calculus ii lecture notespeople.math.umass.edu/~bates/calc_ii_notes.pdf4 matthew bates 2. section...

CALCULUS II LECTURE NOTES

MATTHEW BATES

Contents

1. Notation 3

2. Section 5.2 : The Definite Integral 4

2.1. What is Area? 4

2.2. Riemann Integrals 5

3. Section 5.3 : Fundamental Theorem of Calculus 9

4. Section 5.4 : Indefinite Integrals and Net Change Theorem 18

4.1. Indefinite Integrals 18

4.2. Net change 20

5. Section 5.5 : Substitution Rules 23

5.1. Indefinite Integrals Using Substitution 23

5.2. Definite integrals using substitution 26

5.3. Symmetry 29

6. Section 6.1 : Area between curves 31

7. Section 6.2 : Volume 34

7.1. Revolution 35

8. Section 7.1 : Integration by Parts 40

9. Section 7.2 : Trigonometric Integrals 45

10. Section 7.3 : Trigonometric Substitution 51

11. Section 7.4 : Partial Fractions 55

12. Section 7.8 : Improper Integrals 61

13. Section 11.1 : Sequences 661

2 MATTHEW BATES

14. Section 11.2 : Series 71

15. Section 11.3 : Integral Test and Estimates of Sums 76

15.1. Estimating the Sum of a Series 78

16. Section 11.4 : The Comparison Test 80

17. 11.5 : Alternating Series 84

18. Section 11.6 : Absolute Convergence, Ratio and Root Test 86

18.1. Ratio Test 87

18.2. Root Test 88

18.3. Rearrangements 88

19. Section 11.8 : Power Series 89

20. Section 11.9 : Representing a Function by Power Series 92

21. Section 11.10 : Taylor and Maclaurin Series 94

22. Section 10.1 : Curves Defined by Parametric Equations 99

23. Section 10.2 : Calculus With Parametric Curves 101

24. Section 10.3 - Polar Coordinates 106

25. Section 10.4 : Area and Lengths of Polar Curves 110

CALCULUS II LECTURE NOTES 3

1. Notation

During lectures I will use some notation/shorthand which you may not of seen before. I dothis for two reasons; firstly, it greatly increases the speed at which I can write mathematics,secondly, at some point you will have to learn it if you ever want to read more advanced mathstexts.

- R,Q,Z represent the real numbers, rational numbers and integers respectively.- [a, b] means the set of real numbers, x, such that a ≤ x ≤ b.- “x ∈ R” is short hand for “x is in R, or equivalently, x is a real number”.- “cts” is short hand for, “continuous”.- “s.t.” is short hand for “such that”.- “f : [a, b]→ R” means “f is a real valued function defined on the interval, [a, b]”.- “∀x” is short hand for “for every x”.- “∃x” is short hand for “there exists an x”.- “fn” is short hand for “function”.- “thm” is short hand for “theorem”.- “def” is short hand for “definition”.

Example 1.1.f : R→ R , s.t. f(x) > 0 ∀x

This is short hand for, “f is a function which is defined over the real numbers, such thatf(x) > 0 for every real number, x.

Example 1.2.∃x ∈ [0, 1] s.t. g(x) = 0

This is short hand for, “there is a real number, x between 0 and 1 such that g(x) = 0”.

If you find this stuff a bit confusing, don’t worry. This wont be on any exam or homework, Iincluded it since it is a life skill which everyone should have.

4 MATTHEW BATES

2. Section 5.2 : The Definite Integral

In this chapter I will give a quick review of Riemann sums. All of this stuff should be familiarto you. Make sure you understand it!

2.1. What is Area?

Area is obvious for rectangles, triangles, polygons, etc. But we define the area of other shapesby the limit of approximation.

Definition 2.1.The area of a shape is the limit of the approximating areas.

Definition 2.2 (Riemann Sum).Given a function f : [a, b]→ R, we define its nth left (right, mid) Riemann sum on [a, b] asfollows. Let ∆x = b−a

n ,

Ln =

n−1∑i=0

f(x∗i )∆x, where x∗i = a+i(b− a)

n.

Rn =n∑i=1

f(x∗i )∆x, where x∗i = a+i(b− a)

n.

Mn =

n−1∑i=0

f(x∗i )∆x, where x∗i = a+b− a2n

+i(b− a)

n.

Definition 2.3 (Riemann Integral).Given a function f : [a, b]→mathbbR, we define its Riemann Integral on [a, b] as the limit of any Riemann sum, if sucha limit exists. i.e. ∫ b

af(x)dx

def= lim

n→∞Ln = lim

n→∞Rn = lim

n→∞Mn.

Some notation:

- f is called the integrand- a is the lower bound- b is the upper bound

-∫ ba f is sometimes used as shorthand for

∫ ba f(x)dx.

Definition 2.4 (Area Under a Curve).

We define the area “under” a curve to be∫ ba f(x) dx.


Example 2.5. ∫ 1

0x dx = 1/2.

Example 2.6. ∫ 1

0

√1− x2 dx = π/4.

Example 2.7. ∫ π

0cos(x) dx = 0.

Example 2.8. Find the area under the curve y = ex on the interval [0, 1].

Area = limn→∞

n∑i=1

(ei/n

)( 1

n

)= lim

n→∞

1

n

(1 + e1/n +

(e1/n

)2+ . . .+

(e1/n

)n)= lim

n→∞

1

n

((e1/n

)n − 1

e1/n − 1

)

= limn→∞

e− 1

n(e1/n − 1

)= e− 1.

This is a long and technical process. �

2.2. Riemann Integrals

Remark 2.9. Note, the Riemann sum measures signed area. Consider the following picture:

6 MATTHEW BATES

Theorem 2.10 (Properties of the Integral).

1)

∫ b

ac dx = c(b− a)

2)

∫ b

a(f + g) =

∫ b

af +

∫ b

ag

3)

∫ b

acf = c

∫ b

af

4)

∫ b

a(f − g) =

∫ b

af −

∫ b

ag

Proof.

1) ∫c dx = lim

n→∞

n∑i=1

cb− an

= limn→∞

c(b− a)

= c(b− a).

2) ∫f + g = lim

n→∞

∑(f + g)∆x

= limn→∞

(∑f∆x+

∑g∆x

)= lim

n→∞

∑f∆x+ lim

n→∞

∑g∆x.

3) Exercise.

4) Exercise. �


Example 2.11. ∫ 9

7(3 + 4x2 − x) dx =

∫ 9

73 dx+ 4

∫ 9

7x2 dx−

∫ 9

7x dx.

Theorem 2.12 (Additive property of the integral).∫ b

af(x) dx+

∫ c

bf(x) dx =

∫ c

af(x) dx.

Proof. Proof by picture:

�

The following theorem gives us some comparison properties of the integral.

Theorem 2.13 (Bounding Integrals).

1) If f(x) ≥ 0, then

∫ b

af(x) dx ≥ 0.

2) If f ≥ g, then

∫ b

af(x) dx ≥

∫ b

ag(x) dx.

3) if m ≤ f ≤M , then m(b− a) ≤∫ b

af(x) dx ≤M(b− a).

Example 2.14. Suppose that we wanted to prove that

∫ 1

0e−x dx exists and is finite. We can

do this by an application of the previous theorem.

8 MATTHEW BATES

Notice that e−x is decreasing, so max{e−x} = e−0 = 1. Therefore,∫ 1

0e−x dx ≤ 1(1− 0) = 1.

�

Question: Which functions are integrable?

By definition, the limit must exist. For example, f(x) = 1/x is not integrable on [0, 1] since∫ 1

0

1

xdx is not finite.

Theorem 2.15.If f : [a, b]→ R is continuous, then f is integrable.

Proof. ?? �


3. Section 5.3 : Fundamental Theorem of Calculus

In this section we will talk about what is (in my opinion) the most important (and underappreciated) theorems in this course. I am talking about the fundamental theorem of cal-culus. One way in which the fundamental theorem of calculus (henceforth FTC) is amazingis that it establishes a connection between differential (differentiation) calculus and integral(integration) calculus. Take a moment to think about this; differentiation is concerned withtangent lines, while integration deals with area. Why, a priori, should these two things berelated?

Definition 3.1.Let f : [a, b]→ R be continuous. Define a new function, g : [a, b]→ R by,

g(x) =

∫ x

af(t)dt.

At first, this may look like a strange way to define a function, but such functions are actuallymore common than you might expect. You can think of g as measuring the area under f froma to x.

Example 3.2. Let f(t) = 7, for t ≥ 2. Then

g(x) =

∫ x

27 dx = 7(x− 2).

Example 3.3. Let f : [0, 4]→ R be the function with the following graph...

If g(x) is defined as above, we see that,

• g(0)=• g(1)=• g(2)=• g(3)=• g(4)=

Example 3.4. Let f(t) = t. Is there a nice way to express

g(x) =

∫ x

0f(t)dt?

If we consider g(x) as the area under f from 0 to x we see that this g(x) is actually the areaof a right angle triangle with side length x. Thus,

g(x) =x2

2.

10 MATTHEW BATES

Notice, in the last example, g′(x) = f(x). Why would this be true? Lets think about this fora moment. By definition

g′(x) = limh→0

g(x+ h)− g(x)

h.

Notice that, g(x+ h)− g(x) is the area under the graph of f between x and x+ h. As we leth go to zero, this area starts to look like a rectangle, with height f(x) and width h. Thus,

g(x+ h)− g(x)

h∼=hf(x)

h= f(x).

(NOTE... this is obviously not a proof! It is only included to give you intuition about theproof).

We are now ready for the big theorem, The Fundamental Theorem of Calculus (FTC).

Theorem 3.5 (Fundamental Theorem of Calculus I).Let f : [a, b]→ R be a continuous function. If,

g(x) =

∫ x

af(t)dt,

then g is continuous and

g′ = f.

Proof. Suppose h > 0 and x, x+ h ∈ (a, b), then

g(x+ h)− g(x) =

∫ x+h

af(t)dt −

∫ x

af(t)dt

=

∫ x

af(t)dt +

∫ x+h

xf(t)dt−

∫ x

af(t)dt

=

∫ x+h

xf(t)dt.

Thus,

g(x+ h)− g(x)

h=

1

h

∫ x+h

xf(t)dt.

Since f is cts on [x, x+ h] it must attain a maximum and minimum somewhere in [x, x+ h].i.e.

f(xm) ≤ f(x) ≤ f(xM )

for some xm, xM ∈ [x, x+ h]. Thus,

hf(xm) ≤∫ x+h

xf(t)dt ≤ hf(xM )

so,

f(xm) ≤ 1

h

∫ x+h

xf(t)dt ≤ f(xM ).


We now consider what happens as we let h go to zero. Note,

limh→0

f(xm) = f(x),

since f(x) is continuous. Similarly for f(xM ). Thus, when we take limits of the aboveinequality, we see

f(x) ≤ g′(x) ≤ f(x)

thus, g′ = f

One last thing to check, what happens if x = a or b? In such a case, we consider the onesided limit and follow the same procedure. �

Another way of writing FTC is as

d

dx

∫ x

af(t)dt = f(x).

Lets get a taste of the greatness of the FTC by doing some simple examples.

Example 3.6. Find the derivative, with respect to x, of∫ x

0

√1 + sin2(t)dt.

If we let

f(x) =

√1 + sin2(x)

then, by FTC, we see the answer is f(x).�

Example 3.7. Find g′(x) where g(x) =

∫ x

2

√1 + t2dt.

Let f(t) =√

1 + t2, then by the FTC, we see that g′(x) =√

1 + x2.�

Example 3.8. Find,

d

dx

∫ x3

0cos(u)du.

12 MATTHEW BATES

Let y = x3. Then, using the chain rule we see,

d

dx

∫ x3

0cos(u)du

=dy

dx

d

dy

(∫ y

0cos(u)du

)=dy

dxcos(y)

= 3x2 cos(x3).

�

Example 3.9. Find,

d

dx

∫ x4

0sec(s) ds.

Let y = x4. Using the chain rule we see,

d

dx

∫ x4

0sec sds

=dy

dx

d

dy

∫ y

0sec(s)ds

=dy

dxsec(y)

= 4x3 sec(x4).

�

We can derive a general formula for,

d

dx

∫ b(x)

a(x)f(t)dt.

Note,

d

dx

∫ b(x)

a(x)f(t)dt

=d

dx

(∫ b(x)

0f(t)dt−

∫ a(x)

0f(t)dt

)

=d

dx

∫ b(x)

0f(t)dt− d

dx

∫ a(x)

0f(t)dt

= b′(x)f(b(x))− a′(x)f(a(x)).


Example 3.10. Calculate,

d

dx

∫ 3x2−1

sin(x)tet dt.

Using the above formula we see that the answer is,

6x(3x2 − 1)e3x2−1 − cos(x) sin(x)esin(x).

�

Example 3.11. Find the derivative of the following indefinite integral,∫ 3+x2

7xsin(t)dt

By the above remark, we see the answer is

= (3 + x2)′ sin(3 + x2)− (7x)′ sin(7x)

= 2x sin(3 + x2)− 7 sin(7x).

�

Notice that if f is an integrable function, then

g(x) =

∫ x

af(t)dt,

is a perfectly well defined function. Thus, we should treat it like one. For example,

Example 3.12. Calculate,d

dx

∫ x

0

(∫ t

ah(u)du

)dt.

If we write f(t) :=∫ ta h(u)du, then we can rewrite the question as,

d

dx

∫ x

0f(t)dt,

by FTC I we see that the answer isf(x),

i.e. ∫ x

ah(u)du.

�

The FTC also helps/allows us to calculate integrals.

14 MATTHEW BATES

Theorem 3.13 (Fundamental Theorem of Calculus II).If f is continuous and F ′ = f then,∫ b

af(x)dx = F (b)− F (a).

A function F such that F ′ = f is called an antiderivative of f .

Proof. Let F ′ = f , define

g(x) =

∫ x

af(x)dx.

Note, by FTC I, g′ = f . Thus,(g − F )′ = 0,

and thusF = g + C,

for some constant C. Moreover,

g(a) =

∫ a

af(x)dx = 0.

Thus,

F (b)− F (a) = (g(b) + C)− (g(a) + C)

= g(b)− g(a)

=

∫ b

af(x)dx.

�

Why is this great? It lets us calculate/evaluate integrals! If we want to know∫ b

af

all we need to do is find some F such that

F ′ = f.

Lets introduce some useful notation,

F (x)

∣∣∣∣ba

def= F (b)− F (a).

Example 3.14. Find, ∫ 4

0xdx.

Note,d

dx

x2

2= x.


Thus, by FTC II, the integral isx2

2

∣∣∣∣40

= 8.

�


7exdx.

Note,d

dxex = ex.

Thus, by FTC II, the integral is

ex∣∣∣∣10

7

= e10 − e7.

�

Example 3.16. Find, ∫ π

π/2cos(x)dx.

Note,d

dxsin(x) = cos(x).


sin(x)

∣∣∣∣ππ/2

= 0− 1 = −1.

�


3

1

xdx.

Note,d

dxlog(x) =

1

x.


log(x)

∣∣∣∣53

= log(5)− log(3).

�

These are very standard examples, they should be second nature to you. In a way, all otherexamples are built up from these ones. Lets do some harder examples.

16 MATTHEW BATES


1(t− 1)(3t+ 1)dt.

First we expand the brackets,

=

∫ 2

1(3t2 − 2t− 1)dt

=(t3 − t2 − t

)∣∣∣∣21

= 2− (−1) = 3.

�


15√x dx.

Note,d

dx

(2

3x

32

)=√x.

Thus ∫ 2

15√x dx = 5× 2

3x

32

∣∣∣∣21

=10

3(2)3/2 − 10

3(1)3/2

=10

3(2√

2− 1).

�

Example 3.20. Find, ∫ π4

0sec2(x)dx.

Note,d

dxtan(x) = sec2(x).

Thus ∫ π4

0sec2(x)dx = tan(x)

∣∣∣∣π40

= 1.

�



1

x− 2√xdx.

Note, this is the same as,∫ 2

1x1/2 − 2x−1/2 dx,

=

(2

3x3/2 − 4x1/2

) ∣∣∣∣21

=

(2

3(2)3/2 − 4(2)3/2

)−(

2

3(1)3/2 − 4(1)3/2

)=

2

32√

2− 8√

2− 2

3+ 4

= −20

3

√2 +

10

3.

�

It is not always obvious that a certain question requires integration to solve, for example

Example 3.22. Find,

limn→∞

n∑i=1

i3

n4.

Note, this is the same as

limn→∞

n∑i=1

1

n

(i

n

)3

,

and this is a Riemann sum, i.e. it is equal to∫ 1

0x3dx,

which we can easily solve,x4

4

∣∣∣∣10

=1

4.

�

18 MATTHEW BATES

4. Section 5.4 : Indefinite Integrals and Net Change

Theorem

4.1. Indefinite Integrals

We have seen that evaluating an integral boils down to finding an anti-derivative. Lets givesome notation for a anti-derivative.

Definition 4.1.If F ′ = f , then we say that F is an anti-derivative of f .We use the following notation to denote this,∫

f(x)dx = F (x), or,

∫f = F.

Lets list some easy anti derivatives,

•∫k = kx

•∫xn = xn+1

n+1

•∫ex = ex

•∫

sin(x) = − cos(x)•∫

cos(x) = sin(x)

You have to know other, more complicated trig anti derivatives as well.

•∫

sec2(x) = tan(x)•∫

csc2(x) = cot(x)•∫

sec(x) tan(x) = sec(x)•∫

csc(x) cot(x) = csc(x)

•∫

1√1−x2 = sin−1(x)

•∫

1x2+1

= tan−1(x)

Example 4.2. ∫(7x5 − 2 sec2(x))dx =

7x6

6− 2 tan(x) + C.

Example 4.3. ∫cos(x)

sin2(x)dx =

∫1

sin(x)

cos(x)

sin(x)dx

=

∫csc(x) cot(x)dx

= csc(x) + C.


Example 4.4. ∫(e3x − sin(t))dx =

1

3e3x − x sin(t) + C.

Example 4.5. ∫ (3x2 − 1 +

2

1 + x2

)dx = x3 − x+ 2 tan−1(x) + C.

Example 4.6. ∫(2x− 3)(4x+ 5)dx =

∫(8x2 − 2x− 15)dx

=8

3x3 − x2 − 15x+ C.

Example 4.7. ∫ π/4

0

1 + cos2(θ)

cos2(θ)dθ =

∫ π/4

0sec2(θ) + 1dθ

= (tan(θ) + θ)

∣∣∣∣π/40

= 1 +π

4.

Example 4.8. What happens to the anti-derivative if I translate a function? e.g.∫cos(θ + 3)dθ.

Notice that,

d

dθsin(θ + 3) = (θ + 3)′ cos(θ + 3)

= cos(θ + 3).

We can generalize this example, i.e.∫cos(θ +K) = sin(θ +K) + C,

similarly, ∫sin(θ +K) = − cos(θ +K) + C,∫sec2(θ +K) = tan(θ +K) + C,

... we could go on.

20 MATTHEW BATES

Lets formalise the above observation.

Theorem 4.9.Suppose f : [a, b]→ R is integrable, and

F (x) =

∫f(x)dx,

then,

F (x+K) =

∫f(x+K)dx.

Proof. Differentiate F (x+K) and see what you get,

d

dxF (x+K) = (x+K)′F ′(x+K) = f(x+K)

as required. �

4.2. Net change

Note, there is another way of writing FTC, i.e.∫ b

aF ′(x)dx = F (b)− F (a).

This formalization is useful in many practical applications. The idea is, if you give me the rateat which something changes, then I can tell you how much it has changed during a given timeperiod. These sorts of problems arise naturally in many different areas; population growth,heat diffusion, interest on a savings account, and mechanics.

Example 4.10. Suppose the function s(t) gives the displacement of a particle from a fixedpoint at time t. It is clear that

d

dts(t) = v(t),

where v(t) is the velocity function. Thus, if you are in a situation where you are only giventhe velocity of a given particle, using FTC, you can say something about the distance whichthe particle travels in a given time period. i.e.∫ t2

t1

v(t)dt = s(b)− s(a).

This is the NET change in displacement during the time t1 and t2. Note, this is not thetotal distance travelled by the particle! This is because the integral measures signed distance.In order to find the total distance travelled, we need to integrate the absolute value of thevelocity.


Example 4.11. Thus, if we have the velocity function, v(t) = sin(t), then during the timeperiod 0 to 2π, the net displacement is given by,∫ 2π

0sin(t)dt = − cos(t)|2π0 = 0.

The total distance travelled is,∫ 2π

0| sin(t)|dt =

∫ π

0| sin(t)|dt+

∫ 2π

π| sin(t)|dt

=

∫ π

0sin(t)dt+

∫ 2π

πsin(t)dt

= − cos(t)

∣∣∣∣π0

+− cos(t)

∣∣∣∣2ππ

= 4.

�

Example 4.12. Suppose the velocity of a particle is given by, v(t) = t2−4t−12, for 1 ≤ t ≤ 8.1) Find the net displacement2) Find the total distanced travelled.

1)

Net Displacement = s(8)− s(1)

=

∫ 8

1s′(t)dt

=

∫ 8

1v(t)dt.

=

(1

3t3 − 2t2 − 12t

) ∣∣∣∣81

= ...

2) Total distance travelled. We first find when the velocity is positive and when it is negative.

t2 − 4t− 12 = (t− 6)(t+ 2).

Which is positive when t < −2 or t > 6. So the total distance travelled is,∫ 8

1|t2 − 4t− 12|dt =

∫ 6

1(−t2 + 4t+ 12)dt+

∫ 8

6(t2 − 4t− 12)dt

... �

22 MATTHEW BATES

Example 4.13 (Bacteria Population). Suppose the population of some bacteria at time t = 0is 1000, and that the growth rate at time t is 1000 × 2t. What is the population of bacteriaat time: t = 1, t = 2, t = 10?

Let P (t) be the population of the bacteria at time t. We are given that P ′(t) = 1000 × 2t,and P (0) = 1000. Thus,

P (1)− P (0) =

∫ 1

0P ′(t)dt

=

∫ 1

01000× 2tdt

= 1000

∫ 1

0eln(2)tdt

= 1000

(1

ln(2)eln(2)t

) ∣∣∣∣10

=1000

ln(2)(21 − 20)

=1000

ln(2).

�


5. Section 5.5 : Substitution Rules

5.1. Indefinite Integrals Using Substitution

By now we understand the importance of finding anti-derivatives. The trouble is that this iseasier said than done. For example, find an anti-derivative for each of the following functions.

• 2x√x2 − 1

• x√

5− 3x2

• ye1+2y2

• z3 sin(3z4 − 2)

• 4x4

By the end of this chapter you will be able to find a anti derivative for some of these functions.

Lets examine the first function, 2x√x2 − 1. We know how to integrate

√x, but we don’t

know how to deal with√x2 − 1. Let u = x2 − 1. So du = 2xdx. Substituting this into the

integral, gives ∫x√x2 − 1 dx =

∫ √u du

=2

3u

32

=2

3(x2 − 1)

32 .

In general we have the following result,

Theorem 5.1 (Substitution).Suppose F ′ = f , then ∫

f(g(x))g′(x)dx = F (g(x)).

Another way of saying this is, if u = g(x), then∫f(g(x))g′(x)dx =

∫f(u)du.

Proof. Chain rule. �

Example 5.2. ∫e6xdx.

24 MATTHEW BATES

Use the substitution u = 6x. This gives du = 6xdx. Thus the integral becomes,∫eu

6du =

eu

6+ C

=e6x

6+ C.

�

Example 5.3. ∫x2ex

3dx.

Use the substitution u = x3. This gives, du = 3x2dx. Thus the integral becomes,∫eu

3du =

eu

3+ C

=ex

3

3+ C.

�

Example 5.4. ∫cos(x)esin(x)dx.

Use the substitution u = sin(x). This gives, du = cos(x)dx. Thus the integral becomes,∫eudu = eu + C

= esin(x) + C.

Example 5.5. ∫3x2 cos(x3 + 7)dx.

Use the substitution u = x3 + 7. This gives du = 3x2 dx. Thus the integral becomes,∫cos(u)du = sin(u) + C

= sin(x3 + 7) + C.

�

Example 5.6. ∫cos4(x) sin(x)dx.


Use the substitution u = cos(x). This gives, du = − sin(x)dx. Thus the integral becomes,∫−u4du = −u

5

5+ C

= −cos5(x)

5+ C.

�

Example 5.7. ∫x√

1− 4x2dx.

Use the substitution u = 1− 4x2. This gives du = 8xdx. Thus the integral becomes,∫du

8√u

=2

8

√u+ C

=1

4

√1− 4x2 + C.

�

Example 5.8. ∫x5√x2 + 1dx.

Use the substitution u = x2 + 1. This gives du = 2xdx. Note, x2 = u− 1, thus∫x4x√x2 + 1dx =

∫(u− 1)2x

√udx

=1

2

∫(u− 1)2√udu

=1

2

∫(u2 − 2u+ 1)

√udu

=1

2

∫(u5/2 − 2u3/2 + u1/2)du

�

Example 5.9. ∫tan(x)dx.

Use the substitution u = cos(x). This gives, du = − sin(x)dx. Thus the integral becomes,∫sin(x)

cos(x)dx =

∫−1

udu

= − ln(u) + C

= − ln(cos(x)) + C

= ln(sec(x)) + C.

26 MATTHEW BATES

Example 5.10. ∫(3x− 7)31dx.

Use the substitution u = 3x− 7. This gives, du = 3dx. Thus the integral becomes,∫1

3u31du =

u32

96+ C

=(3x− 7)32

96+ C.

Example 5.11. ∫x7

1 + x16dx.

Let u = x8.

5.2. Definite integrals using substitution

We will now examine methods for calculating a definite integral using substitution.

Theorem 5.12 (Substitution).If f, g are continuous functions, then,∫ b

af(g(x))g′(x)dx =

∫ g(b)

g(a)f(u)du.

Proof. Suppose F ′(x) = f(x). Then,

(F (g(x)))′ = g′(x)F ′(g(x)) = g′(x)f(g(x)).

By FTC, ∫ b

af ′(g)g′dx = F (x)

∣∣∣∣ba

= F (g(a))− F (g(b)).

Similarly, ∫ g(b)

g(a)f(u)du = F (x)

∣∣∣∣g(b)g(a)

= F (g(a))− F (g(b)).

�

Lets do some examples of definite integrals using substitution.


Two ways to do definite integrals using substitution. Either first find the indefinite integral(in terms of the original variable) then substitute in the limits, or do a substitution and plugin the new limits.

Example 5.13. ∫ 1

02x(1 + x2)100dx.

Let u = 1 + x2, then du = 2xdx. The integral becomes,

=

∫ 2

1u100du

=u101

101

∣∣∣∣21

=2101 − 1

101.

�

Example 5.14. ∫ 1

0

√3x+ 2dx.

Let u = 3x+ 2, so du = 3dx. The integral becomes,∫ 5

2

√u

3du =

2

9u3/2

∣∣∣∣52

= ......

Example 5.15. ∫ 2

0

dx

(x+ 1)2.

Let u = x+ 1, so du = dx. The integral becomes,

=

∫ 3

1

du

u2

= −1

u

∣∣∣∣31

=2

3.

�

28 MATTHEW BATES

Example 5.16. ∫ 1

0

ex

1 + exdx.

Let u = ex + 1, so du = exdx. The integral becomes,

=

∫ 2+e

2

1

udu

= ln(x)

∣∣∣∣2+e

2

= ln(2 + e)− ln(2).

�

Example 5.17. ∫ e

1

ln(x)

xdx.

Let u = ln(x), so du = 1xdx. The integral becomes,

=

∫ 1

0udu

=u2

2

∣∣∣∣10

=1

2.

�

Example 5.18. ∫ 1

−1

x√25 + 2x

dx.

Let u = 25 + 2x.

Exercises ∫(ln(x))10

xdx, u = ln(x)∫

tan−1(3θ)

1 + 9θ2dθ, u = tan−1(3θ)

∫sin(ln(3x2 − 1))

6x

3x2 − 1dx, u = ln(3x2 − 1)


5.3. Symmetry

Sometimes we can exploit the symmetry of a given function gain information about an integralwithout actually calculating it.

Example 5.19. ∫ 2π

0sin(θ)dθ = 0.

Example 5.20. ∫ 13

−13x5 = 0.

Example 5.21. ∫ 1

−1|x|dx = 2

∫ 1

0xdx.

Definition 5.22.Let f be a function. We say that f is even if

f(x) = f(−x).

We say that f is odd iff(x) = −f(−x).

Theorem 5.23.If f is an even function, then ∫ a

−af = 2

∫ a

0f.

If f is an odd function then, ∫ a

af = 0.

Proof. Do it yourself �

This simple theorem can save us a lot of work sometimes.

Example 5.24. x2 is even. x3 is odd. sin(x) is odd. cos(x) is even.

30 MATTHEW BATES

Example 5.25. ∫ −8

−8

x2 tan(x)

1 + 2x2 + 4x4dx = 0

Example 5.26. Show, ∫ π

0xf(sin(x))dx =

π

2

∫ π

0f(sin(x))dx

Consider the substitution u = π − x, so du = −dx. The integral becomes,∫ 0

π−(π − u)f(sin(π − u))du

=

∫ π

0(π − u)f(sin(π − u))du

=

∫ π

0πf(sin(u))du−

∫ π

0uf(sin(u))du.

Thus,

2

∫ π

0xf(sin(x))dx =

∫ π

0πf(sin(u))du

Example 5.27. ∫ 1

0xa(1− x)bdx =

∫ 1

0xb(1− x)adx.

Do the substitution, u = 1− x.

Example 5.28. Show, ∫ π/2

0f(cos(x))dx =

∫ π/2

0f(sin(x))dx.

Do the substitution, u = π/2− x.


6. Section 6.1 : Area between curves

We use the same reasoning as we did to calculate the area under a curve. i.e., the signed areabetween two functions, f > g, for x in [a, b], is

Area = limn→∞

n−1∑i=0

(f(x∗i )− g(x∗i ))∆x.

Note, this is the same as

limn→∞

n−1∑i=0

(f(x∗i ))∆x− limn→∞

n−1∑i=0

(g(x∗i ))∆x =

∫ b

af − g

Thus, if f > g then the area between f and g is∫

(f − g). Note, in the special case whereg = 0, we have the standard integral.

Example 6.1. Find area between the curves, f(x) = x2 and g(x) = 2− x2.

First find intersection points. We solve for x,

x2 = 2− x2

so 2x2 = 2, so x = ±1. So, the area is given by,∫ 1

−1(2− x2 − x2)dx

=

(2x− 2

3x3

) ∣∣∣∣1−1

=

(2− 2

3

)−(−2− −2

3

)=

8

3.

�

Example 6.2. Fine the area between ex and x, on [0, 1].First note that ex > x on [0, 1], thus the area is given by,∫ 1

0(ex − x)dx

=

(ex − x2

2

) ∣∣∣∣10

=

(e− 1

2

)− (1− 0)

= e− 3

2.

32 MATTHEW BATES

�

Example 6.3. Fine the area enclosed by, x3 and 3√x on the interval [0, 1].

If f(x) is not always bigger than g(x), then to work out the area enclosed between f and gwe need to split the integral. i.e.

Definition 6.4.The area between f(x) and g(x) on [a, b] is,∫ b

a

∣∣∣∣f(x)− g(x)

∣∣∣∣dx.Example 6.5. Find the area bounded between cos(θ) and sin(θ) on the interval [0, π/2].Note that between [0, π/4] cos(x) is larger than sin(x), and on [π/4, π/2], sin(x) is larger thancos(x). Thus the area is

=

∫ π/4

0(cos(θ)− sin(θ))dθ +

∫ π/2

π/4(sin(θ)− cos(θ))dθ

= (sin(θ) + cos(θ))

∣∣∣∣π/40

+ (− cos(θ)− sin(θ))

∣∣∣∣π/2π/4

= (√

2/2 +√

2/2− 0− 1) + (−0− 1 +√

2/2 +√

2/2)

= 2√

2− 2.

�

Example 6.6. Find the area enclosed between x3 − x and x.First find the intersection points,

x3 − 2 = x,

this gives x = 0 or x = ±√

2. Thus the area is given by,∫ 0

−√

2(x3 − x)− xdx+

∫ √2

0x− (x3 − x)dx

=

Sometimes it is more convenient to think of a curve as a function of y instead of as a functionof x, i.e. x = g(y).


Example 6.7.x = y2.x = sin(y)x = ey

x =√

1− y2.

Now instead of integrating with respect to x, we can integrate with respect to y.∫ d

c(xR − xL)dy

Example 6.8. Find the area enclosed between, y = x− 1 and y2 = 2x+ 6.First find the points of intersection,

(x− 1)2 = 2x+ 4,

sox2 − 4x− 5 = 0,

so x = 5,−1. So the points of intersection are, (5, 4) and (−1,−2). We integrate with respectto y. ∫ 4

−2(y + 1)−

(y2 − 6

2

)dy

=

∫ 4

−2

(−y2

2+ y + 4

)dy

=

(−y3

6+y2

2+ 4y

) ∣∣∣∣4−2

Example 6.9. Fine the area enclosed by 4x+ y2 = 12, and x = y.

Example 6.10. Fine the area enclosed by x = y2 − 4, and x = ey, between y = −1 and y = 1.

34 MATTHEW BATES

7. Section 6.2 : Volume

How do we define volume? We asked this same question for area earlier. We know the volumeof some solids, e.g. cubes, parallelepipeds, cylinders etc...

Given a general solid, S, how do we define its area? Consider a cross section of the S, i.e.the intersection of S and a plane. This is a closed curve, and we know how to find the areaof this. We can slice S into finitely many slices. It is clear the area of S is the sum of theareas of the slices. If we slice thin enough, then each slice is approximately a prism, with baseequal to a cross section and height the width of the slice.

If ∆x is the width of the slice and A(x) is the cross sectional area of S at a given x, then eachslice has volume approximately equal to hA(x). Thus the volume is approximately,∑

A(xi)∆x.

We define the volume to be the limit of the sum as ∆x→ 0, if it exists.

i.e. the volume

Vol(S) = lim∆x→0

∑A(xi)∆x =

∫A(x)dx.

Lets think about cross sections of some common solids.

Sphere. Torus. Cylinder. Cube. Pyramid .

Example 7.1. Let S be a cylinder, radius r, and length h. Then the volume is equal to∫ h

0A(x)dx =

∫ h

0πr2dx

= hπr2.

As expected. �

Example 7.2. Find th volume of the unit sphere.Place the sphere at the origin. What is A(x)? The cross sections are all circles, so we needonly to find the radii. Pythagoras gives us that

x2 + r2 = 1

r =√

1− x2.


So, A(x) = πr2 = π − πx2. So the volume is,∫ 1

−1(π − πx2)dx =

(πx− π

3x3) ∣∣∣∣1−1

= 2π − 2π

3

=4π

3.

Example 7.3. Calculate the volume of a square based pyramid with base length l and height h.

We need to calculate the cross sectional area, A(x). All of the cross sections are squares, sowe need only calculate their side length. Let s be the side length of the square at x unitsfrom the apex of the pyramid. Then,

s

x=l

h,

so s = lxh . So the volume is ∫ h

0

l2x2

h2dx =

l2x3

3h2

∣∣∣∣h0

=l2h

3.

7.1. Revolution

Note, we can think of the sphere are a circle rotated around the x-axis. This is a generalconstruction. Consider the graph of a function y = f(x). We can rotate this around thex-axis to form a solid.

What is the volume of this solid? We need to find the cross sectional area. Note that at eachx, the cross section is a circle, with radius f(x).

Thus, the volume is given by,

π

∫ b

a(f(x))2dx.

Example 7.4. Find the volume of a solid formed by rotating the curve y =√x around the

x-axis, from 0 to 1.Using the above formula, we see the volume is given by,∫ 1

0π(√x)2dx =

∫ 1

0πxdx

=π

2.

�

36 MATTHEW BATES

Example 7.5. Find the volume of revolution (around the x-axis) of the graph of y = 3xbetween x = 1 and x = 2.Answer is

π

∫ 2

1

(9x2)dx = π(3x3)

∣∣∣∣21

= 23π.

�

Example 7.6. Find the volume of revolution (around the x-axis) of the graph of y = sin(3x)between x = 0 and x = π/3.Answer is

π

∫ π/3

0sin2(3x)dx.

To calculate this integral we first do a substitution u = 3x, so the integral becomes,

π

3

∫ π

0sin2(u)du.

Now use the trig identity,

cos(2θ) = 1− 2 sin2(θ).

=π

3

∫ π

0sin2(u)du =

π

3

∫ π

0

1− cos(2u)

2du

=π

3

(x

2− sin(2u)

4

) ∣∣∣∣π0

=π

3

(π2− 0)− π

3(0− 0)

= π2/6.

�

Example 7.7. Calculate the volume of x = tan(y) rotated around the y-axis for y in [0, π/4].∫ π/4

0π tan2(y)dy = π

∫ π/4

0

(sec2(y)− 1

)dy

= π (tan(y)− y)

∣∣∣∣π/40

= π(1− π/4).

�


Example 7.8. Find the volume of the solid formed by rotating the region bounded by y = x3,y = 0, x = 0 and x = 1, around the line x = 2.Consider the cross sectional area, in terms of y. Note that we can write the equation of thecurve as, x = y1/3. Thus

A(y) = π(2− y13 )2 − π(1)2

= π(3− 4y13 + y

23 ).

Thus, the volume is, ∫ 1

0π(3− 4y

13 + y

23 )dy = π

(3y − 3y4/3 +

3

5y

53

) ∣∣∣∣10

= π

(3

5

).

�

We now consider the problem of finding the volume between two solids. For example, considertwo cylinder, one inside the other, what is the volume enclosed between them? It is simplythe volume of the larger minus the volume of the smaller.

Example 7.9. Two cones, each with apex at the origin. Larger has slope = 2, the smaller hasslope = 1. What is the volume enclosed between x = 0 and x = 1?

We have two methods for solving this problem.Method I :Find the cross sectional area of the enclosed area and integrate it. Cross section is concentriccircles, with radii 2x and x. Thus, A(x) = 3πx2.

Integrating this gives, ∫ 1

03πx2dx = π.

Method II :Find volume of smaller and subtract from volume of larger.∫ 1

04πx2dx−

∫ 1

0πx2dx = π.

Example 7.10. Cylinder radius 1, enclosing square prism of side length√

2. Find area enclosedbetween 0 and 1.Find expression for cross sectional area, A(x) = π − 2. Thus, the volume is given by,∫ 1

0(π − 2)dx = π − 2.

38 MATTHEW BATES

Example 7.11. Find the volume of revolution enclosed by y = x2 and y =√x, rotated around

the x-axis.First we must find the points of intersection, x = 1, and x = 1. Thus, the volume is given by,∫ 1

0π(√x)2dx−

∫ 1

0πx4dx = π

(x2

2

) ∣∣∣∣10

− π(x5

5

) ∣∣∣∣10

=π

2− π

5

=3π

10.

�

General formula for the volume enclosed by two surfaces of revolution.

Volume = π

∫(f2(x)− g2(x))dx.

Proof. Do it later. �

NOTE!! This is not the same as

Volume = π

∫(f(x)− g(x))2dx.

This is a very common mistake, don’t make it.

Example 7.12. Fine the volume enclosed between y = x and y = x2, rotated around thex-axis.The points of intersection are obviously x = 0 and x = 1, thus the volume is,∫ 1

0πx2dx−

∫ 1

0(x2)2dx = π

(x3

3

∣∣∣∣10

)− π

(x5

5

∣∣∣∣01

)= π

(1

3− 1

5

).

Example 7.13 (Sphere Inside A Cylinder). Consider the unit sphere, sitting inside a cylinderof radius 1 and height 2. Find the volume enclosed.Consider the graphs, y = 1 and y =

√1− x2. Note, the revolution is exactly what we want.


Thus, the desired volume is

Volume = π

∫ 1

−1

(12 − (

√1− x2)2

)dx

= π

(x− x+

x3

3

) ∣∣∣∣1−1

=2π

3.

�

Example 7.14 (Torus). Fine the volume of a torus, with radius 1, and major radius 3.

Consider the graphs, y = 3 +√

1− x2 and y = 3−√

1− x2. These intersect at x = ±1. Thedesired volume is the volume enclosed by these two functions. i.e.

Volume = π

∫ 1

−1

(3 +

√1− x2

)2−(

3−√

1− x2)2dx

= π

∫ 1

−112√

1− x2dx

=

This is a mess, but you can solve it with the current techniques.

40 MATTHEW BATES

8. Section 7.1 : Integration by Parts

Integration by parts is the integration rule which corresponds to the product rule for differ-entiation. i.e.

d

dx(f(x)g(x)) = f ′(x)g(x)− f(x)g′(x),

thus, ∫d

dx(f(x)g(x))dx =

∫ (f ′(x)g(x) + f(x)g′(x)

)dx

so,

f(x)g(x) =

∫ (f ′(x)g(x)

)dx+

∫ (f(x)g′(x)

)dx.

Rearranging this gives,∫ (f(x)g′(x)

)dx = f(x)g(x)−

∫ (f ′(x)g(x)

)dx.

This is what we call “integration by parts”.

It is sometimes easier to remember when written in shorthand,∫udv = uv −

∫vdu.

Example 8.1. Calculate ∫x sin(x)dx.

Let u = x, dv = sin(x)dx, then du = dx and v = − cos(x). So by integration by parts theintegral is equal to,

−x cos(x)−∫− cos(x)dx = −x cos(x) + sin(x) + C.

Note that the hardest part is knowing what to make u and dv. The general idea is to makeu something which becomes simpler after differentiating. For example, if we chose u = sin(x)and dv = xdx in the previous example, then we would get∫

x sin(x)dx =x2

2sin(x)−

∫x2

2cos(x)dx,

which is perfectly true, but just not helpful.

Example 8.2. ∫tetdt


Example 8.3. ∫t2etdt

Example 8.4. ∫ln(x)dx

Let u = ln(x), dv = dx, then du = 1xdx, and v = x. Thus the integral becomes,

x ln(x)−∫x

1

xdx = x ln(x)− x+ C.

Example 8.5. ∫ex sin(x)dx.

Let u = ex, dv = sin(x)dx, then du = exdx, and v = − cos(x). Thus the integral becomes,∫ex sin(x)dx = −ex cos(x) +

∫ex cos(x)dx,

Now do integration by parts on∫ex cos(x)dx. u = ex, dv = cos(x)dx, and du = exdx,

v = sin(x). Thus,∫ex sin(x)dx = −ex cos(x) +

(ex sin(x)−

∫ex sin(x)dx

).

It may look like we are going in circles but now we can solve for∫ex sin(x)dx. i.e.∫

ex sin(x)dx =1

2(ex sin(x)− ex cos(x)) .

This is a standard trick and it is important that you understand it.

Example 8.6. ∫ln(5x+ 3)dx.

Let u = ln(5x+ 3), dv = dx, then du = 55x+3dx and v = x. The integral becomes,

= x ln(5x+ 3)−∫

5x

5x+ 3dx

= x ln(5x+ 3)−∫

5x+ 3− 3

5x+ 3dx

= x ln(5x+ 3)−∫

1− 3

5x+ 3dx

= x ln(5x+ 3)− x+3

5ln(5x+ 3).

42 MATTHEW BATES

Example 8.7. ∫(x2 + 1)exdx.

Theorem 8.8.Integration by parts for definite integrals,∫ b

af(x)g′(x)dx = f(x)g(x)

∣∣∣∣ba

−∫ b

af ′(x)g(x)dx.

Example 8.9. ∫ 1

0tan−1(x)dx.

Let u = tan−1(x), dv = dx, then du = dx1+x2

, and v = x. Thus,∫ 1

0tan−1(x)dx = x tan−1(x)

∣∣∣∣10

−∫ 1

0

x

1 + x2dx

To evaluate∫ 1

0x

1+x2dx we make the substitution, u = 1 + x2, then du = 2xdx. Thus,∫ 1

0

x

1 + x2dx =

∫ 2

1

1

2udu =

1

2ln(u)

∣∣∣∣21

=1

2ln(2).

Thus, ∫ 1

0tan−1(x)dx = x tan−1(x)

∣∣∣∣10

− 1

2ln(2) =

π

4− 1

2ln(2).

Example 8.10. ∫ 12

0x cos(πx)dx

Let u = x, dv = cos(πx)dx, then du = dx and v = 1π sin(πx). So the integral becomes,

= x1

πsin(πx)

∣∣∣∣ 120

−∫ 1

2

0

1

πsin(πx)dx

=1

2πsin(π/2) +

1

π2cos(πx)

∣∣∣∣ 120

=1

2π− 1

π2.

�


Example 8.11. ∫ 2

1

(ln(x))2

x3dx

Let u = (ln(x))2, dv = 1x3dx, then du = 2 ln(x)

x , and v = −12x2

. So the integral becomes,

=−1

2x2(ln(x))2

∣∣∣∣21

+

∫ 2

1

ln(x)

x3dx,

now do integration by parts again to solve∫ 2

1ln(x)x3

dx. Let u = ln(x), dv = 1x3dx, then du = 1

x ,

and v = −12x2

. So we have that,∫ 2

1

(ln(x))2

x3dx =

−1

2x2(ln(x))2

∣∣∣∣21

+

∫ 2

1

ln(x)

x3dx

=−1

2x2(ln(x))2

∣∣∣∣21

+

(−1

2x2ln(x)

∣∣∣∣21

+

∫ 2

1

1

2x3dx

)

=−1

8(ln(2))2 +

−1

8ln(2) +

−1

4x−2

∣∣∣∣21

= −1

8ln(2) (ln(2) + 1) +

−1

16+

1

4.

�

Example 8.12. ∫ e

1x ln(x)dx.

Let u = ln(x), dv = xdx, thus du = (1/x)dx and v = x2/2. So the integral is,

= (ln(x)x2/2)

∣∣∣∣e1

−∫ e

1

x2

2

1

xdx

= e2/2−∫ e

1

x

2dx

= e2/2− (x2/4)

∣∣∣∣e1

= e2/2− e2/4 + 1/4

= e2/4− 1/4.

�

Example 8.13. ∫ 1

0

y

e2ydy

44 MATTHEW BATES

Let u = y, dv = e−2ydy, then du = dy, and v = −12e−2y. So the integral becomes,

= −1

2ye−2y

∣∣∣∣10

−∫ 1

0−1

2e−2ydy

= −1

2e−2 −

(1

4e−2y

∣∣∣∣10

)

= −1

2e−2 − 1

4e−2 +

1

4

=1

4(1− e−2).

�


9. Section 7.2 : Trigonometric Integrals

This section is all about tricks for integrating trigonometric functions.You should know the following identities,

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

2(1 + cos(2x))

tan2(x) + 1 = sec2(x) sin(2x) = 2 sin(x) cos(x) sin2(x) =1

2(1− cos(2x))

1 + cot2(x) = csc2(x)

Example 9.1. Calculate, ∫cos3(x)dx.

We first use the identity cos2(x) = 1− sin2(x),∫cos3(x)dx =

∫cos(x)(1− sin2(x))dx

=

∫cos(x)dx−

∫sin2(x) cos(x)dx

= sin(x)−∫

sin2(x) cos(x)dx

Use the substitution, u = sin(x), then du = cos(x)dx.

= sin(x)−∫u2du

= sin(x)− u3

3

= sin(x)− sin3(x)

3+ C.

�

Example 9.2. Calculate, ∫sin5(x) cos2(x)dx.

=

∫sin(x)(1− cos2(x))2 cos2(x)dx

=

∫sin(x)(1− 2 cos2(x) + cos4(x)) cos2(x)dx

=

∫sin(x)(cos2(x)− 2 cos4(x) + cos6(x))dx,

46 MATTHEW BATES

Use the substitution, u = cos(x), then du = − sin(x)dx.

=

∫−u2 + 2u4 − u6du

= −1

3u3 +

2

5u5 − 1

7u7 + C

= −1

3cos3(x) +

2

5cos5(x)− 1

7cos7(x) + C.

�

Example 9.3. Calculate, ∫sin2(2x) cos5(2x)dx.

=

∫sin2(2x)(1− sin2(2x))2 cos(2x)dx

=

∫sin2(2x)(1− 2 sin2(2x) + sin4(2x)) cos(2x)dx

=

∫ (sin2(2x)− 2 sin4(2x) + sin6(2x)

)cos(2x)dx

Use the substitution, u = sin(2x), then du = 2 cos(2x)dx.

=1

2

∫(u2 − 2u4 + u6)du

=1

2(1

3u3 − 2

5u5 +

1

7u7)

=1

6sin3(2x)− 2

10sin5(2x) +

1

14sin7(2x) + C.

�

Example 9.4. Calculate, ∫cos2(x)dx.

Use the identity, cos2(x) = 12(1 + cos(2x)),∫

cos2(x)dx =

∫1

2+

1

2cos(2x)dx

=1

2x+

1

4sin(2x) + C.

�


Example 9.5. Calculate, ∫sin4(x)dx.

Use the identity sin2(x) = 12(1− cos(2x)),∫

sin4(x)dx =

∫(sin2(x))2dx

=

∫ (1

2− 1

2cos(2x)

)2

dx

=

∫1

4− 1

2cos(2x) +

1

4cos2(2x)dx

=

∫1

4− 1

2cos(2x) +

1

4

(1

2+

1

2cos(4x)

)dx

=

∫1

4− 1

2cos(2x) +

1

8+

1

8cos(4x)dx

=3

8x− 1

4sin(2x) +

1

32sin(4x) + C.

�

Example 9.6. ∫4 sin2(θ) cos2(θ)dθ.

Example 9.7. Calculate, ∫tan6(θ) sec4(θ)dθ.

Use the identity sec2(x) = tan2(x) + 1,∫tan6(θ) sec4(θ)dθ =

∫tan6(θ)(tan2(θ) + 1) sec2(θ)dθ

=

∫(tan8(θ) + tan6(θ)) sec2(θ)dθ

Use the substitution u = tan(θ), then du = sec2(θ)dθ.

=

∫u8 + u6du

=1

9u9 +

1

7u7 + C

=1

9tan9(θ) +

1

7tan7(θ) + C.

�

48 MATTHEW BATES

Example 9.8. Calculate, ∫tan5(θ) sec7(θ)dθ.

Note, we can’t write this as some function of tan times sec2, since 7 is odd. But we can writethis as some function of sec times tan sec. Use the identity tan2(x) = sec2(x)− 1,∫

tan5(θ) sec7(θ)dθ =

∫tan(θ)(sec2(θ)− 1)2 sec7(θ)dθ

=

∫tan(θ) sec(θ)(sec2(θ)− 1)2 sec6(θ)dθ

Let u = sec(θ), then du = tan(θ) sec(θ)dθ.

=

∫(u2 − 1)2u6du

=

∫(u4 − 2u2 + 1)u6du

=

∫u10 − 2u8 + u6du

=1

11u11 − 2

9u9 +

1

7u7 + C

=1

11sec11(θ)− 2

9sec9(θ) +

1

7sec7(θ) + C

�

Example 9.9. Calculate, ∫ π/4

0tan(θ) sec3(θ)dθ.

Use the substitution, u = sec(θ), so du = tan(θ) sec(θ),∫ π/4

0tan(θ) sec3(θ)dθ =

∫ √2

1u2du

=u3

3

∣∣∣∣√

2

1

=2√

2

3− 1

3.

�

Example 9.10. Calculate, ∫ π/4

0tan(θ) sec4(θ)dθ.


Use the identity sec2(x) = tan2(x) + 1,∫ π/4

0tan(θ) sec4(θ)dθ =

∫ π/4

0tan(θ)(tan2(θ) + 1) sec2(θ)dθ

=

∫ π/4

0(tan3(θ) + tan(θ)) sec2(θ)dθ

Let u = tan(θ), so du = sec2(θ).

=

∫ 1

0u3 + udu

= (u4

4+u2

2)

∣∣∣∣10

=1

4+

1

2

=3

4.

Recall the angle sum formula,

sin(A+B) = sin(A) cos(B) + sin(B) cos(A)

cos(A+B) = cos(A) cos(B)− sin(A) sin(B).

From these equation we can derive,

sin(A) cos(B) =1

2(sin(A−B) + sin(A+B))

sin(A) sin(B) =1

2(cos(A−B)− cos(A+B))

cos(A) cos(B) =1

2(cos(A−B) + cos(A+B)).

Example 9.11. Calculate, ∫sin(4x) cos(5x)dx.

Using the above formulas we see that the integral is equal to,

=

∫1

2(sin(−x) + sin(9x))dx

=1

2(cos(−x)− 1

9cos(9x)) + C.

�

Example 9.12. Calculate, ∫sin(3x) sin(5x)dx.

50 MATTHEW BATES

=

∫1

2(cos(−2x)− cos(8x))dx

=

∫1

2(cos(2x)− cos(8x))dx

=1

2(1

2sin(2x)− 1

8sin(8x)) + C.

�

Example 9.13. ∫tan3(x)dx.

=

∫− tan(x) + tan(x) sec2(x)dx

= − ln(sec(x)) +tan2(x)

2+ C.

Example 9.14. ∫sec(x)dx =

∫sec(x)

sec(x) + tan(x)

sec(x) + tan(x)dx

=

∫sec2(x) + tan(x) sec(x)

sec(x) + tan(x)dx

= ln | sec(x) + tan(x)|+ C.

Example 9.15. ∫x sec(x) tan(x)dx.

Use integration by parts, with u = x, dv = sec(x) tan(x)dx, so du = dx, and v = sec(x).

= x sec(x)−∫

sec(x)dx

= x sec(x)− ln(sec(x) + tan(x)) + C.


10. Section 7.3 : Trigonometric Substitution

Sometimes it is use full to do substitution backwards, i.e. to calculate∫f(x)dx make a

substitution x = g(t), so the integral becomes∫f(g(t))g′(t)dt.

Example 10.1. Calculate, ∫ √a2 − x2dx.

Let x = a sin(t), then dx = a cos(t)dt. Thus the integral becomes,

=

∫ √a2 − a2 sin2(t)

(a cos(t)

)dt

=

∫a

√1− sin2(t)

(a cos(t)

)dt

=

∫a2 cos2(t)dt

=

∫a2

2(1 + cos(2t))dt

=a2

2t+

a2

4sin(2t) + C.

We now need to translate back to the original coordinate, x.

=a2

2sin−1(x/a) +

a2

42 sin(t) cos(t) + C

=a2

2sin−1(x/a) +

a

2x cos(t) + C

=a2

2sin−1(x/a) +

a

2x

√1− sin2(t) + C

=a2

2sin−1(x/a) +

a

2x√

1− (x/a)2 + C.

�

Example 10.2. Calculate, ∫ √9− x2

x2dx.

52 MATTHEW BATES

Use the substitution, x = 3 sin(t), then dx = 3 cos(3t)dt. The integral becomes,

=

∫ √9− 9 sin2(t)

9 sin2(t)3 cos(t)dt

=

∫cos2(t)

sin2(t)dt

=

∫cot2(t)dt

=

∫ (csc2(t)− 1

)dt

= − cot(t)− t+ C.

�

Example 10.3. Calculate, ∫1

x2√x2 + 4

dx.

Let x = 2 tan(t), then dx = 2 sec2(t)dt. The integral becomes,

=

∫1

4 tan2(t)2√

tan2(t) + 12 sec2(t)dt

=

∫1

4 tan2(t) sec(t)sec2(t)dt

=

∫1

4 tan2(t)sec(t)dt

=

∫cos(t)

4 sin2(t)dt

Let u = sin(t), then du = cos(t)dt.

=

∫1

4u2du

= − 1

4u+ C

= − 1

4 sin(t)+ C

= −1

4

√x2 + 4

x+ C.

�

Example 10.4. Calculate, ∫1√

x2 − a2dx.


Let x = a sec(t), then dx = a sec(t) tan(t)dt, and the integral becomes,

=1

a√

sec2(t)− 1a sec(t) tan(t)dt

=

∫sec(t)dt

= ln | sec(t) + tan(t)|+ C

= ln

∣∣∣∣xa +

√x2 − axa

∣∣∣∣+ C.

�

Example 10.5. Calculate, ∫1

u√

1− u2du

Let u = cos(t), then du = − sin(t)dt, and the integral becomes,

=

∫1

cos(t)√

1− cos2(t)

(− sin(t)

)dt

=

∫− 1

cos(t)dt

=

∫sec(t)dt

= ln | sec(t) + tan(t)|+ C

= ln

∣∣∣∣∣1u +

√1− u2

u

∣∣∣∣∣ .�

Example 10.6. Calculate, ∫1√

3− 2x− x2dx.

Complete the square to rewrite this as,∫1√

4− (x+ 1)2dx.

Now use the substitution, x = 2 sin(t)− 1, then dx = 2 cos(t)dt, and the integral becomes,

=

∫1√

4− 4 sin2(t)2 cos(t)dt

=

∫1dt

= t+ C

= sin−1

(x+ 1

2

)+ C.

54 MATTHEW BATES

�

Example 10.7 (Complete the Square).∫ √40− 6x− x2dx.

Complete the square to get, ∫ √72 − (x− 3)2dx,

now do the substitution x− 3 = 7 sin(θ).

Example 10.8 (Trick Question). Calculate,∫2t√

16− t2dt.

Use standard substitution. i.e. u = 16− t2, so du = −2tdt, and the integral is,

=

∫−1√udu

= −2u1/2 + C.

�


11. Section 7.4 : Partial Fractions

Definition 11.1 (Rational Function).A rational function is a ratio of polynomials.

Example 11.2. For example the following are rational functions,3x2 − 1

x2 − 3,

1 + x+ x2 + x3

31x− x4 + 3.

Notice that,1

x− 1+

2

x− 2=x− 2 + 2x− 2

(x− 1)(x− 2)=

3x− 4

x2 − 3x+ 2,

thus, ∫3x− 4

x2 − 3x+ 2dx =

∫1

x− 1+

2

x− 2dx

= ln(x− 1) + 2 ln(x− 2) + C.

Given a rational function, we try to write it as a sum of simple rational functions, ones whichwe can integrate.

Definition 11.3 (Proper Fraction).

A rational function, f(x) = P (x)Q(x) is called proper if deg(P ) ≤ deg(Q), and called improper

otherwise.Where the degree of a polynomial, P (x), is the highest power of x appearing in P (x).

If you have an improper fraction you can do long division to write it as a polynomial plus aproper fraction. i.e.

f(x) =P (x)

Q(x)= S(x) +

R(x)

Q(x).

Example 11.4 (Long Division). We consider the improper rational function,

x3 + x

x− 1.

Do long division,

(x− 1)√x3 + 0x2 + x+ 0...

So,

x3 + x

x− 1= x2 + x+ 2− 2

x− 1.

56 MATTHEW BATES

So, ∫x3 + x

x− 1dx =

∫x2 + x+ 2− 2

x− 1dx

=1

3x3 +

1

2x2 + 2x− 2 ln(x− 1) + C.

�

Example 11.5 (Distinct Linear). Calculate,∫x2 + 2x− 1

2x3 + 3x2 − 2xdx.

First we factorise the denominator,

2x3 + 3x2 − 2x = x(2x2 + 3x− 2) = x(2x− 1)(x+ 2),

We want to writex2 + 2x− 1

2x3 + 3x2 − 2x=A1

x+

A2

2x− 1+

A3

x+ 2,

for some A1, A2, A3 ∈ R. Solve for the Ai,

A1(2x− 1)(x+ 2) +A2(x)(x+ 2) +A3(x)(2x− 1) = x2 + 2x− 1,

a quick way to find the Ai is to cunningly choose x so that most of the terms vanish.x = 0 then the equation becomes, A1(−1)(2) = −1, so A1 = 1

2 .

x = 12 then the equation becomes, A2(1

2)(52) = 1

4 , so A2 = 15 .

x = −2 then the equation becomes, A3(−2)(5) = −1, so A1 = − 110 .

Thus the integral becomes,∫x2 + 2x− 1

2x3 + 3x2 − 2xdx =

∫1

2

1

x+

1

5

1

2x− 1− 1

10

1

x+ 2dx

=1

2ln(x) +

1

10ln(2x− 1)− 1

10ln(x+ 2) + C.

�

Example 11.6 (Distinct Linear). Find∫x4 − 3x3 + 2x2 + 2x− 1

x2 − 3x+ 2dx.

Note, this is an improper fraction, so the first thing we do is some long division,

(x2 − 3x+ 2)√x4 − 3x3 + 2x2 + 2x− 1....

So,x4 − 3x3 + 2x2 + 2x− 1

x2 − 3x+ 2= x2 +

2x− 1

x2 − 3x+ 2.

We now examine 2x−1x2−3x+2

, and try to write it as a sum of simpler fractions. Note, x2−3x+2 =

(x− 1)(x− 2).2x− 1

x2 − 3x+ 2=

A1

x− 1+

A2

x− 2,


then,A1(x− 2) +A2(x− 1) = 2x− 1.

x = 1 : then A1(−1) = 1, so A1 = −1.x = 2 : then A2(1) = 3, so A2 = 3.Thus, the integral becomes,∫

x4 − 3x3 + 2x2 + 2x− 1

x2 − 3x+ 2dx =

∫x2 +

2x− 1

x2 − 3x+ 2dx

= x2 − 1

x− 1+

3

x− 2

=1

3x3 − ln(x− 1) + 3 ln(x− 2) + C.

�

Depending on how the denominator factors changes how we write a rational function as asum of partial fractions.

Theorem 11.7.Any polynomial can be written as a product of linear terms and irreducible quadratic terms.

Theorem 11.8 (Partial Fractions).

Let f(x) = P (x)Q(x) be a proper rational function. There are four possible cases to consider.

Case 1: Q(x) factors into distinct linear terms, i.e.

Q(x) = (a1x− b1)(a2x− b2) . . . (anx− bn).

Then write,P (x)

Q(x)=

A1

a1x− b1+ · · ·+ An

anx− bn.

Case 2 : Q(x) factors into linear factors with some repeated, i.e.

Q(x) = (a1x− b1)r1(a2x− b2)r2 . . . (anx− bn)rn .

Then write,

P (x)

Q(x)=

A1,1

a1x− b1+

A1,2

(a1x− b1)2+ · · ·+ A1,n

(a1x− b1)r1

+A2,1

a2x− b2+

A2,2

(a2x− b2)2+ · · ·+ A2,n

(a2x− b2)r1

...

+An,1

anx− bn+

An,2(anx− bn)2

+ · · ·+ An,n(anx− bn)rn

.

58 MATTHEW BATES

Case 3 : Q(x) has a non-repeated quadratic term, i.e.

Q(x) = Q̃(x)(ax2 + bx+ c).

Then use,Ax+B

ax2 + bx+ c,

for the quadratic term.

Case 4 : Q(x) has a repeated quadratic term. Then do the analogous thing to what happenedin case 2.

Example 11.9 (Repeated Linear). Write the form of the partial fraction decomposition of thefunction

x3 − x+ 1

x2(x− 1)3=A

x+B

x2+

C

x− 1+

D

(x− 1)2+

E

(x− 1)3.

�

Example 11.10 (Repeated Linear). Write the form of the partial fraction decomposition ofthe function

x3

x2 + 5x+ 4.

First do some long division to get that,

x3

x2 + 5x+ 4= x− 5 +

21x+ 20

x2 + 5x+ 4.

Now factorise the denominator, x2 + 5x+ 4 = (x+ 1)(x+ 4). Thus the answer is,

x− 5 +A

x+ 1+

B

x+ 4.

�

Example 11.11 (Repeated Linear). Write the form of the partial fraction decomposition ofthe function

8x+ 1

(x+ 1)3(x2 + 3)2.

The answer is,A

x+ 1+

B

(x+ 1)2+

C

(x+ 1)3+Dx+ E

x2 + 3+

Fx+G

(x2 + 3)2.

�


Example 11.12 (Repeated Linear). Calculate,∫x4 − 2x2 + 4x+ 1

x3 − x2 − x+ 1dx.

First do long division,

(x3 − x2 − x+ 1)√x4 + 0x3 − 2x2 + 4x+ 1...

So the integral becomes, ∫x+ 1 +

4x

x3 − x2 − x+ 1dx.

We now concentrate on the fraction. Factorise the denominator,

x3 − x2 − x+ 1 = (x− 1)(x2 − 1) = (x− 1)2(x+ 1).

Thus we consider partial fractions of the form,

4x

x3 − x2 − x+ 1=

A

x− 1+

B

(x− 1)2+

C

x+ 1.

Putting everything over a common denominator gives that

4x = A(x− 1)(x+ 1) +B(x+ 1) + C(x− 1)2,

Letting x = 1 we see that 4 = 2B, so B = 2.Letting x = −1 we see that −4 = 4C, so C = −1Thus, 4x = A(x− 1)(x+ 1) + 4(x+ 1)− (x− 1)2, so A = 1.So the integral becomes,∫

x4 − 2x2 + 4x+ 1

x3 − x2 − x+ 1dx =

∫x+ 1 +

1

x− 1+

2

(x− 1)2− 1

x+ 1dx

= x2/2 + x+ ln(x− 1)− 2

x− 1− ln(x+ 1) + C.

�

Example 11.13 (Distinct Quadratic). Calculate,∫2x2 − x+ 4

x3 + 4xdx.

Factorise the denominator, x3 + 4x = x(x2 + 4), so we need to consider partial fractions ofthe form,

2x2 − x+ 4

x3 + 4x=A

x+Bx+ C

x2 + 4.

So,2x2 − x+ 4 = A(x2 + 4) +Bx2 + Cx = (A+B)x2 + Cx+ 4A.

So A = 1, C = −1, and B = 1. The integral becomes,∫2x2 − x+ 4

x3 + 4xdx =

∫1

x+

x− 1

x2 + 4dx

= ln(x) +

∫x

x2 + 4dx−

∫1

x2 + 4dx

= ln(x) +1

2ln(x2 + 4)− 1

2arctan

(x2

)+ C.

60 MATTHEW BATES

�

Exercise 11.14. Calculate the following,∫1

(x− 3)(x+ 2)dx∫

2x

(x+ 4)(2x− 1)dx∫

x

x− 17dx

Example 11.15. Calculate, ∫ 16

9

√x

x− 4dx.

First we make the substitution, u =√x, so x = u2 and dx = 2udu. So the integral becomes,∫ 16

9

√x

x− 4dx =

∫ 4

3

u

u2 − 42udu

= 2

∫ 4

3

(1 +

4

u2 − 4

)du

= 2u

∣∣∣∣43

+

∫ 4

3

4

u2 − 4du

= 8− 6 +

∫ 4

3

1

u− 2− 1

u+ 2du

= 2 + (ln(u− 2) + ln(u+ 2))

∣∣∣∣43

= 2 + ln

(2

6

)− ln

(1

5

).

�


12. Section 7.8 : Improper Integrals

Does it make sense to talk about the area of an unbounded shape?

Example 12.1. Calculate the area under the graph of f(x) = 1x2

form x = 1 to x =∞.Infinity isn’t a real number, so we can’t just calculate the antiderivate then plug in infinity.What we can do is estimate the area by considering∫ t

1

1

x2dx,

for larger and larger values of t. ∫ t

1

1

x2dx =

−1

x

∣∣∣∣t1

= 1− 1

t.

So for larger and larger values of t, this estimate gets closer and closer to 1. So it is reasonablyto say, ∫ ∞

1

1

x2dx = 1.

�

Definition 12.2 (Type 1 Improper Integrals).∫∞a f(x)dx = lim

t→∞

∫ ta f(x)dx, if it exists.∫ b

−∞ f(x)dx = limt→−∞

∫ bt f(x)dx, if it exists.∫∞

−∞ f(x)dx =∫∞a f(x)dx+

∫ b−∞ f(x)dx, if it exists.

Example 12.3. Calculate, ∫ ∞1

1

xdx.


1

x3dx.


1

xpdx,

for a general p ∈ R.

62 MATTHEW BATES

Theorem 12.6. ∫ ∞1

1

xpdx,

converges if p > 1, and diverges otherwise.


e−xdx.


xe−x2dx.


xe−xdx.

Example 12.10. Calculate, ∫ ∞−∞

1

x2 + 1dx = π.


sin(x)dx.

There are other ways in which the graph of a function can be unbounded, i.e. infinite discon-tinuities.

Definition 12.12 (Type 2 Improper Integrals).If f(x) has a vertical asymptote at x = c where a < c < b, then∫ c

af(x)dx

def= lim

t→c−

∫ t

af(x)dx.

Sinilarly, ∫ b

cf(x)dx

def= lim

t→c+

∫ b

tf(x)dx.

And, ∫ b

af(x)dx

def=

∫ c

af(x)dx+

∫ b

cf(x)dx.

When such limits exist.



0

1√xdx.


3

1√x− 3

dx.


0

1

x− 1dx.

This function has a asymptote at x = 1, so we have to split this into two integrals,∫ 3

0

1

x− 1dx =

∫ 1

0

1

x− 1dx+

∫ 3

1

1

x− 1dx.

Lets concentrate on the first term,∫ 1

0

1

x− 1dx = lim

t→1−

∫ t

0

1

x− 1dx

= limt→1−

ln |x− 1|∣∣∣∣t0

= limt→1−

ln |t− 1| − ln | − 1|

= limt→1−

ln(1− t)

= −∞.

Thus the integral doesn’t exist.

Note that in the previous example it is important to split the integral, if you didn’t realisethat it was an improper integral and just tried to integrate it naively you would get,∫ 3

0

1

x− 1dx = ln |x− 1|

∣∣∣∣30

= ln(2)− ln(1) = ln(2).

Which is incorrect.


0ln(x)dx.

64 MATTHEW BATES

This has an vertical asymptote at x = 0, so the integral equals,∫ 1

0ln(x)dx = lim

t→0+

∫ 1

tln(x)dx

= limt→0+

(x ln(x)

∣∣∣∣1t

−∫ 1

t1dx

)= lim

t→0+(−t ln(t)− 1 + t)

= limt→0+

ln(t)

1/t− 1

= limt→0+

1/t

−1/t2− 1

= limt→0+

−t− 1

= −1.


−1

e1x

x2dx

First find an antiderivative for e1x

x2. Use the substitution u = 1/x.∫

e1x

x2dx = −

∫eudu = −eu = −e1/x.

So, ∫ 0

−1

e1x

x2dx = lim

t→0−

∫ t

−1

e1x

x2dx

= limt→0−

−e1/x

∣∣∣∣t−1

= limt→0−

e1/t − e−1

= lims→−∞

es − e−1

= e−1.

Theorem 12.18 (Comparison Theorem for Improper Integrals).If f(x) ≥ g(x) for all x ≥ a, then if

∫∞a f(x)dx exists then so does

∫∞a g(x)dx, and∫ ∞

af(x)dx ≥

∫ ∞a

g(x)dx.


Example 12.19 (Comparison Theorem). The integral∫ ∞4

ln(x)

xdx,

diverges.

66 MATTHEW BATES

13. Section 11.1 : Sequences

Definition 13.1.A Sequence is any order list of numbers, {a1, a2, . . . }. Sometimes denoted,

{an}∞n=0, {an}.

One could think of a sequence as a function f : N→ R.

It is important to note that a sequence doesn’t need to follow any sort of pattern.

Example 13.2 (Sequences as lists).

{3, 1, 4, 1, 5, . . . }.

an = nth digit of π.

Example 13.3 (Sequences as functions).

{n}∞n=1 an = n {1, 2, 3, 4, . . . }{1/n}∞n=1 an = 1/n {1, 1/2, 1/3, 1/4, . . . }

{sin(nπ/3)}∞n=1 an = sin(nπ/3) {√

3/2,√

3/2, 0,−√

3/2,−√

3/2, 0, . . . }.

Example 13.4 (Sequences recursively defined). The sequence defined by a1 = a2 = 1, andan = an−1 + an−2 is called the Fibonacci sequence. It has lots of strange properties.

{1, 1, 2, 3, 5, 8, 13, 21, 34, . . . }

For example, not that

22 − 1× 3 = 1,

32 − 2× 5 = −1,

52 − 3× 8 = 1,

82 − 5× 13 = −1

. . .

Example 13.5 (Pictures of sequences). One can think of sequences pictorially, this can behelpful to understand convergence.


Definition 13.6 (Limit of a sequence).We define the limit of the sequence, {an} as n→∞ to be,

limn→∞

an = l

if

∀ε > 0,∃N > 0 such that, n > N =⇒ |an − l| < ε.

We also sometimes write this as, an → l.If a sequence has a limit, we say that it converges, else we say that it diverges.If limn→∞ an =∞, then we say that the sequence diverges to infinity.

You should think of the sequence an converging to l to mean that no matter how close youwant the sequence to be to l, there is some number, N > 0, such that if n > N then thesequence is always at least that close.

Theorem 13.7.If f(x) is some function, and an = f(n) for all n, then

limn→∞

an = limx→∞

f(x).

Example 13.8. If an = 1n , then

limn→∞

1

n= lim

x→∞

1

x= 0.

Example 13.9. Again, let an = 1n . We will show that an → 0 using the ε−N definition.

Fix some ε > 0. We want to find some N > 0, such that

n > N =⇒ | 1n− 0| < ε.

This is true when n > 1ε . So let N = [1

ε ].

Theorem 13.10.If f(x) is a continuous function then,

limn→∞

f(an) = f(

limn→∞

an).

68 MATTHEW BATES

Example 13.11. Calculate the limit of the following sequence,

an =4 + 5n2

n− 3n2.

Divide the numerator and denominator by n2,

4 + 5n2

n− 3n2=

4n2 + 51n − 3

→ −5/3.


bn = 21/n − 1.

Use the function f(x) = 21/x − 1.


cn = (−1)n.

The limit doesn’t exist. To see this not that if the limit did exist, (−1)n → l for some l, thenfor ε = 1/2, there must exist some N > 0 such that

n > N =⇒ |(−1)n − l| < 1/2.

So in particular,

|(−1)2N − l| < 1/2,

and

|(−1)2N+1 − l| < 1/2.

So, 1/2 < l < 3/2 and −3/2 < l < −1/2, which is impossible.


dn = sin(nπ/2).

Note, that

dn = {1, 0,−1, 0, 1, 0,−1, 0, . . . }so the limit doesn’t exist for the same reason as the previous example.

Definition 13.15.We say that a sequence diverges to infinity, an →∞, if

∀M > 0, ∃N > 0, such that, n > N =⇒ an > M.


Example 13.16.

limn→∞

(2 + n3) =∞.

To show this let M > 0, we want to find N > 0, such that,

n > N =⇒ 2 + n3 > M

i.e. n3 > M − 2, so n > 3√M − 2.

Theorem 13.17 (Limit Laws).If lim an <∞ and lim bn <∞, then,

lim(an + bn) = lim(an) + lim(bn),

lim(can) = c lim(an),

lim(anbn) = lim(an) lim(bn),

lim(an/bn) = lim(an)/ lim(bn).

Note, it is very important that both lim(an) <∞, and lim(bn) <∞, other wise these equationsare not true. For example,

1 = lim(1) = lim(n× 1

n) 6= lim(n) lim(1/n).

Theorem 13.18 (Squeeze Rule).If an ≤ bn ≤ cn for all n > n0 for some n0 > 0, then,

lim(an) ≤ lim(bn) ≤ lim(cn).


an =sin(nπ/2)

n.

Note,

−1 ≤ sin(nπ/2) ≤ 1,

so,

−1/n ≤ sin(nπ/2)/n ≤ 1/n.

So by the Squeeze rule,

0 = lim(−1/n) ≤ lim(sin(nπ/2)/n) ≤ lim(1/n) = 0.

Thus,

lim(sin(nπ/2)/n) = 0.

70 MATTHEW BATES

Example 13.20. Calculate the limit of,n!

nn.

Note that,

0 ≤ n!

nn≤ 1

n→ 0.

Theorem 13.21 (Absolute Convergence Theorem for Sequences).If limn→∞ |an| = 0, then limn→∞ an = 0.

Definition 13.22.A sequence an is called...Increasing : if an+1 ≥ an for all n.Decreasing : if an+1 ≤ an for all n.Monotonic : if it is increasing or decreasing.Bounded : if there exists some K > 0 such that −K < an < K for all n.

Example 13.23. Determine whether the sequence is bounded, increasing, decreasing, or notmonotonic.

an =1

4n+ 1

Example 13.24. Determine whether the sequence is bounded, increasing, decreasing, or notmonotonic.

an =

(−1

2

)n

Theorem 13.25 (Monotone Convergence Theorem).Every bounded monotonic sequence converges.

Theorem 13.26 (Geometric Sequence).

limn→∞

rn =

∞ if r > 1

0 if − 1 < r ≤ 1

DNE if r ≤ −1.

Example 13.27. Does the following sequence converge?

an =(−2)3n+1

32n


14. Section 11.2 : Series

What does it mean when we write a decimal number?

0.125 =0

100+

1

101+

2

102+

5

103.

So what does it mean when we write an infinite decimal?

0.33333 =3

101+

3

102+

3

103+ . . . .

This is an infinite sum. How do we make sense of infinite sums in general?

Definition 14.1 (Partial Sums).Given a sequence an, we define

sk =k∑i=1

aidef= a1 + a2 + · · ·+ ak.

We denote this sum by sk, and call it the kth partial sum of the sequence.

Example 14.2.n∑i=1

i = 1 + 2 + 3 + 4 + · · ·+ n =1

2n(n+ 1).

Definition 14.3 (Series).Given a sequence, the associated series is defined to be,

∞∑i=1

aidef= lim

n→∞

n∑i=1

ai = limn→∞

sn.

You can think of this as an infinite sum.

Example 14.4. Calculate,∞∑i=1

(1

2

)n.

Draw a square.


n.

Look at partial sums.

72 MATTHEW BATES


(−1)n.

Look at partial sums. They don’t converge.

Definition 14.7 (Geometric Series).A series of the form

a+ ar + ar2 + ar3 + ar4 + · · · =∞∑i=1

ari−1,

is called a geometric series.

Theorem 14.8 (Geometric Series).

∞∑i=1

ari−1 =

{Diverges if |r| ≥ 1a

1−r if |r| < 1

Proof. Note,

sn = a+ ar + ar2 + · · ·+ arn−2 + arn−1,

rsn = ar + ar2 + ar3 + · · ·+ arn−1 + arn,

So,

sn − rsn = a− arn,and thus,

sn =a(1− rn)

1− r.

�

Example 14.9. Show that∞∑i=1

(1

2

)i= 1

Example 14.10. Calculate∞∑i=1

2

(1

3

)i



22i31−i


sini(x).

Example 14.13. For what values of x does the following converge? And when it does converge,what is it’s limit?

∞∑n=0

xn =

1

1−x if |x| < 1,

∞ if x ≥ 1,

DNE if x ≤ −1.

Example 14.14. For what values of x does the following converge? And when it does converge,what is it’s limit?

∞∑n=0

(x− 2)n

3n.

This is a geometric series, with common ration x−23 . Thus

∞∑n=0

(x− 2)n

3n=

1

1−x−23

if |x−23 | < 1,

∞ if x−23 ≥ 1,

DNE if x−23 ≤ −1.

Simplifying gives,

∞∑n=0

(x− 2)n

3n=

3

5−x if x ∈ (1, 5),

∞ if x ≥ 5,

DNE if x ≤ 1.

Theorem 14.15 (Series Laws).If∑an and

∑bn are both convergent series, then,∑

(an + bn) =∑

(an) +∑

(bn)∑(can) = c

∑(an)

Example 14.16. Calculate∞∑n=1

1 + 3n

5n.

74 MATTHEW BATES

Example 14.17 (Partial Fractions / Telescoping Series). Calculate∞∑n=1

1

n(n+ 1).

Use partial fractions to write this as,∞∑n=1

(1

n− 1

n+ 1

)Now calculate the partial sums,

sk =

(1

1− 1

2

)+

(1

2− 1

3

)· · ·+

(1

k− 1

k + 1

)= 1− 1

k + 1

Thus,∞∑n=1

(1

n− 1

n+ 1

)= lim(sk) = lim

(1− 1

k + 1

)= 1.

Example 14.18 (Harmonic Series). The Harmonic series is defined as, an = 1n . We will show

that the associated series is divergent.

s1 = 1

s2 = 1 +1

2

s4 = 1 +1

2+

1

3+

1

4≥ 1 +

1

2+

1

4+

1

4= 1 +

2

2

s8 = 1 +1

2+ · · ·+ 1

8≥ 1 +

3

2...

...

s2k ≥ 1 +k

2Thus,

∞∑n=1

1

n= lim(sk) =∞.

This is a very important example since it is an example of a sequence, an, where an → 0, but∑an does not converge.

Theorem 14.19 (Divergence Test).If an 6→ 0, then

∑an diverges.

Equivalently, if∑an converges then an → 0.

Proof. If∑an = l <∞, then lim sk = l. Now note that, an = sn − sn−1 → l − l = 0. �


Note the direction of the implication! The above theorem says that if∑an <∞ then an → 0.

It does not say an → 0 then∑an <∞!!!

Example 14.20. Calculate the following limits,∞∑n=1

n2

3n2 + 1

∞∑n=1

(−3

π

)n∞∑n=1

2

n2 + 2n

∞∑n=1

ln

(n

n+ 1

)∞∑n=2

2

n2 − 1

Example 14.21. Write 2.317 as a fraction.

2.317 = 2.3 +17

103+

17

105+

17

107. . .

=23

10+

17

103

(1

1− 1102

)

=23

10+

17

103

(102

99

)=

23

10+

17

990.

76 MATTHEW BATES

15. Section 11.3 : Integral Test and Estimates of Sums

It is very hard to calculate the exact value of a general series. So on settles for estimates.Moreover, it is often useful to know that a series converges, even if you don’t know it’s exactvalue.

Example 15.1. We will show that∞∑n=1

1

n2,

converges.There is no simple formula for the partial sums of this series, but we can still prove that itconverges by thinking of it geometrically. Consider the function,

f(x) =1

x2.

Thinking of the series as the area of a bunch of rectangles, we see that,

∞∑n=1

1

n2≤ 1 +

∫ ∞1

1

x2dx = 2.

Thus the series converges. (The exact value is actually π2

6 , but this is very hard to show.)

Example 15.2. We will show that∞∑n=1

1√n,

diverges.Consider the function,

f(x) =1√x.

Thinking of the series as the area of a bunch of rectangles, we see that,

∞∑n=1

1

n2≥∫ ∞

1

1

x2dx =∞.

Thus the series diverges.

Theorem 15.3 (Integral Test).

If f(x) is a positive, continuous, decreasing function, and an = f(n), then the series∞∑n=1

an

converges iff the integral

∫ ∞1

f(x)dx, converges.


Note, we need decreasing so that the rectangle either all lie above or all lie below the graph.The integral test doesn’t tell us what the actual value of a series is if it converges, it only tellsus that it exists.

Example 15.4. Show that the following series converges,

∞∑n=1

1

n2 + 1.

Note the function, f(x) = 1x2+1

is positive, continuous and decreasing.

∫ ∞1

1

x2 + 1dx = lim

t→∞tan−1(x)

∣∣∣∣t1

= π/2− π/4= π/4.

Thus, by the integral test, the series converges.

Example 15.5 (p-text). For what values of p does the following converge?

∞∑n=1

1

np,

Example 15.6. Does the following series converge?

∞∑n=1

ne−n

The function f(x) = xe−x is obviously continuous, and positive. To show that it is decreasingwe consider the derivative,

f ′(x) = e−x − xe−x = e−x(1− x).

78 MATTHEW BATES

Which is negative for all x > 1. So the function satisfies the hypothesis of the integral test.∫ ∞1

xe−xdx = limt→∞

∫ t

1xe−xdx

= limt→∞−xe−x

∣∣∣∣t1

+ limt→∞

∫ t

1e−xdx

= limt→∞

(−te−t) +1

e+ limt→∞

(−e−t +1

e)

=2

e+ limt→∞

−te−t

=2

e+ limt→∞

−1

−e−t

=2

e.

Thus, the series converges.

Example 15.7. Does the following series converge?

∞∑n=2

1

n ln(n).

The function, f(x) = 1x ln(x) is positive and continuous for x ≥ 2. To show that it is decreasing

we consider it’s derivative,

f ′(x) =− ln(x)− 1

(x ln(x))2< 0

so it is also decreasing. ∫ ∞2

1

x ln(x)= lim

t→∞

∫ t

2

1

x ln(x)

= limt→∞

ln(ln(x))

∣∣∣∣t2

= limt→∞

ln(ln(t))− ln(ln(2))

=∞.

Thus, the series diverges.

15.1. Estimating the Sum of a Series

Definition 15.8.If∑∞

n=1 an = s, then we define the nth remainder to be,

Rndef= s− sn = an+1 + an+2 + an+3 + . . . .


You should think of the remainder as the error in the approximation, sn.

Theorem 15.9.Suppose f(x) is positive, decreasing, and continuous, and that f(n) = an. If

∑an <∞, then∫ ∞

n+1f(x)dx ≤ Rn ≤

∫ ∞n

f(x)dx.

Example 15.10. How close is the 10th partial sum to the exact value of∑∞

n=11n2 ?

How many terms do you need to calculate so that the error is less than 0.01?From the above theorem we know that the error is bounded by,

R10 ≤∫ ∞

10

1

x2dx =

1

10.

For the error to be less than 0.01 we require Rn ≤ 0.01, i.e.

Rn ≤∫ ∞n

1

x2dx =

1

n≤ 0.01 =

1

100.

So n = 100 will do.

80 MATTHEW BATES

16. Section 11.4 : The Comparison Test

The idea is to compare a given series to an easier one, one which you understand.

Example 16.1 (Comparison Test). Consider the series,

∞∑n=1

1

2n + 1.

This is not a geometric series, but it is pretty close to one. Note that,

N∑n=1

1

2n + 1≤

N∑n=1

1

2n≤∞∑n=1

1

2n= 1.

So the partial sums are bounded and increasing, thus they converge. i.e.∞∑n=1

1

2n + 1,

converges.

Theorem 16.2 (Comparison Test).If an, bn ≥ 0 for all n, and an ≤ bn for all n, then:

If∞∑n=1

bn converges then∞∑n=1

an converges.

If

∞∑n=1

an diverges then

∞∑n=1

bn diverges.

Some series which are commonly used for comparisons,∞∑n=1

1

np,

∞∑n=1

arn−1.

Example 16.3 (Comparison Test). Show that the following series converges,

∞∑n=1

2

3n2 + 1.

Compare with∑ 1

n2 . Note,

2

3n2 + 1≤ 1

3n2 + 1≤ 1

3n2≤ 1

n2.


By the p-test, ∑ 1

n2<∞.

Thus∞∑n=1

2

3n2 + 1,

converges.

Example 16.4 (Comparison Test). Does the following series converge?

∞∑n=1

n+ 1

n2√n.

Use,n+ 1

n2√n≤ 2n

n5/2≤ 2

n3/2.


∞∑n=1

1

n!.

Use,1

n!≤ 1

2n−1.


∞∑n=1

3n

2 + 5n.

Use,3n

2 + 5n≤ 3n

5n.


∞∑n=1

2 + sin(n)

3n.

Use,2 + sin(n)

3n≤ 3

3n.

82 MATTHEW BATES

Example 16.8 (Comparison Test). Does the following series converge?∞∑n=1

e−n

n.

Use,e−n

n≤ e−n.

Note, we only actually need that an is bounded by bn, eventually. This is because the con-vergence of a series only depends on what happens eventually.

Theorem 16.9 (Limit Comparison Test).If, an, bn ≥ 0 for all n, and

limn→∞

anbn

= c 6= 0,

then either both∞∑n=1

an and∞∑n=1

bn converge, or they both diverge.

Note, it doesn’t matter whether you use anbn

or bnan

, since if one limit exists and is non-zerothen so it the other.

Note that when the limit equals zero, you can’t say anything. For example,

1/n2

1/n→ 0,

and∑ 1

n2 <∞ but∑ 1

n =∞.

Example 16.10 (Limit Comparison Test). Does the following series converge?∞∑n=1

1

2n − 1.

Use limit comparison test with, bn = 12n .

limn→∞

2n − 1

2n= 1.

Since,∑

1/2n converges, so does∑

1/(2n − 1).

Example 16.11 (Limit Comparison Test). Does the following converge?∞∑n=1

n2 + n√n5 + 7n− 17

.


Compare with∑ 1√

n.


1

2n + sin(n).

Compare with∑ 1

2n .


n+ 2n

n+ 3n.

Compare with∑ 2n

3n .n3n + 2n3n

n2n + 3n2n=n/2n + 1

n/3n + 1→ 1.


√n

2n− 1.

Compare with∑ 1√

n.


1

4n − 3n.

Compare with∑ 1

4n .

84 MATTHEW BATES

17. 11.5 : Alternating Series

Definition 17.1 (Alternating Series).An alternating series is the series associated to a sequence whose terms alternate in sign. i.e.one of the form,

∞∑n=1

(−1)nan, or

∞∑n=1

(−1)n+1an,

where an ≥ 0 for all n.

Theorem 17.2 (Alternating Series Test).If an is a non-negative, decreasing sequence, then

∞∑n=1

(−1)n−1an <∞ ⇐⇒ limn→∞

an = 0.

Example 17.3 (Alternating Series). Does the following converge?∞∑n=1

(−1)n

n.

It is clear that 1/n is non-negative and decreasing, and that lim 1/n = 0. Thus, by the

alternating series test,∑∞

n=1(−1)n

n converges.

Example 17.4. Does the following converge?∞∑n=1

(−1)n+3 n

3n− 1.

By the divergence test it diverges.


(−1)nln(n)

n.

It is clear that ln(n)/n is non-negative. We now show that it is decreasing.

d

dx

ln(x)

x=

1− ln(x)

x2< 0

when x > e.

limln(n)

n= lim

1/n

1= 0.

So, by the alternating series test,∑∞

n=1(−1)n ln(n)n converges.



(−1)nn!

nn.


cos(nπ)ln(n)

n.

Theorem 17.8 (Alternating Series Estimation Theorem).If∑∞

n=1(−1)nan is an alternating series, with an+1 ≤ an, and limn→∞ an = 0, then

Rn = |s− sn| ≤ an+1.

86 MATTHEW BATES

18. Section 11.6 : Absolute Convergence, Ratio and Root

Test

Given any sequence, an, we can form a new one |an|.

Definition 18.1 (Absolutely Convergent, and Conditionally Convergent).A series

∑an is called absolutely convergent if

∑|an| converges. If

∑an converges, but∑

|an| diverges, then the series is called conditionally convergent.

Note, it is not true that if∑an converges then

∑|an| also converges. For example,

∑(−1)n 1

n .

Example 18.2. ∑(−1)n

1

n2∑(−1)n

1

2n∑(−1)n

1

n!∑(−1)n

1√n

Theorem 18.3.If∑an is absolutely convergent, then it is convergent.

Proof. Note,

0 ≤ an + |an| ≤ 2|an|,

for all n. So, ∑(an + |an|) <∞,

by the comparison test. So, ∑an =

∑(an − |an|)−

∑|an|,

converges. �


18.1. Ratio Test

The general idea is to compare your series with a geometric series.

Theorem 18.4 (Ratio Test).If

limn→∞

∣∣∣∣an+1

an

∣∣∣∣ = L,

then,i) L < 1 implies

∑an is absolutely convergent.

ii) L > 1 implies∑an is divergent.

iii) L = 1 inconclusive.

Example 18.5. Show that the following is absolutely convergent,

∞∑n=1

(−1)nn4

2n.

Example 18.6. Note that if L = 1 then we can’t say anything. For example, both∑

1/n and∑1/n2 have L = 1, but the first diverges while the second converges.

Example 18.7.∞∑n=1

2n

n!

∞∑n=1

(n!)2

(2n)!

∞∑n=1

(−1)n2n

n2

∞∑n=1

(−1)nn!

nn

∞∑n=1

(−1)n7nn!

(n+ 2)!

88 MATTHEW BATES

18.2. Root Test

Theorem 18.8 (Root Test).If

limn→∞

n√|an| = L,

then,i) L < 1 implies

∑an is absolutely convergent.

ii) L > 1 implies∑an is divergent.

iii) L = 1 inconclusive.


(5n+ 7

7n+ 5

)n


(n2 + 7

2n2 + 1

)3n


n2n

3n

18.3. Rearrangements

One big reason why one cares whether a series is absolutely convergent is that being absolutelyconvergent allows one to rearrange the order in which the terms are added without changingthe final sum. This is note true for conditionally convergent series.


(−1)n

n= l <∞.

Now rearrange the order of the sum,

(1− 1

2)− 1

4+ (

1

3− 1

6)− 1

8+ (

1

5− 1

10)− 1

12+ . . .

=1

2− 1

4+

1

6− 1

8+

1

10− 1

12+ . . .

= 2l


19. Section 11.8 : Power Series

Definition 19.1 (Power Series).A power series is a formal sum/series of the form,

∞∑n=0

cnxn,

where the cn are real numbers and x is a variable. We call the cn the coefficients.

Note that we can try to evaluate a power series at specific values of x. But we have to becareful, because it might not converge for every x.

Definition 19.2.For which values of x do the following power series converge?

∞∑n=0

xn

∞∑n=0

nxn

∞∑n=0

xn

n!

∞∑n=0

(−1)nxn

2n

∞∑n=0

n!xn

We can consider a power series∑cnx

n as a function, with domain the set of values of x suchthat it converges.

Definition 19.3 (Power Series Centred at x = a).A series of the form,

∞∑n=0

cn(x− a)n,

is called a power series centred at x = a.

90 MATTHEW BATES

Note that our first definition of a power series is a special case of this one, it is a power seriescentred around x = 0.

Example 19.4. For which values of x does the following converge?

∞∑n=1

(x− 2)n

n

∞∑n=1

(3x+ 5)n

∞∑n=1

(x− 7)n

n!

Notice that the set of values of x such that a power series converges is an interval, this turnsout to always be true.

Theorem 19.5 (Radius of Convergence).Given a power series,

∑cn(x− a)n, one of the following must be true,

i) The series only converges at x = a.ii) The series converges for all x.iii) There exists some R > 0, such that the series converges for all |x− a| < R, and divergesfor all |x− a| > R.

R is called the radius of convergence. In case i) we say R = 0, in case ii) we say R =∞.The set of all x for which it converges is called the interval of convergence.

Note that the third case says nothing about the convergence of the power series when x = ±a.You need to check these separately.

The interval of convergence must be one of the following,{a}∞[a−R, a+R], [a−R, a+R), (a−R, a+R], or (a−R, a+R).


Example 19.6. Find the radius and interval of convergence for the following,∞∑n=0

(−3)n(x− 2)n

∞∑n=1

(−1)n(x− 5)n

n

∞∑n=0

n!(x+ 3)n.

92 MATTHEW BATES

20. Section 11.9 : Representing a Function by Power

Series

There are many benefits of begin able to write a function as a power series.Easy to integrate and differentiate.Easy to approximate.

Example 20.1. How do you, or your computer, calculate sin(1)? or e4?

Example 20.2. The function f(x) = 11−x can be represented by the power seres,

∞∑n=0

xn,

for |x| < 1.

Example 20.3. Represent the function, f(x) = 11+x2

as a power series.

Example 20.4. Represent the function, f(x) = 2x7+3x2

as a power series.

Theorem 20.5 (Differentiating and Integrating Power Series).If∑cn(x− a)n is a power series, with radius of convergence, R, then

f(x) =∑

cn(x− a)n,

is differentiable and integrable on |x− a| < R, and

f ′(x) =∑

ncn(x− a)n−1,

and ∫f(x)dx =

∑ cnn+ 1

(x− a)n+1.

And both have radius of convergence R.

One way to think of this is that power series can be differentiated and integrated term-wise.A nice way to write this is,

d

dx

(∑cn(x− a)n+1

)=∑(

d

dxcn(x− a)n+1

),


and ∫ (∑cn(x− a)n+1

)dx =

∑(∫cn(x− a)n+1dx

).

Example 20.6. Represent the function, f(x) = 1(1−x)2

as a power series.

1

(1− x)2=

d

dx

(1

1− x

)= ...

Example 20.7. Represent the function, f(x) = ln(1− x) as a power series.

ln(1− x) =

∫1

1− xdx = . . .

Example 20.8. Represent the function, f(x) = tan−1(x) as a power series.

tan−1(x) =

∫1

1 + x2dx =

∫(1− x2 + x4 − x6 + ...)dx = x− x3

3+x5

5− x7

7. . .

One can actually show that this converges for all x ∈ [−1, 1]. Thus,

π

4= tan−1(1) = 1− 1

3+

1

5− 1

7. . .

Example 20.9. Represent the function, f(x) = ln(1 + 2x) as a power series.

Example 20.10. Represent the function, f(x) = 1(1−x)3

as a power series.

94 MATTHEW BATES

21. Section 11.10 : Taylor and Maclaurin Series

We describe a method for calculating a power series expansions of a general function. Supposef(x) is some function with a power series expansion, i.e.

f(x) = c0 + c1(x− a) + c2(x− a)2 + c3(x− a)3 + c4(x− a)4 + . . .

then

f(a) = c0.

Moreover,

f ′(x) = c1 + 2c2(x− a) + 3c3(x− a)2 + 4c4(x− a)3 + . . .

so,

f ′(a) = c1.

Similarly,

f ′′(x) = 2c2 + 3× 2c3(x− a) + 4× 3c4(x− a)2 + 5× 4c5(x− a)3 + . . .

so

f ′′(a) = 2c2

In general we have that

cn =f (n)(a)

n!

Theorem 21.1 (Taylor Series).If f(x) has a power series expansion about x = a, then

cn =f (n)(a)

n!.

i.e.

f(x) =∞∑n=0

f (n)(a)

n!(x− a)n.

The series∞∑n=0

f (n)(a)

n!(x− a)n.

is called the Taylor series (or Taylor expansion) of f at a.

The special case that a = 0, i.e.∞∑n=0

f (n)(0)

n!xn.

is called the Maclaurin series.


It is important to note that this theorem does not say that f has a power series expansion, itonly says that if it did then it must be this one. Note every function has is equal to a powerseries.

Example 21.2. Find the Maclaurin series for the function,

f(x) = ex,

and calculate it’s radius of convergence.Note that in this case, f (n)(x) = ex for all n. Thus, f (n)(0) = e0 = 1 for all n. So,

f(x) =∞∑n=0

f (n)(0)

n!xn =

∞∑n=0

1

n!xn.

We also want to calculate it’s radius of convergence. To do this we use the ration test.∣∣∣∣ xn+1

(n+ 1)!

/xn

n!

∣∣∣∣ =

∣∣∣∣ x

n+ 1

∣∣∣∣→ 0

thus the radius of convergence is R =∞. �

This last example shows that if ex has a power series expansion, then

ex =∞∑n=0

xn

n!.

We have not actually shown that the equality holds. We will come back to this later.

Example 21.3 (sin(x)). Find the Maclaurin series for the following function,

f(x) = sin(x).

We calculate the derivatives,

f(0) = sin(0) = 0

f (1)(0) = cos(0) = 1

f (2)(0) = − sin(0) = 0

f (3)(0) = − cos(0) = −1

f (4)(0) = sin(0) = 0

f (5)(0) = cos(0) = 1

f (6)(0) = − sin(0) = 0

f (7)(0) = − cos(0) = −1

f (8)(0) = sin(0) = 0

Thus, the Maclaurin series is,∞∑n=0

(−1)nx2n+1

(2n+ 1)!.

96 MATTHEW BATES

Example 21.4 (log(1 + x)).ln(1 + x).

We calculate the derivatives,

f(x) = log(1 + x)

f (1)(x) =1

1 + x

f (2)(x) =−1

(1 + x)2

f (3)(x) =−1×−2

(1 + x)3

f (4)(x) =−1×−2×−3

(1 + x)4

f (5)(x) =−1×−2×−3×−4

(1 + x)5

Thus,

f (n)(x) =(−1)(n+1)(n− 1)!

(1 + x)n

Thus,

f (n)(0) = (−1)(n+1)(n− 1)!

so the Maclaurin series is,∞∑n=1

(−1)n−1xn

n.

�

Example 21.5.3x2 + 7x+ 1.

Example 21.6. Find the Taylor series for f(x) = cos(x), centred at a = π.

Example 21.7. Use the Maclaurin series for ex to calculate the Maclaurin series for

f(x) = xe2x + e−2x.

We now ask when does a function equal it’s Taylor series? i.e. when is

f(x) =

∞∑n=0

f (n)(a)

n!(x− a)n ?


First recall that an infinite series is equal to the limit of it’s partial sums. In the case of Taylorseries, we give the partial sums a special name.

Definition 21.8 (Taylor Polynomials).The N th-degree Taylor polynomial, TN (x) is defined to be,

TN (x) =

N∑n=0

f (n)(a)

n!(x− a)n.

Thus, saying that a function equals it’s own Taylor series is the same as saying,

f(x) = limTN (x).

Definition 21.9 (Taylor Remainder).The N th Taylor remainder, RN (x), is defined to be,

RN (x) = f(x)− TN (x).

So if we can show that RN (x)→ 0, then we have shown that f(x) is equal to it’s Taylor series.

Theorem 21.10 (Taylor Inequality).

If |f (n)(x)| < M for all |x− a| < d, then

|Rn(x)| ≤ M

(n+ 1)!|x− a|n+1,

for al |x− a| < d.

Proof. Suppose n = 1, so f ′′(x) ≤M . Thus if a ≤ x ≤ a+ d then∫ x

af ′′(t)dt ≤

∫ x

aMdt = (x− a)M.

But note that f ′(x) is an antiderivative of f ′′(x), thus∫ x

af ′′(t)dt = f ′(x)− f ′(a) ≤M(x− a).

Thus, ∫ x

af ′(t)dt ≤

∫ x

a(f ′(a) +M(x− a))dx = f ′(a)(x− a) +

M(x− a)2

2.

So,

f(x)− f(a) ≤ f ′(a)(x− a) +M(x− a)2

2.

98 MATTHEW BATES

Similar arguments work for n > 1. �

Example 21.11. Let f(x) = ex. Note that if |x− 0| < d, then |f (n)(x)| = |ex| < ed. Thus,

|Rn(x)| < ed

(n+ 1)!|x|n+1 → 0.

Thus, ex is equal to it’s Taylor series. �


22. Section 10.1 : Curves Defined by Parametric

Equations

Note that not every curve is the graph of some function y = f(x). But we can always describea curve if we describe the x and y coordinates separately. i.e.

x = f(t) y = g(t).

You should think of this as describing the position of a particle in the plane at time t.

Definition 22.1 (Parametric Equations and Parametric Curves).The equations, x = f(t), y = g(t), are call parametric equations. The curve they define inthe plane is called a parametric curve.

Example 22.2. x = t, y = t2, for t ∈ [0,∞).

Example 22.3. x = cos(t), y = sin(t), for t ∈ [0, 2π].

Example 22.4. x = t2, y = t3, for t ∈ R.

Example 22.5. x = cos(2t), y = sin(2t), for t ∈ [0, 2π].

The above example shows that different parametric equations can describe the same para-metric curve. (Think of the particle moving at a different speed)

Example 22.6. Find a parametric equation for a circle of radius r, centred at (p, q).

Note that we can always write a standard Cartesian equations, y = f(x), as a parametricequation as follows, x = t, y = f(t). Sometimes we can go the other direction as well, or atleast write it in terms of x and y only.

Example 22.7. x = t, y = t2, for t ∈ [0,∞).This is equivalent to, y = x2.

100 MATTHEW BATES

Example 22.8. x = cos(t), y = sin(t), for t ∈ [0, 2π].This is equivalent to y2 + x2 = 1

Example 22.9. x = t2, y = t3, for t ∈ R.This is equivalent to y2 = x3.

Example 22.10. x = t2 − 1, y = t2 − 1, for t ∈ R.This is equivalent to y = (x− 1)2 + (x− 1) = x2 − x.

Example 22.11. x = t−1, y = t(t2 + t), for t ∈ R.This is equivalent to y = x3 − x2.

Example 22.12. x = 2t3 − 3, y = 2t, for t ∈ R.This is equivalent to 2(y2 )3 − 3.

Example 22.13. x = 2 cos(t), y = sin(t), for t ∈ R.

This is equivalent to y =√

1− cos2(t) =√

1− (x/2)2.

Example 22.14 (Cycloid). x = t− sin(t), y = 1− cos(t).


23. Section 10.2 : Calculus With Parametric Curves

We first examine the tangent line to a parametric curve.

Theorem 23.1 (Derivative of Parametric Equation).If y = f(t) and x = g(t), then by the chain rule,

dy

dt=dy

dx

dx

dt.

So,

dy

dx=

dydtdxdt

=f ′(t)

g′(t).

You should think of dydt , and dx

dt as the vertical and horizontal velocities of the particle.

Theorem 23.2.

d2y

dx2=

d

dx

(dy

dx

)=

ddt

(dydx

)dxdt

.

Example 23.3. Find the tangent to the curve, x = cos(t), y = sin(t), at time t = π/4.

dy

dx=

dydtdxdt

=cos(t)

− sin(t)

thus the slope at t = π/4 is −1.So the tangent line is given by,

y − y(π/4) = −1(x− x(π/4)),

i.e.

y − 1√2

= −x+1√2.

Example 23.4. Find the tangent to the curve, x = 2t3, y = t+ t2, at time t = 1.

dy

dx=

dydtdxdt

=1 + 2t

6t2

thus the slope at t = 1 is 1/2.So the tangent line is given by,

y − y(1) = 1/2(x− x(1)),

i.e.y − 2 = 1/2(x− 2).

102 MATTHEW BATES

Example 23.5. Find the tangent to the curve, x = t2, y = t3 + t, at the point (1, 2).First we must find the corresponding value of t which gives the point, (1, 2). This is clearlyt = 1.

dy

dx=

dydtdxdt

=3t2 + 1

2t

thus the slope at t = 1 is 2.So the tangent line is given by,

y − 2 = 2(x− 1),

Example 23.6. Find all the times at which the tangent to the curve, x = t3 − 3, y(t) = t2 − 2is vertical or horizontal.

Example 23.7. Sometimes parametric curves can have more than one tangent line at a givenpoint, for example,

x(t) = t2, y(t) = t3 − t = t(t2 − 1).

Find the equations of the tangent lines at (1, 0).The corresponding values of t are t = ±1. Which give slopes, ±1. Thus the tangent lines aregiven by,

y = x− 1, y = −x+ 1.

We now examine how to calculate the arc length of a parametric curve. Recall that if y = f(x),then the arc length is given by,

L =

∫ b

a

√1 +

(dy

dx

)dx.

We can also talk about the area under a parametric curve. Recall that if y = F (x), andF (x) ≥ 0, then the area under the graph is∫ b

aF (x)dx.

Definition 23.8 (Area).If x(t) = f(t), and y(t) = g(t), then the area under the graph from α = f(a) to β = f(b), is∫ β

αg(t)f ′(t)dt.


Example 23.9 (Area of Cycloid). Find the area under one arch of the cycloid.Recall that the equation for a cycloide is,

x(t) = t− sin(t), y(t) = 1− cos(t).

So the area is,

Area =

∫ 2π

0(1− cos(t))(1− cos(t))dt

=

∫ 2π

01− 2 cos(t) + cos2(t)dt

=

∫ 2π

01− 2 cos(t) +

1

2+

1

2cos(2t)dt

=

(3

2t− 2 sin(t) +

1

4sin(2t)

) ∣∣∣∣2π0

= 3π.

�

Definition 23.10 (Arc Length).If x = f(t), and y = g(t), then

L =

∫ b

a

√√√√1 +

(dydtdxdt

)dx

=

∫ b

a

√√√√1 +

(dydtdxdt

)dx

dtdt

=

∫ b

a

√(dx

dt

)+

(dy

dt

)dt

This can be seen by taking polygonal approximations to the curve then taking the limit.

Example 23.11. Find the length of a circle of radius r, centred at the origin.First we find a parametric equation for such a circle.

x(t) = r cos(t), y(t) = r sin(t).

104 MATTHEW BATES

Now we calculate the integral,

Arc Length =

∫ 2π

0

√(−r sin(t))2 + (r cos(t))2dt

=

∫ 2π

0r

√sin2(t) + cos2(t)dt

=

∫ 2π

01dt

= 2πr.

Example 23.12. Find the arc length of,

x(t) = et cos(t), y(t) = et sin(t), t ∈ [0, π].

The arc length is,

Length =

∫ π

0

√(et(cos(t)− sin(t)))2 + (et(cos(t) + sin(t)))2dt

=

∫ π

0et√

2 cos2(t) + 2 sin2(t)dt

=

∫ π

0

√2etdt

=√

2(eπ − 1).

Example 23.13. Find the arc length of

x = et + e−t, y = 1− 2t, t ∈ [0, 2].

=

∫ 2

0

√(et − e−t)2 + (−2)2dt

=

∫ 2

0

√e2t − 2− e−2t + 4dt

=

∫ 2

0

√e2t + 2− e−2tdt

=

∫ 2

0

√(et − e−t)2dt

=

∫ 2

0et − e−tdt

= (et − e−t)∣∣∣∣20

= e2 − e−2 − (1− 1)

= e2 − e−2.

�


Example 23.14. Find the exact length of the curve.

y = 5 + 2x3/2, x ∈ [0, 1].

Example 23.15 (Arch Length of Cycloid). Calculate the length of one arch of the cycloid.Recal that the equation for a cycloid is,

x(t) = t− sin(t), y(t) = 1− cos(t).

So the length is,

=

∫ 2π

0

√(1− cos(t))2 + (sin(t))2dt

=

∫ 2π

0

√1− 2 cos(t) + cos2(t) + sin2(t)dt

=

∫ 2π

0

√2− 2 cos(t)dt

Now recall that,

sin2(θ) =1

2(1− cos(2θ)).

Thus2(1− cos(t)) = 4 sin2(t/2).

So the integral becomes,

=

∫ 2π

0

√4 sin2(t/2)dt

=

∫ 2π

02 sin(t/2)dt

= −4 cos(t/2)

∣∣∣∣2π0

= −4(−1)−−4(1)

= 8.

�

106 MATTHEW BATES

24. Section 10.3 - Polar Coordinates

Polar coordinates are simply another way of describing the position of a point in the plane.Until now we have been using Cartesian coordinates, (x, y). The first term represents thehorizontal distance and the second term represents the vertical distance.

Definition 24.1 (Polar Coordinates).The polar coordinate (r, θ), represents the point in the plane which is r-units distance fromthe origin, at an angle of θ with the positive x-axis.

Theorem 24.2 (Converting between Polar and Cartesian).If (x, y) = (r, θ), then

r =√x2 + y2

θ = tan−1(y/x).

and

x = r cos(θ)

y = r sin(θ).

Example 24.3.Cartesian Polar

(1, 0) (1, 0)(0, 1) (1, π/2)

(−12 ,√

32 ) (1, 2π

3 )

(1,√

3) (2, π/3)

(−1,−1) (√

2, 5π/4)(0, 3) (3, 7π)

(sin(θ), cos(θ)) (1, θ)

We now examine curves written in terms of polar coordinates. i.e. equations of the form,

F (r, θ) = 0.

Example 24.4.r = 2.

Example 24.5.θ = π/2.


Example 24.6.

r = 2 cos(θ).

Convert this into a Cartesian equation.

r = 2 cos(θ) =2x

r.

So,

2x = r2 = x2 + y2,

Completing the square gives,

(x− 1)2 + y2 = 1.

So it is a circle of radius 1, centred at (1, 0).

Example 24.7. Write the line, x = 1, in polar coordinates.We solve,

1 = x = r cos(θ).

So,

r = sec(θ).

�

Example 24.8 (A flower).

r = cos(2θ).

This is a four leaved flower.

We can also talk about tangents to polar curves.

If r = f(θ), then

dy

dx=

dydθdxdθ

=ddθr sin(θ)ddθr cos(θ)

=f ′(θ) sin(θ) + f(θ) cos(θ)

f ′(θ) cos(θ)− f(θ) sin(θ).

108 MATTHEW BATES

Example 24.9 (Circle Tangent). Fine the tangent line to the polar curve,

r = sin(θ),

at θ = π/2.

dy

dx=

cos(θ) sin(θ) + sin(θ) cos(θ)

cos2(θ)− sin2(θ)

=sin(2θ)

cos(2θ)

= tan(2θ).

So, at θ = π/2, the slope is, dydx = tan(2× π/2) = tan(π) = 0.

Now we find the x and y coordinates when θ = π/2.

x = r cos(θ) = sin(π/2) cos(π/2) = 0,

y = r sin(θ) = sin(π/2) sin(π/2) = 1.

So the equation of the tangent line is,

y − 1 = 0(x− 0) = 0.

i.e.

y = 1.

�

Example 24.10 (Flower Tangent). Find the tangent to the curve,

r = cos(2θ),

at θ = π/6.First we calculate the slope,

dy

dx=

(−2 sin(2θ) sin(θ) + cos(2θ) cos(θ)

(−2 sin(2θ) cos(θ)− cos(2θ) sin(θ).

Plug in θ = π/6.

=−2√

32

12 + 1

2

√3

2

−2√

32

√3

2 −12

12

=

√3

7.

So the tangent line is given by,

y − r sin(π/6) =

√3

7(x− r cos(π/7)).

i.e.

y − cos(π/3) sin(π/6) =

√3

7(x− cos(π/3) cos(π/7)).

y − 1

4=

√3

7(x−

√3

4).

�


Example 24.11 (Tangent to a Spiral). Find the tangent line to the polar curve,

r = θ,

at θ = π/4.The slope is given by,

dy

dx=

dydθdxdθ

=sin(θ) + θ cos(θ)

cos(θ)− θ sin(θ)

=

1√2

+ π4

1√2

1√2− π

41√2

=4 + π

4− π.

So the tangent line is given by,

y − π

4

1√2

=4 + π

4− π(x− π

4

1√2

).

�

110 MATTHEW BATES

25. Section 10.4 : Area and Lengths of Polar Curves

Recall that the area of a sector of a circle of radius r, subtended by the angle, θ, is 12r

2θ.Given a general polar curve, r = f(θ), one can approximate the area bouned between θ = aand θ = b by dividing it up into lots of little sectors then adding the area of all the sectors.This sum has the form, ∑ 1

2f(θi)

2θi.

Taking the limit as the number of approximations approaches infinity, we get a the true area.

Definition 25.1 (Area of Polar Curves).The area of a polar region is,

A =

∫ b

a

1

2f(θ)2dθ

=

∫ b

a

1

2r2dθ.

Example 25.2 (Area of a Petal of a Flower). Calculate the area of one petal of the flower,

r = cos(2θ).

First we need to calculate the bounds on integration. −π/4 to π/4 will do. Thus the area isgiven by,

A =

∫ π/4

−π/4

1

2(cos(2θ))2dθ

=

∫ π/4

−π/4

1

4

1

4cos(4θ)dθ

=

(1

4θ +

1

16sin(4θ)

) ∣∣∣∣π/4−π/4

=

(1

4

π

4+ 0

)−(−1

4

π

4+ 0

)=π

8.

�

Example 25.3 (Area of a Spiral). Sketch and find the area of the polar curve,

r = θ,

for θ ∈ [0, 2π].


The area is given by,

A =1

2

∫ 2π

0θ2dθ

=1

2

(2π)2

3

=2π2

3.

�

Example 25.4. Sketch and calculate the area enclosed in the polar curve,

r = 2 cos(θ),

for θ ∈ [−π/4, π/4]. The area is given by,

A =

∫ π/4

−π/4

1

2(2 cos(θ))2dθ

=

∫ π/4

−π/42 cos2(θ)dθ

=

∫ π/4

−π/41 + cos(2θ)dθ

=

(θ +

1

2sin(2θ)

) ∣∣∣∣π/4−π/4

=

(pi

4+

1

21

)−(−π

4+

1

2(−1)

)=π

2+ 1.

�

Example 25.5 (Area Between Two Polar Curves). Sketch the area bounded by the polarcurves,

r = 2 cos(θ), and, r =√

2,

and find it’s area. The area bounded is given by,

A =

∫ π/4

−π/4

1

2(2 cos(2θ))2dθ −

∫ π/4

−π/4

1

2

√2

2dθ

=(π

2+ 1)−∫ π/4

−π/4dθ

=(π

2+ 1)−(π

4− −π

4

)= 1.

�

calculus ii lecture notespeople.math.umass.edu/~bates/calc_ii_notes.pdf4 matthew bates 2. section...

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