of the area of the region:
1. β« β9 β π₯2 ππ₯3
0
We would first need to sketch the region to determine whether it is a recognizable region that we
easily find the area of.
The function is
π¦ = β9 β π₯2
Squaring both sides and simplifying, we get
π¦2 = 9 β π₯2
π₯2 + π¦2 = 9
This is a circle center the origin, radius 3.
But π¦ = β9 β π₯2 is positive so it represents the part of the circle lying above the x-axis.
23
Moreover, from the limits of integration, we see that x goes from 0 to 3. This means that the
region is a quarter of a circle as seem in the graph below:
So β« β9 β π₯2 ππ₯3
0 is the area of a quarter circle with radius 3 which is
1
4 . π . 32 =
9
4 π
2. β« |π₯| ππ₯1
β3
Again, we first sketch the region:
24
The area of the region is the sum of the areas of the two triangles:
(1
2 Γ 3 Γ 3) + (
1
2Γ 1 Γ 1) = 5
Some properties of the definite integral
1. β« π π(π₯) ππ₯ = π β« π(π₯) ππ₯π
π
π
π where k is a constant.
2. β« [π(π₯) Β± π(π₯)] ππ₯ = β« π(π₯) ππ₯ Β± β« π(π₯) ππ₯π
π
π
π
π
π
3. If π < π < π, then
β« π(π₯) ππ₯ = β« π(π₯)ππ₯ + β« π(π₯) ππ₯π
π
π
π
π
π
4. β« π(π₯) ππ₯ = β β« π(π₯) ππ₯π
π
π
π
25
HOMEWORK ON CHAPTER 2
1. Use Riemann Sums to find:
a) β« π₯21
0 ππ₯
b) β« 2π₯ ππ₯5
0
c) β« 4π₯32
0 ππ₯
d) β« (1 β 2π₯) ππ₯5
2
2. Find the following integrals by first sketching the region and interpreting the area as the
area of a known region:
a) β« β4 β π₯2 ππ₯2
0
b) β« β9 β π₯2 ππ₯3
β3
c) β« β25 β π₯2 ππ₯0
β5
d) β« |π₯ β 3| ππ₯5
0
e) β« (2 β π₯) ππ₯3
β1
26
CHAPTER 3: THE FUNDAMENTAL THEOREM OF CALCULUS
THE FUNDAMENTAL THEOREM OF CALCULUS
The fundamental theorem of Calculus is a very important theorem. It shows us how the processes
of differentiation and integration are related. There are two parts to this theorem.
You will need to recall some of the basic rules of differentiation below:
y π π
π π
π₯π, n is a constant
ππ₯πβ1
ππ₯ ππ₯
ln π₯
1
π₯
sin π₯ cos π₯
cos π₯ β sin π₯
tan π₯ π ππ2π₯
sec π₯ sec π₯ tan π₯
csc π₯ β csc π₯ cot π₯
cot π₯ βππ π2π₯
27
y π π
π π
sinβ1 π₯ 1
β1 β π₯2
cosβ1 π₯ β
1
β1 β π₯2
tanβ1 π₯ 1
1 + π₯2
cscβ1 π₯ β
1
π₯βπ₯2 β 1
secβ1 π₯ 1
π₯βπ₯2 β 1
cotβ1 π₯ β
1
1 + π₯2
Part 1 of the Fundamental theorem of Calculus
The first part enables us to compute the derivative of a definite integral.
Theorem
Suppose f is continuous on an interval. Let
πΉ(π₯) = β« π(π‘) ππ‘π₯
π
where a is a number in that interval. Then
πΉβ²(π₯) =π
ππ₯(β« π(π‘) ππ‘
π₯
π
) = π(π₯)
Recall that β« π(π‘) ππ‘π₯
π is the area under the graph of π¦ = π(π‘) from a to x. Thus this area is
denoted by πΉ(π₯). This theorem enables us to find the derivative of πΉ(π₯), that is, the rate of
28
change of the area, which is equal to the value of the function at x. Thus, we are able to evaluate
the derivative of a definite integral without evaluating that integral.
Note that the lower limit a can be any constant as long as f is continuous at a.
Example 3.1
Find the derivatives of the following integrals:
1. β« sin π‘ ππ‘π₯
π where a is an arbitrary constant.
πΉ(π₯) = β« sin π‘ ππ‘π₯
π
We note that the sine function is continuous everywhere. So
πΉβ²(π₯) = sin π₯
2. β« (1 + π‘3)β4 ππ‘π₯
0
πΉ(π₯) = β« (1 + π‘3)β4 ππ‘π₯
0
πΉβ²(π₯) = (1 + π₯3)β4
3. β« 2π‘ sin π‘ ππ‘1
π₯
We first note that the limits of integration are reversed in that the constant 1 is now the
upper limit. So we would need to change the order of the limits so that 1 is now the lower
limit. We will make use of this theorem:
β« π(π₯) ππ₯ = β β« π(π₯) ππ₯π
π
π
π
Thus
29
πΉ(π₯) = β« 2π‘ sin π‘ ππ‘ = β β« 2π‘ sin π‘ ππ‘π₯
1
1
π₯
πΉβ²(π₯) = β2π₯ sin π₯
4. β« (1 + π‘3)β3 ππ‘π₯2
1
Notice that in this case the upper limit is a function of x. It is not simply x. We would
therefore need to replace it with a single variable like u. Thus,
πΉ(π₯) = β« (1 + π‘3)β3 ππ‘ = β« (1 + π‘3)β3 ππ‘π’
1
π₯2
1
πΉβ²(π’) = (1 + π’3)β3
However, we need to find πΉβ²(π₯). Further work is necessary.
Recall the chain rule from differentiation:
If π = π(π) and π = π(π) are both differentiable functions, then
π π
π π=
π π
π π .
π π
π π
So from above, let π¦ = πΉ(π₯). We want to find ππ¦
ππ₯= πΉβ²(π₯).
Also ππ¦
ππ’= πΉβ²(π’)
And since π’ = π₯2,ππ’
ππ₯= 2π₯
So
ππ¦
ππ₯=
ππ¦
ππ’ .
ππ’
ππ₯
πΉβ²(π₯) = πΉβ²(π’) . 2π₯
30
πΉβ²(π₯) = (1 + π’3)β3 . 2π₯
We then rewrite this result in terms of x by substituting π’ = π₯2.
πΉβ²(π₯) = (1 + (π₯2)3)β3 . 2π₯ = 2π₯(1 + π₯6)β3
This brings us to the following rule for using the first part of the fundamental theorem of
calculus when the upper limit is a function of x.
Let
π(π) = β« π(π) π ππ(π)
π
where π(π) is a function of x. Then
πβ²(π) = πβ²(π) . π(π(π))
5. β« cos(π‘5) ππ‘π₯3
1
πΉ(π₯) = β« cos(π‘5) ππ‘π₯3
1
πΉβ²(π₯) =π
ππ₯(π₯3) . cos((π₯3)5)
πΉβ²(π₯) = 3π₯2 cos(π₯15)
Part 2 of the Fundamental theorem of Calculus
This theorem shows that integration is the inverse operation of differentiation and allows us to
evaluate definite integrals without using Riemann Sums.
31
Theorem
Let f be continuous on [a, b]. If πΉβ²(π‘) = π(π‘), then
β« π(π‘) ππ‘ = πΉ(π) β πΉ(π)π
π
The function F is called the antiderivative of the function f.
Thus, to evaluate a definite integral of f we would first need to find the function F whose
derivative is the function f.
Example 3.2
Evaluate the definite integrals by using the fundamental theorem of calculus:
1. β« 2π₯ ππ₯3
1
The derivative of π₯2 is 2π₯. This means that the antiderivative of 2π₯ is π₯2. So
πΉ(π₯) = π₯2
Thus
β« 2π₯ ππ₯3
1
= πΉ(3) β πΉ(1)
= 9 β 1
= 8
2. β« 3 sin π₯ ππ₯π
20
The antiderivative of 3 sin π₯ is β3 cos π₯. So πΉ(π₯) = β3 cos π₯. So
32
β« 3 sin π₯ ππ₯
π2
0
= πΉ (π
2) β πΉ(0)
= 0 β β3
= 3
3. β« (1 + 4π₯3) ππ₯3
1
The antiderivative of 1 + 4π₯3 is π₯ + π₯4. So πΉ(π₯) = π₯ + π₯4. Thus
β« (1 + 4π₯3)ππ₯3
1
= πΉ(3) β πΉ(1)
= (3 + 81) β (1 + 1)
= 82
4. β« (π₯2 β 4ππ₯)ππ₯3
0
The antiderivative of π₯2 β 4ππ₯ is 1
3π₯3 β 4ππ₯. So πΉ(π₯) =
1
3π₯3 β 4ππ₯. Thus
β« (π₯2 β 4ππ₯)ππ₯3
1
= πΉ(3) β πΉ(1)
= (9 β 4π3) β (1
3β 4π)
=26
3β 4π3 + 4π
33
HOMEWORK ON CHAPTER 3
1. Find the derivatives of the following integrals:
a) β« (1 + 3π‘5)β4π₯
0ππ‘
b) β« βπ‘ β π‘33
π₯ππ‘
c) β« sin(π‘2) ππ₯π₯4
1
d) β« βπ‘0
π₯2 tan(π‘3) ππ₯
e) β« (1 + π‘4)20 ππ‘π₯
2
f) β« cosβ1(2π‘) ππ‘0
π₯
g) β« cos(βπ‘) ππ‘π₯4
3
2. Evaluate the definite integrals by using the fundamental theorem of calculus:
a) β« (6π₯5 + 2ππ₯)1
0 ππ₯
b) β« (1 β 3π₯ + 2π₯2)2
1 ππ₯
c) β« cos π₯ ππ₯π
20
d) β« (2π₯ + 2π₯3 β 3) ππ₯2
1
e) β«1
1+π₯2 ππ₯1
0
34
f) β« 5π₯4 ππ₯1
0
g) β« (2π₯ β π₯3) ππ₯2
1
h) β« cos π₯ ππ₯π
0
i) β« 3π₯5 + π₯2
1 ππ₯
j) β« (2ππ₯ + 3π₯2 β 1) ππ₯2
0
k) β« π ππ2π₯ ππ₯π
4π
6
35
CHAPTER 4: THE INDEFINITE INTEGRAL
In the previous chapters, you learned how to evaluate definite integrals by using Riemann sums
and the fundamental theorem of calculus. Recall that definite integrals have limits of integration.
Integrals without such limits are called indefinite integrals. In this chapter, you will learn how to
evaluate them.
Observe:
The derivative of π₯2 is 2π₯.
The derivative of π₯2 + 1 is 2π₯.
The derivative of π₯2 β 500 is 2π₯.
The derivative of π₯2 + 0.009 is also 2π₯.
So the derivative of π₯2 + πππ¦ ππππ π‘πππ‘ is 2π₯
In the last chapter we saw that the π₯2 is the antiderivative of 2π₯. Another way of expressing the
antiderivative is by using the indefinite integral.
Since the derivative of π₯2 + πππ¦ ππππ π‘πππ‘ is 2π₯, we say that the indefinite integral of 2π₯ is
π₯2 + πππ¦ ππππ π‘πππ‘. This leads us to the following definition:
Definition
If πΉβ²(π₯) = π(π₯) for all x on an interval [a, b], then F is called an antiderivative or indefinite
integral of f :
β« π(π₯)ππ₯ = πΉ(π₯) + πΆ
where C is an arbitrary constant.
36
To find the indefinite integral, we will need to apply some basic integration formulas. This will
also enable us to evaluate the definite integral easily.
Some basic integration formulas
1. β« π ππ₯ = ππ₯ + πΆ, where k is a constant.
For example:
β« 2 ππ₯ = 2π₯ + π
2. β« π₯πππ₯ =π₯π+1
π+1+ πΆ, π β β1
For example:
β« π₯4 ππ₯ = π₯4+1
4 + 1+ πΆ =
π₯5
5+ πΆ
β« 3π₯2 ππ₯ = 3π₯3
3+ πΆ = π₯3 + πΆ
β« βπ₯ ππ₯ = β« π₯12 ππ₯ =
π₯32
32
+ πΆ =2
3π₯
32 + πΆ
3. β«1
π₯ππ₯ = ln |π₯| + πΆ
The absolute value is necessary since by using the chain rule:
π
ππ₯(ln π₯) =
1
π₯
π
ππ₯(ln(βπ₯)) = β
1
π₯ . β1 =
1
π₯
37
For example:
β«3
π₯ππ₯ = 3 ln|π₯| + πΆ
β«1
5π₯ ππ₯ = β«
1
5 .
1
π₯ ππ₯ =
1
5 β«
1
π₯ ππ₯ =
1
5ln|π₯| + πΆ
4. β« ππ₯ππ₯ = ππ₯ + πΆ
For example:
β« 2ππ₯ ππ₯ = 2ππ₯ + πΆ
5. β« sin π₯ ππ₯ = β cos π₯ + πΆ
For example:
β« β3 sin π₯ ππ₯ = 3 cos π₯ + πΆ
6. β« cos π₯ ππ₯ = sin π₯ + πΆ
For example:
β«1
2cos π₯ ππ₯ =
1
2sin π₯ + πΆ
7. β« π ππ2π₯ ππ₯ = tan π₯ + πΆ
For example:
38
β« 5 π ππ2π₯ ππ₯ = 5 tan π₯ + πΆ
8. β« ππ π2π₯ ππ₯ = βcot π₯ + πΆ
For example:
β« 4 ππ π2π₯ ππ₯ = β4 cot π₯ + πΆ
9. β« sec π₯ tan π₯ ππ₯ = sec π₯ + πΆ
For example:
β«1
3 sec π₯ tan π₯ ππ₯ =
1
3 sec π₯ + πΆ
10. β« csc π₯ cot π₯ ππ₯ = βcsc π₯ + πΆ
For example:
β« 0.6 csc π₯ cot π₯ ππ₯ = β 0.6 csc π₯ + πΆ
11. β«1
1+π₯2 ππ₯ = tanβ1 π₯ + πΆ
For example:
β«7
1 + π₯2ππ₯ = 7 β«
1
1 + π₯2 ππ₯ = 7 tanβ1 π₯ + πΆ
12. β«1
β1βπ₯2ππ₯ = sinβ1 π₯ + πΆ
For example:
39
β«3
β1 β π₯2ππ₯ = 3 β«
1
β1 β π₯2ππ₯ = 3 sinβ1 π₯ + πΆ
Please note that it is important to add the constant C after evaluating indefinite integrals.
Now, let us try some examples. In each example, note that we integrate term by term.
Example 4.1
Find:
1. β«(π₯3 β 5π₯ + 2) ππ₯ = π₯4
4β
5π₯2
2+ 2π₯ + πΆ =
1
4π₯4 β
5
2π₯2 + 2π₯ + πΆ
2. β« (5π‘4 +π‘
2+
5
π‘β 3 cos π‘) ππ‘ =
5π‘5
5+
π‘2
2 . 2+ 5 ln|π‘| β 3 sin π‘ + πΆ
= π‘5 +1
4π‘2 + 5 ln|π‘| β 3 sin π‘ + πΆ
3. β«(4 sin π₯ β1
7 π ππ2π₯ β
2
βπ₯ + 5ππ₯ + βπ₯
3) ππ₯
= β« (4 sin π₯ β1
7 π ππ2π₯ β 2π₯β
12 + 5ππ₯ + π₯
13) ππ₯
= β4 cos π₯ β1
7tan π₯ β
2π₯12
12
+ 5ππ₯ +π₯
43
43
+ πΆ
= β4 cos π₯ β 1
7tan π₯ β 4π₯
12 + 5ππ₯ +
3
4π₯
43 + πΆ
4. β«(π₯ β 1)(π₯2 + 5) ππ₯ = β«(π₯3 + 5π₯ β π₯2 β 5) ππ₯
= 1
4 π₯4 +
5
2π₯2 β
1
3π₯3 β 5π₯ + πΆ
40
5. β« π₯β3(2π₯ + 7) ππ₯ = β«(2π₯β2 + 7π₯β3) ππ₯
=2π₯β1
β1+
7π₯β2
β2+ πΆ
= β2π₯β1 β7
2π₯β2 + πΆ
6. β«2+βπ₯+4π₯2
π₯3 ππ₯ = β« π₯β3 (2 + π₯1
2 + 4π₯2) ππ₯
= β« (2π₯β3 + π₯β52 + 4π₯β1) ππ₯
=2π₯β2
β2+
π₯β32
β32
+ 4 ln|π₯| + πΆ
= βπ₯β2 β2
3π₯β
32 + 4 ln|π₯| + πΆ
Evaluating definite integrals
We can now use the basic rules of integration to evaluate definite integrals. Pay attention to the
set-up of these solutions.
Example 4.2
1. β« (π₯2 β 2π₯ β 1) ππ₯2
1
β« (π₯2 β 2π₯ β 1) ππ₯ = [1
3π₯3 β π₯2 β π₯]
1
22
1
= [8
3β 4 β 2] β [
1
3β 1 β 1]
= β5
3
41
2. β« (ππ₯ β cos π₯) ππ₯π
0
β« (ππ₯ β sin π₯) ππ₯
π2
0
= [ππ₯ + cos π₯]0
π2
= [ππ2 + cos
π
2] β [π0 + cos 0]
= ππ2 β 2
42
HOMEWORK ON CHAPTER 4
Evaluate the following indefinite integrals:
1. β« (7π₯11 β 1000 +55
π₯β 11ππ₯) ππ₯
2. β« (8 csc π₯ cot π₯ + 1
9sin π₯ β βπ₯) ππ₯
3. β« (2
π₯4 + 5βπ₯3
+6
7π₯3) ππ₯
4. β«(5π₯2 β 2π₯ + 1)(π₯ β 3) ππ₯
5. β«(π₯4 + 7)(2π₯3 β 5π₯) ππ₯
6. β« 4βπ₯3 (π₯2 β 3π₯ + 1) ππ₯
7. β« (2 +βπ₯
5) (βπ₯ β
4
βπ₯) ππ₯
8. β«2π₯+3π₯2β5
βπ₯ ππ₯
9. β«5π₯18+ βπ₯34
β4
2π₯3 ππ₯
10. β«(π‘2+4π‘)(π‘β1)
2π‘5 ππ‘
11. β« (5π₯3 +7
π₯β 9) ππ₯
12. β«(β2 ππ π2π₯ + 3ππ₯ β βπ₯4
) ππ₯
13. β« (π₯2 β1
π₯) (2π₯3 + 4) ππ₯
14. β«5π₯7+6π₯β2βπ₯+4
π₯2 ππ₯
43
CHAPTER 5: TECHNIQUES OF INTEGRATION I
So far, you have learned how to integrate simple functions like π₯2,5
π₯, etc. by using the basic rules
of integration seen in the last chapter. Now, we will begin to learn techniques of integration that
we can use for integrating more intricate functions like β4π₯ + 7, π₯ππ₯2, etc. In order to employ
such techniques, you will need to learn the basic rules of integration.
In this chapter, you will learn the techniques for integrating a function of a linear function, and
the technique of integration by substitution.
Integrating a function of a linear function
A function of linear function is a composite function π(π(π₯)) whose inner function π(π₯) is a
linear function of the form ππ₯ + π, where m and b are constants. Examples of such functions
are: (3π₯ + 5)7, π5π₯β9, 6
5β4π₯ , sin (2π‘ β π), etc.
To integrate a function of a linear function: Apply the basic rules of integration and divide
by the coefficient of x, the variable within the linear function.
Recall the basic rules of integration:
1. β« π ππ₯ = ππ₯ + πΆ
2. β« π₯πππ₯ =π₯π+1
π+1+ πΆ, π β β1
3. β«1
π₯ππ₯ = ln |π₯| + πΆ
4. β« ππ₯ππ₯ = ππ₯ + πΆ
5. β« sin π₯ ππ₯ = β cos π₯ + πΆ
6. β« cos π₯ ππ₯ = sin π₯ + πΆ
7. β« π ππ2π₯ ππ₯ = tan π₯ + πΆ
8. β« ππ π2π₯ ππ₯ = βcot π₯ + πΆ
9. β« sec π₯ tan π₯ ππ₯ = sec π₯ + πΆ
44
10. β« csc π₯ cot π₯ ππ₯ = βcsc π₯ + πΆ
11. β«1
1+π₯2 ππ₯ = tanβ1 π₯ + πΆ
12. β«1
β1βπ₯2ππ₯ = sinβ1 π₯ + πΆ
Example 5.1
1. β«(3π₯ β 7)4 ππ₯
We use the basic rule #1 above and divide by 3, the coefficient of x, to get:
β«(3π₯ β 7)4 ππ₯ = 1
3 .
1
5 (3π₯ β 7)5
= 1
15 (3π₯ β 7)5 + πΆ
2. β«1
β1β5π₯3 ππ₯
We would first need to write the integrand in exponential form, then apply basic rule #1,
and divide by the coefficient of x.
β«1
β1 β 5π₯3 ππ₯ = β«(1 β 5π₯)β
13 ππ₯
= β1
5 .
3
2 (1 β 5π₯)
23 + πΆ
= β3
10 (1 β 5π₯)
23 + πΆ
3. β« cos 9π₯ ππ₯
Here we use basic rule #6:
β« cos 9π₯ ππ₯ = 1
9sin 9π₯ + πΆ
45
4. β« πβ2π₯+3 ππ₯
We use basic rule #4:
β« πβ2π₯+3 ππ₯ = β1
2 πβ2π₯+3 + πΆ
5. β«4
7π₯+11 ππ₯
We use basic rule #3:
β«4
7π₯ + 11 ππ₯ =
4
7ln|7π₯ + 11| + πΆ
6. β«1
1+9π₯2 ππ₯
We use basic rule #11:
β«1
1 + 9π₯2 ππ₯ = β«
1
1 + (3π₯)2 ππ₯
=1
3 π‘ππβ1(3π₯) + πΆ
Integration by Substitution
This technique of integration is used for integrating some products and quotients, and also
functions of a linear function. We cannot use the method of integration by substitution to
integrate all products and quotients. Thus, we would first need to decide when such a technique
is appropriate.
Under what conditions do we use this technique for integrating products and quotients?
We examine the inner function. If the derivative of the inner function is a multiple of the other
function, then we use this method. For example:
46
Consider the product 3π₯(1 β 6π₯2)5. The inner function is (1 β 6π₯2). Its derivative is β12π₯
which is a multiple of 3π₯. So we can use integration by substitution to integrate this product.
The goal of this technique is to rewrite the integrand in a simpler form so that it is easier to
integrate. This is accomplished through a change of variables.
The method of integration by substitution
1. If the integrand is a product or quotient, let u be the inner function. If the integrand
is a function of a linear function, let u be the linear function.
2. Find du by differentiating both sides and solve for dx.
3. Using substitution, rewrite the integrand in terms of u and du.
4. Integrate the now-simpler integrand.
5. Rewrite the answer in terms of x.
Example 5.2
1. β« 3π₯2(4 + π₯3)5 ππ₯
The derivative of the inner function (4 + π₯3) is 3π₯2 which is the other function 3π₯2. So
we can use integration by substitution.
π’ = 4 + π₯3
ππ’ = 3π₯2 ππ₯
ππ₯ =ππ’
3π₯2
We now substitute into the integral so that the variables are changed from x to u.
β« 3π₯2(4 + π₯3)5 ππ₯ = β« 3π₯2 . π’5 .ππ’
3π₯2
47
= β« π’5 ππ’
= 1
6π’6 + πΆ
= 1
6(4 + π₯3)6 + πΆ
2. β«3π₯
βπ₯2β1 ππ₯
This integral can be written as β« 3π₯(π₯2 β 1)β1
2 ππ₯.
The inner function is (π₯2 β 1) and its derivative is 2π₯ which is a multiple of 3π₯. So again
we use integration by substitution.
π’ = π₯2 β 1
ππ’ = 2π₯ ππ₯
ππ₯ =ππ’
2π₯
We substitute into the integral:
β« 3π₯(π₯2 β 1)β12 ππ₯ = β« 3π₯ . π’β
12 .
ππ’
2π₯
= β«3
2 π’β
12 ππ’
= 2 .3
2 . π’
12 + πΆ
= 3βπ₯2 β 1 + πΆ
3. β« π₯π5π₯2β3 ππ₯
The derivative of the inner function 5π₯2 β 3 is 10π₯ which is a multiple of π₯.
π’ = 5π₯2 β 3
ππ’ = 10π₯ ππ₯
48
ππ₯ =ππ’
10π₯
Substitute:
β« π₯π5π₯2β3 ππ₯ = β« π₯ . ππ’ . ππ’
10π₯
= β«1
10ππ’ ππ’
=1
10ππ’ + πΆ
=1
10π5π₯2β3 + πΆ
4. β« 7π₯4 cos(2π₯5) ππ₯
The derivative of the inner function 2π₯5 is 10π₯4 which is a multiple of 7π₯4.
π’ = 2π₯5
ππ’ = 10π₯4 ππ₯
ππ₯ =ππ’
10π₯
So
β« 7π₯4 cos(2π₯5) ππ₯ = β« 7π₯4 cos π’ . ππ’
10π₯4
= β«7
10cos π’ ππ’
=7
10sin π’ + πΆ
=7
10sin(2π₯5) + πΆ
49
5. β« π ππ6π₯ cos π₯ ππ₯
The integral can be written as β«(sin π₯)6 cos π₯ ππ₯.
Again we use integration by substitution with our inner function sin π₯.
π’ = sin π₯
ππ’ = cos π₯ ππ₯
ππ₯ =ππ’
cos π₯
So
β« π ππ6π₯ cos π₯ ππ₯ = β« π’6 . cos π₯ . ππ’
cos π₯
= β« π’6 ππ’
=1
7π’7 + πΆ
=1
7π ππ7π₯ + πΆ
6. β« π‘ππ53π₯ π ππ23π₯ ππ₯
In this case, what would be our inner function? Here we rely on our knowledge of
derivatives. Since the derivative of tan π₯ is π ππ2π₯ we would need to choose the tangent
function as our inner function. In this example, the inner function is tan 3π₯. Its derivative
(by the chain rule) is 3 π ππ2 3π₯. Recall the chain rule: we apply the basic rules of
differentiation and then multiply by the derivative of the inner function 3π₯.
Again, we use integration by substitution.
π’ = tan 3π₯
ππ’ = 3 π ππ2 3π₯ ππ₯
ππ₯ =ππ’
3 π ππ2 3π₯
50
So
β« π‘ππ53π₯ π ππ23π₯ ππ₯ = β« π’5 . π ππ23π₯ .ππ’
3 π ππ2 3π₯
= β«1
3 π’5 ππ’
=1
18π’6 + πΆ
=1
18(tan 3π₯)6 + πΆ
Integration by substitution for definite integrals
We have seen how to apply integration by substitution to indefinite integrals. How do we apply it
to definite integrals where there are upper and lower limits of integration?
We follow the same procedure we used before but before we make the substitution we would
need to first change the limits to limits with respect to the new variable of integration. That is,
suppose that we are integrating with respect to x, then the limits are the values of x. We then
change the limits from x-limits to u-limits by finding the value of u at these x-values. Only then,
do we make the substitution. Thus, the definite integral will have a new variable of integration u
and new limits for u.
The Method of integration by substitution for definite integrals
1. If the integrand is a product or quotient, let u be the inner function. If the integrand
is a function of a linear function, let u be the linear function.
2. Find du by differentiating both sides and solve for dx.
3. Change limits.
51
4. Using substitution, rewrite the integrand in terms of u and du.
5. Integrate and evaluate.
Example 5.3
1. β«3π₯3
βπ₯4+9 ππ₯
2
0
We rewrite the integral as β« 2π₯3(π₯4 + 9)β1
22
0 ππ₯
π’ = π₯4 + 9
ππ’ = 4π₯3 ππ₯
ππ₯ =ππ’
4π₯3
Now change limits:
When π₯ = 0, π’ = 0 + 9 = 9
When π₯ = 2, π’ = 16 + 9 = 25
Now we substitute:
β« 2π₯3(π₯4 + 9)β12
2
0
ππ₯ = β« 2π₯3 . π’β12 .
ππ’
4π₯3
25
9
=1
2 β« π’β
12
25
9
ππ’
=1
2 . 2 [π’
12]
9
25
= 2512 β 9
12
= 5 β 3
= 2
52
2. β«π₯
1+π₯4 ππ₯1
0
In order to use integration by substitution, we rewrite the integral as β«π₯
1+(π₯2)2 ππ₯1
0
π’ = π₯2
ππ’ = 2π₯ ππ₯
ππ₯ =ππ’
2π₯
Change limits:
When π₯ = 0, π’ = 0
When π₯ = 1, π’ = 1
β«π₯
1 + (π₯2)2 ππ₯
1
0
= β«π₯
1 + π’2 .
ππ’
2π₯
1
0
=1
2β«
1
1 + π’2 ππ’
1
0
=1
2[tanβ1 π’]0
1
=1
2[tanβ1 1 β tanβ1 0]
=1
2[π
4β 0]
=π
8
Harder Integration by Substitution
Sometimes we are presented with an integral and we are unable to find the technique for
integration because the established rules for using the technique do not apply. In such cases, it is
advisable to apply the technique of integration by substitution. However, we would need to
rewrite all terms in the integrand in terms of u since cancelling will not occur as in previous
examples.
53
Example 5.4
1. β« π₯2βπ₯ + 3 ππ₯
We rewrite the integral as β« π₯2(π₯ + 3)1
2 ππ₯
π’ = π₯ + 3
ππ’ = ππ₯
We now rewrite π₯2 in terms of u.
π’ = π₯ + 3
π₯ = π’ β 3
π₯2 = (π’ β 3)2
So
β« π₯2(π₯ + 3)12 ππ₯ = β«(π’ β 3)2 . π’
12 ππ₯
= β« π’12(π’2 β 6π’ + 9) ππ’
= β« (π’52 β 6π’
32 + 9π’
12) ππ’
=2
7 π’
72 β
12
5π’
52 + 6π’
32 + πΆ
=2
7 (π₯ + 3)
72 β
12
5(π₯ + 3)
52 + 6(π₯ + 3)
32 + πΆ
54
HOMEWORK ON CHAPTER 5
Evaluate the following integrals:
1. β«(8π₯ + 3)5 ππ₯
2. β«(1 + 11π₯)6 ππ₯
3. β« β2 β 3π₯ ππ₯
4. β« π4π₯ ππ₯
5. β« sin(7π₯ + 3π) ππ₯
6. β« 2 π ππ2(10 π‘) ππ‘
7. β«1
1β9π₯ ππ₯
8. β« β4 β 5π₯3
ππ₯
9. β« π ππ2 (11π₯) ππ₯
10. β«7
3π₯+1 ππ₯
11. β« β2π3π₯+2 ππ₯
12. β« sec 5π₯ tan 5π₯ ππ₯
13. β«1
β1β4π₯2 ππ₯
14. β« 4 sin(2π β 9π₯) ππ₯
15. β«3π₯3
(1+π₯4)2 ππ₯
16. β« 5π₯2β2π₯3 + 7 ππ₯
17. β« πππ 3π₯ sin π₯ ππ₯
18. β«π₯
1+3π₯2
1
0 ππ₯
19. β« π₯3ππ₯4 ππ₯
20. β«2π₯4
π₯5β2 ππ₯
21. β« 3π₯(7π₯2 + 2)4 ππ₯
22. β« πππ 35π₯ sin 5π₯ ππ₯
55
23. β«1
βπ₯πβπ₯ ππ₯
24. β« π ππ32π₯ tan 2π₯ ππ₯
25. β« π₯(1 + π₯2)32
1
26. β« 2π₯ (π₯2 + 1)1
3 ππ₯β7
0
27. β« π‘ππ3π₯ π ππ2π₯ ππ₯π
30
28. β« 2π₯βπ₯ β 1 ππ₯
29. β« π₯βπ₯ + 1 ππ₯
56
CHAPTER 6: TECHNIQUES OF INTEGRATION II
Integration by Parts
Integration by parts is another technique of integration that can be used to integrate products
when other techniques (like substitution and distribution) cannot be used. For example, to find
β«(π₯2 β 4)(2π₯3 + π₯) ππ₯
we first distribute and then integrate. Also, to find
β« π₯(7π₯2 + 5)11 ππ₯
we use integration by substitution. However, consider the following integrals:
β« π₯2ππ₯ ππ₯, β« π₯3 ln π₯ ππ₯, β« π₯ sin π₯ ππ₯, β« sinβ1 π₯ ππ₯
Integration by substitution cannot be used in these cases, neither will distribution work. So use
the technique of integration by parts.
Formula for integration by parts
β« π πβ² π π = ππ β β« πβ²π π π
The left hand side of this formula represents the product to be integrated and the right hand side
is the result upon application of integration by parts. Note that the right hand side also contains
an integral which, in most cases, will be a simpler integral to evaluate.
In examining the left hand side, notice that one of the functions in the product to be integrated
should be u and the other π£β². We would need to wisely choose the function representing u. How
do we choose u?
57
In most cases, u is the simpler function, that is, the function which evaluates to 0 when
differentiated repeatedly. However, if the integrand contains the natural logarithmic or
inverse trigonometric function, then u = natural logarithmic function or the inverse
trigonometric function.
Example 6.1
1. β« π₯ cos π₯ ππ₯
In this case the simpler function is x. So π’ = π₯ and π£β² = cos π₯. From the formula
β« π πβ² π π = ππ β β« πβ²π π π
we would need to find π’β² by differentiating u, and v by integrating π£β². We set up in this
way:
π’ = π₯ π£β² = cos π₯
π’β² = 1 π£ = sin π₯
We now substitute these terms into the formula to get:
β« π₯ cos π₯ ππ₯ = π₯ sin π₯ β β« 1 . sin π₯ ππ₯
Simplify the integrand on the right hand side and then integrate to get:
β« π₯ cos π₯ ππ₯ = π₯ sin π₯ β β« sin π₯ ππ₯
β« π₯ cos π₯ = π₯ sin π₯ β (β cos π₯) + πΆ
β« π₯ cos π₯ ππ₯ = π₯ sin π₯ + cos π₯ + πΆ
58
2. β« 2π₯ π5π₯ ππ₯
π’ = 2π₯ π£β² = π5π₯
π’β² = 2 π£ =1
5 π5π₯
β« π πβ² π π = ππ β β« πβ²π π π
So
β« 2π₯ π5π₯ ππ₯ = 2π₯ .1
5 π5π₯ β β« 2 .
1
5 π5π₯ ππ₯
= 2
5 π₯ π5π₯ β β«
2
5 π5π₯ ππ₯
= 2
5 π₯ π5π₯ β
2
25 π5π₯ + πΆ
3. β« π₯2 ln π₯ ππ₯
In this example, since the integrand contains the natural logarithm function, then
π’ = ln π₯.
π’ = ln π₯ π£β² = π₯2
π’β² =1
π₯ π£ =
1
3 π₯3
So
β« π₯2 ln π₯ ππ₯ = ln π₯ . 1
3 π₯3 β β«
1
π₯ .
1
3 π₯3 ππ₯
= 1
3 π₯3 ln π₯ β β«
1
3 π₯2 ππ₯
= 1
3 π₯3 ln π₯ β
1
9 π₯3 + πΆ
59
4. β« tanβ1 π₯ ππ₯
Any integral containing an inverse trigonometric function should be integrated using
integration by parts and u must be the inverse trigonometric function. In this example, the
integrand can be written as a product of tanβ1 π₯ . 1. Thus
π’ = tanβ1 π₯ π£β² = 1
π’β² =1
1 + π₯2 π£ = π₯
So
β« tanβ1 π₯ ππ₯ = tanβ1 π₯ . π₯ β β«1
1 + π₯2 . π₯ ππ₯
= π₯ tanβ1 π₯ β β«π₯
1 + π₯2 ππ₯
Note that to integrate the remaining integral we will need to use integration by
substitution. So to find
β«π₯
1 + π₯2 ππ₯
π’ = 1 + π₯2
ππ’ = 2π₯ ππ₯
ππ₯ =ππ’
2π₯
β«π₯
1 + π₯2 ππ₯ = β«
π₯
π’ .
ππ’
2π₯
= β«1
2 .
1
π’ ππ’
= 1
2 ln|π’|
= 1
2ln |1 + π₯2|
60
Thus
β« tanβ1 π₯ ππ₯ = π₯ tanβ1 π₯ β 1
2ln|1 + π₯2| + πΆ
Repeated integration by parts
In the last example, after applying integration by parts, we saw that we had to apply integration
by substitution to the integral in the result. Sometimes, we may have to apply integration by parts
again to the integral in the result. This technique is called repeated integration by parts.
Example 6.2
1. β« π₯2 π2π₯ ππ₯
π’ = π₯2 π£β² = π2π₯
π’β² = 2π₯ π£ =1
2π2π₯
So
β« π₯2 π2π₯ ππ₯ = 1
2π₯2π2π₯ β β« π₯π2π₯ ππ₯ β¦ β¦ β¦ β¦ β¦ β¦ β¦ . . (1)
Notice that the only technique that we can apply to β« π₯π2π₯ ππ₯ is integration by parts.
β« π₯π2π₯ ππ₯
π’ = π₯ π£β² = π2π₯
π’β² = 1 π£ = 1
2π2π₯
So
β« π₯π2π₯ ππ₯ = 1
2π₯π2π₯ β β«
1
2π2π₯ ππ₯
61
=1
2π₯π2π₯ β
1
4 π2π₯
Now we substitute this into (1) above:
β« π₯2 π2π₯ ππ₯ = 1
2π₯2π2π₯ β (
1
2π₯π2π₯ β
1
4 π2π₯) + πΆ
=1
2π₯2π2π₯ β
1
2π₯π2π₯ +
1
4 π2π₯ + πΆ
2. β« π₯3ππ₯ ππ₯
In this example, we apply integration by parts 3 times.
π’ = π₯3 π£β² = ππ₯
π’β² = 3π₯2 π£ = ππ₯
So
β« π₯3ππ₯ ππ₯ = π₯3ππ₯ β β« 3π₯2ππ₯ ππ₯ β¦ β¦ β¦ β¦ β¦ β¦ β¦ β¦ (1)
We apply integration by parts to β« 3π₯2ππ₯ ππ₯:
β« 3π₯2ππ₯ ππ₯
π’ = 3π₯2 π£β² = ππ₯
π’β² = 6π₯ π£ = ππ₯
So
β« 3π₯2ππ₯ ππ₯ = 3π₯2ππ₯ β β« 6π₯ππ₯ ππ₯ β¦ β¦ β¦ β¦ β¦ β¦ β¦ β¦ β¦ (2)
We apply integration by parts to β« 6π₯ππ₯ ππ₯:
β« 6π₯ππ₯ ππ₯
π’ = 6π₯ π£β² = ππ₯
62
π’β² = 6 π£ = ππ₯
So
β« 6π₯ππ₯ ππ₯ = 6π₯ππ₯ β β« 6ππ₯ ππ₯
= 6π₯ππ₯ β 6ππ₯
Substitute in (2):
β« 3π₯2ππ₯ ππ₯ = 3π₯2ππ₯ β (6π₯ππ₯ β 6ππ₯)
= 3π₯2ππ₯ β 6π₯ππ₯ + 6ππ₯
Now substitute in (1):
β« π₯3ππ₯ ππ₯ = π₯3ππ₯ β (3π₯2ππ₯ β 6π₯ππ₯ + 6ππ₯) + πΆ
= π₯3ππ₯ β 3π₯2ππ₯ + 6π₯ππ₯ β 6ππ₯ + πΆ
3. β« ππ₯ cos π₯ ππ₯
In this example, it appears that we may have to apply integration by parts multiple times,
but this is not so. Here, you will learn a useful trick.
In this case, none of the functions obey our definition of what a simple function is. So we
can choose any function as u.
π’ = ππ₯ π£β² = cos π₯
π’β² = ππ₯ π£ = sin π₯
β« ππ₯ cos π₯ ππ₯ = ππ₯ sin π₯ β β« ππ₯ sin π₯ ππ₯ β¦ β¦ β¦ β¦ β¦ β¦ . . (1)
63
We again apply integration by parts and choose u to be the same type of function we
chose it to be previously:
β« ππ₯ sin π₯ ππ₯
π’ = ππ₯ π£β² = sin π₯
π’β² = ππ₯ π£ = β cos π₯
β« ππ₯ sin π₯ ππ₯ = βππ₯ cos π₯ + β« ππ₯ cos π₯ ππ₯
It would be futile to apply integration by parts again since β« ππ₯ cos π₯ ππ₯ is the integral
we started with in the first place. So instead we substitute our result into (1):
β« ππ₯ cos π₯ ππ₯ = ππ₯ sin π₯ β (βππ₯ cos π₯ + β« ππ₯ cos π₯ ππ₯)
β« ππ₯ cos π₯ ππ₯ = ππ₯ sin π₯ + ππ₯ cos π₯ β β« ππ₯ cos π₯ ππ₯
Notice that the same integral appears on both sides of the integration. We can therefore
collect like terms:
β« ππ₯ cos π₯ ππ₯ + β« ππ₯ cos π₯ ππ₯ = ππ₯ sin π₯ + ππ₯ cos π₯
2 β« ππ₯ cos π₯ ππ₯ = ππ₯ sin π₯ + ππ₯ cos π₯
So
β« ππ₯ cos π₯ ππ₯ = 1
2(ππ₯ sin π₯ + ππ₯ cos π₯ ) + πΆ
Definite integration by parts
Here is one example of definite integration by parts. We use the same procedure for integration
by parts, but after integrating we substitute the limits of integration as we did before.
64
Example 6.3
1. β« π₯ πβπ₯ ππ₯3
0
π’ = π₯ π£β² = πβπ₯
π’β² = 1 π£ = βπβπ₯
β« π₯ πβπ₯ ππ₯ = [βπ₯πβπ₯]03 + β« πβπ₯ ππ₯
3
0
3
0
= [βπ₯πβπ₯ β πβπ₯]03
= [β3πβ3 β πβ3] β [β1]
= β3πβ3 β πβ3 + 1
65
HOMEWORK ON CHAPTER 6
Evaluate the following integrals:
1. β« π₯ππ₯ ππ₯
2. β« π₯2 cos π₯ ππ₯
3. β« ln π₯ ππ₯
4. β« 3π₯ sin π₯ ππ₯
5. β« 7π₯ ln π₯ ππ₯
6. β« sinβ1 π₯ ππ₯
7. β« π₯5 ln π₯ ππ₯
8. β« 5π₯ cos 2π₯ ππ₯π
0
9. β«(π₯2 β 3π₯)π2π₯ ππ₯
10. β« π₯2 sin π₯ ππ₯
11. β« π‘3πβπ‘ ππ‘1
0
12. β« π3π₯ sin 5π₯ ππ₯
13. β« π₯ . β3 β π₯3
ππ₯
14. β« 7π₯ tanβ1(π₯2) ππ₯
15. β« π₯5ππ₯3 ππ₯ (Hint: express the integral as a different product: π₯3 . π₯2ππ₯3
66
CHAPTER 7: TECHNIQUES OF INTEGRATION III
INTEGRATING RATIONAL FUNCTIONS
In this chapter you will learn how to integrate a rational function using other methods not
previously taught. You already know how to integrate some rational functions like π₯2β5
π₯3 or
π₯
(4π₯2β5)4. In this chapter we will examine other rational functions and learn other techniques to
integrate them.
Integrating Improper Rational functions
A rational function π(π₯)
π(π₯) is said to be improper if the degree of π(π₯) is greater than or equal to the
degree of π(π₯). Examples of improper rational functions are π₯β1
π₯+3 and
π₯2
π₯β5.
To integrate an improper rational function π(π₯)
π(π₯) , we first divide the numerator by the
denominator to obtain a result of the form π(π) +π(π)
π(π). Then, integrate the result.
Example 7.1
1. β«π₯+1
π₯β5 ππ₯
We divide π₯ + 1 by π₯ β 5 to get:
1 +6
π₯ β 5
So
β«π₯ + 1
π₯ β 5 ππ₯ = β« (1 +
6
π₯ β 5) ππ₯ = π₯ + 6 ln|π₯ β 5| + πΆ
67
2. β«π₯2+4π₯β2
π₯2+1 ππ₯
Divide π₯2 + 4π₯ β 2 by π₯2 + 1 to get 1 +4π₯β3
π₯2+1 . So
β«π₯2 + 4π₯ β 2
π₯2 + 1 ππ₯ = β« (1 +
4π₯ β 3
π₯2 + 1) ππ₯
= β« (1 +4π₯
π₯2 + 1β
3
π₯2 + 1 ) ππ₯
= π₯ + 2 ln|π₯2 + 1| β 3 tanβ1 π₯ + πΆ
Note that integration by substitution was used to integrate the second term.
Integrating Rational functions whose denominators contain irreducible quadratic factors
An irreducible quadratic factor is a quadratic expression that cannot be factored. For example,
π₯2 + 3, π₯2 β π₯ β 1 are irreducible quadratic factors.
To integrate such rational functions, we first complete the square of the denominator of the
rational function, and then integrate the resulting rational function by using the integration by
substitution technique.
Review of completing the square
To complete the square of a quadratic expression ππ₯2 + ππ₯ + π, first factor out the coefficient of
π₯2 so that the coefficient of π₯2 is 1. Then add and subtract from the expression, the square of half
the coefficient of x. Factor the trinomial. That is,
ππ₯2 + ππ₯ + π = π (π₯2 +π
ππ₯ +
π
π)
= π (π₯2 +π
ππ₯ + (
π
2π)
2
+π
πβ (
π
2π)
2
)
= π [(π₯ +π
2π)
2
+π
πβ
π2
4π2]
68
= π [(π₯ +π
2π)
2
βπ2 β 4ππ
4π2]
= π (π₯ +π
2π)
2
βπ2 β 4ππ
4π
When integrating the resulting expression, you will find these formulas useful:
β«π
π + ππ π π = ππ«ππππ§ π + πͺ
β«π
ππ + ππ π π =
π
πππ«ππππ§
π
π+ πͺ
β«π
βπ β ππ π π = ππ«ππ¬π’π§ π + πͺ
β«π
βππ β ππ π π = ππ«ππ¬π’π§
π
π+ πͺ
Example 7.2
1. β«1
π₯2+2π₯+5 ππ₯
Since π₯2 + 2π₯ + 5 is irreducible, first complete the square.
π₯2 + 2π₯ + 5 = π₯2 + 2π₯ + 1 + 5 β 1
= (π₯ + 1)2 + 4
So
β«1
π₯2 + 2π₯ + 5 ππ₯ = β«
1
(π₯ + 1)2 + 4 ππ₯
Now use integration by substitution:
π’ = π₯ + 1
ππ’ = ππ₯
69
β«1
π₯2 + 2π₯ + 5 ππ₯ = β«
1
(π₯ + 1)2 + 4 ππ₯
= β«1
π’2 + 4 ππ’
To integrate this expression, use the formula:
β«π
ππ + ππ π π =
π
πππ«ππππ§
π
π+ πͺ
In this case π2 is 4. So
β«1
π₯2 + 2π₯ + 5 ππ₯ = β«
1
(π₯ + 1)2 + 4 ππ₯
= β«1
π’2 + 4 ππ’
= β«1
π’2 + 22 ππ’
=1
2tanβ1
π’
2+ πΆ
=1
2tanβ1
π₯ + 1
2+ πΆ
2. β«1
4π₯2+4π₯+5 ππ₯
Since 4π₯2 + 4π₯ + 5 is irreducible, first complete the square.
4π₯2 + 4π₯ + 5 = 4 (π₯2 + π₯ +5
4)
= 4 (π₯2 + π₯ +1
4+
5
4β
1
4)
= 4 (π₯2 + π₯ +1
4+ 1)
= 4 [(π₯ +1
2)
2
+ 1]
70
= 4 (π₯ +1
2)
2
+ 4
= 22 (π₯ +1
2)
2
+ 4
= (2π₯ + 1)2 + 4
So
β«1
4π₯2 + 4π₯ + 5 ππ₯ = β«
1
(2π₯ + 1)2 + 4 ππ₯
Now use integration by substitution:
π’ = 2π₯ + 1
ππ’ = 2 ππ₯
ππ₯ =ππ’
2
β«1
4π₯2 + 4π₯ + 5 ππ₯ = β«
1
(2π₯ + 1)2 + 4 ππ₯
=1
2β«
1
π’2 + 4 ππ’
To integrate this expression, use the formula:
β«π
ππ + ππ π π =
π
πππ«ππππ§
π
π+ πͺ
In this case π2 is 4. So
β«1
4π₯2 + 4π₯ + 5 ππ₯ = β«
1
(2π₯ + 1)2 + 4 ππ₯
=1
2β«
1
π’2 + 4 ππ’
71
=1
2β«
1
π’2 + 22 ππ’
=1
2 .
1
2tanβ1
π’
2+ πΆ
=1
4tanβ1
2π₯ + 1
2+ πΆ
So far, you have learned how to integrate rational expressions whose denominators are
irreducible quadratic factors and numerators are constants. You will now learn how to integrate
rational expressions whose denominators are irreducible quadratic factors and numerators are
functions of π₯. To integrate such rational functions, you will again complete the square of the
denominator of the rational function. Then express the numerator in terms of the term which is
squared in the denominator. Finally, integrate the resulting rational function by using the
integration by substitution technique.
You will find this formula useful:
β«πβ²(π)
π(π) π π = π₯π§|π(π)| + πͺ
For example, since the derivative of π₯2 + 5 is 2π₯, then
β«2π₯
π₯2 + 5 ππ₯ = ln|π₯2 + 5| + πΆ
Also
β«6π₯
π₯2 + 5 ππ₯ = 3 β«
2π₯
π₯2 + 5 ππ₯ = 3 ln|π₯2 + 5| + πΆ
β«π₯
π₯2 + 5 ππ₯ =
1
2β«
2π₯
π₯2 + 5 ππ₯ =
1
2ln|π₯2 + 5| + πΆ
72
Example 7.3
1. β«3π₯+1
π₯2+4π₯+5 ππ₯
Since π₯2 + 4π₯ + 5 is irreducible, first complete the square.
π₯2 + 4π₯ + 5 = π₯2 + 4π₯ + 4 + 5 β 4
= (π₯ + 2)2 + 1
We now express 3π₯ + 1 in terms of π₯ + 2. We replace x with π₯ + 2 and write an expression in
terms of π₯ + 2 that is equal to 3π₯ + 1.
3π₯ + 1 = 3(π₯ + 2) β 5
β«3π₯ + 1
π₯2 + 4π₯ + 5 ππ₯ = β«
3(π₯ + 2) β 5
(π₯ + 2)2 + 1 ππ₯
We now use integration by substitution:
π’ = π₯ + 2
ππ’ = ππ₯
β«3π₯ + 1
π₯2 + 4π₯ + 5 ππ₯ = β«
3(π₯ + 2) β 5
(π₯ + 2)2 + 1 ππ₯
= β«3π’ β 5
π’2 + 1 ππ’
= β«3π’
π’2 + 1 ππ’ β β«
5
π’2 + 1 ππ’
=3
2β«
2π’
π’2 + 1 ππ’ β β«
5
π’2 + 1 ππ’
=3
2ln|π’2 + 1| β 5 tanβ1 π’ + πΆ
=3
2ln|(π₯ + 2)2 + 1| β 5 tanβ1(π₯ + 2) + πΆ
73
2. β«2π₯+5
π₯2β6π₯+25 ππ₯
π₯2 β 6π₯ + 25 = π₯2 β 6π₯ + 9 + 25 β 9
= (π₯ β 3)2 + 16
We now express 2π₯ + 5 in terms of π₯ β 3. We replace x with π₯ β 3 and write an expression in
terms of π₯ β 3 that is equal to 2π₯ + 5.
2π₯ + 5 = 2(π₯ β 3) + 11
β«2π₯ + 5
π₯2 β 6π₯ + 25 ππ₯ = β«
2(π₯ β 3) + 11
(π₯ β 3)2 + 16 ππ₯
Integrate by substitution:
π’ = π₯ β 3
ππ’ = ππ₯
β«2π₯ + 5
π₯2 β 6π₯ + 25 ππ₯ = β«
2(π₯ β 3) + 11
(π₯ β 3)2 + 16 ππ₯
= β«2π’ + 11
π’2 + 16 ππ’
= β«2π’
π’2 + 16 ππ’ + β«
11
π’2 + 16 ππ’
= ln|π’2 + 16| + 11
4 tanβ1
π’
4+ πΆ
= ln|(π₯ β 3)2 + 16| + 11
4 tanβ1
π₯ β 3
4+ πΆ
Integrating Rational Functions whose denominators contain reducible quadratic factors
If the denominator π(π₯) of the rational function π(π₯)
π(π₯) can be factored, we first decompose the
rational function into partial fractions, and then integrate. You will now learn how to decompose
a rational function into partial fractions.
74
Partial Fractions
We know how to simplify 1
π₯β1+
1
π₯+1 to get
2π₯
(π₯β1)(π₯+1).
Now given 2π₯
(π₯β1)(π₯+1) and expressing it in the form:
2π₯
(π₯ β 1)(π₯ + 1)=
1
π₯ β 1+
1
π₯ + 1
is known as decomposing 2π₯
(π₯β1)(π₯+1) into partial fractions.
There are various rules for decomposing a rational function into partial fractions. This depends
on the factors of the denominator of the rational function.
Rule 1
Consider the rational function π(π₯)
π(π₯).
Suppose π(π₯) can be factored into linear factors, then the partial fraction decomposition is
π(π₯)
π(π₯)=
π΄
ππ₯ + π+
π΅
ππ₯ + π+ β―
where A and B are constants.
How to find the values of the constants
1. Rewrite the R.H.S. (right hand side) as one fraction by finding the LCD β least common
denominator.
2. Compare the numerators.
3. Find the constants by choosing appropriate values for x.
75
Example 7.4
Decompose into partial fractions:
1. 4
(π₯+3)(π₯β2)
Since the denominator is made up of linear factors we decompose as follows, then
simplify the R.H.S:
4
(π₯ + 3)(π₯ β 2)=
π΄
π₯ + 3+
π΅
π₯ β 2
4
(π₯ + 3)(π₯ β 2)=
π΄(π₯ β 2) + π΅(π₯ + 3)
(π₯ + 3)(π₯ β 2)
Equate numerators:
4 = π΄(π₯ β 2) + π΅(π₯ + 3)
To find the values of the constants, we choose values for x which will make each of the
terms on the left equal to 0.
When π₯ = 2:
4 = 5π΅
π΅ =4
5
When π₯ = β3:
4 = β5π΄
π΄ = β4
5
So
4
(π₯ + 3)(π₯ β 2)=
β45
π₯ + 3+
45
π₯ β 2
2. 3π₯+1
π₯2β6π₯+5
3π₯ + 1
π₯2 β 6π₯ + 5=
3π₯ + 1
(π₯ β 5)(π₯ β 1)=
π΄
π₯ β 5+
π΅
π₯ β 1
76
3π₯ + 1
(π₯ β 5)(π₯ β 1)=
π΄(π₯ β 1) + π΅(π₯ β 5)
(π₯ β 5)(π₯ β 1)
3π₯ + 1 = π΄(π₯ β 1) + π΅(π₯ β 5)
When π₯ = 1:
3 + 1 = β4π΅
π΅ = β1
When π₯ = β5:
β15 + 1 = β6π΄
π΄ =7
3
So
3π₯ + 1
π₯2 β 6π₯ + 5=
73
π₯ β 5β
1
π₯ β 1
Rule 2
Suppose Q(x) has repeated linear factors of the form (ππ₯ + π)π, use the decomposition:
π(π₯)
π(π₯)=
π΄1
ππ₯ + π+
π΄2
(ππ₯ + π)2+ β― +
π΄π
(ππ₯ + π)π
where the π΄πβ²π are constants.
How to find the values of the constants
1. Rewrite the R.H.S. as one fraction by finding the LCD.
77
2. Compare the numerators.
3. Find the constants by choosing appropriate values for x.
4. Equate coefficients of π₯2 or any other variable to find the last constant.
Example 7.5
1. 3π₯
(π₯+2)(π₯β1)2
3π₯
(π₯ + 2)(π₯ β 1)2=
π΄
π₯ + 2+
π΅
π₯ β 1+
πΆ
(π₯ β 1)2
3π₯
(π₯ + 2)(π₯ β 1)2=
π΄(π₯ β 1)2 + π΅(π₯ + 2)(π₯ β 1) + πΆ(π₯ + 2)
(π₯ + 2)(π₯ β 1)2
3π₯ = π΄(π₯ β 1)2 + π΅(π₯ + 2)(π₯ β 1) + πΆ(π₯ + 2)
When π₯ = 1:
3 = 3πΆ
πΆ = 1
When π₯ = β2:
β6 = 9π΄
π΄ = β2
3
We now equate coefficients of π₯2 on both sides of the equation to find B:
0 = π΄ + π΅
0 = β2
3+ π΅
π΅ =2
3
3π₯
(π₯ + 2)(π₯ β 1)2=
β23
π₯ + 2+
23
π₯ β 1+
1
(π₯ β 1)2
78
2. π₯2+4π₯+3
2π₯3βπ₯2
π₯2 + 4π₯ + 3
2π₯3 β π₯2=
π₯2 + 4π₯ + 3
π₯2(2π₯ β 1)=
π΄
π₯+
π΅
π₯2+
πΆ
2π₯ β 1
π₯2 + 4π₯ + 3
π₯2(2π₯ β 1)=
π΄π₯(2π₯ β 1) + π΅(2π₯ β 1) + πΆπ₯2
π₯2(2π₯ β 1)
π₯2 + 4π₯ + 3 = π΄π₯(2π₯ β 1) + π΅(2π₯ β 1) + πΆπ₯2
When π₯ = 0:
3 = βπ΅
π΅ = β3
When π₯ =1
2:
1
4+ 2 + 3 =
1
4πΆ
πΆ = 21
Equate coefficients of π₯2:
1 = 2π΄ + πΆ
1 = 2π΄ + 21
β20 = 2π΄
π΄ = β10
So
π₯2 + 4π₯ + 3
2π₯3 β π₯2=
β10
π₯β
3
π₯2+
21
2π₯ β 1
Rule 3
Suppose Q(x) has an irreducible quadratic factor ππ₯2 + ππ₯ + π, then use the decomposition
π΄π₯ + π΅
ππ₯2 + ππ₯ + π
79
How to find the values of the constants
1. Rewrite the R.H.S. as one fraction by finding the LCD.
2. Compare the numerators.
3. Find the constants by choosing appropriate values for x.
4. Substitute π₯ = 0.
5. Equate coefficients of π₯2 to find the last constant.
Example 7.6
1. 4
(π₯+1)(π₯2+3)
4
(π₯ + 1)(π₯2 + 3)=
π΄
π₯ + 1+
π΅π₯ + πΆ
π₯2 + 3
4
(π₯ + 1)(π₯2 + 3)=
π΄(π₯2 + 3) + (π΅π₯ + πΆ)(π₯ + 1)
(π₯ + 1)(π₯2 + 3)
4 = π΄(π₯2 + 3) + (π΅π₯ + πΆ)(π₯ + 1)
When π₯ = β1:
4 = 4π΄
π΄ = 1
When π₯ = 0:
4 = 3π΄ + πΆ
4 = 3 + πΆ
πΆ = 1
Equate coefficients of π₯2:
80
0 = π΄ + π΅
0 = 1 + π΅
π΅ = β1
So
4
(π₯ + 1)(π₯2 + 3)=
1
π₯ + 1+
1 β π₯
π₯2 + 3
2. π₯2
(π₯β2)(π₯2+π₯+1)
π₯2
(π₯ β 2)(π₯2 + π₯ + 1)=
π΄
π₯ β 2+
π΅π₯ + πΆ
π₯2 + π₯ + 1
π₯2
(π₯ β 2)(π₯2 + π₯ + 1)=
π΄(π₯2 + π₯ + 1) + (π΅π₯ + πΆ)(π₯ β 2)
(π₯ β 2)(π₯2 + π₯ + 1)
π₯2 = π΄(π₯2 + π₯ + 1) + (π΅π₯ + πΆ)(π₯ β 2)
When π₯ = 2:
4 = 7π΄
π΄ =4
7
When π₯ = 0:
0 = π΄ β 2πΆ
0 =4
7β 2πΆ
81
πΆ =2
7
Equate coefficients of π₯2:
1 = π΄ + π΅
1 =4
7+ π΅
π΅ =3
7
So
π₯2
(π₯ β 2)(π₯2 + π₯ + 1)=
47
π₯ β 2+
37 π₯ +
27
π₯2 + π₯ + 1
Once you have resolved a rational function into its partial fractions, it is now easy to
integrate. You will find these formulas useful:
β«1
1 + π₯2 ππ₯ = arctan π₯ + πΆ
β«1
π2 + π₯2 ππ₯ =
1
πarctan
π₯
π+ πΆ
β«πβ²(π₯)
π(π₯) ππ₯ = ln|π(π₯)| + πΆ
Example 7.7
1. β«4
(π₯+3)(π₯β2) ππ₯
Earlier on, we decomposed this rational function and found that
82
4
(π₯ + 3)(π₯ β 2)=
β45
π₯ + 3+
45
π₯ β 2
So
β«4
(π₯ + 3)(π₯ β 2) ππ₯ = β« (
β45
π₯ + 3+
45
π₯ β 2) ππ₯
= β4
5ln|π₯ + 3| +
4
5ln|π₯ β 2| + πΆ
2. β«3π₯+1
π₯2β6π₯+5 ππ₯
We found that
3π₯ + 1
π₯2 β 6π₯ + 5=
73
π₯ β 5β
1
π₯ β 1
So
β«3π₯ + 1
π₯2 β 6π₯ + 5 ππ₯ = β« (
73
π₯ β 5β
1
π₯ β 1) ππ₯
=7
3ln|π₯ β 5| β ln|π₯ β 1| + πΆ
Example 7.7
1. β«3π₯
(π₯+2)(π₯β1)2 ππ₯
3π₯
(π₯ + 2)(π₯ β 1)2=
β23
π₯ + 2+
23
π₯ β 1+
1
(π₯ β 1)2
So
β«3π₯
(π₯ + 2)(π₯ β 1)2 ππ₯ = β« (
β23
π₯ + 2+
23
π₯ β 1+
1
(π₯ β 1)2) ππ₯
83
= β« (β
23
π₯ + 2+
23
π₯ β 1+ (π₯ β 1)β2) ππ₯
= β2
3ln|π₯ + 2| +
2
3ln|π₯ β 1| β
1
π₯ β 1+ πΆ
2. β«π₯2+4π₯+3
2π₯3βπ₯2 ππ₯
π₯2 + 4π₯ + 3
2π₯3 β π₯2=
β10
π₯β
3
π₯2+
21
2π₯ β 1
So
β«π₯2 + 4π₯ + 3
2π₯3 β π₯2 ππ₯ = β« (
β10
π₯β
3
π₯2+
21
2π₯ β 1) ππ₯
= β10 ln|π₯| +3
π₯+
21
2ln|2π₯ β 1| + πΆ
Example 7.8
1. β«4
(π₯+1)(π₯2+3) ππ₯
4
(π₯ + 1)(π₯2 + 3)=
1
π₯ + 1+
1 β π₯
π₯2 + 3
So
β«4
(π₯ + 1)(π₯2 + 3) ππ₯ = β« (
1
π₯ + 1+
1 β π₯
π₯2 + 3) ππ₯
= β« (1
π₯ + 1+
1
π₯2 + 3β
π₯
π₯2 + 3) ππ₯
= ln|π₯ + 1| + 1
β3tanβ1
π₯
β3β
1
2ln|π₯2 + 3| + πΆ
84
HOMEWORK ON CHAPTER 7
Evaluate the following rational integrals:
1. β«π₯
π₯β3 ππ₯
2. β«π₯
π₯+2 ππ₯
3. β«π₯2+5
π₯β1 ππ₯
4. β«π₯3+π₯+1
π₯2+4 ππ₯
5. β«1
π₯2β6π₯+13 ππ₯
6. β«1
π₯2+4π₯+13 ππ₯
7. β«5
9π₯2+12π₯+5 ππ₯
8. β«3π₯β1
π₯2+4π₯+13 ππ₯
9. β«π₯+3
π₯2+2π₯+10 ππ₯
10. β«2π₯β5
π₯2β8π₯+41 ππ₯
11. β«5π₯
(π₯+4)(π₯+2) ππ₯
12. β«6
3π₯2+2π₯β1 ππ₯
13. β«π₯2
π₯2β1 ππ₯
14. β«1
(π₯+2)(π₯β1)2 ππ₯
15. β«2π₯2+25
π₯3+5π₯ ππ₯
16. β«1
(π₯β2)(π₯2+1) ππ₯
17. β«π₯2+5π₯+3
(π₯β1)(π₯2+2) ππ₯
18. β«2π₯β1
π₯2+3π₯+2 ππ₯
19. β«2π₯β1
π₯ (π₯+3)2 ππ₯
20. β«10π₯
(π₯+3)(π₯2+1) ππ₯
85
CHAPTER 8: TECHNIQUES OF INTEGRATION IV
TRIGONOMETRIC INTEGRALS AND TRIGONOMETRIC SUBSTITUTIONS
In this chapter, you will learn how to solve more complicated trigonometric integrals, as well as
integrate expressions using trigonometric substitutions.
Trigonometric Integrals
To evaluate trigonometric integrals, you will need to use some of the trigonometric identities you
learned in a trigonometry course. Recall these trigonometric identities:
πππππ + πππππ = π β¦ β¦ β¦ β¦ β¦ . (π)
Dividing each term in equation (1) by πππ 2π₯ we get:
π + πππππ = πππππ β¦ β¦ β¦ β¦ . (π)
The double angle formula for cosine is:
πππ ππ = πππππ β πππππ β¦ β¦ . . (π)
Replacing πππ 2π₯ with 1 β π ππ2π₯ and solving for π ππ2π₯ we get:
πππππ =π
π(π β ππ¨π¬ ππ) β¦ β¦ β¦ β¦ (π)
Similarly, replacing π ππ2π₯ with 1 β πππ 2π₯ and solving for πππ 2π₯ we get:
πππππ =π
π(π + ππ¨π¬ ππ) β¦ β¦ β¦ β¦ (π)
Equations (4) and (5) are called half-angle identities.
Note that equations (4) and (5) enable us to write any squared cosine and sine function in terms
of its double angle. For example,
86
π ππ23π₯ =1
2(1 β cos 6π₯)
πππ 2π₯
2=
1
2(1 + cos π₯)
Integrating even powers of sine and cosine
To integrate even powers of sine and cosine, replace the even power with its half-angle identity.
Then simplify the result and integrate. For even powers higher than 2, you would need to use the
half angle identities more than once.
Example 8.1
1. β« π ππ22π₯ ππ₯
β« π ππ22π₯ ππ₯ = β«1
2(1 β cos 4π₯) ππ₯
= β« (1
2β
1
2cos 4π₯) ππ₯
=1
2π₯ β
1
8sin 4π₯ + πΆ
2. β« πππ 43π₯ ππ₯
β« πππ 43π₯ ππ₯ = β«(πππ 23π₯)2 ππ₯
= β« [1
2(1 + cos 6π₯)]
2
ππ₯
=1
4β«(1 + 2 cos 6π₯ + πππ 26π₯) ππ₯
=1
4β« [1 + 2 cos 6π₯ +
1
2(1 + cos 12π₯)] ππ₯
=1
4β« (1 + 2 cos 6π₯ +
1
2+
1
2cos 12π₯) ππ₯
87
= β« (3
8+ 2 cos 6π₯ +
1
2cos 12π₯) ππ₯
=3
8π₯ +
1
3sin 6π₯ +
1
24sin 12π₯ + πΆ
Integrating odd powers of sine and cosine
To integrate odd powers of sine and cosine:
β’ Factor out πππ 2π₯ or π ππ2π₯
β’ Substitute for πππ 2π₯ or π ππ2π₯ by using the identity π ππ2π₯ + πππ 2π₯ = 1
β’ Use integration by substitution to then evaluate the integral.
Example 8.2
1. β« πππ 32π₯ ππ₯
β« πππ 32π₯ ππ₯ = β« cos 2π₯ . πππ 22π₯ ππ₯
= β« cos 2π₯(1 β π ππ22π₯) ππ₯
π’ = sin 2π₯
ππ’ = 2 cos 2π₯ ππ₯
ππ₯ =ππ’
2 cos 2π₯
Substituting:
β« cos 2π₯ (1 β π’2) . ππ’
2 cos 2π₯= β«
1
2(1 β π’2) ππ’
=1
2(π’ β
1
3π’3) + πΆ
=1
2sin 2π₯ β
1
6 π ππ32π₯ + πΆ
88
Integrating π πππ π₯ πππ π π₯ where m and n are positive integers
Follow the rules above for odd and even powers of sine and cosine. However, it is wise to first
manipulate the odd power of sine or cosine, if m or n is odd.
Example 8.3
1. β« π ππ4π₯ πππ 3π₯ ππ₯
Since the power of cosine is odd we will use the rule for odd powers of cosine.
β« π ππ4π₯ πππ 3π₯ ππ₯ = β« π ππ4π₯ cos π₯ . πππ 2π₯ ππ₯
= β« π ππ4π₯ cos π₯ (1 β π ππ2π₯) ππ₯
= β« cos π₯ (π ππ4π₯ β π ππ6π₯) ππ₯
π’ = sin π₯
ππ’ = cos π₯ ππ₯
Substituting:
β«(π’4 β π’6) ππ’ = 1
5π’5 β
1
7π’7 + πΆ
=1
5π ππ5π₯ β
1
7π ππ7π₯ + πΆ
2. β« π ππ2π₯ πππ 2π₯ ππ₯
β« π ππ2π₯ πππ 2π₯ ππ₯ = β«1
2(1 β cos 2π₯) .
1
2 (1 + cos 2π₯) ππ₯
=1
4β«(1 β πππ 22π₯) ππ₯
=1
4β« [1 β
1
2(1 + cos 4π₯)] ππ₯
= β« (1
8β
1
8cos 4π₯) ππ₯
89
=1
8π₯ β
1
32sin 4π₯ + πΆ
Integrating powers of tangent
Use the identity 1 + π‘ππ2π₯ = π ππ2π₯ and substitute π‘ππ2π₯ with π ππ2π₯ β 1.
Example 8.4
1. β« π‘ππ25π₯ ππ₯
β« π‘ππ25π₯ ππ₯ = β«(π ππ25π₯ β 1) ππ₯
=1
5tan 5π₯ β π₯ + πΆ
2. β« π‘ππ4π₯ ππ₯
β« π‘ππ4π₯ ππ₯ = β« π‘ππ2π₯ . π‘ππ2π₯ ππ₯
= β« π‘ππ2π₯ (π ππ2π₯ β 1) ππ₯
= β«(π‘ππ2π₯ π ππ2π₯ β π‘ππ2π₯) ππ₯
= β«(π‘ππ2π₯ π ππ2π₯ β π ππ2π₯ + 1) ππ₯
=1
3π‘ππ3π₯ β tan π₯ + π₯ + πΆ
Note that we used integration by substitution to integrate the first term.
Integrating π‘πππ π₯ π πππ π₯ where m and n are positive integers
In this case it would be better to manipulate the power of the secant function.
For even powers of secant:
90
β’ Factor out π ππ2π₯
β’ Use the substitution π ππ2π₯ = 1 + π‘ππ2π₯
β’ To integrate, use integration by substitution by letting π’ = tan π₯
For odd powers of secant:
β’ Factor out sec π₯ tan π₯
β’ Use the substitution π‘ππ2π₯ = π ππ2π₯ β 1
β’ To integrate, use integration by substitution by letting π’ = sec π₯
Example 8.5
1. β« π‘ππ5π₯ π ππ3 π₯ ππ₯
β« π‘ππ5π₯ π ππ3 π₯ ππ₯ = β« π‘ππ4π₯ π ππ2 π₯ . sec π₯ tan π₯ ππ₯
= β«(π‘ππ2π₯)2 π ππ2π₯ . sec π₯ tan π₯ ππ₯
= β«(π ππ2π₯ β 1)2 π ππ2π₯ . sec π₯ tan π₯ ππ₯
π’ = sec π₯
ππ’ = sec π₯ tan π₯ ππ₯
Substituting:
β«(π’2 β 1)2. π’2 ππ’ = β«(π’4 β 2π’2 + 1)π’2 ππ’
= β«(π’6 β 2π’4 + π’2) ππ’
=1
7π’7 β
2
5π’5 +
1
3π’3 + πΆ
=1
7π ππ7π₯ β
2
5π ππ5π₯ +
1
3π ππ3π₯ + πΆ
91
2. β« π‘ππ5π₯ π ππ4π₯ ππ₯
β« π‘ππ5π₯ π ππ4π₯ ππ₯ = β« π‘ππ5π₯ π ππ2π₯ π ππ2π₯ ππ₯
= β« π‘ππ5π₯ (π‘ππ2π₯ + 1 )π ππ2π₯ ππ₯
π’ = tan π₯
ππ’ = π ππ2π₯ ππ₯
Substituting:
β« π’5(π’2 + 1) ππ’ =1
8π’8 +
1
6π’6 + πΆ
=1
8π‘ππ8π₯ +
1
6π‘ππ6π₯ + πΆ
Trigonometric substitutions
Earlier one, we used these integration formulas:
β«1
1 + π₯2 ππ₯ = arctan π₯ + πΆ
β«1
π2 + π₯2 ππ₯ =
1
πarctan
π₯
π+ πΆ
The results of these integrals were found by making a trigonometric substitution in the integral.
We use the same method for integration by substitution to evaluate such integrals. Then to
simplify the resulting integral we use trigonometric identities, then integrate.
Example 8.6
1. Use the substitution π₯ = 3 tan π to find
β«ππ₯
(π₯2 + 9)
π₯ = 3 tan π
92
ππ₯ = 3 π ππ2π ππ
Substituting:
β«ππ₯
(π₯2 + 9) = β«
3 π ππ2π
9 π‘ππ2π + 9 ππ
= β«3 π ππ2π
9 (π‘ππ2π + 1) ππ
= β«3 π ππ2π
9 π ππ2π ππ
= β«1
3 ππ
=1
3π + πΆ
Now
π₯ = 3 tan π
π₯
3= tan π
π = tanβ1π₯
3
So
β«ππ₯
(π₯2 + 9) =
1
3tanβ1
π₯
3+ πΆ
2. Use the substitution 2π₯ = 3 sin π to find
β«π₯2
β9 β 4π₯2 ππ₯
2π₯ = 3 sin π
93
π₯ =3
2sin π
ππ₯ =3
2cos π ππ
Substituting:
β«π₯2
β9 β 4π₯2 ππ₯ = β«
(32
sin π)2
β9 β 4 (32
sin π)2
. 3
2cos π ππ
= β«
94
π ππ2π
β9 β 4 .94
π ππ2π
. 3
2cos π ππ
= β«
94
π ππ2π
β9 β 9π ππ2π .
3
2cos π ππ
= β«
94
π ππ2π
β9(1 β π ππ2π) .
3
2cos π ππ
=9
4 .
1
3.
3
2β«
π ππ2π
β(1 β π ππ2π) . cos π ππ
=9
8 β«
π ππ2π
cos π . cos π ππ
=9
8β« π ππ2π ππ
=9
8β«
1
2(1 β cos 2π) ππ
=9
16(π β
1
2sin 2π) + πΆ
Now
2π₯ = 3 sin π
94
2π₯
3= sin π
ΞΈ = sinβ12π₯
3
How do we write sin 2π in terms of x. We use the double angle formula:
sin 2π = 2 sin π cos π
We then rewrite sin π and cos π in terms of x. We already know this for sin π but we will need to
calculate this for cos π. We can do this in several ways. One way is by using the identity
πππ 2π₯ + π ππ2π₯ = 1
Then substitute 2π₯
3 for sin π thus enabling us to find cos π. Another way is to use a right-angled
triangle which I will illustrate here.
We know that
sin π = 2π₯
3=
πππππ ππ‘π
βπ¦πππ‘πππ’π π
Now draw a right-angled triangle whose opposite is 2x and hypotenuse 3.
3
2x
We use the Pythagorean theorem to find the base which is
β9 β 4π₯2
So from the triangle
cos π =β9 β 4π₯2
3
Thus
95
9
16(π β
1
2sin 2π) + πΆ =
9
16(π β sin π cos π) + πΆ
= 9
16(sinβ1
2π₯
3β
2π₯
3 .
β9 β 4π₯2
3) + πΆ
β«π₯2
β9 β 4π₯2 ππ₯ =
9
16(sinβ1
2π₯
3β
2π₯β9 β 4π₯2
9) + πΆ
Evaluating definite integrals using trigonometric substitution
We use the same method we learned in Chapter 5 for evaluating definite integrals by integration
by substitution. That is, follow the same procedure above with one difference: change the limits
of integration.
Example 8.7
1. Use the substitution 3π₯ = 4 sin π to evaluate
β«ππ₯
β16 β 9π₯2
43
0
3π₯ = 4 sin π
π₯ =4
3sin π
ππ₯ =4
3cos π ππ
Change limits:
π₯ =4
3sin π
When π₯ =4
3:
96
4
3=
4
3sin π
sin π = 1
π =π
2
When π₯ = 0:
0 =4
3sin π
π = 0
Thus
β«ππ₯
β16 β 9π₯2
43
0
= β«
43 cos π
β16 β 9(43
sin π)2
π2
0
ππ
= β«
43 cos π
β16 β 9 . 169
π ππ2π
π2
0
ππ
= β«
43 cos π
β16(1 β π ππ2π)
π2
0
ππ
=4
3 .
1
4β«
cos π
cos π
π2
0
ππ
=1
3β« ππ
π2
0
=1
3[π]
0
π2
=1
3 .
π
2
=π
6
97
Trigonometric substitutions with no given substitutions
How do you know that you would need to use a trigonometric substitution to evaluate an
integral, and how would you know what substitution to make?
Previously, you learned how to evaluate an integral using a given trigonometric substitution. In
the following examples you will have to determine what trigonometric substitution to make by
following the following rules:
Expression contains Trigonometric substitution Use Identity
π2 β π₯2
π₯ = a sin π 1 β π ππ2π = πππ 2π
π2 + π₯2
π₯ = π tan π 1 + π‘ππ2π = π ππ2π
π₯2 β π2
π₯ = π sec π π ππ2π β 1 = π‘ππ2π
For example, suppose the integral contains (16 β π₯2)3
2, then the trigonometric substitution
should be π₯ = 4 sin π.
If the integral contains β16 + 25π₯2, then the trigonometric substitution should be 5π₯ = 4 tan π.
Example 8.8
1. β«π₯
β4β25π₯2 ππ₯
Use the substitution:
5π₯ = 2 sin π
π₯ =2
5sin π
98
ππ₯ =2
5cos π ππ
β«π₯
β4 β 25π₯2 ππ₯ = β«
25
sin π
β4 β 25 (25
sin π)2
. 2
5cos π ππ
= β«
25
sin π
β4 β 25.4
25π ππ2π
. 2
5cos π ππ
= β«
25
sin π
β4(1 β π ππ2π) .
2
5cos π ππ
=2
5.1
2.
2
5β«
sin π
cos π . cos π ππ
=4
25β« sin π ππ
= β4
25cos π + πΆ
Now
sin π =5π₯
2=
πππππ ππ‘π
βπ¦πππ‘πππ’π π
ππππππππ‘ = β4 β 25π₯2
5x 2
So
cos π =β4 β 25π₯2
2
Thus
99
β«π₯
β4 β 25π₯2 ππ₯ = β
4
25 .
β4 β 25π₯2
2+ πΆ
= β 2β4 β 25π₯2
25+ πΆ
2. β«1
β9+π₯2 ππ₯
3
β3
Use the substitution
π₯ = 3 tan π
ππ₯ = 3 π ππ2π ππ
Change limits:
When π₯ = 3:
3 = 3 tan π
tan π = 1
π =π
4
When π₯ = β3:
β3 = 3 tan π
tan π = 1
β3
π =π
6
β«1
β9 + π₯2 ππ₯ = β«
1
β9 + 9 π‘ππ2π
π4
π6
3
β3
. 3 π ππ2π ππ
100
= β«1
3 sec π . 3 π ππ2π ππ
π4
π6
= β« sec π ππ
π4
π6
= [ln | sec π + tan π|]π6
π4
= ln | secπ
4+ tan
π
4| β ln | sec
π
6+ tan
π
6|
= ln |β2 + 1| β ln |2
β3+
1
β3|
= ln|β2 + 1| β ln β3
= ln (β2 + 1
β3)
101
HOMEWORK ON CHAPTER 8
1. β« πππ 26π₯ ππ₯
2. β« π ππ27π₯ ππ₯
3. β« πππ 25π₯ ππ₯
4. β« π ππ45π₯ ππ₯
5. β« π ππ3π₯ ππ₯
6. β« π ππ34π₯ ππ₯
7. β« πππ 36π₯ ππ₯
8. β« π ππ3π₯ πππ 3π₯ ππ₯
9. β« π ππ3π₯ πππ 2π₯ ππ₯
10. β« π‘ππ26π₯ ππ₯
11. β« π‘ππ3π₯ π ππ3π₯ ππ₯
12. Use the substitution π₯ = 5 sin π to evaluate β«π₯3
β25βπ₯2 ππ₯.
13. Use the substitution 3π₯ = 7 tan π to evaluate β«ππ₯
9π₯2+49
7
30
.
14. Use an appropriate substitution to evaluate:
a) β«1
β16β25π₯2 ππ₯
b) β«1
(4+π₯2)32
2
0 ππ₯
102
CHAPTER 9: REVIEW OF TECHNIQUES OF INTEGRATION
In this chapter we will summarize the techniques of integration and review when each technique
should be used to evaluate an integral.
First, you must know the basic integration formulas:
1. β« π ππ₯ = ππ₯ + πΆ, where k is a constant.
2. β« π₯πππ₯ =π₯π+1
π+1+ πΆ, π β β1
3. β«1
π₯ππ₯ = ln |π₯| + πΆ
4. β« ππ₯ππ₯ = ππ₯ + πΆ
5. β« sin π₯ ππ₯ = β cos π₯ + πΆ
6. β« cos π₯ ππ₯ = sin π₯ + πΆ
7. β« π ππ2π₯ ππ₯ = tan π₯ + πΆ
8. β« ππ π2π₯ ππ₯ = βcot π₯ + πΆ
9. β« sec π₯ tan π₯ ππ₯ = sec π₯ + πΆ
10. β« csc π₯ cot π₯ ππ₯ = βcsc π₯ + πΆ
11. β« sec π₯ ππ₯ = ln | sec π₯ + tan π₯|
12. β«1
1+π₯2 ππ₯ = tanβ1 π₯ + πΆ
13. β«1
β1βπ₯2ππ₯ = sinβ1 π₯ + πΆ
14. β«1
π2+π₯2 ππ₯ =1
πtanβ1 π₯
π+ πΆ
15. β«1
βπ2βπ₯2ππ₯ = sinβ1 π₯
π+ πΆ
16. β«πβ²(π₯)
π(π₯) ππ₯ = ln|π(π₯)| ππ₯
Here is a step by step approach which should enable you to determine when a particular
technique should be used.
A) Check whether the basic integration formulas can be applied without using any
technique. It may be necessary to first simplify the integrand. For example, consider
103
β«(π₯2 + 4)(π₯ β 1) ππ₯
We can easily distribute the function and then use the basic integral formulas to evaluate
the integral.
B) If the integrand is a function of a linear function, then apply the basic integration
formulas and divide the result by the coefficient of π₯. For example,
β« π3π₯+2 ππ₯
C) If the integrand is a quotient or product, look for an obvious substitution and use the
technique of integration by substitution. For example, consider
β«π₯
π₯2 β 1 ππ₯, β« 4π₯3(5π₯4 + 10)5 ππ₯
We use integration by substitution to evaluate both integrals: in the first one, since x is a
multiple of the derivative of π₯2 β 1, and in the second, since 4π₯3 is a multiple of the
derivative of 5π₯4 + 10.
Note also, that we could have used another technique to integrate the first. Since π₯2 β 1
can be factored, we could have first resolved the rational function into partial fractions
and then integrate. However, this latter technique may take a longer time so even though
the denominator of a rational function can be factored, it is advisable to first check
whether integration by substitution can be used.
D) If the integrand is a product and integration by substitution does not work, use integration
by parts. For example, consider
β« π₯ ππ₯ ππ₯, β« ln π₯ ππ₯, β« π₯2 sinβ1 π₯ ππ₯
104
Each of these integrals can be evaluated using integration by parts. Recall that the
formula is:
β« π πβ² π π = ππ β β« πβ²π π π
Remember the general rule is to let the simpler function be u except when the integrand
contains the natural logarithm or inverse trigonometric function, in which case u would
be that function.
E) Suppose the integrand is a rational function:
β’ If the integrand is an improper rational function, then perform long division and
apply the techniques for integrating the result. For example,
β«2π₯ β 1
5π₯ + 3 ππ₯
β’ If the rational function is proper and the denominator can be factored, decompose
it into partial fractions, then integrate. For example,
β«π₯
π₯2 + 4π₯ + 3 ππ₯
β’ If the rational function is proper and the quadratic denominator cannot be
factored, use the concept of completing the square on the denominator, then
integrate by using integration by substitution. For example,
β«3
π₯2 + 2π₯ + 10 ππ₯, β«
2π₯ + 5
π₯2 + 6π₯ + 34 ππ₯
F) Check for and apply other methods:
β’ Trigonometric integrals
β’ Trigonometric substitution
105
Example 9.1
1. β«2π₯+1
π₯2+π₯β30 ππ₯
β«2π₯ + 1
π₯2 + π₯ β 30 ππ₯ = β«
2π₯ + 1
(π₯ + 6)(π₯ β 5) ππ₯
We first decompose the rational function into partial fractions:
2π₯ + 1
(π₯ + 6)(π₯ β 5)=
π΄
π₯ + 6+
π΅
π₯ β 5
2π₯ + 1
(π₯ + 6)(π₯ β 5)=
π΄(π₯ β 5) + π΅(π₯ + 6)
(π₯ + 6)(π₯ β 5)
2π₯ + 1 = π΄(π₯ β 5) + π΅(π₯ + 6)
When π₯ = 5:
11 = 11π΅
π΅ = 1
When π₯ = β6:
β11 = β11π΄
π΄ = 1
2π₯ + 1
(π₯ + 6)(π₯ β 5)=
1
π₯ + 6+
1
π₯ β 5
β«1
π₯ + 6+
1
π₯ β 5 ππ₯ = ln|π₯ + 6| + ln|π₯ β 5| + πΆ
= ln |π₯ + 6
π₯ β 5| + πΆ
106
2. β«π₯β1
π₯2β8π₯+41 ππ₯
Since the denominator of the rational function is irreducible, we complete the square:
π₯2 β 8π₯ + 41 = π₯2 β 8π₯ + 16 + 41 β 16
= (π₯ β 4)2 + 25
β«π₯ + 1
π₯2 β 8π₯ + 41 ππ₯ = β«
(π₯ β 4) + 5
(π₯ β 4)2 + 25 ππ₯
π’ = π₯ β 4
ππ’ = ππ₯
β«π’ + 5
π’2 + 25 ππ’ = β«
π’
π’2 + 25 ππ’ + β«
5
π’2 + 25 ππ’
=1
2ln|π’2 + 25| +
5
5 tanβ1
π’
5+ πΆ
=1
2ln|(π₯ β 4)2 + 25| + tanβ1 (
π₯ β 4
5) + πΆ
3. β«(π₯3 β 5)(π₯2 + 1) ππ₯
We apply the basic integration formulas after distributing:
β«(π₯3 β 5)(π₯2 + 1) ππ₯ = β«(π₯5 + π₯3 β 5π₯2 β 5) ππ₯
=1
5π₯6 +
1
4π₯4 β
5
3π₯3 β 5π₯ + πΆ
4. β«(3π₯ + 7)11 ππ₯
The integrand is a function of a linear function.
β«(3π₯ + 7)11 ππ₯ =1
3 .
1
12 (3π₯ + 7)12 + πΆ
107
=1
36 (3π₯ + 7)12 + πΆ
5. β« 3π₯2π4π₯3β5 ππ₯
We use integration by substitution here.
π’ = 4π₯3 β 5
ππ’ = 12π₯2 ππ₯
ππ₯ =ππ’
12π₯2
β« 3π₯2π4π₯3β5 ππ₯ = β« 3π₯2ππ’ . ππ’
12π₯2
=1
4β« ππ’ ππ’
=1
4ππ’ + πΆ
=1
4π4π₯3β5 + πΆ
6. β« π₯5 ln 3π₯ ππ₯
We use integration by parts here.
π’ = ln 3π₯ ππ£ = π₯5
π’β² =3
3π₯=
1
π₯ π£ =
1
6π₯6
β« π₯5 ln 3π₯ ππ₯ =1
6π₯6 ln 3π₯ β β«
1
π₯ .
1
6π₯6 ππ₯
=1
6π₯6 ln 3π₯ β β«
1
6π₯5 ππ₯
=1
6π₯6 ln 3π₯ β
1
36π₯6 + πΆ
108
HOMEWORK ON CHAPTER 9
Evaluate the following integrals:
1. β« π₯5ππ₯6+2 ππ₯
2. β«3π₯
π₯2+3π₯+2 ππ₯
3. β«π₯2β4π₯+1
π₯2+9 ππ₯
4. β«(π₯4 β 2π₯2 + 5)(π₯3 β 1) ππ₯
5. β«2
π₯2+4π₯+40 ππ₯
6. β« 3π₯π2π₯+1 ππ₯
7. β«5π₯+1
π₯2+2π₯+82 ππ₯
8. β«(ln π₯)2 ππ₯
9. β« β7 β 4π₯ ππ₯
10. β«11π₯
(π₯2β9)2 ππ₯
109
CHAPTER 10: APPROXIMATING DEFINITE INTEGRALS
Sometimes it is difficult or impossible to find the exact value of a definite integral. For example,
β« ππ₯2 ππ₯
1
0 cannot be integrated by conventional means. In these cases, we use numerical
methods to find approximate values for the definite integrals. We approximate these values by
finding an approximation for the required area under the curve. We will examine the following
methods:
1. Midpoint Rule
2. Trapezoidal Rule
3. Simpsonβs Rule
Midpoint Rule
Here, we divide the required area into intervals. Then, approximate the area of the region by
drawing rectangles whose total area is close to the actual area. We will choose the midpoint of
each interval to draw the rectangle.
Letβs derive the formula. We will choose 6 rectangles to illustrate. Denote the midpoints of the
intervals π₯1β, π₯2
β, π₯3β, π₯4
β, π₯5β, π₯6
β . Let the width of each rectangle be βπ₯. Then the area of the
region is approximately equal to
110
βπ₯[π(π₯1β) + π(π₯2
β) + π(π₯3β) + π(π₯4
β) + π(π₯5β) + π(π₯6
β)]
= βπ₯ [π (π₯0 + π₯1
2) + π (
π₯1 + π₯2
2) + π (
π₯2 + π₯3
2) + π (
π₯3 + π₯4
2) + π (
π₯4 + π₯5
2)
+ π (π₯5 + π₯6
2)]
Extend this to n rectangles and we obtain the midpoint rule.
Midpoint rule (formula)
β« π(π₯) ππ₯ β ππ = βπ₯[π(π₯1β) + π(π₯2
β) + π(π₯3β) + π(π₯4
β) + π(π₯5β) + π(π₯6
β)]π
π
where π₯πβ is the midpoint of the ππ‘β interval.
βπ₯ =π β π
π
π₯πβ =
π₯πβ1 + π₯π
2
Example 10.1
1. Use the midpoint rule with π = 4 to approximate β«1
π₯ ππ₯
3
1. Then find the exact value of
the definite integral and the error in using the approximation.
π(π₯) =1
π₯
βπ₯ =3 β 1
4= 0.5
The end-points of the intervals are: 1, 1.5, 2, 2.5, 3. So the midpoints of the intervals are:
1.25, 1.75, 2.25, 2.75
β«1
π₯ ππ₯ β 0.5[π(1.25) + π(1.75) + π(2.25) + π(2.75)]
3
1
111
= 0.5 [1
1.25+
1
1.75+
1
2.25+
1
2.75]
= 1.08975469
The exact value of the integral is:
β«1
π₯ ππ₯
3
1
= [ln|π₯|]13
= ln 3 β ln 1
= 1.098612289
Error = 1.098612289 β 1.08975469 = 0.00886
Trapezoidal Rule
We approximate the area under the curve by using trapezoids.
Letβs derive the formula.
Suppose that we have again divided the area of the region into 6 intervals. We find the area of
the region by summing the areas of the trapezoids. Thus the area of the region is approximately
equal to
112
βπ₯ [(π(π₯0) + π(π₯1)
2) + (
π(π₯1) + π(π₯2)
2) + (
π(π₯2) + π(π₯3)
2) + (
π(π₯3) + π(π₯4)
2)
+ (π(π₯4) + π(π₯5)
2) + (
π(π₯5) + π(π₯6)
2)]
= βπ₯
2[π(π₯0) + π(π₯1) + π(π₯1) + π(π₯2) + π(π₯2) + π(π₯3) + π(π₯3) + π(π₯4) + π(π₯4) + π(π₯5)
+ π(π₯5) + π(π₯6)]
= βπ₯
2[π(π₯0) + 2{π(π₯1) + π(π₯2) + π(π₯3) + π(π₯4) + π(π₯5)} + π(π₯6)]
Extend this to n rectangles and we obtain the trapezoidal rule:
β« π(π₯) ππ₯ β ππ = βπ₯
2[π(π₯0) + 2{π(π₯1) + π(π₯2) + β― + π(π₯πβ1)} + π(π₯π)]
π
π
βπ₯ =π β π
π
Example 10.2
1. Use the trapezoidal rule with π = 4 to approximate β«1
π₯ ππ₯
3
1. Then find the exact value of
the definite integral and the error in using the approximation.
(π₯) =1
π₯
βπ₯ =3 β 1
4= 0.5
β«1
π₯ ππ₯ β
0.5
2[π(1) + 2{π(1.5) + π(2) + π(2.5)} + π(3)]
3
1
=0.5
2[1 + 2 (
1
1.5+
1
2+
1
2.5) +
1
3]
= 1.116666667
113
From the last example,
β«1
π₯ ππ₯ = 1.098612289
3
1
Error = 1.098612289 β 1.116666667 = β0.01805437767
Some observations
1. Approximations may overestimate or underestimate the exact area. This depends on the
concavity of the curve.
For a concave down curve, ππ, the mid-point rule formula, gives an overestimation and
ππ, the trapezoidal rule, gives an underestimation.
For a concave up curve, ππ, the mid-point rule formula, gives an underestimation and
ππ, the trapezoidal rule, gives an overestimation.
Note that in the example above, the defined region corresponds to a concave up curve.
2. We get more accurate approximations when we increase the value of n.
3. The errors in ππ and ππ are opposite in sign.
Simpsonβs Rule
This is a more accurate approximation. We approximate the definite integral by using parabolas
instead of line segments.
114
The formula is:
β« π(π₯) ππ₯ β ππ = βπ₯
3[π(π₯0) + 4π(π₯1) + 2π(π₯2) + β― + 2π(π₯πβ2) + 4π(π₯πβ1) + π(π₯π)]
π
π
One difference with this approximation is that n must be even.
The approximation can also found by taking a weighted average of ππ and ππ.
β« π(π₯) ππ₯ β ππ = 2ππ + ππ
3
π
π
Example 10.3
1. Use Simpsonβs rule with π = 4 to approximate β«1
π₯ ππ₯
3
1. Then find the exact value of the
definite integral and the error in using the approximation.
β«1
π₯ ππ₯ β
0.5
3[π(1) + 4π(1.5) + 2π(2) + 4π(2.5) + π(3)]
3
1
=0.5
3[1 +
4
1.5+
2
2+
4
2.5+
1
3]
= 1.1
From the last example,
β«1
π₯ ππ₯ = 1.098612289
3
1
Error = 1.098612289 β 1.1 = β0.001387711
Note that using the weighted average formula gives:
β« π(π₯) ππ₯ β ππ = 2ππ + ππ
3
π
π
116
HOMEWORK ON CHAPTER 10
Approximate the integral
β« π₯3 ππ₯3
0
with π = 6 by using:
1. The midpoint rule
2. The trapezoidal rule
3. Simpsons rule
Then find the exact value of the definite integral and in each case, the error in using the
approximation.
117
CHAPTER 11: IMPROPER INTEGRALS
The integral β« π(π₯) ππ₯π
π is said to be an improper integral if a or b is infinite or if π(π₯) becomes
infinite within or at an extremity of the range of integration. For example,
β« πβ3π₯ ππ₯β
2 is an improper integral because its upper limit is infinite.
β«1
π₯4 ππ₯5
0 is an improper integral because π(π₯) =
1
π₯4 is infinite when π₯ = 0.
β«1
π₯ ππ₯
5
β2 is an improper integral because π(π₯) =
1
π₯ is infinite when π₯ = 0 and β2 β€ 0 β€ 5.
To evaluate such integrals, please review finding limits at infinity which can be found in Chapter
2.
We will first learn how to evaluate improper integrals in which the upper and /or lower limits of
integration are infinite.
Improper integrals with infinite limits
β« π(π₯) ππ₯ = limπ‘ ββ
β« π(π₯) ππ₯π‘
π
β
π
β« π(π₯) ππ₯ = limπ‘ β ββ
β« π(π₯) ππ₯π
π‘
π
ββ
β« π(π₯) ππ₯ = β« π(π₯) ππ₯ + β« π(π₯) ππ₯β
π
π
ββ
β
ββ
where b is any real number at which π(π₯) is defined.
So, to evaluate such integrals, we write them in the appropriate format above. We then evaluate
the definite integral and then find the limit at infinity.
118
An improper integral is said to converge if the limit exists, otherwise it diverges.
Example 11.1
Evaluate the following improper integrals, that is, investigate their convergence.
1. β« πβ2π₯ ππ₯β
1
β« πβ2π₯ ππ₯β
1
= limπ‘ββ
β« πβ2π₯ ππ₯π‘
1
= limπ‘ββ
[β1
2 πβ2π₯]
1
π‘
= limπ‘ββ
[β1
2 πβ2π‘ +
1
2πβ2]
= limπ‘ββ
[β1
2π2π‘ +
1
2πβ2]
= 0 +1
2πβ2
=1
2πβ2
This improper integral converges.
2. β«ππ₯
π₯3
β
2
β«ππ₯
π₯3=
β
2
β« π₯β3 ππ₯β
2
= limπ‘ββ
β« π₯β3 ππ₯π‘
2
= limπ‘ββ
[β2π₯β2]2π‘
= limπ‘ββ
[β2
π₯2]
2
π‘
= limπ‘ββ
[β2
π‘2+
2
4]
119
=1
2
This improper integral converges.
3. β« ln π₯ ππ₯1
ββ
β« ln π₯ ππ₯ = limπ‘βββ
β« ln π₯ ππ₯1
π‘
1
ββ
We use integration by parts to find β« ln π₯ ππ₯1
π‘
π’ = ln π₯ π£β² = 1
π’β² =1
π₯ π£ = π₯
β« ln π₯ ππ₯1
π‘
= [π₯ ln π₯]π‘1 β β« ππ₯
1
π‘
= [π₯ ln π₯ β π₯]π‘1
= (0 β 1) β (π‘ ln π‘ β π‘)
= β1 β π‘ ln π‘ + π‘
So
limπ‘βββ
(β1 β π‘ ln π‘ + π‘)
= β
This improper integral diverges.
120
4. β«ππ₯
1+π₯2 β
ββ
β«ππ₯
1 + π₯2
β
ββ
= β«ππ₯
1 + π₯2+ β«
ππ₯
1 + π₯2
β
0
0
ββ
β«ππ₯
1 + π₯2= lim
π‘ββββ«
ππ₯
1 + π₯2
0
π‘
0
ββ
= limπ‘βββ
[tanβ1 π₯]π‘0
= limπ‘βββ
(tanβ1 0 β tanβ1 π‘)
= 0 β (βπ
2)
=π
2
β«ππ₯
1 + π₯2= lim
π‘βββ«
ππ₯
1 + π₯2
π‘
0
β
0
= limπ‘ββ
[tanβ1 π₯]0π‘
= limπ‘ββ
(tanβ1 π‘ β tanβ1 0)
=π
2
So
β«ππ₯
1 + π₯2=
π
2+
π
2
β
ββ
= π
This improper integral converges.
121
We now evaluate improper integrals whose limit(s) of integration make the integrand infinite.
These are called infinite integrands.
Infinite integrands
Suppose f is continuous and positive on [π, π). If limπ₯ βπ
π(π₯) is infinite, then
β« π(π₯) ππ₯ = limπ‘ βπβ
β« π(π₯) ππ₯π‘
π
π
π
Similarly, if limπ₯ βπ
π(π₯) is infinite, then
β« π(π₯) ππ₯ = limπ‘ βπ+
β« π(π₯) ππ₯π
π‘
π
π
Example 11.2
1. β«1
βπ₯β1 ππ₯
5
1
Note that the lower limit of integration 1 makes the integrand infinite. So
β«1
βπ₯ β 1 ππ₯ = lim
π‘β1+β« (π₯ β 1)β
12 ππ₯
5
π‘
5
1
= limπ‘β1+
[2(π₯ β 1)12]
π‘
5
= 2 limπ‘β1+
[(5 β 1)12 β (π‘ β 1)
12]
= 2(2 β 0)
= 4
This improper integral converges.
122
2. β«π₯
π₯2β9 ππ₯
3
0
The upper limit of 3 makes the integrand infinite. So
β«π₯
π₯2 β 9 ππ₯ = lim
π‘β3ββ«
π₯
π₯2 β 9 ππ₯
π‘
0
3
0
= limπ‘β3β
[1
2ln |π₯2 β 9|]
0
π‘
= limπ‘β3β
1
2 [ln|π‘2 β 9| β ln |0 β 9|]
which does not exist since the natural logarithm of a negative number is undefined.
This improper integral diverges.
3. β«1
π₯13
ππ₯8
β1
We note that 0 makes the integrand infinite. Since 0 is within the range of integration we
must split up the integral as follows:
β«1
π₯13
ππ₯8
β1
= β« π₯β13 ππ₯ + β« π₯β
13 ππ₯
8
0
0
β1
β« π₯β13 ππ₯ = lim
π‘β0ββ« π₯β
13 ππ₯
π‘
β1
0
β1
= limπ‘β0β
[3
2π₯
23]
β1
π‘
=3
2lim
π‘β0β [π‘
23 β 1]
=3
2
123
β« π₯β13 ππ₯ = lim
π‘β0+β« π₯β
13 ππ₯
8
π‘
8
0
= limπ‘β0+
[3
2π₯
23]
π‘
8
=3
2lim
π‘β0β [8
23 β π‘
23]
=3
2 . 4
= 6
β«1
π₯13
ππ₯ = 3
2+ 6
8
β1
= 15
2
This improper integral converges.
124
HOMEWORK ON CHAPTER 11
Evaluate the following improper integrals. State whether the integral converges or diverges.
1. β«ππ₯
βπ₯
β
9
2. β« π₯πβπ₯ ππ₯β
1
3. β«ππ₯
9+π₯2
β3
ββ
4. β«ππ₯
β1βπ₯2
0
β1
5. β«ππ₯
π₯β3
5
2
6. β«5
π₯ ππ₯
β
ββ
125
CHAPTER 12: AREA
Recall:
The definite integral
β« π(π) π ππ
π
gives the area of the region bounded by the graph of π¦ = π(π₯), the x-axis and the lines π₯ = π
and π₯ = π.
Let us now apply this to finding the area of regions. When finding the area of such a region, it is
advisable to make a sketch of that region.
Example 12.1
1. Find the area bounded by π¦ = π₯2, the x-axis, and the line π₯ = 3.
126
Area =
β« π₯2 ππ₯ = [1
3π₯3]
0
33
0
= 9
2. Find the area of the region bounded by the curve π¦ = 9 β π₯2 and the x-axis.
127
In this example, we would need to find the limits of integration which are the x-intercepts
of the graph. We therefore solve
9 β π₯2 = 0
π₯2 = 9
π₯ = Β±3
Area =
β« (9 β π₯2) ππ₯ = [9π₯ β1
3π₯3]
β3
33
β3
= (27 β 9) β (β27 + 9)
= 36
Note that since π¦ = π₯2 is an even function, the region is symmetrical about the y-axis.
Thus, we could have found the area of the region in this way:
Area =
2 β« (9 β π₯2) ππ₯ = 2 [9π₯ β1
3π₯3]
0
33
0
= 2(27 β 9)
= 36
3. Find the area of the region bounded by the curve π¦ = π₯(π₯2 β 4) and the x-axis.
128
We find the limits of integration:
π₯(π₯2 β 4) = 0
π₯(π₯ β 2)(π₯ + 2) = 0
π₯ = 0, π₯ = 2, π₯ = β2
We will evaluate the area of each region separately.
β« π₯(π₯2 β 4) ππ₯0
β2
= β« (π₯3 β 4π₯) ππ₯ 0
β2
= [1
4π₯4 β 2π₯2]
β2
0
= [0 β (4 β 8)]
= 4
And
β« π₯(π₯2 β 4) ππ₯ = β« (π₯3 β 4π₯) ππ₯2
0
2
0
129
= [1
4π₯4 β 2π₯2]
0
2
= (4 β 8)
= β4
Recall that this definite integral is negative since the region lies below the x-axis. We
therefore take the absolute value to find the area. Thus
The area of the required region = 4 + 4 = 8
Again, note that since the graph of π¦ = π₯(π₯2 β 4) is an odd function (symmetric about
the origin), we could have found the area in this way.
Area =
2 β« π₯(π₯2 β 4) ππ₯0
β2
= 2 β« (π₯3 β 4π₯) ππ₯ 0
β2
= 2 [1
4π₯4 β 2π₯2]
β2
0
= 2[0 β (4 β 8)]
= 8
Now consider the following region:
y x=f(y)
d
R
c
0 x
130
The area of the region R bounded by the curve π₯ = π(π¦), the y-axis and the lines π¦ = π and
π¦ = π is given by:
β« π(π) π ππ
π
Example 12.2
1. Find the area between the curve π₯ = 9 β π¦2 and the y-axis.
We first find the limits of integration by calculating the y-intercepts.
0 = 9 β π¦2
π¦2 = 9
π¦ = Β±3
Area =
β« (9 β π¦2) ππ¦ = [9π¦ β1
3π¦3]
β3
33
β3
= (27 β 9) β (β27 + 9)
131
= 36
2. Find the area of the region between the curve π¦ = π₯2, the y-axis, and the lines π¦ = 1 and
π¦ = 9.
We would first need to rewrite the equation for the curve π¦ = π₯2 as:
π₯ = βπ¦
Area =
β« βπ¦9
1
ππ¦ = β« π¦12
9
1
ππ¦
= [2
3π¦
32]
1
9
= 2
3(9
32 β 1
32)
132
= 2
3(27 β 1)
= 52
3
Area between two curves
Let us find the area enclosed between two curves.
y y = g(x)
R
y = f(x)
x
Let the functions f and g be continuous with π(π₯) β₯ π(π₯) for all x in [π, π]. Let π₯ = π and π₯ = π
be the x-intercepts of the points of intersection of the graphs of π¦ = π(π₯) and π¦ = π(π₯). Then
the area of the region R bounded by the curves π¦ = π(π₯) and π¦ = π(π₯) is:
β« [π(π₯) β π(π₯)] ππ₯π
π
Note that in the integrand, π(π₯) appears first since π(π₯) β₯ π(π₯). That is, the graph of π(π₯) lies
above the graph of π(π₯).
Note
β’ It does not matter whether the curves are below or above the x-axis. The same formula
applies.
133
β’ Always sketch the region.
β’ Sometimes the area of the region comprises of the sum of more than one integral. In such
a case the region is not a vertically simple region. Picture a vertically simple region in
this way: take a vertical strip and move it throughout the region. If the top of that strip
always touches π¦ = π(π₯) and the bottom always touches π¦ = π(π₯) as the strip moves
throughout the region, then the region is called vertically simple. Otherwise, the region is
not vertically simple and you would have to divide it into more than one simple region
before using the formula above. An example of this is illustrated in Example 3 below.
Example 12.3
1. Find the area bounded by the curves π¦ = π₯2 and π¦ = π₯ .
Note that this region is vertically simple.
We first find the x-coordinates of the points of intersection.
π₯2 = π₯
π₯2 β π₯ = 0
134
π₯(π₯ β 1) = 0
π₯ = 0, π₯ = 1
Area of the region =
β« (π₯ β π₯2) ππ₯ = [1
2π₯2 β
1
3π₯3]
0
11
0
= 1
2β
1
3
= 1
6
2. Find the area of the region enclosed by the curves π¦ = π₯2 and π¦ = βπ₯.
This is another vertically simple region.
First find the points of intersection:
π₯2 = βπ₯
135
π₯4 = π₯
π₯4 β π₯ = 0
π₯(π₯3 β 1) = 0
π₯ = 0, π₯ = 1
Area =
β« (βπ₯ β π₯2) ππ₯ = [2
3π₯
32 β
1
3π₯3]
0
11
0
= 2
3β
1
3
= 1
3
3. Find the area of the region enclosed by the parabola π₯ = π¦2 and the line π¦ = π₯ β 2.
This is not a vertically simple region. We would therefore need to divide the region into
two. We then find the points of intersection:
136
π₯ = (π₯ β 2)2
π₯ = π₯2 β 4π₯ + 4
π₯2 β 5π₯ + 4 = 0
(π₯ β 4)(π₯ β 1) = 0
π₯ = 1, π₯ = 4
Area of the region =
β« (βπ₯ β (ββπ₯)) ππ₯ + β« (βπ₯ β (π₯ β 2)) ππ₯4
1
1
0
= β« 2βπ₯ ππ₯ + β« (βπ₯ β π₯ + 2) ππ₯4
1
1
0
= [4
3π₯
32]
0
1
+ [2
3π₯
32 β
1
2π₯2 + 2π₯]
1
4
= 4
3+ [(
16
3β 8 + 8) β (
2
3β
1
2+ 2)]
= 9
2
This could have been solved in a simpler manner by making it a horizontally simple region. This
would necessitate the use of the following formula:
137
y
d
R x = f(y)
x = g(y)
c
0 x
Let f and g be continuous with π(π¦) β₯ π(π¦) for all y in [π, π]. Let π¦ = π and π¦ = π be the y-
intercepts of the points of intersection of the graphs of π₯ = π(π¦) and π₯ = π(π¦), or simply the
bounds of the region. Then the area of the region bounded by the curves π₯ = π(π¦) and π₯ = π(π¦)
is:
β« [π(π¦) β π(π¦)] ππ¦π
π
Note that in the integrand, π(π¦) appears first since π(π¦) β₯ π(π¦). That is, the graph of π(π¦) lies
to the right of the graph of π(π¦).
Again, sometimes the area of the region comprises of the sum of more than one integral. In such
a case the region is not a horizontally simple region. Picture a horizontally simple region in this
way: take a horizontal strip and move it throughout the region. If the right of that strip always
touches π₯ = π(π¦) and the left always touches π₯ = π(π¦) as the strip moves throughout the region,
then the region is called horizontally simple. Otherwise, the region is not horizontally simple and
you would have to divide it into more than one simple region before using the formula above.
138
Example 12.4
1. Find the area bounded by the curves π¦ = π₯2 and π¦ = π₯ .
Note that this region is horizontally simple.
The equations of both functions must be expressed in the form π₯ = π(π¦). π¦ = π₯ is
already in this form. So we now do the same to π¦ = π₯2. Note that the portion of this
curve being considered corresponds to positive values of x. So
π₯ = βπ¦
We now find the y-coordinates of the points of intersection.
βπ¦ = π¦
π¦ = π¦2
π¦ β π¦2 = 0
π¦(1 β π¦) = 0
π¦ = 0, π¦ = 1
Area of the region =
139
β« (βπ¦ β π¦) ππ¦ = [2
3π¦
32 β
1
2π¦2]
0
11
0
= 2
3β
1
2
= 1
6
2. Find the area of the region enclosed by the parabola π₯ = π¦2 and the line π¦ = π₯ β 2.
Earlier on we saw that this region was not vertically simple. However, it is horizontally
simple.
π¦ = π₯ β 2 β π₯ = π¦ + 2
We find the y-coordinates of the points of intersection.
π¦2 = π¦ + 2
π¦2 β π¦ β 2 = 0
(π¦ β 2)(π¦ + 1) = 0
π¦ = 2, π¦ = β1
140
Area of the region =
β« (π¦ + 2 β π¦2) ππ¦ = [1
2π¦2 + 2π¦ β
1
3π¦3]
β1
22
β1
= (2 + 4 β8
3) β (
1
2β 2 +
1
3)
= 9
2
141
HOMEWORK ON CHAPTER 12
1. Find the area of the region bounded by the curve π¦ = π₯3, the x-axis, and π₯ = 0 and
π₯ = 2.
2. Find the area of the region bounded by the curve π¦ = π₯3, the y-axis, and π¦ = 1 and
π¦ = 8.
3. Find the area of the region enclosed by the parabola π¦ = 6 β π₯2 and the line π¦ = βπ₯.
4. Find the area of the region enclosed by the curves π¦ = π₯3 and π₯ = π¦2.
5. Find the area of the region enclosed by the curves π¦ = π₯2 β 16 and π¦ = βπ₯2 β 4π₯.
6. Find the area of the region enclosed by the curve π¦2 β 4π₯ = 4 and 4π₯ β π¦ = 16.
7. Find the area of the region between the curve π¦ = π₯2 + 5 and the line π¦ = 3π₯ + 9.
8. Find the area of the region enclosed by the curves π¦ = π₯2 and π¦ = βπ₯2 + 6π₯.
142
CHAPTER 13: VOLUME
We will examine two methods for finding the volume of a solid:
β’ By slicing
β’ By rotation about an axis
Volume by slicing
Here we define volumes of solids whose cross sections are plane regions. A cross section of a solid
S is a plane region formed by intersecting S with a plane.
The idea behind finding the volume of the solid by slicing is to cut the solid into slices, find the
volume of each slice and add up these volumes.
Consider the solid lying on an interval [π, π]
A x b
Partition [π, π] into subintervals. Slice the solid. Consider the ππ‘β slice with base area π΄(π₯π) and
height βπ.
Volume of the ππ‘β slice = ππ = π΄(π₯π)βπ
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So the volume of the solid V =
limπ β β
β π΄(π₯π)βπ = β« π΄(π₯) ππ₯π
π
π
π=1
The volume of a solid of known integrable cross-sectional area π¨(π) from π = π to π = π
is:
π½ = β« π¨(π) π ππ
π
Steps for calculating volume by slicing
1. Sketch the solid and a typical cross section
2. Find a formula for π΄(π₯), the area of a typical cross section
3. Find the limits of integration
4. Integrate
Example 13.1
1. Find the volume of a sphere of radius r.
We make a sketch of the sphere with radius r and center it at the origin. We now draw a cross-
section (slice) on the y-axis, positioned y units from the origin. This cross-section is a circle nor
disk. Let the radius of the circle be π₯.
144
We need to find an expression for the radius π₯ of the cross-section. From the diagram, knowing
that the radius of the sphere is r, we use the Pythagorean theorem to find π₯. Thus
π₯2 + π¦2 = π2
π₯ = βπ2 β π¦2
Therefore, the area of the slice is given by:
π΄(π¦) = ππ₯2 = π(π2 β π¦2)
Since the radius of the sphere is r, then the limits of integration will go from -r to r. So
Volume =
β« π(π2 β π¦2) ππ¦ = π [π2π¦ β1
3π¦3]
βπ
ππ
βπ
= π [(π3 β1
3π3) β (βπ3 +
1
3π3)]
= 4
3ππ3
2. Find the volume of a right circular cone of base radius 2 and height 6.
y
145
We sketch the solid with its height on the y-axis and the origin as the center of the base.
At a particular value of y on the y-axis, we sketch a slice with radius x on the y-axis,
positioned y units from the origin. Note that the slice is a circle. We need to find the area
of that slice. To do so we will use the concept of similar triangles
6 β π¦
π₯=
6
2
6 β π¦ = 3π₯
π₯ = 2 βπ¦
3
So, the area of the slice =
π (2 βπ¦
3)
2
Since the height of the cone is 6, the limits of integration are from 0 to 6.
Volume =
β« π (2 βπ¦
3)
2
ππ¦ = π [β3 . 1
3 (2 β
π¦
3)
3
]0
66
0
= π(0 + 23) = 8π
146
Volume of solids of revolution
Another way of finding the volume of solids is to rotate some plane region about an axis. The
solid generated is called a solid of revolution.
The shaded regions below are rotated about the x-axis to form the solids.
We can find the volume of these solids by slicing. This will enable us to arrive at a formula to
find the volume of solids of revolution.
Suppose these solids are bounded on the x-axis by the lines π₯ = π and π₯ = π. We slice the solid
at some point x. This cross-section is a circle with radius [π(π₯)]2. So the area of the cross-section
is:
π[π(π₯)]2
The volume of the solid is therefore:
π β« [π(π₯)]2 ππ₯π
π
Thus, we can state the formula as follows:
147
Consider the region bounded by the curve π¦ = π(π₯), the x-axis, and the lines π₯ = π and π₯ = π.
We rotate that region about the x-axis to obtain the following solid of revolution:
The volume of the solid obtained by rotating the shaded region about the x-axis is:
π β« ππ π ππ
π
= π β« [π(π)]π π ππ
π
Similarly, consider the region bounded by the curve π₯ = π(π¦), the y-axis, and the lines π¦ = π
and π¦ = π
The volume of the solid obtained by rotating this region about the y-axis is:
π β« ππ π ππ
π
= π β« [π(π)]π π ππ
π
Example 13.2
1. The region bounded by π¦ = π₯3, the x-axis and the line π₯ = 1 is rotated about the x-axis.
Calculate the volume of the solid generated.
148
Volume =
π β« (π₯3)2 ππ₯ = 1
0
π β« π₯6 ππ₯1
0
= [1
7π₯7]
0
1
= 1
7π
2. Find the volume of the solid obtained by rotating about the y-axis the region bounded by
π¦ = π₯2 in the first quadrant, the line π¦ = 4 and y-axis.
π¦ = π₯2
π₯ = βπ¦
Volume =
π β« (βπ¦)2
ππ¦ = π β« π¦ ππ¦4
0
4
0
= π [1
2π¦2]
0
4
= 8π
149
Rotation of a plane region about other horizontal and vertical lines
Example 13.3
1. Find the volume of the solid generated by revolving the region bounded by π¦ = π₯3 and the
lines π¦ = 1 and π₯ = 2 about the line π¦ = 1.
When we rotate the solid and find a cross-section, observe that the cross-section is a
circle with radius π₯3 β 1. So the area of the cross-section is
π(π₯3 β 1)2
Volume of the generated solid =
π β« (π₯3 β 1)2 ππ₯ = π β« (π₯6 β 2π₯3 + 1) ππ₯2
1
2
1
= π [1
7π₯7 β
1
2π₯4 + π₯]
1
2
= π [(128
7β 8 + 2) β (
1
7β
1
2+ 1)]
= 163
14π
150
2. Find the volume of the solid generated by revolving the region between the parabola
π₯ = π¦2 and the line π₯ = 4 about the line π₯ = 4.
The radius of the cross-section (a circle) is 4 β π¦2.
Area of the cross-section is
π(4 β π¦2)2
To find the limits of integration we solve
π¦2 = 4
π¦ = Β±2
Volume of the solid =
π β« (4 β π¦2)2 ππ¦ = π β« (16 β 8π¦2 + π¦4) ππ¦2
β2
2
β2
= π [16π¦ β8
3π¦3 +
1
5π¦5]
β2
2
151
= π [(32 β64
3+
32
5) β (β32 +
64
3β
32
5)]
= 512
15π
The washer method
If the region we revolve to generate a solid does not border on or cross the axis of revolution, the
solid has a hole in it.
Consider the region lying between the curves π(π₯) =π₯2
8+ 1 and π(π₯) =
π₯2
16+
π₯
8+
1
2. The solid
formed when this region is rotated about the x-axis, denoted by S, can be seen in the following
diagram.
π1 and π2 are the solids formed when the regions under the curves π¦ = π(π₯) and π¦ = π(π₯)
respectively are rotated about the x-axis, as seen in the following diagrams:
152
π1:
π2:
The cross-section of solid S is the region between two concentric circles and is called an annulus
or washer. The area of the annulus is the difference between the areas of the circles, which are the
cross-sections of solids π1 and π2. So the area of the annulus is:
153
π[π(π₯)]2 β π[π(π₯)]2
Our limits of integration are a and b. Thus,
The volume of the solid obtained by rotating the shaded region R about the x-axis is:
π β« [π(π)]π β [π(π)]π π ππ
π
Note that π¦ = π(π₯) is the curve lying above the region, and π¦ = π(π₯) is the curve below the
region.
Example 13.4
1. Find the volume of the solid generated when the region bounded by π¦ = 4 β π₯2 and
π¦ = π₯ + 2 is rotated about the x-axis.
Let us first find the limits of integration by finding the points of intersection of the
graphs.
π₯ + 2 = 4 β π₯2
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π₯2 + π₯ β 2 = 0
(π₯ + 2)(π₯ β 1) = 0
π₯ = β2, π₯ = 1
Volume =
π β« [(4 β π₯2)2 β (π₯ + 2)2] ππ₯ = π β« (16 β 8π₯2 + π₯4 β π₯2 β 4π₯ β 4) ππ₯1
β2
1
β2
= π β« (π₯4 β 9π₯2 β 4π₯ + 12) ππ₯1
β2
= π [1
5π₯5 β 3π₯3 β 2π₯2 + 12π₯]
β2
1
= π [(1
5β 3 β 2 + 12) β (β
32
5+ 24 β 8 β 24)]
= 108
5π
2. The plane region bounded by the curves π¦ = π₯3 and π₯ = π¦2 is rotated about the y-axis.
Calculate the volume of the solid generated.
155
We find the limits of integration:
π¦ = (π¦2)3
π¦ = π¦6
π¦6 β π¦ = 0
π¦(π¦5 β 1) = 0
π¦ = 0, π¦ = 1
Note that we are rotating the region about the y-axis. So the formula for volume would be
something like this:
π β« (πππβπ‘ ππ’ππ£π)2 β (ππππ‘ ππ’ππ£π)2 ππ¦1
0
Volume =
π β« [(π¦13)
2
β (π¦2)2 ] ππ¦ = π β« (π¦23 β π¦4)
1
0
1
0
ππ¦
= π [3
5π¦
53 β
1
5π¦5]
0
1
= 2
5π
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HOMEWORK ON CHAPTER 13
1. Find the volume (by slicing) of a pyramid with height 5 cm and whose base is an
equilateral triangle with side 2 cm.
2. Find the volume of a right circular cone of base radius 6 and height 8.
3. Find the volume of the solid obtained by rotating the region bounded by the curve
π¦ = 9 β π₯2 and the x-axis about the x-axis.
4. Consider the curve π¦ = βπ₯ β 4. Find the volume of the solid obtained by rotating about
the x-axis, the region bounded by this curve, the line π₯ = 5 and the x-axis.
5. Find the volume of the solid obtained by rotating the region bounded by the curve
π₯ = βπ¦, the lines π¦ = 1 and π¦ = 3, and the y-axis about the y-axis.
6. Find the volume of the solid obtained by rotating the region bounded by the curve
π¦ = ππ₯, the lines π¦ = 1 and π¦ = 2, and the y-axis about the y-axis.
7. The plane region in the first quadrant bounded by the curve π¦ = π₯3 and the line π¦ = 4π₯
is rotated about the x-axis. Calculate the volume of the solid generated.
8. The plane region bounded by the curve π₯ = π¦2 and the line π¦ = 3π₯ is rotated about the y-
axis. Calculate the volume of the solid generated.
157
CHAPTER 14: ARC LENGTH
In this chapter we will learn how to find the length of a curve using calculus.
Consider the curve π¦ = π(π₯). We want to find the length of the curve between π₯ = π and π₯ = π.
Just as we did when we used Riemann sums to find the area under a curve, we will subdivide the
curve into sub-curves as follows:
a ______________________________________________________________b
We now find the length of each line segment connecting each of the points ππ. So denote the
length of the curve by L. Then
πΏ β |π0π1| + |π1π2| + |π2π3| + |π3π4|
This can be written as
πΏ β β|ππβ1ππ|
4
π=1
Recall that the greater the number of intervals, the more accurate the approximation. So as the
number of intervals approach infinity the sum above approaches the exact length of the curve.
Thus
πΏ = limπββ
β|ππβ1ππ|
π
π=1
Let us now express |ππβ1ππ| in another way by using the Pythagorean theorem.
158
a___________________________________________________________ b
Consider the line segment ππβ1ππ which is the hypotenuse of a right-angled triangle whose base
is βπ₯ and height βπ¦.
ππβ1
βπ¦
βπ₯ ππ
βπ₯ = π₯π β π₯πβ1
βπ¦ = π¦π β π¦πβ1 = π(π₯π) β π(π₯πβ1)
So by the Pythagorean theorem:
|ππβ1ππ| = β(βπ₯)2 + (βπ¦)2
Now recall the Mean Value Theorem from Calculus I β Differential Calculus:
Let f be a function that satisfies the following hypotheses:
1. f is continuous on the closed interval [a, b].
2. f is differentiable on the open interval (a, b).
Then there is a number c in (a, b) such that
159
πβ²(π) =π(π) β π(π)
π β π
So choose a number π₯πβ within the interval [π₯πβ1, π₯π]. Then from the Mean Value Theorem
πβ²(π₯πβ ) =
π(π₯π) β π(π₯πβ1)
π₯π β π₯πβ1=
βπ¦
βπ₯
βπ¦ = πβ²(π₯πβ )βπ₯
So
|ππβ1ππ| = β(βπ₯)2 + (βπ¦)2
= β(βπ₯)2 + (πβ²(π₯πβ )βπ₯)2
= β(βπ₯)2[1 + (πβ²(π₯πβ ))2]
= β[1 + (πβ²(π₯πβ ))2] . βπ₯
So
πΏ = limπββ
(β β[1 + (πβ²(π₯πβ ))2] . βπ₯
π
π=1
)
Recall the definition of the definite integral from Chapter 2:
β« π(π₯) ππ₯π
π
= limπββ
(β π(π₯π) βπ₯
π
1=1
)
This enables us to write the formula for the arc length or length of a curve:
The length of the curve π = π(π) from π = π to π = π is given by:
π³ = β« βπ + (πβ²(π))π π
π
π π = β« βπ + (π π
π π)
π
π
π
π π
160
Similarly, the length of the curve π = π(π) from π = π to π = π is given by:
π³ = β« βπ + (πβ²(π))π π
π
π π = β« βπ + (π π
π π)
π
π
π
π π
Example 14.1
1. Find the arc length of π¦ = 2π₯ + 1 from π₯ = β1 to π₯ = 0.
πΏ = β« β1 + (πβ²(π₯))2 π
π
ππ₯ = β« β1 + (ππ¦
ππ₯)
2
π
π
ππ₯
π¦ = 2π₯ + 1
ππ¦
ππ₯= 2
πΏ = β« β1 + 22 ππ₯0
β1
= β5 β« ππ₯0
β1
= β5(0 β (β1))
= β5
2. Find the length of the curve π¦ =1
12π₯3 +
1
π₯ from π₯ = 2 to π₯ = 4.
ππ¦
ππ₯=
π₯2
4β
1
π₯2
(ππ¦
ππ₯)
2
= (π₯2
4β
1
π₯2)
2
= π₯4
16β
1
2+
1
π₯4
πΏ = β« β1 +π₯4
16β
1
2+
1
π₯4
4
2
ππ₯
161
= β« βπ₯4
16+
1
2+
1
π₯4
4
2
ππ₯
= β« β(π₯2
4+
1
π₯2)
2
4
2
ππ₯
= β« (π₯2
4+
1
π₯2) ππ₯
4
2
= [1
12π₯3 β
1
π₯]
2
4
= (16
3β
1
4) β (
2
3β
1
2)
= 59
12
3. Find the length of the curve π₯ = 2π¦3
2 from π₯ = 0 to π₯ = 3.
πΏ = β« β1 + (πβ²(π¦))2 π
π
ππ¦ = β« β1 + (ππ₯
ππ¦)
2
π
π
ππ¦
ππ₯
ππ¦= 3π¦
12
(ππ₯
ππ¦)
2
= 9π¦
πΏ = β« β1 + 9π¦ ππ¦3
0
= β« (1 + 9π¦)12 ππ¦
3
0
= [2
3 .
1
9 (1 + 9π¦)
32]
0
3
=2
27(28)
32
162
HOMEWORK ON CHAPTER 14
1. Find the arc length of π¦ = 2π₯ + 1 from π₯ = β3 to π₯ = 2.
2. Find the length of the curve π₯ =π¦4
8+
1
4π¦2 from π¦ = 1 to π¦ = 3.
3. Find the length of the curve π¦ =1
3π₯
3
2 β π₯1
2 from π₯ = 4 to π₯ = 9.
4. Find the length of the curve π¦ = ln (cos π₯) from π₯ = 0 to π₯ =π
6.
163
CHAPTER 15: HYDROSTATIC PRESSURE AND FORCE
In this chapter you will learn how to find hydrostatic force, the force due to some fluid pressure.
For the most part, we will consider water pressure. This is yet another application of integral
calculus.
The basic strategy for finding hydrostatic force is to break physical quantity into smaller parts,
approximate each small part, add results, take limits and evaluate the resulting integrals. This is
again similar to the technique of Riemann sums. To understand the idea behind the method for
finding hydrostatic force, we should note that
1. Water pressure increases with depth.
2. At any point in a liquid, the pressure is the same in all directions
Suppose that a thin horizontal plate with area A square meters is submerged in a fluid of mass
density π ππ/π3 at a depth of d meters below the surface of the fluid.
The volume of the fluid directly above the plate is:
π = π΄π
The mass of the fluid directly above the plate is:
π = ππππ ππ‘π¦ ππ πππ’ππ Γ π£πππ’ππ
= ππ΄π
where π is the Greek letter rho used here to denote the mass density of the fluid.
The force exerted by the fluid on the plate is hydrostatic force and it is given by:
π = ππ = πππ¨π
where π = 9.8 π/π 2 is acceleration due to gravity.
The hydrostatic pressure on the plate is the force per unit area and is given by:
164
π =πΉ
π΄
π· = πππ
This formula for hydrostatic pressure can also be written as:
π· = πΉπ
Similarly, the formula for hydrostatic force can be written as:
π = πΉπ¨π
where πΏ = ππ is the Greek letter delta used to denote the weight density of the fluid.
The mass density of water is: π = 1000 ππ/π3
The weight density of water is: πΏ = 62.5 ππ/ππ‘3
Steps for finding hydrostatic force
1. Divide submerged region into horizontal strips.
2. Find area A and depth d of the ππ‘β horizontal strip.
3. Find the force F acting on the ππ‘β horizontal strip:
F = pressure x Area
= πππ΄π
= πΏπ΄π
4. Integrate the force expression over the interval of depths occupied by the object.
Example 15.1
1. The end of a certain tank is in the shape of an isosceles triangle. The top of the triangle is
24 feet across. The tank is 8 feet deep at the tip of the triangle and 30 feet long. If the
165
tank is full of water, find the hydrostatic force on the end of the tank. (Assume water
weighs 62.5 πππ /ππ‘3).
We need to set up an axis system. Let π¦ = 0 correspond to the water surface and π¦ = 8
the depth.
Now break up the triangle into horizontal strips of width βπ¦. We assume that βπ¦ is small
enough so that the hydrostatic pressure on each strip is constant. Place the ππ‘β strip at a
depth of y. So the remaining depth is 8 β π¦.
We will use similar triangles to find the length of the strip which will allow us to find its
area. Thus we would need to draw a vertical line midway through the triangle.
Let a be half the length of the strip. Using similar triangles we get
π
8 β π¦=
12
8
8π = 12(8 β π¦)
π =3
2(8 β π¦)
So the length of the strip is
2π = 3(8 β π¦) = 24 β 3π¦
166
Since the width of the strip is βπ¦, then the area of the strip is
(24 β 3π¦)βπ¦
Hydrostatic force on the strip is:
πΉ = πΏπ΄π
πΉ = 62.5(24 β 3π¦)π¦ βπ¦
Hydrostatic force on the end of the tank =
β« 62.5(24 β 3π¦)π¦ ππ¦ = 62.5 β« (24π¦ β 3π¦2) ππ₯8
0
8
0
= 62.5 [12π¦2 β π¦3]08
= 62.5 (768 β 512)
= 16000 πππ
2. A dam has the shape of the trapezoid shown below. The height is 10 meters and the width
is 30 meters at the top and 20 meters at the bottom. Find the force on the dam due to the
hydrostatic pressure if
a) the dam is full
b) the water level is 2 meters from the top of the dam.
167
a) Again, we need to set up an axis system. Let π¦ = 0 correspond to the water surface and
π¦ = 10 the depth.
Break up the trapezoid into horizontal strips of width βπ¦. We assume that βπ¦ is small
enough so that the hydrostatic pressure on each strip is constant. Place the ππ‘β strip at a
depth of y. So the remaining depth is 10 β π¦.
We will use similar triangles to find the length of the strip which will allow us to find its
area. Thus we would need to draw a vertical line as shown in the diagram below:
Using similar triangles, we get
π
10 β π¦=
5
10=
1
2
2π = 10 β π¦
π =1
2(10 β π¦)
So the length of the strip is
20 + 2π = 20 + 10 β π¦ = 30 β π¦
168
Since the width of the strip is βπ¦, then the area of the strip is
(30 β π¦)βπ¦
Hydrostatic force on the strip is:
πΉ = πππ΄π
where π = 1000 ππ/π3 and π = 9.8 π/π 2. So
πΉ = 1000 . 9.8 (30 β π¦)π¦ βπ¦
Hydrostatic force on the end of the tank =
β« 980(30 β π¦)π¦ ππ¦ = 980 β« (30π¦ β π¦2) ππ¦10
0
10
0
= 980 [15π¦2 β1
3π¦3]
0
10
= 980 (1500 β1000
3)
=3430000
3 π
a) We do this again with the water level 2 meters from the top of the dam.
169
Again, using similar triangles we get
π
8 β π¦=
5
10=
1
2
2π = 8 β π¦
π =1
2(8 β π¦)
So the length of the strip is
20 + 2π = 20 + 8 β π¦ = 28 β π¦
Since the width of the strip is βπ¦, then the area of the strip is
(28 β π¦)βπ¦
Hydrostatic force on the strip is:
πΉ = πππ΄π
where π = 1000 ππ/π3 and π = 9.8 π/π 2. So
πΉ = 1000 . 9.8 (28 β π¦)π¦ βπ¦
Note that the upper limit of integration is now 8.
Hydrostatic force on the end of the tank =
β« 980(28 β π¦)π¦ ππ¦ = 980 β« (28π¦ β π¦2) ππ¦8
0
8
0
= 980 [14π¦2 β1
3π¦3]
0
8
= 980 (896 β512
3)
=2132480
3 π
170
HOMEWORK ON CHAPTER 15
1. A trough whose cross-section is a trapezoid, measures 6 meters across the bottom and 8
meters across the top, and is 3 meters deep. If the trough is filled with a liquid of mass
density π, what is the force due to hydrostatic pressure on one end of the trough? Write
your answer in terms of π.
2. A trough whose cross section is a trapezoid is 4 meters across the bottom, 10 meters
across the top, and 4 m deep. If the trough is filled to 3 meters with water, what is the
force due to water pressure on one end of the trough?
171
3. A trough whose cross section is an equilateral triangle with side 6 meters long is filled
with water. What is the force due to water pressure on one end of the trough?
172
CHAPTER 16: MOMENTS AND CENTER OF MASS
In this chapter, you will learn yet another application of integral calculus β the calculation of
moments and centers of mass. Our goal is to find the point P on which a thin plate of any shape
balances horizontally.
P is called the plateβs centroid or center of mass.
Moment is the tendency of an object to rotate about an axis.
Simple example
2 masses π1 and π2 are attached to a rod of negligible mass at distances π1 and π2 from the
fulcrum.
Under what conditions will the rod balance?
π1 π2
π1 π2
Moment of each mass is the product of the mass and its distance from the fulcrum, that is, π1π1
and π2π2.
The rod will balance when the moments are equal, that is, when
π1π1 = π2π2
This is the center of mass condition for such a system.
173
Now suppose that the rod lies along the x-axis with π1 at π₯1 and π2 at π₯2. Denote the center of
mass by οΏ½Μ οΏ½.
y
π₯1 οΏ½Μ οΏ½ π₯2 x
π1 π2
Then
π1(οΏ½Μ οΏ½ β π₯1) = π2(π₯2 β οΏ½Μ οΏ½)
π1οΏ½Μ οΏ½ β π1π₯1 = π2π₯2 β π2οΏ½Μ οΏ½
π1οΏ½Μ οΏ½ + π2οΏ½Μ οΏ½ = π1π₯1 + π2π₯2
(π1 + π2)οΏ½Μ οΏ½ = π1π₯1 + π2π₯2
οΏ½Μ οΏ½ =π1π₯1 + π2π₯2
π1 + π2
Thus the center of mass οΏ½Μ οΏ½ is given by
π π’π ππ π‘βπ ππππππ‘π ππ π‘βπ πππ π ππ
π π’π ππ π‘βπ πππ π ππ
In general, if there is a system of n masses π1, π2, β¦ , ππ located at the points π₯1, π₯2, β¦ , π₯π on
the x-axis, the center of mass is:
οΏ½Μ οΏ½ =π1π₯1 + π2π₯2 + β― + πππ₯π
π1 + π2 + β― + ππ=
β πππ₯πππ=1
β ππππ=1
= β πππ₯π
ππ=1
π
where m is the sum of the masses and πππ₯π denotes the moment of respective mass.
174
π = β πππ₯πππ=1 is called the moment of the system about the origin.
Now consider a system of n particles with masses π1, π2, β¦ , ππ located at points
(π₯1, π¦1), β¦ , (π₯π, π¦π).
y
β π1 π¦1
β π2 π¦2
π¦π β ππ
π₯1 π₯2 π₯π x
Let us find a formula for the moment of the system about the y-axis. To do this, we multiply each
mass by its distance from the y-axis and add up the products. Denote the moment about the y-
axis by ππ¦. We define ππ¦ to be
π΄π = ππππ + ππππ + β― + ππππ = β ππππ
π
π=π
Similarly, we define ππ₯, the moment of the system about the x-axis to be:
π΄π = ππππ + ππππ + β― + ππππ = β ππππ
π
π=π
Denote the center of mass to be the point (οΏ½Μ οΏ½, οΏ½Μ οΏ½). Then
175
οΏ½Μ οΏ½ = ππππ + ππππ + β― + ππππ
ππ + ππ + β― + ππ=
β ππππππ=π
π =
π΄π
π
οΏ½Μ οΏ½ =ππππ + ππππ + β― + ππππ
ππ + ππ + β― + ππ=
β ππππππ=π
π=
π΄π
π
Example 16.1
Find the moments and center of mass of the system of objects that have masses 2, 5 and 6 at the
points (β1,1), (2,0) and (3, β1).
ππ¦ = β πππ₯π
π
π=1
So the moment of the system about the y-axis is:
ππ¦ = 2(β1) + 5(2) + 6(3) = 26
ππ₯ = β πππ¦π
π
π=1
So the moment of the system about the x-axis is:
ππ₯ = 2(1) + 5(0) + 6(β1) = β4
Sum of the masses = 2 + 5 + 6 = 13
οΏ½Μ οΏ½ = ππ¦
π
οΏ½Μ οΏ½ =26
13= 2
οΏ½Μ οΏ½ = ππ₯
π
οΏ½Μ οΏ½ = β4
13
176
So the center of mass is (2, β4
13)
Now consider a lamina with uniform density π that occupies a region R in the plane. We want to
locate the center of mass (or centroid) of the plate.
If the region R is symmetric about a line, then we can use the symmetry principle to find the
centroid:
Symmetry principle: If R is symmetric about a line L, then the centroid of R lies on L.
If the region R has 2 lines of symmetry, then the centroid is the point of intersection of those
lines. For example, the centroid of a circle is its center.
Now suppose that R is not a symmetric region. Let us find a formula for the centroid.
We divide R into rectangles like we did when finding Riemann sums:
Divide [π, π] into n sub-intervals π₯1, π₯2, β¦ , π₯π with width βπ₯ where
βπ₯ =π β π
π
Let π₯οΏ½Μ οΏ½ be the midpoint of each interval.
Recall that the centroid of a rectangle is the point of intersection of the two lines of symmetry.
So the centroid of the ππ‘β rectangle π π is (π₯οΏ½Μ οΏ½ ,1
2π(π₯οΏ½Μ οΏ½ )).
To find moments, we would need to find the mass of the system. Let us find the mass of the ππ‘β
rectangle.
177
πππ π = ππππ ππ‘π¦ Γ ππππ
Area of the ππ‘β rectangle = π(π₯οΏ½Μ οΏ½)βπ₯
Mass of the ππ‘β rectangle = ππ(π₯οΏ½Μ οΏ½)βπ₯
Moment of π π about y-axis = Mass Γ Distance of centroid to y-axis
That is, ππ¦(π π) = [ ππ(π₯οΏ½Μ οΏ½)βπ₯]π₯οΏ½Μ οΏ½
So
π΄π = π₯π’π¦πββ
β πποΏ½Μ οΏ½π(ποΏ½Μ οΏ½)βπ = π β« ππ(π) π ππ
π
π
π=π
Similarly
π΄π = π β«π
π[π(π)]π
π
π
π π
Mass is given by:
π = π β« π(π) π ππ
π
So
οΏ½Μ οΏ½ = π΄π
π=
π β« ππ(π) π ππ
π
π β« π(π) π ππ
π
= β« ππ(π) π π
π
π
β« π(π) π ππ
π
οΏ½Μ οΏ½ = π΄π
π=
π β«ππ
[π(π)]ππ
π π π
π β« π(π) π ππ
π
= β«
ππ
[π(π)]ππ
π π π
β« π(π) π ππ
π
Note that the centroid is not dependent on the density π.
178
Example 16.2
1. Find the center of mass of a semicircular plate of radius 1 cm.
The plate is symmetrical about the y-axis so οΏ½Μ οΏ½ = 0.
Let us now find οΏ½Μ οΏ½.
Equation of circle is π₯2 + π¦2 = 1.
Solving for y we get the equation of the semi-circle:
π¦ = β1 β π₯2
This is our function π(π₯).
οΏ½Μ οΏ½ = ππ₯
π=
β«12
[π(π₯)]2π
π ππ₯
β« π(π₯) ππ₯π
π
For the semi-circle
β«1
2[π(π₯)]2
π
π
ππ₯ = β«1
2
1
β1
(1 β π₯2) ππ₯
=1
2[π₯ β
1
3π₯3]
β1
1
179
=1
2[(1 β
1
3) β (β1 +
1
3)]
=2
3
Now β« π(π₯) ππ₯π
π is the area of the semi-circle which can be easily found without
integrating.
Area of the semicircle =
1
2 . π . 12 =
π
2
So
οΏ½Μ οΏ½ = 2
3βπ
2β=
4
3π
Center of mass = (0,4
3π).
2. Find the centroid of the lamina occupying the region bounded by π¦ = π₯2, π¦ = 0, π₯ = 0,
and π₯ = 1.
180
οΏ½Μ οΏ½ = ππ¦
π=
β« π₯π(π₯) ππ₯π
π
β« π(π₯) ππ₯π
π
οΏ½Μ οΏ½ = ππ₯
π=
β«12
[π(π₯)]2π
π ππ₯
β« π(π₯) ππ₯π
π
π = β« π₯2 ππ₯1
0
= [1
3π₯3]
0
1
= 1
3
ππ¦ = β« π₯ . π₯2 ππ₯1
0
= β« π₯3 ππ₯1
0
= [1
4π₯4]
0
1
= 1
4
οΏ½Μ οΏ½ = 1
4β
13β
= 3
4
ππ₯ = β«1
2 (π₯2)2 ππ₯
1
0
181
= β«1
2π₯4 ππ₯
1
0
= [1
10π₯5]
0
1
= 1
10
οΏ½Μ οΏ½ = 1
10β
13β
= 3
10
Centroid = (3
4,
3
10)
3. Find the centroid of the region bounded by the line π¦ = βπ₯ and π¦ = π₯2.
To find the centroid of regions bounded by two curves we apply the same principles
when we found area between two curves. Thus
182
οΏ½Μ οΏ½ = ππ¦
π=
β« π₯[π(π₯) β π(π₯)] ππ₯π
π
β« π(π₯) β π(π₯) ππ₯π
π
οΏ½Μ οΏ½ = ππ₯
π=
β«12
{[π(π₯)]2 β [π(π₯)]2}π
π ππ₯
β« π(π₯) β π(π₯) ππ₯π
π
where π(π₯) is the function above the region and π(π₯) is the function above the region.
We first find the limits of integration:
π₯2 = βπ₯
π₯2 + π₯ = 0
π₯(1 + π₯) = 0
π₯ = 0, π₯ = β1
So
ππ¦ = β« π₯(βπ₯ β π₯2)0
β1
ππ₯
= β« (βπ₯2 β π₯3)0
β1
ππ₯
= [β1
3π₯3 β
1
4π₯4]
β1
0
= 0 β (1
3β
1
4)
= β1
12
π = β« (βπ₯ β π₯2)0
β1
ππ₯
183
= [β1
2π₯2 β
1
3π₯3]
β1
0
= 0 β (β1
2+
1
3)
= 1
6
οΏ½Μ οΏ½ = β 1
12β
16β
= β 1
2
ππ₯ = β«1
2[(βπ₯)2 β (π₯2)2] ππ₯
0
β1
= β«1
2(π₯2 β π₯4) ππ₯
0
β1
= 1
2[1
3π₯3 β
1
5π₯5]
β1
0
= 0 β1
2(β
1
3+
1
5)
= 1
15
οΏ½Μ οΏ½ = 1
15β
16β
= 2
5
Centroid = (β1
2,
2
5)
184
HOMEWORK ON CHAPTER 16
1. Find the centroid of the region bounded by π¦ = 9 β π₯2 and the x-axis.
2. Find the centroid of the lamina occupying the region bounded by π¦ = π₯3, π¦ = 0,
π₯ = 0, and π₯ = 1.
3. Find the centroid of the region bounded by π¦ = 6 β π₯2 and π¦ = π₯.
4. Find the centroid of the region bounded by π¦ = π₯2 and π¦ = βπ₯.
185
CHAPTER 17: POLAR COORDINATES
To understand the concepts in this chapter, you will need to review the unit circle.
The Unit Circle
From this unit circle, you can easily find the sine and cosine of some angles given in radians and
degrees. For example,
sin 210Β° = β1
2
186
cos 210Β° = ββ3
2
sin7π
4= β
β2
2
cos7π
4=
β2
2
Recall that
tan π₯ =sin π₯
cos π₯
So, for example
tan 210Β° =β 1
2β
β β32
β=
1
β3
Polar Coordinates
So far we have dealt with points in the Cartesian plane (that is, x-y plane) which are specified
using Cartesian or rectangular coordinates (π₯, π¦). However, in some situations, it would be better
to specify a point in another way for ease of calculation. In this chapter, you will learn about
writing a point using polar coordinates (π, π).
r is the length of the line from the origin to the point (π₯, π¦)
π is the angle the line makes with the positive x-axis. When we move in an counterclockwise
direction, π is positive and π is negative when we move in a clockwise direction.
187
π can be negative. To plot a point with a negative r-coordinate, we plot the point with the
positive r-coordinate and then extend it in the opposite direction to get the negative r-coordinate.
Thus if (π, π), π > 0 is in the third quadrant, then (βπ, π) is in the first quadrant.
Example 17.1
Plot the following points:
1. (3,2π
3)
189
Converting from polar to rectangular coordinates
Consider the diagram:
From the right angled triangle and basic trigonometric formulas, recall that
sin π =πππππ ππ‘π
βπ¦πππ‘πππ’π π
cos π = ππππππππ‘
βπ¦πππ‘πππ’π π
So
sin π =π¦
π
cos π = π₯
π
Thus to convert a point from polar to rectangular coordinates, use the formulas:
π₯ = π cos π, π¦ = π sin π
Example 17.2
Convert to rectangular coordinates:
1. (2,π
2)
π₯ = 2 cosπ
2= 0
190
π¦ = 2 sinπ
2= 2
(2,π
2) = (0,2)
2. (β2,7π
6)
π₯ = β2 cos7π
6= β2 . β
β3
2= β3
π¦ = β2 sin7π
6= β2 . β
1
2= 1
(β2,7π
6) = (β3, 1)
Converting from rectangular to polar coordinates
Again, from the diagram:
π2 = π₯2 + π¦2 from the Pythagorean theorem, and tan π = π¦
π₯.
Thus to convert a point from rectangular coordinates to polar coordinates, first illustrate the point
and to find π and π, use the formulas
π2 = π₯2 + π¦2
tan π = π¦
π₯
191
But exercise caution! Consider which quadrant the point lies in. It is important to refer to the unit
circle. Also recall that tangent is positive in the first and third quadrants, and negative in the
second and fourth quadrants.
Example 17.3
Convert to polar coordinates. Find 4 different sets of points such that the angle satisfies
β2π β€ π β€ 2π
1. (1, ββ3)
We first plot the point:
π = β1 + 3 = 2
tan π = ββ3
192
But π is in the 4th quadrant. From the unit circle,
π =5π
3
But recall π can be negative when moving in a clockwise direction. Thus
π =5π
3β 2π = β
π
3
This gives us these 2 points:
(2,5π
3) , (2, β
π
3)
Now π can be negative. So extend the line in the opposite direction so that we are now in the 2nd
quadrant.
From the unit circle
π =2π
3
That is,
π = π βπ
3=
2π
3
Again, π can be negative. So
193
π =2π
3β 2π = β
4π
3
We now have 2 more points:
(β2,2π
3) , (β2, β
4π
3)
2. (2, 2)
We first plot the point:
The point is in the first quadrant.
π = β4 + 4 = β8 = 2β2
tan π = 1
π =π
4
194
We calculate a negative value for π:
π =π
4β 2π = β
7π
4
This gives us these 2 points:
(2β2,π
4) , (2β2, β
7π
4)
π can be negative. So extend the line in the opposite direction so that we are now in the
3rd quadrant.
From the unit circle,
π =5π
4
That is,
π = π +π
4=
5π
4
For negative π:
195
π =5π
4β 2π = β
3π
4
We now have 2 more points:
(2β2,5π
4) , (2β2, β
3π
4)
Graphs of polar equations
So far, you know how to draw graphs in the Cartesian plane. Now, you will learn how to draw
polar curves.
Simple example
1. π = 1 is the unit circle.
2. π = 3 is a circle centered at the origin with radius 3.
Now, you will learn to sketch more complicated curves by hand. Again, refer to the position of
the angles on the unit circle to help you draw the graphs. You first construct a table of values.
Example 17.4
Graph
1. π = 2 cos π
Here we will use intervals of π
2 for π.
π 0 π
2 π
π 2 0 -2
197
3. π = β3 sin π for 0 β€ π β€π
2
π 0 π
4
π
2
π 0 β
3
β2
-3
4. π = cos 2π
π 0 π
4
π
2
3π
4
π
π 1 0 -1 0 1
198
Points of intersection of polar curves
We find points of intersection the usual way by solving the two equations simultaneously. This is
useful when finding the area between two polar curves.
Example 17.5
Find the point(s) of intersection the curves π = sin π and π = cos π
We graph the two functions:
199
Solving the two equations we get:
sin π = cos π
Dividing by cos π, cos π β 0:
tan π = 1
π =π
4
Exercise caution when using the graphs of the function to find the points of intersection.
From the graph it appears that (0,0) is indeed a point of intersection. But this is false
since π = 0 does not satisfy
sin π = cos π
Area in polar coordinates
Consider the shaded region R below bounded by the curve π = π(π) and the lines representing
π = πΌ and π = π½.
200
The area of R is given by:
1
2β« [π(π)]2 ππ
π
π
Let us look at some simple examples. Note that you should always make a sketch of the curve
and shade the region. You can use your calculator to graph the curves.
How to graph polar curves on the TI-84
β’ Press the MODE key.
β’ Use the ARROW keys and the ENTER key to select Pol.
β’ Press the Y= key to enter the function. If there are two functions, then enter the second
one below the first.
β’ Use the WINDOW key to change the domain and range of the function.
β’ Press GRAPH.
201
Example 17.6
In the following examples, only find the integrals, but do not evaluate them.
1. Find the integral which gives the area inside π = 2 cos π.
Recall: we had graphed this function as follows:
π 0 π
2 π
π 2 0 -2
Notice that 0 β€ π β€ π maps out the entire circle. So these would be the limits of
integration.
Thus, to find the area, evaluate:
1
2β« (2 cos π)2 ππ = β« 2 πππ 2π ππ
π
0
π
0
202
2. Identify the integral which gives area inside π = 1 + sin π
Here this region is defined for 0 β€ π β€ 2π. So to find the area, evaluate:
1
2β« (1 + sin π)2 ππ
2π
0
Now let us examine some other examples in which we will find the area between curves.
Steps to finding the area between two curves
1. Graph the curves
2. Label the curves
3. Shade the region
4. Find π from the point of intersection
5. Find the limits of integration
6. It may be necessary to draw a line from the origin to the point of intersection. The
importance of this will be revealed shortly.
203
7. Use the concept of symmetry when finding area of symmetrical regions.
Example 17.7
Set up, but do not evaluate the integrals:
1. Find the area of the region outside the circle π = 3 cos π but inside the cardioid
π = 1 + cos π.
We graph, label the curves and shade the region.
Notice that the region is symmetrical so it suffices to take either the lower or upper region
and multiply its area by 2 to find the area of the entire region. Let us take the upper
region.
We now find the points of intersection:
1 + cos π = 3 cos π
2 cos π = 1
204
cos π =1
2
π =π
3
Now draw a line from π
3 to the origin. Notice that in order to find the required area we
will have to subtract the area for the region bounded by the circle from the region
bounded by the cardioid. Let us set up these integrals separately:
Circle: The curve for that particular region is defined for π
3β€ π β€
π
2. So the integral to be
evaluated is:
1
2β« (3 cos π)2 ππ
π2
π3
Cardioid: The curve for that particular region is defined for π
3β€ π β€ π. So the integral to
be evaluated is:
1
2β« (1 + cos π)2 ππ
π
π3
Thus the integral for evaluating the required region is:
2 [1
2β« (1 + cos π)2 ππ β
1
2
π
π3
β« (3 cos π)2 ππ
π2
π3
]
= β« (1 + cos π)2 ππ β π
π3
β« (3 cos π)2 ππ
π2
π3
2. Find the area of the region outside the circle π = cos π but inside the circle
π = sin π.
We graph, label the curves and shade the region.
205
We now find the points of intersection:
sin π = cos π
tan π = 1
π =π
4
Now draw a line from π
4 to the origin. Notice that in order to find the required area we
will have to subtract the area for the region bounded by π = cos π from the region
bounded by π = sin π. Let us set up these integrals separately:
π = sin π: The curve for that particular region is defined for π
4β€ π β€ π. So the integral to
be evaluated is:
1
2β« (sin π)2 ππ
π
π4
π = cos π: The curve for that particular region is defined for π
4β€ π β€
π
2. So the integral to
be evaluated is:
206
1
2β« (cos π)2 ππ
π2
π4
Thus the integral for evaluating the required region is:
=1
2β« (sin π)2 ππ
π
π4
β1
2β« (cos π)2 ππ
π
π3
3. Find the area of the region inside π = 2 + sin π and inside π = β3 sin π.
We graph, label the curves and shade the region.
Notice that the region is symmetrical so it suffices to take either the left or right region
and multiply its area by 2 to find the area of the entire region. Let us take the left region.
We now find the points of intersection:
2 + sin π = β3 sin π
207
β4 sin π = 2
sin π = β1
2
π =7π
6
But this value of π applies to π = 2 + sin π and not to π = β3 sin π since the latter is
negative.
For π = β3 sin π, π =π
6.
Now draw a line from 7π
6 to the origin. Notice that in order to find the required area we
will have to add the area for the region bounded by π = 2 + sin π to the region bounded
by π = β3 sin π. Let us set up these integrals separately:
π = β3 sin π: The curve for that particular region is defined for 0 β€ π β€π
6. So the
integral to be evaluated is:
1
2β« (β3 sin π)2 ππ
π6
0
π = 2 + sin π: The curve for that particular region is defined for 7π
6β€ π β€
3π
2. So the
integral to be evaluated is:
1
2β« (2 + sin π)2 ππ
3π2
7π6
Thus the integral for evaluating the required region is:
2 [1
2β« (β3 sin π)2 ππ
π6
0
β1
2β« (2 + sin π)2 ππ
3π2
7π6
]
= β« (β3 sin π)2 ππ
π6
0
β β« (2 + sin π)2 ππ
3π2
7π6
208
HOMEWORK ON CHAPTER 17
Set up, but do not evaluate the integrals:
1. Find the area of the region inside the circle π = cos π but outside the circle
π = sin π.
2. Find the area of the region inside the circle π = 2 but outside π = 3 + 2 sin π.
3. Find the area of the region enclosed by the curve π = 2 sin π.
4. Find the area of the region that lies inside the curves π = 2 cos π and π = 1.
5. Find the area of the region that lies inside π = 2 + sin π and outside π = 3 sin π.
6. Find the area of the region inside the cardioid π = 3 β 3 sin π and outside the
cardioid π = 1 + sin π.
209
CHAPTER 18: SEQUENCES
In this chapter, you will learn about sequences which is an important pre-requisite for learning
about series, which will follow subsequently.
Definition
A sequence is a list of numbers written in a definite order.
e.g. {1, 3, 5, 7, β¦}
The numbers 1, 3, 5, 7 are called the terms of the sequence. For example, 1 is the first term of the
sequence.
Notation
The sequence {π1, π2, β¦ , ππ, β¦ } is denoted by {ππ}
π1 is the first term
π2 is the second term
ππ is the ππ‘β term. It is also called the general term or general formula for the sequence.
Listing the terms of a sequence
To list the terms of a sequence, we substitute appropriate values for n. For example, to find the
2nd term of a sequence, evaluate the ππ‘β term at π = 2.
Example 18.1
List the first 4 terms in the sequences:
1. ππ = 2π + 1
210
π1 = 2(1) + 1 = 3
π2 = 2(2) + 1 = 5
π3 = 2(3) + 1 = 7
π4 = 2(4) + 1 = 9
So
{ππ} = {3, 5, 7, 9}
2. ππ = (β1)π π+3
2π
π1 = (β1)11 + 3
21= β2
π2 = (β1)22 + 3
22=
5
4
π3 = (β1)33 + 3
23= β
3
4
π4 = (β1)44 + 3
24=
7
16
So
{ ππ} = {β2,5
4, β
3
4,
7
16}
Notice that when the ππ‘β term of a sequence contains (β1)π or similar, the terms of the sequence
alternate in sign.
211
Finding the formula for ππ, the general term of a sequence
Example 18.2
1. {ππ} = {1
3,
1
6,
1
9,
1
12, β¦ }
What general formula can be used to define the sequence? Let us look at the terms to see
whether we can find a pattern:
π1 =1
3=
1
3(1)
π2 =1
6=
1
3(2)
π3 =1
9=
1
3(3)
π4 =1
12=
1
3(4)
So
ππ =1
3π
2. {ππ} = {1
2,
1
4,
1
8,
1
16, β¦ }
π1 =1
2=
1
21
π2 =1
4=
1
22
π3 =1
8=
1
23
π4 =1
16=
1
24
So
212
ππ =1
2π
3. {ππ} = [2, 5, 8, 11, β¦ }
Note that terms increase by 3. This means that ππ will be of the form:
ππ = 3π + π₯
We find x by evaluating the terms in the sequence. For example:
π1 = 2 = 3(1) β 1
π2 = 5 = 3(2) β 1
So
ππ = 3π β 1
4. {ππ} = {β7
5,
11
25, β
15
125,
19
625, β
23
3125, β¦ }}
Note the terms have alternating signs which means that ππ contains a power of (-1).
Since the first term is negative and every odd numbered term is negative, then ππ
contains (β1)π.
The denominators of the terms are powers of 5 so the denominator of ππ is 5π.
Now consider the numerators 7, 11, 19, 19, 23.
The terms increase by 4. This is similar to the previous example. So the numerator of ππ
is 4π + 3.
Thus
ππ = (β1)π 4π + 3
5π
213
Recursively defined sequences
Consider the sequence
{ππ} = [2, 5, 8, 11, β¦ }
We found that the general term of this sequence is:
ππ = 3π β 1
However, this sequence could also have been defined as follows:
π1 = 2
ππ+1 = ππ + 3, π β₯ 1
This is a recursive definition of the sequence where one term is given (usually the first term), and
subsequent terms are defined in terms of the preceding ones.
Note that ππ+1 is the term following ππ.
So in the example above, since we are given π1, we can now find π2. We are not able to find π3
unless we know π2. Similarly, we are not able to find π31 unless we know π30. This is how a
recursively defined sequence is evaluated, which is unlike what we have done before. Thus
π2 = π1 + 3 = 2 + 3 = 5
π3 = π2 + 3 = 5 + 3 = 8
π4 = π3 + 3 = 8 + 3 = 11
A famous recursively defined sequence is the Fibonacci sequence:
{1,1,2,3,5,8,13, β¦ }
It is defined recursively as follows:
π1 = 1, π2 = 1
ππ = ππβ1 + ππβ2, π β₯ 3
214
Example 18.3
Find the first 5 terms of the following sequences:
1. π1 = 1, ππ+1 = 3ππ β 1
π1 = 1
π2 = 3π1 β 1 = 3(1) β 1 = 2
π3 = 3π2 β 1 = 3(2) β 1 = 5
π4 = 3π3 β 1 = 3(5) β 1 = 14
π5 = 3π4 β 1 = 3(14) β 1 = 41
2. π1 = 2, ππ+1 = ππ + 3π
π1 = 2
π2 = π1 + 3(1) = 2 + 3 = 5
π3 = π2 + 3(2) = 5 + 6 = 11
π4 = π3 + 3(3) = 11 + 9 = 20
π5 = π4 + 3(4) = 20 + 12 = 32
3. ππ = ππβ1 β ππβ2, π1 = 5, π2 = 3
π1 = 5
π2 = 3
π3 = π2 β π1 = 3 β 5 = β2
π4 = π3 β π2 = β2 β 3 = β5
215
π5 = π4 β π3 = β5 β (β2) = β5 = β3
Convergence of sequences
A sequence {ππ} has a limit L, if limπββ
ππ = πΏ.
If L exists, then the sequence converges; otherwise it diverges.
Note that you will need to review finding limits at infinity. Recall, this was reviewed in Chapter
2. Please return to that chapter and review that topic if necessary.
Example 18.4
1. Consider the sequence whose ππ‘β term is given by:
ππ =1
3π
limπββ
1
3π= 0
.
So the sequence
{ππ} = {1
3,1
6,1
9,
1
12, β¦ }
converges to 0.
2. Consider the sequence whose ππ‘β term is given by:
ππ = π
limπββ
π = β
.
So the sequence {ππ} = π diverges.
216
Helpful hints for testing the convergence of sequences
1. Consider a sequence {ππ} whose terms alternate in sign. As we have seen before, these
are sequences whose ππ‘β term has a form containing (β1)π.
If limπββ
|ππ| = 0, then limπββ
ππ = 0; otherwise the sequence diverges.
For example, the sequence given by
ππ = (β1)π1
3π
converges to 0 since
limπββ
1
3π= 0
But the sequence given by
ππ = (β1)ππ
3π + 1
diverges since
limπββ
π
3π + 1= β
1
3
The following points give other techniques for evaluating limits at infinity which are useful when
testing for convergence of sequences.
2. limπββ
ππ = {0, β 1 < π < 11, π = 1β, ππ‘βπππ€ππ π
3. limπββ
1
ππ = 0 if p > 0.
4. The laws of limits apply to sequences.
217
5. Use the squeeze theorem where applicable. That is:
If ππ β€ ππ β€ ππ for all n and limπββ
ππ = limπββ
ππ = πΏ, then limπββ
ππ = πΏ.
6. LβHοΏ½ΜοΏ½pitalβs rule may need to be used.
Review of LβHοΏ½ΜοΏ½pitalβs rule
Suppose f and g are differentiable functions and πβ²(π₯) β 0 on an open interval containing c
except possibly at c. Suppose that
limπ₯βπ
π(π₯) = 0 and limπ₯βπ
π(π₯) = 0
or that
limπ₯βπ
π(π₯) = Β±β and limπ₯βπ
π(π₯) = Β±β
Then
limπ₯βπ
π(π₯)
π(π₯)= lim
π₯βπ
πβ²(π₯)
πβ²(π₯)
A second application of LβHοΏ½ΜοΏ½pitalβs rule is sometimes necessary if upon the first application, the
result is still in indeterminate form.
For example:
limπββ
ln π
π= lim
π ββ
1πβ
1= 0
limπββ
ππ
π2= lim
π ββ
ππ
2π= lim
π ββ
ππ
2= β
Example 18.5
Determine whether the following sequences converge. If a sequence converges, what is its limit?
218
1. ππ = 3π β 1
limπββ
(3π β 1) = β
The sequence diverges.
2. ππ =3π3+5π2β1
4π3+7
limπββ
3π3 + 5π2 β 1
4π3 + 7=
3
4
The sequence converges to 3
4.
3. ππ = (1
2)
π
Since β1 <1
2< 1,
limπββ
(1
2)
π
= 0
The sequence converges to 0.
4. ππ = (β1)π 3π
π5+2
limπββ
3π
π5 + 2= 0
The sequence converges to 0.
219
5. ππ = (7
3)
π
Since 7
3> 1,
limπββ
(7
3)
π
= β
The sequence diverges.
6. ππ =3π2
π2π
limπββ
3π2
π2π= lim
πββ
6π
2π2π= lim
πββ
6
4π2π= 0
The sequence converges to 0.
220
HOMEWORK ON CHAPTER 18
1. List the first 5 terms of the sequence:
a) ππ =2π
3π+1
b) ππ = (β1)π+1 π+3
π!
c) π1 = 2, ππ+1 = 3 β 5ππ
d) π1 = 1, π2 = 3, ππ = 2ππβ1 β ππβ2
2. Find a formula for the general term ππ of the sequence:
a) {5, 7, 9,11, β¦ . }
b) {1, 4,16, 64, 256, β¦ }
c) {β5
3,
10
9, β
15
27,
20
81, β¦ }
d) {1, β1, 1, β1, 1, β1 β¦ }
3. Determine whether the sequence converges or diverges. If it converges, find its limit:
a) ππ = 5 β1
π2
b) ππ = π4
2π3β1
c) ππ = (0.4)π
d) ππ = 3π5β5
11π5β4π3β1
e) ππ = π(π β 1)
f) ππ = (9
8)
π
g) ππ = π3πβπ
h) ππ = ln π
ln 2π
i) ππ = (β1)π 1
2π3β1
j) ππ = (β1)π+1 2π3
3π3β5
k) ππ = π sin1
π
221
CHAPTER 19: SERIES I β GEOMETRIC SERIES
Review of summation notation
The summation symbol, denoted by β , tells us to add up its elements.
Here are some important terms associated with the summation symbol:
Example 19.1
1. β π = 1 + 2 + 3 + 4 + 5 β¦5π=1
2. β π2 = 12 + 22 + 32 + 42 + β―βπ=1
3. β 3π β 1 = 5 + 8 + 11 + 14 + β―βπ=2
Definition of a series
An expression of the form π1 + π2 + β― + ππ + β¦ is called an infinite series (or series) and is
denoted by β ππβπ=1 or simply β ππ.
The ππβ²π are the terms of the series.
For example:
β 2π = 2 + 4 + 6 + 8 + β―
β
π=1
An important example of an infinite series is the geometric series which can take the following
forms:
β πππβ1 = π + ππ + ππ2 + ππ3 + β― .
β
π=1
222
β πππ = π + ππ + ππ2 + ππ3 + β― .
β
π=0
In the geometric series, each term is obtained from the preceding one by multiplying it by a
constant called the common ratio. The common ratio is denoted by r and the first term of the
geometric series is denoted by a.
For example, consider the geometric series:
β 5 (1
2)
πβ
π=1
=5
2+
5
4+
5
8+
5
16+ β―
π =5
2, π =
1
2
Example 19.2
Decide which of the following are geometric series. For those which are, find a and r.
1. 2 + 10 + 50 + 250 + β―
This is a geometric series with π = 5, π = 2
2. 3 + 12 + 16 + 64 + 256 + β―
This is not a geometric series because there is no common ratio.
3. 1 + 4 + 7 + 10 + 13 + β―
This is not a geometric series.
223
4. 3 +3
2+
3
3+
3
4+ β―.
This is not a geometric series.
5. 5 β10
3+
20
9β
40
27+ β―
This is a geometric series with π = β2
3, π = 5
Sum of a finite geometric series
We want a formula for finding ππ, the sum of the first n terms of the finite geometric series:
ππ = π + ππ + ππ2 + ππ3 + β― πππβ1 β¦ β¦ β¦ β¦ β¦ β¦ β¦ β¦ . (1)
Multiply this equation by r to get (2):
ππ = π + ππ + ππ2 + ππ3 + β― πππβ1 β¦ β¦ β¦ β¦ β¦ β¦ β¦ β¦ . (1)
πππ = ππ + ππ2 + ππ3 + β― πππβ1 + πππ β¦ β¦ β¦ β¦ β¦ β¦ β¦ β¦ . (2)
Subtracting the 2 equations we get:
ππ β πππ = π β πππ
ππ(1 β π) = π(1 β ππ)
So
πΊπ =π(π β ππ)
π β π, π β π
Thus to find the sum of a finite geometric series, you would need the common ration r, the first
term a, and the number of terms n.
224
Example 19.3
Find the sum of the finite geometric series:
1. 2 + 1 +1
2+
1
4+
1
8+ β― +
1
26
π = 2, π =1
2, π = 8
ππ =π(1 β ππ)
1 β π
So
π8 =2(1 β (0.5)8)
1 β 0.5
= 3.984375
2. 3 + 3(0.2) + 3(0.2)2 + β― + 3(0.2)9
π = 3, π = 0.2, π = 10
π20 =3(1 β (0.2)10)
1 β 0.2
β 3.75
Sum of an infinite geometric series
Using the same approach for finding ππ:
πβ = π + ππ + ππ2 + ππ3 + β― β¦ β¦ β¦ β¦ β¦ β¦ β¦ β¦ . (1)
πππ = ππ + ππ2 + ππ3 + β― β¦ β¦ β¦ β¦ β¦ β¦ β¦ β¦ . (2)
Subtracting the 2 equations we get:
ππ β πππ = π
ππ(1 β π) = π
So
225
ππ =π
1 β π
We can find this formula in another way:
Denote the partial sum, the sum of the first n terms of an infinite geometric series, by ππ .
Denote the sum of an infinite series by S. Then
π = limπββ
ππ
So, to find the sum of an infinite geometric series, we find:
π = limπββ
ππ = limπββ
π(1 β ππ)
1 β π
which depends on the value of r.
Recall: limπββ
ππ = {0, β 1 < π < 11, π = 1β, ππ‘βπππ€ππ π
So
π = limπββ
π(1 β ππ)
1 β π=
π
1 β π
for β1 < π < 1.
That is, the sum of an infinite geometric series is given by
πΊ =π
π β π
for |π| < π.
We say that the infinite geometric series β ππ converges to π
1βπ for |π| < 1; otherwise it diverges.
226
Example 19.4
1. Find the sum of the geometric series:
a) 5 β5
4+
5
16β
5
64+ β―
π = 5, π = β1
4
π =π
1 β π
So
π =5
1 β (β14
)
= 4
The series converges to 4.
b) β (2
5)
πβπ=2
π = (2
5)
2
=4
25
π =2
5
So
π =
425
1 β25
=4
15
The series converges to 4
15.
227
Applications
Example 19.5
1. Write the number 1.345Μ Μ Μ Μ = 1.3454545 β¦ as a ratio of integers.
1.345Μ Μ Μ Μ = 1.3454545 β¦
= 1.3 + 0.045 + 0.00045 + 0.0000045 + β―.
=13
10+
45
103+
45
105+
45
107+
Observe that if the first term is excluded then, the infinite series is a geometric series with
π =45
103=
45
1000
π =1
102=
1
100
The sum of this infinite geometric series is:
451000
1 β1
100
=45
990
So
1.345Μ Μ Μ Μ = 1.3454545 β¦
=13
10+
45
990
=1332
990
228
2. How much money will you have at the end of 25 years if you deposit $100 at 5%
compounded annually?
To find the amount at the end of each year we would add the interest accrued to the
principal.
Interest = 5% of principal or 0.05 Γ principal
Principal +Interest = 1.05 Γ principal
Amount at the end of year 1 = $1.05(100) = $105
Amount at the end of year 2 = $1.05(105) = 1.05(1.05)(100) = 1.052(100)
Amount at the end of year 3 = $1.05(1.05)(105) = 1.05(1.05)(1.05)(100) = 1.053(100)
So, the amount at the end of 25 years is:
$1.0525(100) = $338.64
229
HOMEWORK ON CHAPTER 19
1. Find the sum of the finite geometric series:
a) β (0.8)π10π=2
b) 1 +1
5+
1
52 + β― +1
513
2. Determine whether the infinite geometric series converges or diverges. If it converges,
find its sum.
a) β 3(0.4)πβπ=0
b) β (1.6)πβπ=2
c) 2 β4
5+
8
25β
16
125+ β―.
3. Write the number 1.72Μ as a ratio of integers.
4. Sam takes 200 mg of a drug at 6:00 pm every day. 3% of the drug remains in the body
after the drug is taken.
a) How much of the drug is in the body after the 2nd day?
b) How much of the drug is in the body after the 5th day?
c) How much of the drug is in the body after the nth day?
d) How much of the drug remains in the body in the long run?
230
CHAPTER 20: SERIES II β CONVERGENCE OF SERIES
So far you have learned how to test the geometric series for convergence. In this chapter, you
will learn some techniques for testing series for convergence
The partial sum technique
Recall that the ππ‘β partial sum ππ is the sum of the first n terms of an infinite series. So {ππ} is
the sequence of partial sums. Consider the infinite series
β ππ = π1 + π2 + π3 + β―
β
π=1
Then
π1 = π1
π2 = π1 + π2
π3 = π1 + π2 + π3
ππ = π1 + π2 + π3 + β― ππ
The sequence of partial sums is then given by:
{ππ} = {π1, π2, π3, β¦ }
If the sequence {ππ} converges, then the series β ππ converges. Recall, that to test a sequence
{ππ} for convergence, we find limπββ
ππ. Thus,
If π₯π’π¦πββ
πΊπ exists, then β ππ converges. Moreover, the sum of the series is π₯π’π¦πββ
πΊπ.
This technique for testing convergence of a series and finding its sum is called the partial sum
technique.
231
Example 20.1
1. Show that the series β1
(π+1)(π+2)βπ=1 converges, and find its sum.
We write out ππ and try to find an expression for the sum so that we can find the limit at
infinity.
ππ =1
6+
1
12+
1
20+ β― +
1
(π + 1)(π + 2)
This expression is not helpful to us so we try another technique.
We first decompose 1
(π+1)(π+2) into partial fractions:
1
(π + 1)(π + 2)=
π΄
π + 1+
π΅
π + 2
1 = π΄(π + 2) + π΅(π + 1)
When π = β2:
1 = βπ΅
π΅ = β1
When π = β1:
1 = π΄
So
1
(π + 1)(π + 2)=
1
π + 1β
1
π + 2
We again write out ππ and try to find an expression for the sum:
ππ = (1
2β
1
3) + (
1
3β
1
4) + (
1
4β
1
5) + β― (
1
πβ
1
π + 1) + (
1
π + 1β
1
π + 2)
232
Notice that the middle terms cancel out. Thus, we can simplify this sum to:
ππ =1
2β
1
π + 2
limπββ
ππ = limπββ
1
2β
1
π + 2
=1
2
So, the series converges to 1
2. The sum of the series is
1
2.
2. Investigate the convergence of the series β ππ (π+4
π+3)β
π=1
We write out ππ and try to find an expression for the sum so that we can find the limit at
infinity.
ππ = ln (5
4) + ln (
6
5) + ln (
7
6) + ln (
8
7) + β¦ + ln (
π + 3
π + 2) + ππ (
π + 4
π + 3)
= ln (5
4 .
6
5.
7
6.
8
7. β¦.
π + 3
π + 2 .
π + 4
π + 3)
= ln (π + 4
4)
limπββ
ππ = limπββ
ln (π + 4
4)
= β
The series diverges.
The πππ term test for divergence
The following test only proves that the series β ππ diverges.
If π₯π’π¦πββ
ππ β π, then β ππ diverges
233
Note: If π₯π’π¦πββ
ππ = π, it does not necessarily mean that the series converges. In this case,
further testing is necessary.
Example 20.2
Test for convergence:
1. β3π2+1
2π2β5βπ=1
limπββ
3π2 + 1
2π2 β 5=
3
2β 0
The series diverges.
2. β ln 3πβπ=1
limπββ
ln 3π = β
The series diverges.
3. βπ2
π3+1βπ=1
limπββ
π2
π3 + 1= 0
This test fails. We will then need to use another test for convergence which you will learn
about in the next chapter.
234
Popular series test
There are some well-known series whose convergence is known (one we have already learned
about from the last chapter). Thus, another technique used for testing an infinite series for
convergence is to determine whether it is a popular series whose convergence is known. Here are
some of these known series:
(i) The Geometric series
β πππβ1 = π + ππ + ππ2 + ππ3 + β― .
β
π=1
The geometric series converges for |π| < π with a sum of π
πβπ.
(ii) The Harmonic series
β1
π= 1 +
1
2+
1
3+ β―
β
π=1
The harmonic series diverges.
For example,
β2
πβπ=1 is a harmonic series and so diverges.
(iii) The p-series
β1
ππ=
1
1π+
1
2π+
1
3π+ β―
β
π=1
The p-series converges for π > π and diverges for π β€ π.
For example,
235
a) β1
π4βπ=1 converges since it is a p-series with p = 4 >1.
b) β2
π1
2ββπ=1 diverges since it is a p-series with π =
1
2< 1
Example 20.3
Test for convergence:
1. β3
π7βπ=1
The series converges because it is a p-series with π = 7 > 1.
2. β4
πβπ=1
The series diverges because it is a harmonic series.
3. β (1
5)
πβπ=1
The series converges because it is a geometric series with π =1
5.
236
HOMEWORK ON CHAPTER 20
1. Find the ππ‘β partial sum of the series and determine whether the series converges or
diverges. If the series converges, find its sum.
a) β1
(π+3)(π+4)βπ=1
b) β4
π2β1βπ=2
c) β ln (π+1
π+2)β
π=1
2. Test the series for convergence:
a) βπβ1
3π+1βπ=1
b) β (1.3)πβπ=1
c) β1
3πβπ=1
d) β1
2π5βπ=1
e) β1
βπβπ=1
f) β 5 (1
3)
πβπ=1
g) β11
πβπ=1
237
CHAPTER 21: SERIES III β CONVERGENCE OF SERIES
So far, we have examined the following tests for convergence of series:
1. The ππ‘β partial sum test
2. The ππ‘β term test for divergence
3. The popular series test.
Let us examine more tests for convergence of series.
The Integral Test
To test β ππβπ=1 for convergence we apply the following theorem:
Suppose ππ = π(π), where π(π) is decreasing and positive.
β’ If β« π(π) ππβ
1 converges, then β ππ converges
β’ If β« π(π) ππβ
1 diverges, then β ππ diverges
To use the integral test, we must first show that ππ = π(π) is a decreasing, positive function.
Only then can we apply the integral test. Note that to show convergence or divergence, you will
need to evaluate an improper integral. You can review this in Chapter 11.
Let us now use the integral test to show why the p-series converges for π > 1 and diverges for
π β€ 1.
Example 21.1
We will show that the p-series β1
ππβπ=1 is convergent for π > 1 and divergent for π β€ 1 by using
the integral test.
π(π) =1
ππ= πβπ
238
Now π(π) > 0 for π β₯ 1.
For π > 0:
πβ²(π) = βππβπβ1 < 0
So π(π) is decreasing.
We can now use the integral test since both conditions hold.
β«1
ππ ππ = lim
π‘βββ« πβπ ππ
π‘
1
β
1
= limπ‘ββ
[1
1 β π π1βπ]
1
π‘
=1
1 β π limπ‘ ββ
[π‘1βπ β 1]
=1
1 β π limπ‘ ββ
[1
π‘πβ1β 1]
If π = 1, the integral diverges, so the p-series diverges.
If π < 1, the integral diverges, so the p-series diverges.
If π > 1, the integral converges to 1
πβ1, so the p-series converges.
Example 21.2
Use the integral test to test the following series for convergence:
1. The harmonic series β1
πβπ=1
π(π) =1
π
π(π) > 0
Also
239
πβ²(π) = β1
π2< 0
So π(π) is decreasing.
β«1
π ππ = lim
π‘βββ«
1
π ππ
π‘
1
β
1
= limπ‘ββ
[ln π]1π‘
= limπ‘ ββ
[ln π‘]
= β
The integral diverges so the harmonic series diverges.
2. βπ
π2+1βπ=1
π(π) =π
π2 + 1
π(π) > 0
πβ²(π) =1. (π2 + 1) β 2π . π
(π2 + 1)2
=βπ2 + 1
(π2 + 1)2< 0
So π(π) is decreasing.
β«π
π2 + 1 ππ = lim
π‘βββ«
π
π2 + 1 ππ
π‘
1
β
1
= limπ‘ββ
[1
2ln|π2 + 1|]
1
π‘
=1
2 limπ‘ ββ
[ln(π‘2 + 1) β ln 2]
= β
240
The integral diverges, so the series diverges.
The following test allow us to test positive term series for convergence by finding other series to
compare them to. There are two comparison tests. Please be careful in using both tests. Ensure
that the conditions hold for convergence and divergence.
The Comparison Test
Let β ππ and β ππ be positive term series. Then
1. If ππ β€ ππ for all n and β ππ converges, then β ππ converges.
2. If ππ β₯ ππ for all n and β ππ diverges, then β ππ diverges.
The Limit Comparison Test
If limπββ
ππ
ππ exists and is positive, then either both series converge or both diverge.
So, to test β ππ for convergence, we will first need to find a known series β ππ for comparison.
To find such a series, we usually find the quotient of the dominant terms in the numerator and
denominator of the rational function represented by ππ.
Example 21.3
Use the comparison test, to test the series for convergence:
1. β1
π(π+1)βπ=1
ππ =1
π(π + 1)
241
Our comparing series is:
ππ =1
π. π=
1
π2
This comparing series converges because it is a p-series with π = 2 > 1. Thus we will
use the first condition in the comparison test and check if it holds.
1
π(π + 1)<
1
π2
since they both have the same numerator, and the denominator of 1
π(π+1) is greater than
that of 1
π2.
So by the comparison test, β1
π(π+1)βπ=1 also converges.
2. β1
π 3πβπ=1
ππ =1
π 3π
Our comparing series is:
ππ =1
3π
This comparing series converges because it is a geometric series with π =1
3. Thus we will
again use the first condition in the comparison test and check if it holds.
1
π 3π<
1
3π
So by the comparison test, β1
π 3πβπ=1 also converges.
242
3. β1
5πβ1βπ=1
ππ =1
5π β 1
Our comparing series is:
ππ =1
5π
This comparing series diverges because it is a harmonic series. Thus we will use the
second condition in the comparison test and check if it holds.
1
5π β 1>
1
5π
So by the comparison test, β1
5πβ1βπ=1 diverges.
4. βπ2
π5βπβ2βπ=1
ππ =π2
π5 β π β 2
Our comparing series is:
ππ =π2
π5=
1
π3
This comparing series converges because it is a p-series with π = 3 > 1. Thus we will
use the first condition in the comparison test and check if it holds.
However, observe that
π2
π5 β π β 2>
π2
π5=
1
π3
243
This means that in this case, the comparison test fails. We will therefore use the limit
comparison test.
limπββ
ππ
ππ= lim
πββ
π2
π5 β π β 2 . π3
limπββ
π5
π5 β π β 2
= 1
This limit exists and is positive. So by the limit comparison test, both series converge.
That is, βπ2
π5βπβ2βπ=1 converges.
244
HOMEWORK ON CHAPTER 21
1. Use the integral test to test the series for convergence:
a) β1
π2βπ=1
b) β1
π+5βπ=1
c) βln π
π2βπ=1
2. Use the comparison test or the limit comparison test to test the series for convergence:
a) β1
βπ2πβπ=1
b) βπ2
π5+3π+5βπ=1
c) β1
βπβ1βπ=2
d) βπ3
π5β3πβ5βπ=1
245
CHAPTER 22: SERIES IV β CONVERGENCE OF SERIES
In the last three chapters, you learned how to test the convergence of positive term series. In this
chapter, you will learn how to determine the convergence of general series, that is, series with
positive and/or negative terms.
Alternating Series
An alternating series is an infinite series of the form
β(β1)π+1ππ = π1 β π2 + π3 β π4 + β―
β
π=1
Alternating series Test
If ππ > ππ+1 > 0 for all n and limπββ
ππ = 0 then the alternating series
β(β1)π+1ππ = π1 β π2 + π3 β π4 + β―
β
π=1
converges.
Thus, to test an alternating series for convergence, we must show that:
1. ππ is decreasing
2. limπββ
ππ = 0
Example 22.1
Test the alternating series for convergence:
1. β(β1)π
βπ3
βπ=1
246
ππ =1
βπ3
ππ+1 =1
βπ + 13
1
βπ3 >
1
βπ + 13 > 0
limπββ
1
βπ3 = 0
So by the alternating series test, β(β1)π
βπ3
βπ=1 converges.
2. β(β1)π+1
π2βπ=1
ππ =1
π2
ππ+1 =1
(π + 1)2
1
π2>
1
(π + 1)2> 0
limπββ
1
π2= 0
So by the alternating series test, β(β1)π+1
π2βπ=1 converges.
Absolute Convergence
The series β ππ is said to converge absolutely if the series β |ππ | converges, where
β |ππ | = |π1| + |π2| + β―
247
Theorem
If β ππ converges absolutely, that is, if β |ππ | converges, then so does the series β ππ.
However, if β |ππ | diverges, it does not necessarily mean that β ππ diverges.
Conditional Convergence
Earlier on, we saw that the series β(β1)π
βπ3
βπ=1 converges.
But β |(β1)π
βπ3
βπ=1 | = β
1
βπ3
βπ=1 diverges since it is a p-series with π =
1
3< 1.
The series β(β1)π
βπ3
βπ=1 is called conditionally convergent β it converges but it is not absolutely
convergent.
Example 22.2
Show that the series β(β1)π+1
πβπ=1 is conditionally convergent.
ππ =1
π
ππ+1 =1
π + 1
1
π>
1
π + 1> 0
limπββ
1
π= 0
So by the alternating series test, β(β1)π+1
πβπ=1 converges.
But β |(β1)π+1
π|β
π=1 = β1
πβπ=1 which diverges since it is a harmonic series.
248
The Ratio Test
For the series β ππ, suppose limπββ
|ππ+1
ππ| = πΏ.
1. If L < 1, then β ππ converges.
2. If L > 1 or πΏ = Β±β, then β ππ diverges.
3. If L = 1, this test is inconclusive. Use a different test.
This section uses the concept of factorial notation.
Review of factorial notation
π! = π(π β 1)(π β 2)(π β 3) β¦ 3 . 2 . 1
So
5! = 5 . 4 . 3 . 2 . 1
Factorials can also be defined recursively. For example,
(π + 1)! = (π + 1) π!
Example 22.3
Test the following series for convergence by using the ratio test:
1. β1
2πβπ=1
ππ =1
2π ππ+1 =
1
2π+1
ππ+1
ππ=
1
2π+1 Γ·
1
2π
=2π
2π+1
249
=2π
2π. 2
=1
2
limπββ
1
2=
1
2< 1
By the ratio test, β1
2πβπ=1 converges.
2. β(β5)π
π!βπ=1
We use absolute value here because this series contains negative terms.
|ππ| =5π
π! |ππ+1| =
5π+1
(π + 1)!
|ππ+1
ππ| =
5π+1
(π + 1)! Γ·
5π
π!
=5π+1
(π + 1)! .
π!
5π
=5π. 5
(π + 1)π! .
π!
5π
= 5
π + 1
limπββ
5
π + 1= 0 < 1
By the ratio test, β(β5)π
π!βπ=1 converges.
250
3. β2π
πβπ=1
ππ =2π
π ππ+1 =
2π+1
π + 1
ππ+1
ππ=
2π+1
π + 1 Γ·
2π
π
=2π+1
π + 1 .
π
2π
=2π
π + 1
limπββ
2π
π + 1= 2 > 1
By the ratio test, β2π
πβπ=1 diverges.
The Root Test
For the series β ππ, suppose limπββ
β|ππ|π= πΏ.
1. If L < 1, then β ππ converges.
2. If L > 1 or πΏ = Β±β, then β ππ diverges.
3. If L = 1, this test is inconclusive. Use a different test.
Example 22.4
Test for convergence:
1. β (β1)π+1 (2π3β5
3π3+1)
πβπ=1
|ππ| = (2π3 β 5
3π3 + 1)
π
251
β|ππ|π
= β(2π3 β 5
3π3 + 1)
ππ
= (2π3 β 5
3π3 + 1)
limπββ
2π3 β 5
3π3 + 1=
2
3< 1
By the root test, β (β1)π+1 (2π3β5
3π3+1)
πβπ=1 converges.
2. β (3π+1
5πβ4)
β2πβπ=1
ππ = (3π + 1
5π β 4)
β2π
βπππ = β(
3π + 1
5π β 4)
β2ππ
= (3π + 1
5π β 4)
β2
limπββ
(3π + 1
5π β 4)
β2
= (3
5)
β2
= 25
9> 1
By the root test, β (3π+1
5πβ4)
β2π β
π=1 diverges.
Strategies for testing series for convergence
You have learned various tests for investigating the convergence of the series β ππ, however,
you will need to know when to apply each test. In this section we review each test and examine
ways of deciding which test to use.
1. The ππ‘β term test for divergence:
If limπββ
ππ β 0, then β ππ diverges
252
Try this test first but remember it only proves divergence and not convergence.
2. The popular series test:
(i) The Geometric series: β πππβ1 = π + ππ + ππ2 + ππ3 + β― .βπ=1
The geometric series converges for |π| < 1 with a sum of π
1βπ.
(ii) The Harmonic series β1
π= 1 +
1
2+
1
3+ β―β
π=1
The harmonic series diverges.
(iii) The p-series β1
ππ =1
1π +1
2π +1
3π + β―βπ=1
The p-series converges for π > 1 and diverges for π β€ 1.
This is a relatively straightforward test and should be used if the series is recognizable.
3. If the series is an alternating series, use the Alternating series theorem. The ratio test may
also work here.
Alternating series Test:
If ππ > ππ+1 > 0 for all n and limπββ
ππ = 0 then the alternating series
β(β1)π+1ππ = π1 β π2 + π3 β π4 + β―
β
π=1
converges.
4. The Root test:
Suppose limπββ
β|ππ|π= πΏ.
(i) If L < 1, then β ππ converges.
(ii) If L > 1 or πΏ = Β±β, then β ππ diverges.
(iii) If L = 1, this test is inconclusive. Use a different test.
253
If ππ is of the form (π(π))π(π), use the root test.
5. The Ratio test
Suppose limπββ
|ππ+1
ππ| = πΏ.
(i) If L < 1, then β ππ converges.
(ii) If L > 1 or πΏ = Β±β, then β ππ diverges.
(iii)If L = 1, this test is inconclusive. Use a different test.
For series containing factorials or terms with constants raised to some ππ‘β power, use the
ratio test.
6. The Comparison test and Limit Comparison test.
The Comparison Test:
Let β ππ and β ππ be positive term series. Then
3. If ππ β€ ππ for all n and β ππ converges, then β ππ converges.
4. If ππ β₯ ππ for all n and β ππ diverges, then β ππ diverges.
The Limit Comparison Test:
If limπββ
ππ
ππ exists and is positive, then either both series converge or both diverge.
Note that these tests are only valid for positive term series. Use this test when the general
term of the series is a rational function and all other tests fail for proving convergence.
7. The Integral Test:
Suppose ππ = π(π), where π(π) is decreasing and positive.
β’ If β« π(π) ππβ
1 converges, then β ππ converges
254
β’ If β« π(π) ππβ
1 diverges, then β ππ diverges
Again, this test is valid only for positive term series. Use this test when all other tests fail
or when the improper integral can be easily evaluated.
Example 22.5
Test the series for convergence:
1. β1
π5+ππβπ=1
Use the comparison test:
ππ =1
π5 + ππ
Our comparing series is:
ππ =1
π5
This comparing series converges because it is a p-series with π = 5 > 1.
1
π5 + ππ<
1
π5
So by the comparison test, β1
π5+ππβπ=1 also converges.
2. β (π3+1
3π3+5)
3πβπ=1
Use the root test:
ππ = (π3 + 1
3π3 + 5)
3π
255
βπππ = β(
π3 + 1
3π3 + 5)
3ππ
= (π3 + 1
3π3 + 5)
3
limπββ
(π3 + 1
3π3 + 5)
3
= (1
3)
3
= 1
27< 1
By the root test, β (π3+1
3π3+5)
3π
βπ=1 converges.
3. β (β1)π 3π . π
(2π)!βπ=1
Use the ratio test:
|ππ| =3π . π
(2π)! |ππ+1| =
3π+1 . (π + 1)
[2(π + 1)]!=
3π+1 . (π + 1)
(2π + 2)!
|ππ+1
ππ| =
3π+1 . (π + 1)
(2π + 2)! Γ·
3π . π
(2π)!
=3π+1 . (π + 1)
(2π + 2)! .
(2π)!
3π . π
=3π . 3 . (π + 1)
(2π + 2)(2π + 1)(2π)! .
(2π)!
3π . π
= 3(π + 1)
(2π + 2)(2π + 1)π
limπββ
3(π + 1)
(2π + 2)(2π + 1)π= 0 < 1
By the ratio test, β (β1)π 3π . π
(2π)!βπ=1 converges.
256
4. β (β1)π 3π+1
π+2βπ=1
Use the ππ‘β test for divergence:
limπββ
3π + 1
π + 2= 3 β 0
By the ππ‘β test for divergence, β (β1)π 3π+1
π+2βπ=1 diverges.
5. β5
11πβπ=1
Use the popular series test:
This is a harmonic series, so β5
11πβπ=1 diverges.
257
HOMEWORK ON CHAPTER 22
1. Use the ratio test or the root test to investigate the convergence of the series
a) β (β1)π π2
5πβπ=1
b) β 1
ππβπ=1
c) β (β1)π (4π3+πβ1
3π3+2)
πβπ=1
d) β π!
7πβπ=1
e) β (1 +1
π)
π2
βπ=1
2. Determine whether the series converges:
a) β π3
π!βπ=1
b) β 3π5+2π+7
5π5+1βπ=1
c) β 1
π2πβπ=1
d) β (π
2π+5)
βπβπ=1
e) β 3π
π5βπ=1
f) β 1
2π+πβπ=1
g) β π2β2πβ1
π3+5π+4βπ=1
h) β 1
π32
βπ=1
i) β 1
π ln πβπ=2
258
CHAPTER 23: POWER SERIES
In this chapter you will learn about a special type of series called a power series and how to
determine the conditions under which such a series would converge.
Definition
A power series is a series of the form
β πππ₯π = π0 + π1π₯ + π2π₯2 + π3π₯3 + β―
β
π=0
where the ππβ²π are constants called the coefficients of the series.
A power series about π₯ = π is a series of the form:
β ππ(π₯ β π)π = π0 + π1(π₯ β π) + π2(π₯ β π)2 + π3(π₯ β π)3 + β―
β
π=0
For example,
β(β1)ππ₯π = 1 β π₯ + π₯2 β π₯3 + β―
β
π=0
is a power series that is centered at 0.
An important question is: βWhen does a power series converge?β When we test a power series
for convergence, we determine the values of x for which the series converges.
In previous chapters, we learned various tests for determining the convergence of series. Any of
these can be used to test power series for convergence, but the most popular test to use in these
cases is the ratio test.
259
The following examples makes use of another test, other than the ratio test.
Example
1. β (2π₯)π = 1 + 2π₯ + 4π₯2 + 8π₯3 + β―βπ=0
This power series is a geometric series with common ratio π = 2π₯.
Recall that a geometric series converges for |π| < 1. So the series converges for
|2π₯| < 1
β1 < 2π₯ < 1
β1
2< π₯ <
1
2
2. 1 β1
3(π₯ β 1) +
1
9(π₯ β 1)2 β
1
27(π₯ β 1)3 + β―.
This is another geometric series with common ratio β1
3(π₯ β 1).
The series converges for
|β1
3(π₯ β 1)| < 1
β1 <1
3(π₯ β 1) < 1
β3 < π₯ β 1 < 1
β2 < π₯ < 2
260
Not all power series are popular series like the geometric series. Generally, to test a power series
for convergence, use the ratio test. Recall the ratio test:
For the series β ππ, suppose π₯π’π¦πββ
|ππ+π
ππ| = π³.
4. If L < 1, then β ππ converges.
5. If L > 1 or π³ = Β±β, then β ππ diverges.
6. If L = 1, this test is inconclusive. Use a different test.
Note that if πΏ = 1, the test is inconclusive. This means that we would need to do some further
testing at these x-values.
Example
For what values of x do the following power series converge?
1. β ππ₯πβπ=1
|ππ| = | ππ₯π| |ππ+1| = |(π + 1)π₯π+1|
|ππ+1
ππ| = |
(π + 1)π₯π+1
ππ₯π|
=(π + 1)|π₯|
π
limπ ββ
|ππ+1
ππ| = lim
π ββ
(π + 1)|π₯|
π
= |π₯| limπ ββ
π + 1
π
= |π₯|. 1
= |π₯|
By the ratio test, this series converges for |π₯| < 1, that is for β1 < π₯ < 1.
But because the test is inconclusive for |π₯| = 1, we would need to do further testing at
π₯ = 1 and at π₯ = β1, that is, at the end-points of the interval. We check these, by
substituting these values of x, into the actual series and then test for convergence.
261
Check at π₯ = 1:
β ππ₯π
β
π=1
= β π(1)π
β
π=1
= β π
β
π=1
= 1 + 2 + 3 + β―
This series clearly diverges. But we can test it by using the ππ‘β term test for divergence:
limπ ββ
π = β β 0
Check at π₯ = β1:
β ππ₯π
β
π=1
= β π(β1)π
β
π=1
= β1 + 2 β 3 + β―
Again, this series clearly diverges. We can test it by using the ππ‘β term test for
divergence:
limπ ββ
|π(β1)π| = limπ ββ
π = β β 0
Thus, the power series β ππ₯πβπ=1 converges for β1 < π₯ < 1.
2. βπ₯2π
π!βπ=1
|ππ| = | π₯2π
π!| |ππ+1| = |
π₯2(π+1)
(π + 1!| = |
π₯2π+2
(π + 1!|
|ππ+1
ππ| = |
π₯2π+2
(π + 1! .
π!
π₯2π|
=π₯2
π + 1
limπ ββ
|ππ+1
ππ| = lim
π ββ
π₯2
π + 1
= π₯2 limπ ββ
1
π + 1
= π₯2 . 0
262
= 0 < 1
By the ratio test, the series βπ₯2π
π!βπ=1 converges for all values of π₯.
3. βπ₯π
3ππ2βπ=1
|ππ| = | π₯π
3ππ2| |ππ+1| = |
π₯π+1
3π+1(π + 1)2|
|ππ+1
ππ| = |
π₯π+1
3π+1(π + 1)2 .
3ππ2
π₯π|
=|π₯|π2
3(π + 1)2
limπ ββ
|ππ+1
ππ| = lim
π ββ
|π₯|π2
3(π + 1)2
= |π₯| limπ ββ
|π2
3(π + 1)2
=|π₯|
3
By the ratio test, this series converges for |π₯|
3< 1, that is for β3 < π₯ < 3.
We would need to do further testing at the end-points π₯ = 3 and at π₯ = β3.
Check at π₯ = 3:
βπ₯π
3ππ2
β
π=1
= β3π
3ππ2
β
π=1
= β1
π2
β
π=1
This is a p-series with π = 2. So the series converges at π₯ = 3.
Check at π₯ = β3:
263
βπ₯π
3ππ2
β
π=1
= β(β3)π
3ππ2
β
π=1
= β(β1)π
π2
β
π=1
Now
β |(β1)π
π2|
β
π=1
= β1
π2
β
π=1
which converges. So the series converges at π₯ = β3.
Thus, the power series π₯π
3ππ2 converges for β3 β€ π₯ β€ 3.
We will now define two concepts: the radius of convergence and the interval of convergence.
Theorem
For a given power series
β ππ(π₯ β π)π = π0 + π1(π₯ β π) + π2(π₯ β π)2 + π3(π₯ β π)3 + β―
β
π=0
There are only 3 possibilities:
1. The series converges only when π₯ = π
2. The series converges for all x
3. There is a positive number R such that the series converges if |π₯ β π| < π
R is called the radius of convergence. In the first two possibilities above:
1. π = 0
2. π = β
Thus in the examples above:
Example 1: R = 1
Example 2: π = β
264
Example 3: π = 3
The interval of convergence
The interval of convergence of a power series is the interval that consists of all values of x for
which the series converges. Again in the possibilities above:
1. Interval of convergence is a single point a.
2. Interval of convergence = (ββ, β)
3. Interval of convergence can be one of the following:
a) (a β R, a + R)
b) [a β R, a + R]
c) (a β R, a + R]
d) [a β R, a + R)
Example
Find the radius and interval of convergence of the series:
1. β(β1)π(π₯β3)π
πβπ=0
|ππ| = | (β1)π(π₯ β 3)π
π| |ππ+1| = |
(β1)π+1(π₯ β 3)π+1
π + 1|
|ππ+1
ππ| = |
(π₯ β 3)π+1
π + 1 .
π
(π₯ β 3)π|
=|π₯ β 3|π
π + 1
limπ ββ
|ππ+1
ππ| = lim
π ββ
|π₯ β 3|π
π + 1
= |π₯ β 3| limπ ββ
π
π + 1
265
= |π₯ β 3|
By the ratio test, this series converges for |π₯ β 3| < 1, that is for
β1 < π₯ β 3 < 1
2 < π₯ < 4
The radius of convergence is 1.
.
We would need to do further testing at the end-points π₯ = 2 and at π₯ = 4.
Check at π₯ = 2:
β(β1)π(π₯ β 3)π
π
β
π=1
= β(β1)π. (β1)π
π
β
π=1
= β1
π
β
π=1
This is the harmonic series which diverges. So the series diverges at π₯ = 2.
Check at π₯ = 4:
β(β1)π(π₯ β 3)π
π
β
π=1
= β(β1)π. (1)π
π
β
π=1
= β(β1)π
π
β
π=1
We use the alternating series test to test this series for convergence.
ππ =1
π
ππ+1 =1
π + 1
1
π>
1
π + 1> 0
limπββ
1
π= 0
266
So by the alternating series test, β(β1)π
πβπ=1 converges. So the series converges for
π₯ = 4.
Thus, the power series β(β1)π(π₯β3)π
πβπ=0 converges for 2 < π₯ β€ 4.
The interval of convergence is π < π β€ π.
267
HOMEWORK ON CHAPTER 23
Find the radius of convergence and interval of convergence of the series:
1. βπ₯2π
(2π)!βπ=1
2. β2ππ₯π
π!βπ=1
3. β π3(π₯ β 1)πβπ=1
4. βπ₯π3π
ππβπ=1
5. β(β1)π(π₯+1)π
5πβπβπ=1
6. β 5ππ₯πβπ=1
7. βπ!π₯π
7πβπ=1
268
CHAPTER 24: SERIES EXPANSIONS
This chapter teaches you a general method for writing a function as a series. You will learn how
to approximate functions by polynomials. For example,
ππ₯ = 1 + π₯ +π₯2
2!+
π₯3
3!+ β―
Taylor Polynomials
Taylorβs polynomial gives the expansion of π(π₯) in powers of π₯ β π, with π being a constant.
Letβs derive the Taylor polynomial expansion:
Suppose
π(π₯) = π0 + π1(π₯ β π) + π2(π₯ β π)2 + π3(π₯ β π)3 + β― + ππ(π₯ β π)π β¦ . . (1)
Substitute π₯ = π. We get:
ππ = π(π)
We differentiate (1):
πβ²(π₯) = π1 + 2π2(π₯ β π) + 3π3(π₯ β π)2 + β― + πππ(π₯ β π)πβ1
Substitute π₯ = π. We get:
ππ = πβ²(π)
Differentiate again:
πβ²β²(π₯) = 2π2 + 3 β 2π3(π₯ β π) + β― + π(π β 1)ππ(π₯ β π)πβ2
Substitute π₯ = π. We get:
ππ =πβ²β²(π)
π!
269
Continuing we get:
ππ =πβ²β²β²(π)
π!
ππ =π(π)(π)
π!
Substituting these values for the ππβ²π in π(π₯) gives us:
Taylor polynomial of degree n approximating π(π) near π = π denoted by π·π(π)
π(π) β π·π(π)
where
π·π(π) = π(π) + πβ²(π)(π β π) +πβ²β²(π)
π!(π β π)π +
πβ²β²β²(π)
π!(π β π)π + β― +
π(π)(π)
π!(π β π)π
πππ degree Taylor polynomial about π = π is:
π(π) β π(π) + πβ²(π)π +πβ²β²(π)
π!ππ +
πβ²β²β²(π)
π!ππ + β― +
π(π)(π)
π!ππ
Example 24.1
1. Find the 5π‘β degree Taylor polynomial of π(π₯) = π2π₯ at π₯ = 0.
We find up to the 5π‘β derivative of π(π₯) at π₯ = 0:
π(π₯) = π2π₯ π(0) = 1
πβ²(π₯) = 2π2π₯ πβ²(0) = 2
πβ²β²(π₯) = 4π2π₯ πβ²β²(0) = 4
270
πβ²β²β²(π₯) = 8π2π₯ πβ²β²β²(0) = 8
π(4)(π₯) = 16π2π₯ π(4)(0) = 16
π(5)(π₯) = 32π2π₯ π(5)(0) = 32
π(π₯) β π(0) + πβ²(0)π₯ +πβ²β²(0)
2!π₯2 +
πβ²β²β²(0)
3!π₯3 +
π(4)(0)
4!π₯4 +
π(5)(0)
5!π₯5
So
π2π₯ β 1 + 2π₯ +4
2!π₯2 +
8
3!π₯3 +
16
4!π₯4 +
32
5!π₯5
π2π₯ β 1 + 2π₯ + 2π₯2 +4
3π₯3 +
2
3π₯4 +
4
15π₯5
2. Find the 3ππ degree Taylor polynomial of π(π₯) = cos π₯ at π₯ =π
3.
We find up to the 3ππ derivative of π(π₯) at π₯ =π
3:
π(π₯) = cos π₯ π (π
3) =
1
2
πβ²(π₯) = βsin π₯ πβ² (π
3) = β
β3
2
πβ²β²(π₯) = βcos π₯ πβ²β² (π
3) = β
1
2
πβ²β²β²(π₯) = sin π₯ πβ²β²β² (π
3) =
β3
2
π(π₯) β π(π) + πβ²(π)(π₯ β π) +πβ²β²(π)
2!(π₯ β π)2 +
πβ²β²β²(π)
3!(π₯ β π)3
So
271
cos π₯ β π (π
3) + πβ² (
π
3) (π₯ β
π
3) +
πβ²β² (π3
)
2!(π₯ β
π
3)
2
+ πβ²β²β² (
π3
)
3!(π₯ β
π
3)
3
cos π₯ β 1
2β
β3
2(π₯ β
π
3) β
1
2(π₯ β
π
3)
2
+β3
2(π₯ β
π
3)
3
3. Find the ππ‘β degree Taylor polynomial of π(π₯) = ln(1 + π₯) at π = 0.
π(π₯) = ln(1 + π₯) π(0) = 0
πβ²(π₯) =1
1 + π₯ πβ²(0) = 1
πβ²β²(π₯) = β1
(1 + π₯)2 πβ²β²(0) = β1
πβ²β²β²(π₯) =2
(1 + π₯)3 πβ²β²β²(0) = 2
π(4)(π₯) = β6
(1 + π₯)4 π(4)(0) = β6
π(5)(π₯) =24
(1 + π₯)5 π(5)(0) = 24
π(π₯) β π(0) + πβ²(0)π₯ +πβ²β²(0)
2!π₯2 +
πβ²β²β²(0)
3!π₯3 + β― +
π(π)(0)
π!π₯π
So
ln(1 + π₯) β π₯ β1
2!π₯2 +
2
3!π₯3 β
6
4!π₯4 +
24
5!π₯5 β¦ +
π(π)(0)
π!π₯π
ln(1 + π₯) β π₯ βπ₯2
2+
π₯3
3β
π₯4
4+
π₯5
5β β― + (β1)π+1
π₯π
π
272
Taylor series
The Taylor series for π(π₯) about π₯ = π is an infinite series given by:
π(π) = π(π) + πβ²(π)(π β π) +πβ²β²(π)
π!(π β π)π +
πβ²β²β²(π)
π!(π β π)π + β― +
π(π)(π)
π!(π β π)π+..
The Taylor series for π(π₯) about π₯ = 0 is called the Maclaurin series and is given by:
π(π) = π(π) + πβ²(π)π +πβ²β²(π)
π!ππ +
πβ²β²β²(π)
π!ππ + β― +
π(π)(π)
π!ππ + β―
The following are common Maclaurin series:
ππ₯ = 1 + π₯ +π₯2
2!+
π₯3
3!+ β― +
π₯π
π!+ β―
sin π₯ = π₯ βπ₯3
3!+
π₯5
5!β
π₯7
7!+ β― + β― (β1)π
π₯2π+1
(2π + 1)!+ β―
cos π₯ = 1 βπ₯2
2!+
π₯4
4!β
π₯6
6!+ β― + β― (β1)π
π₯2π
(2π)!+ β―
ln(1 + π₯) β π₯ βπ₯2
2+
π₯3
3β
π₯4
4+
π₯5
5β β― + (β1)π+1
π₯π
π+ β―
arctan π₯ = π₯ βπ₯3
3+
π₯5
5β
π₯7
7+ β― + β― (β1)π
π₯2π+1
(2π + 1)+ β―
1
1 β π₯= 1 + π₯ + π₯2 + π₯3 + π₯4 + β― + π₯π + β―
We can use these common series to write out series expansions for related functions without
using the Taylor formula.
273
Example 24.2
Find the Taylor series for:
1. π3π₯
We know that
ππ₯ = 1 + π₯ +π₯2
2!+
π₯3
3!+ β― +
π₯π
π!+ β―
So to find the series expansion for π3π₯, we replace π₯ with 3π₯ to get:
π3π₯ = 1 + 3π₯ +(3π₯)2
2!+
(3π₯)3
3!+ β― +
(3π₯)π
π!+ β―
π3π₯ = 1 + 3π₯ +9π₯2
2!+
27π₯3
3!+ β― +
3ππ₯π
π!+ β―
π3π₯ = 1 + 3π₯ +9π₯2
2+
9π₯3
2+ β― +
3ππ₯π
π!+ β―
2. cos (π₯2)
We know that
cos π₯ = 1 βπ₯2
2!+
π₯4
4!β
π₯6
6!+ β― + β― (β1)π
π₯2π
(2π)!+ β―
So
cos π₯2 = 1 β(π₯2)2
2!+
(π₯2)4
4!β
(π₯2)6
6!+ β― + β― (β1)π
(π₯2)2π
(2π)!+ β―
cos π₯2 = 1 βπ₯4
2!+
π₯8
4!β
π₯12
6!+ β― + β― (β1)π
π₯4π
(2π)!+ β―
274
3. ππ₯ sin π₯
Find the series expansion up to the term in π₯3.
ππ₯ = 1 + π₯ +π₯2
2!+
π₯3
3!= 1 + π₯ +
π₯2
2+
π₯3
3
sin π₯ = π₯ βπ₯3
3!= π₯ β
π₯3
6
So
ππ₯ sin π₯ = (1 + π₯ +π₯2
2+
π₯3
3) (π₯ β
π₯3
6)
= π₯ βπ₯3
6+ π₯2 +
π₯3
2
= π₯ + π₯2 +π₯3
3
Binomial Series
An important series which often appears in applications is the binomial series. This series is the
Maclaurin series expansion of (1 + π₯)π and is given by:
(π + π)π = π + ππ +π(π β π)
π!ππ +
π(π β π)(π β π)
π!ππ + β―
This expansion is valid for βπ < π < π or |π| < π. That is, the series converges for |π| < π.
Example 24.3
1. Expand (1 + π₯)4 as a power series up to the 5th term.
(1 + π₯)4 = 1 + ππ₯ +π(π β 1)
2!π₯2 +
π(π β 1)(π β 2)
3!π₯3 +
π(π β 1)(π β 2)(π β 3)
4!π₯4
275
= 1 + 4π₯ +4(4 β 1)
2!π₯2 +
4(4 β 1)(4 β 2)
3!π₯3 +
4(4 β 1)(4 β 2)(4 β 3)
4!π₯4
= 1 + 4π₯ + 6π₯2 + 4π₯3 + π₯4
2. Expand (1 β 2π₯)1
2β as a power series up to the 4th term.
In the binomial expansion formula, we would need to let π =1
2 and replace π₯ with β2π₯.
(1 β 2π₯)1
2β = 1 +1
2(β2π₯) +
12
(12
β 1)
2!(β2π₯)2 +
12
(12
β 1) (12
β 2)
3!(β2π₯)3
= 1 β π₯ +
12
. β12
2!4π₯2 +
12
. β12
. β32
3!. β8π₯3
= 1 β π₯ β1
2π₯2 β
1
2π₯3
3. Find the series expansion up to the 4th term for 1
3+π₯.
1
3 + π₯= (3 + π₯)β1
In order to use the binomial series expansion, we would need to first factor out 3. So
(3 + π₯)β1 = [3 (1 +π₯
3)]
β1
= 3β1 (1 +π₯
3)
β1
=1
3[1 + (β1) (
π₯
3) +
β1 . (β1 β 1)
2!(
π₯
3)
2
+β1 . (β1 β 1)(β1 β 2)
3!(
π₯
3)
3
]
=1
3[1 β
π₯
3+
π₯2
9β
π₯3
27]
276
=1
3β
π₯
9+
π₯2
27β
π₯3
81
We can use the binomial series expansion for estimation.
Example 24.4
1. Expand β1 + π₯ as a power series. Hence estimate β1.01 to 3 decimal places.
β1 + π₯ = (1 + π₯)12
= 1 +1
2π₯ +
12
(12
β 1)
2!π₯2 +
12
(12
β 1) (12
β 2)
3!π₯3 + β―
= 1 +1
2π₯ +
12
. β12
2!π₯2 +
12
. β12
. β32
3!. π₯3 + β―
= 1 +1
2π₯ β
1
8π₯2 +
1
16π₯3 + β―
It was necessary to expand only up to the term in π₯3 since the estimation must be
rounded to 3 decimal places.
To find β1.01 we express it as β1 + 0.01 and substitute 0.01 for π₯ in the series
expansion. So
β1.01 = β1 + 0.01
= 1 +1
2(0.01) β
1
8(0.01)2 +
1
16(0.01)3 + β―
= 1 + 0.005 β 0
= 1.005
277
Finding series expansions through differentiation and integration of other series
Example 24.5
1. Find the Taylor series expansion for 1
(1βπ₯)2 from the series expansion for 1
1βπ₯.
1
1 β π₯= 1 + π₯ + π₯2 + π₯3 + π₯4 + β― + π₯π + β―
Observe that
(1
1 β π₯)
β²
=1
(1 β π₯)2
So to find the series expansion of 1
(1βπ₯)2 we differentiate the series expansion of 1
1βπ₯. So
1
(1 β π₯)2=
π
ππ₯(1 + π₯ + π₯2 + π₯3 + π₯4 + β― + π₯π + β― )
= 1 + 2π₯ + 3π₯2 + 4π₯3 + β― + ππ₯πβ1 + β―
2. Find the Taylor series expansion for tanβ1 π₯ from the series expansion for 1
1+π₯2.
1
1 + π₯2= (1 + π₯2)β1 = 1 β π₯2 +
β1 . β2
2! (π₯2)2 +
β1 . β2 . β3
3! (π₯2)3 +
= 1 β π₯2 + π₯4 β π₯6 + π₯8 β β―
Observe that
β«1
1 + π₯2 ππ₯ = tanβ1 π₯ + πΆ
So to find the series expansion for tanβ1 π₯ we integrate the series expansion for 1
1+π₯2. So
tanβ1 π₯ = β«(1 β π₯2 + π₯4 β π₯6 + π₯8 β β― ) ππ₯
278
π₯ β1
3π₯3 +
1
5π₯5 β
1
7π₯7 +
1
9π₯9 β β― + πΆ
To find C, substitute π₯ = 0 on both sides of the equation to get
πΆ = 0
So
tanβ1 π₯ = π₯ β1
3π₯3 +
1
5π₯5 β
1
7π₯7 +
1
9π₯9 β β―
279
HOMEWORK ON CHAPTER 24
1. Use the formula for the binomial series expansion to find the series expansion for the
following functions:
a) π(π₯) =1
β1+3π₯
b) π(π₯) =2
4βπ₯
c) π(π₯) = (1 + 5π₯)4
2. Find the Taylor series expansion for the function about the given value of π up to the
term in π₯4:
a) π(π₯) = π₯β3 about π = 2
b) π(π₯) = π₯4 + 3π₯3 β 1 about π = β1
3. Find the Maclaurin series expansion up to the term in π₯4 for:
a) πβ2π₯
b) cos(π₯3)
c) πβπ₯ sin 2π₯
4. Expand β1 β π₯. Hence estimate β0.999 up to 4 decimal places.
5. Find the Taylor series expansion for 1
(1+π₯)2 from the series expansion for 1
1+π₯.
6. Find the Taylor series expansion for ln(1 + π₯) from the series expansion for 1
1+π₯.
280
CHAPTER 25: AN INTRODUCTION TO DIFFERENTIAL EQUATIONS
Suppose that we need to determine the rate of change of some quantity with respect to time (for
example, the growth or decay of a population, the cooling temperature of a hot cup of tea), we
would study differential equations. The importance of differential equations can be observed in
disciplines such as pure and applied mathematics, physics, engineering, biology, and economics.
A differential equation is an equation that contains an unknown function and one or more of its
derivatives. The order of a differential equation is the order of the highest derivative that occurs
in the equation. For example,
ππ¦
ππ₯= π₯2 is a first order differential equation
π₯π2π¦
ππ₯2 =1
π¦ is a second order differential equation.
The solution of a differential equation
π¦ = π(π₯) is a solution of a differential equation if it satisfies the differential equation.
For example, consider the differential equation ππ¦
ππ₯= 3π₯2 + 2π₯. The following are solutions for
this differential equation:
π¦ = π₯3 + π₯2
π¦ = π₯3 + π₯2 β 1
π¦ = π₯3 + π₯2 + 5
π¦ = π₯3 + π₯2 + πΆ
Observe that the derivative of each of these solutions is the differential equation. When we solve
a differential equation, we are finding all possible solutions of the equation. We usually solve
differential equations by integration.
281
Example 25.1
Solve the differential equations:
1. ππ¦
ππ₯= 4π₯3 β 2π₯ + 5
π¦ = β«(4π₯3 β 2π₯ + 5) ππ₯
= π₯4 β π₯2 + 5π₯ + πΆ
2. . ππ¦
ππ₯= π₯2 +
1
π₯
π¦ =1
3π₯3 + ln π₯ + πΆ
This solution of a differential equation is called the general solution because it contains C, a
constant.
When we graph the general solution we obtain a family of solutions.
Consider the differential equation
ππ¦
ππ₯= 2π₯
The general solution is:
π¦ = π₯2 + πΆ
Substituting values for C would give a family of solutions. The following graph represents such a
family:
282
Usually, when we solve a differential equation we are interested in finding a particular solution,
i.e. a solution that satisfies some requirement.
A differential equation is usually presented with an initial condition, that is, the value of the
dependent variable at some initial value of the independent variable. For example,
ππ¦
ππ₯= π₯2, π¦(0) = 1
The initial condition π¦(0) = 1 means that when π₯ = 0, π¦ = 1. This problem is called an initial
value problem and its solution will be a particular solution for the differential equation.
Some differential equation models
1. Population growth:
Model: The rate of change of a population is proportional to the size of the population
π π·
π π= ππ·
283
where π is the population size, π‘ is time, and k is a constant of proportionality
This is a simple model and it does not take into consideration factors that influence
population growth.
The solution of this differential equation is
π = π0πππ‘
where π0 is the initial population.
If π > 0, the population grows exponentially. Such a model does not take into account
that resources are limited.
A better model for population growth is:
π π·
π π= π (π β
π·
π΅) π·
where N is the carrying capacity of the population.
2. Newtonβs law of cooling:
This law states that the rate at which the temperature of a body changes is proportional to
the difference between the temperature of the body and the temperature of the
surroundings. The differential equation modeling that law is:
π π»
π π= π(π» β π»π)
where π is the temperature of the body, and ππ is the temperature of the surroundings.
Slope fields
Slope fields enable us to answer qualitative questions about the solution of a differential
equation. For example, how does the solution behave near a certain point?
Consider the differential equation
284
ππ¦
ππ₯= π₯ + π¦
How do we sketch the graph of its solution without solving the differential equation? We do so
by plotting a slope field.
Consider the differential equation
ππ¦
ππ₯= π(π₯, π¦)
A slope field is a collection of short straight line segments with slope π(π₯, π¦) at several points.
For example, in the differential equation
ππ¦
ππ₯= π₯ + π¦
ππ¦
ππ₯|
(1,3)= 1 + 3 = 4
This means that the slope of the function (which is the solution of the differential equation) at the
point (1,3) is 4. Thus, the slope field would contain a short line segment at (1,3) with slope 4.
ππ¦
ππ₯|
(4,β1)= 4 β 1 = 3
This means that the slope of the function (which is the solution of the differential equation) at the
point (4, β1) is 3, and the slope field would contain a short line segment at (4, β1) with slope 3.
Example 25.1
The following is the slope field for
ππ¦
ππ₯= π¦ β π₯
285
Here is the slope field with solution curves:
Eulerβs Method
This method enables us to find a numerical approximation to the solution of a differential
equation.
286
Euler used a slope field as a road map to finding an approximate solution. The idea is to start
with a first approximation (π₯0, π¦0). To find the next approximation we draw the tangent line to
the curve π¦ = π(π₯) at (π₯0, π¦0), move a short distance away to (π₯1, π¦1). Draw a tangent line to the
curve at this new approximation (π₯1, π¦1) and find another approximation (π₯2, π¦2). Continue this
procedure.
Let us find a formula for Eulerβs method by using the first two approximations: (π₯0, π¦0) and
(π₯1, π¦1).
Finding Eulerβs formula
Given
ππ¦
ππ₯= π(π₯, π¦), π¦(π₯0) = π¦0
Recall that ππ¦
ππ₯= π(π₯, π¦) represents the slope of the function. Also recall that we drew a tangent
line to the curve at (π₯0, π¦0) to find (π₯1, π¦1). The slope of that tangent line is therefore π(π₯0, π¦0).
But the slope of the line joining (π₯0, π¦0) to (π₯1, π¦1) is also
π¦1 β π¦0
π₯1 β π₯0
So
π¦1 β π¦0
π₯1 β π₯0= π(π₯0, π¦0)
Let the step size
π₯1 β π₯0 = β
Then
π¦1 β π¦0 = βπ(π₯0, π¦0)
π¦1 = π¦0 + βπ(π₯0, π¦0)
287
Extending this to arbitrary points (π₯π, π¦π) and (π₯π+1, π¦π+1), we obtain Eulerβs formula:
Eulerβs formula:
ππ+π = ππ + ππ(ππ, ππ)
where h is the step size.
Example 25.2
1. Use Eulerβs method to find 3 more approximate solutions to the initial value problem:
ππ¦
ππ₯= 2π₯ + π¦, π¦(1) = 3
Use the step size β = 0.1.
Since we want 3 more approximations, we will first write 4 approximations for π₯π
starting with π₯0.
π₯0 = 1 π₯1 = 1.1 π₯2 = 1.2 π₯2 = 1.3
Also
π¦0 = 3
Eulerβs formula is:
π¦π+1 = π¦π + βπ(π₯π, π¦π)
where
π(π₯π, π¦π) = 2π₯ + π¦
So
π¦1 = π¦0 + 0.2(2π₯0 + π¦0)
= 3 + 0.2[2(1) + 3]
ππ = π
288
π¦2 = π¦1 + 0.2(2π₯1 + π¦1)
= 4 + 0.2[2(1.1) + 4]
ππ = π. ππ
π¦3 = π¦2 + 0.2(2π₯2 + π¦2)
= 5.24 + 0.2[2(1.2) + 5.24]
ππ = π. πππ
2. Use Eulerβs method to find 3 more approximate solutions to the initial value problem:
ππ¦
ππ‘= π¦ β π‘2, π¦(0) = 2, β = 0.5
Since we want 3 more approximations, we will first write 4 approximations for π‘π
starting with π‘0.
π‘0 = 0 π‘1 = 0.5 π‘2 = 1 π‘3 = 1.5
Also
π¦0 = 2
Eulerβs formula is:
π¦π+1 = π¦π + βπ(π‘π, π¦π)
where
π(π₯π, π¦π) = π¦ β π‘2
So
π¦1 = π¦0 + 0.5(π¦0 β π‘02)
= 2 + 0.5[2 β 0]
ππ = π
289
π¦2 = π¦1 + 0.5(π¦1 β π‘12)
= 3 + 0.5[3 β 0.52]
ππ = π. πππ
π¦3 = π¦2 + 0.5(π¦2 β π‘22)
= 4.375 + 0.5[4.375 β 12]
ππ = π. ππππ
290
HOMEWORK ON CHAPTER 25
1. Solve the following differential equations:
a) ππ¦
ππ₯= 2π₯4 +
1
3π₯β 1
b) ππ¦
ππ₯= π₯7 + 3π₯5 β 6π₯ β 4
2. Use Eulerβs method with step size 0.4 to compute the approximate values π¦0, π¦1, π¦2, and
π¦3 of the solution to the initial value problem
ππ¦
ππ‘= π¦2 β 2π‘, π¦(0) = 1
3. Use Eulerβs method with step size 0.5 to compute the approximate values π¦0, π¦1, π¦2, and
π¦3 of the solution to the initial value problem
ππ¦
ππ₯= π₯2π¦ β 3π₯, π¦(1) = 4
291
CHAPTER 26: SOLVING DIFFERENTIAL EQUATIONS
In this chapter we will learn to solve specific types of differential equations using the method of
separation of variables.
Separation of variables
The differential equation
ππ¦
ππ₯= π(π₯)β(π¦)
is said to be separable because we can basically separate the variables by placing the dependent
variable π¦ with the differential ππ¦ on the left side of the equation, and placing the independent
variable π₯ with the differential ππ₯ on the right side of the equation. We achieve this by
multiplication and/or division.
Steps to solving a differential equation by separation of variables
1. Separate the variables in the differential equation ππ¦
ππ₯= π(π₯)β(π¦) to get
1
β(π¦) ππ¦ = π(π₯) ππ₯
2. Integrate both sides
β«1
β(π¦) ππ¦ = β« π(π₯) ππ₯
3. Solve by integrating both sides of the equation and putting the constant C on the right
hand side:
π»(π¦) = πΊ(π₯) + πΆ
4. If possible, solve for π¦.
292
Example 26.1
Solve:
1. ππ¦
ππ₯= 2π₯2π¦
We separate variables by dividing by π¦ and multiplying by ππ₯.
1
π¦ ππ¦ = 2π₯2 ππ₯
Integrate:
β«1
π¦ ππ¦ = β« 2π₯2 ππ₯
ln|π¦| = 2
3π₯3 + πΎ
We solve for π¦ by exponentiating both sides of the equation:
πln |π¦| = π23π₯3+πΎ
|π¦| = π23π₯3+πΎ
π¦ = Β±π23π₯3+πΎ
π¦ = Β±ππΎ . π23π₯3
Now Β±ππΎ is a constant which we can replace with the constant πΆ. Thus the solution of
the differential equation is:
π¦ = πΆπ23π₯3
293
2. ππ¦
ππ₯= 3βπ¦, π¦(2) = 4
This is an initial value problem so we will obtain a particular solution upon solving. Let
us first find the general solution.
ππ¦
ππ₯= 3βπ¦
Divide by βπ¦ and integrate:
β«1
βπ¦ ππ¦ = β« 3 ππ₯
β« π¦β12 ππ¦ = β« 3 ππ₯
2π¦12 = 3π₯ + πΆ
π¦12 =
3
2π₯ + πΆ
Again C is a constant, so even though we have divided both sides of the equation by 2, it
will remain a constant. Instead of writing C, you may choose to write K instead and
replace πΎ
2 with C.
Solving for π¦ we get:
π¦ = (3
2π₯ + πΆ)
2
We find C by substituting the values in the initial condition:
4 = (3
2(2) + πΆ)
2
2 = 3 + πΆ
πΆ = β1
So
π¦ = (3
2π₯ β 1)
2
294
3. ππ₯
ππ‘=
π‘π₯
2+π₯
β«2 + π₯
π₯ππ₯ = β« π‘ ππ‘
β« (2
π₯+
π₯
π₯) ππ₯ =
1
2π‘2 + πΆ
β« (2
π₯+ 1) ππ₯ =
1
2π‘2 + πΆ
2 ln|π₯| + π₯ =1
2π‘2 + πΆ
It would be non-trivial to solve for π₯ so we leave the solution as is.
4. 2π₯ ππ¦
ππ₯β π¦ ππ₯ = 0
We rewrite the equation as:
2π₯ ππ¦
ππ₯= π¦ ππ₯
We then separate variables:
2
π¦ ππ¦ =
1
π₯ ππ₯
2 ln|π¦| = ln|π₯| + ln πΆ
When both sides of the solution contain a natural logarithm function, it would be better to
write ln πΆ instead of πΆ.
Simplifying both sides by using the properties of logarithms we get:
ln π¦2 = ln πΆπ₯
295
So
π¦2 = πΆπ₯
π¦ = βπΆπ₯
Applications of the method of separation of variables:
Population Growth
Example 26.2
1. The rate at which a herd of cattle grows is proportional to the number of animals present
at time t. At a certain instant the number of animals is 3000. After 10 days, the number of
animals is found to be 4000. How many animals are there after 20 days?
Let P be the number of animals at time t.
We first set up the differential equation
ππ
ππ‘= ππ
We solve the differential equation by separation of variables:
β«1
π ππ = β« π ππ‘
ln|π| = ππ‘ + πΆ
|π| = πππ‘+πΆ
π = πΆπππ‘
At time π‘ = 0, π = 3000. So substituting in the general solution, we get
3000 = πΆπ0
πΆ = 3000
296
So
π = 3000πππ‘
When π‘ = 10, π = 4000. Substituting we get
4000 = 3000π10π
We solve for π. To retain accuracy we will leave our solution in terms of ππ.
4000
3000= π10π
ππ = (4
3)
110
Substituting this value into our formula for π we get
π = 3000 (4
3)
π‘10
We want to find π when π‘ = 20.
π = 3000 (4
3)
2010
π β 5333
2. A radioactive isotope decays at a rate proportional to the amount of the isotope present.
The half-life of the isotope is 8 days. After how much time will 10% of the isotope be
left?
Let A be the amount of the isotope present at time t. Then
ππ΄
ππ‘= βππ΄
Solving:
297
β«1
π΄ ππ΄ = β« βπ ππ‘
ln|π΄| = βππ‘ + πΆ
|π΄| = πβππ‘+πΆ
π΄ = πΆπβππ‘
Suppose that at time π‘ = 0, π΄ = π΄0. So substituting in the general solution, we get
π΄0 = πΆπ0
πΆ = π΄0
So
π΄ = π΄0πβππ‘
The half-life of the isotope is 8 days. That is, when π‘ = 8, π΄ =π΄0
2. Substituting:
π΄0
2= π΄0πβππ‘
1
2= πβ8π
πβπ = (1
2)
18
So
π΄ = π΄0 (1
2)
π‘8
When π΄ = 0.1π΄0:
0.1π΄0 = π΄0 (1
2)
π‘8
0.1 = (0.5)π‘8
298
ln 0.1 =π‘
8ln 0.5
8 ln 0.1 = π‘ ln 0.5
π‘ =8 ln 0.1
ln 0.5β 26.6 πππ¦π
Newtonβs Law of Cooling
Recall:
Newtonβs law if cooling states that the rate at which the temperature of a body changes is
proportional to the difference between the temperature of the body and the temperature of the
surroundings.
Example 26.3
1. Hot tea in a cup has an initial temperature of 170β. The cup is placed on a table to cool
in a room maintained at a temperature of 70β. After 20 minutes, the temperature of the
tea is 90β. Find the temperature of the tea after 30 minutes.
Let T be the temperature of the tea at time t.
We first set up the differential equation for Newtonβs law of cooling:
ππ
ππ‘= π(π β 70)
We solve the differential equation by separation of variables:
β«1
π β 70 ππ = β« π ππ‘
ln|π β 70| = ππ‘ + πΆ
|π β 70| = πππ‘+πΆ
299
π = 70 + πΆπππ‘
At time π‘ = 0, π = 170. So substituting in the general solution, we get
170 = 70 + πΆπ0
πΆ = 100
So
π = 70 + 100πππ‘
When π‘ = 20, π = 90. Substituting we get
90 = 70 + 100π20π
20 = 100π20π
We solve for π. To retain accuracy we will leave our solution in terms of ππ.
20
100= π20π
ππ = (0.2)1
20
Substituting this value into our formula for π we get
π = 70 + 100(0.2)π‘
20
We want to find π when π‘ = 30.
π = 70 + 100(0.2)3020
π β 78.9 β
300
Mixing Problems
Example 26.3
1. A large tank contains 80 kg of sucrose dissolved in 500 liters of water. Water with a
sucrose concentration of 0.2 kg of sucrose per liter of water enters the tank at a rate of
50π/πππ. The water mixes and exits the tank at the same rate. Find an expression for the
amount of sucrose in that tank at time t.
Let S be the amount of sucrose in the tank at time t.
ππ
ππ‘= ππ’ππππ π πππ‘π ππ β ππ’ππππ π πππ‘π ππ’π‘
Rate in is:
0.2 Γ 50 = 10
The amount of sucrose leaving the tank at any time depends on the concentration of the
sucrose in the tank at that time.
Concentration is:
π
500
Rate out is:
π
500 . 50 =
π
10
So
ππ
ππ‘= 10 β
π
10=
100 β π
10
Solving:
β«1
100 β π ππ = β«
1
10 ππ‘
301
β ln|100 β π| =1
10π‘ + πΆ
ln|100 β π| = β1
10π‘ + πΆ
100 β π = πΆπβπ‘
10
π = 100 β πΆπβπ‘
10
When π‘ = 0, π = 80
80 = 100 β πΆπ0
20 = πΆ
So
π = 100 β 20πβπ‘
10
2. A tank contains 120 lbs. of salt dissolved in 300 gallons of water. Brine that contains 4
lbs. of salt per gallon of water enters the tank at the rate of 5 gallons per minute. The
mixture leaves the tank at the same rate. How much salt is in the tank after 60 minutes?
Let S be the amount of salt in the tank at time t.
ππ
ππ‘= πππ‘π ππ β πππ‘π ππ’π‘
Rate in is:
4 Γ 5 = 20
The amount of salt leaving the tank at any time depends on the concentration of the salt
in the tank at that time.
Concentration is:
302
π
300
Rate out is:
π
300 . 5 =
π
60
So
ππ
ππ‘= 20 β
π
60=
1200 β π
60
Solving:
β«1
1200 β π ππ = β«
1
60 ππ‘
β ln|1200 β π| =1
60π‘ + πΆ
ln|1200 β π| = β1
60π‘ + πΆ
1200 β π = πΆπβπ‘
60
π = 1200 β πΆπβπ‘
60
When π‘ = 0, π = 120
120 = 1200 β πΆπ0
1080 = πΆ
So
π = 1200 β 1080πβπ‘
60
When π‘ = 60:
π = 1200 β 1080πβ1 = 802.7 πππ
303
HOMEWORK ON CHAPTER 26
1. A certain colony of bacteria grows at a rate proportional to the number of bacteria
present. At π‘ = 0, there are 10 bacteria. At π‘ = 4, there are 200 bacteria. At what time
will the colony have 3000 bacteria?
2. The amount of a drug present in the body decays at a rate proportional to the amount of
the drug present in the body. At time π‘ = 0, there is 25 mg of the drug in the body. The
half-life of the drug is 4 hours. How much of the drug is in the body after 10 hours?
3. A hard-boiled egg at 96β is put in a sink of 20β water. After 5 minutes, the eggβs
temperature is 36β. Assuming that the water has not warmed appreciably, how much
longer will it take for the egg to reach 22β?
4. A detective is investigating a murder. He finds the body at 6 p.m. in a room maintained at
a temperature of 68 β. At that time, the temperature of the body is 86 β. Thirty minutes
later, the temperature of the body is 85.4 β. Assuming that the body had a normal
temperature of 98.6 β, estimate the time of death.
5. A tank contains 30 kg of salt dissolved in 4000 liters of water. Brine that contains 0.02 kg
of salt per liter of water enters the tank at a rate of 25π/πππ. The solution is kept
thoroughly mixed and drains from the tank at the same rate. How much salt remains in
the tank after half an hour?
304
REFERENCES
Stewart, James. Single Variable Calculus: Early Transcendentals 7th ed. USA: Brooks/Cole,
2012.
Rogawski, Jon & Adams, Colin. Calculus 3rd ed. USA: W.H. Freeman and Company, 2015
Taalman, Laura & Kohn, Peter. Calculus. USA: W.H. Freeman and Company, 2014
Thomas, George B., Weir, Maurice & Haas, Joel. Thomasβ Calculus Early Transcendentals 11th
ed. USA: Pearson Education, Inc. 2008
www.tutorial.math.lamar.edu
www.columbia.edu/itc/sipa/math/summation.html
www.17calculus.com/infinite-series/radent-convergence
www.shmoop.com/differential-equations/slope-fields.html
www.calculuslab.deltacollege.edu/ODE/7-C-1/7-C-1-h-a.html
www.vias.org/calculus
www.cf.linnbenton.edu
www.csun.edu