Math 1114: Introductory Calculus II Lecture Notes Mount ...€¦ · Math 1114: Introductory Calculus II Lecture Notes Mount Saint Vincent University Darien DeWolf [email protected]

Math 1114: Introductory Calculus II

Lecture Notes

Mount Saint Vincent University

Darien DeWolf

[email protected]

Winter, 2015

Contents

1 Riemann Sums 3

1.1 Area Under Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.2 Riemann Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2 Definite Integrals 9

2.1 Definite Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

2.2 Properties of Definite Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

3 The Fundamental Theorem of Calculus 12

3.1 Antiderivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

3.2 Fundamental Theorem of Calculus, Part 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

3.3 Fundamental Theorem, Part I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

3.4 Fundamental Theorem, Part II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

3.5 Application: Probability Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

4 The Substitution Rule 21

4.1 More Complicated Integrands . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

4.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

4.3 The Substitution Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23


5 Integration by Parts 26

5.1 Integration by Parts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30


5.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

6 Trigonometric Integrals 32

6.1 Powers of Sine and Cosine . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

6.2 Powers of Tangent and Cotangent . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

7 Trigonometric Substitutions 37

7.1 Inverse Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

7.2 Techniques for√a2 + x2,

√a2 − x2 and

√x2 − a2. . . . . . . . . . . . . . . . . . . . . . . . . 38

7.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

8 Partial Fractions 40

8.1 Finding Partial Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

8.2 Larger Degree Numerators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

9 Improper Integrals 45

1

10 Areas Between Curves 48

11 Volumes 52

12 Arc Length 55

2

Chapter 1

Riemann Sums

1.1 Area Under Curves

The content of this course is most directly motivated by the area problem.

Problem: How much area is contained in the region a function f(x) and the x-axis between x = a and

x = b? That is, what is the area of the following shaded region?

To develop some intuition for what the area of this region should be, we will first estimate it using rectangles

whose area function is quite well known.

Example 1.1.1. Estimate the area bounded by f(x) = x2 and the x-axis between x = 0 and x = 1. That

is, estimate the area of the following region:

3

We don’t know yet how to calculate this area exactly, but I will tell you that its area is1

3. Let us estimate

this area using 2 rectangles and then re-estimate it using 5 rectangles. For simplicity sake, we are going to

use rectangles of equal width.

Two Rectangles: The interval [0, 1] is 1 wide so that two equally wide rectangle placed in this inter-

val would have width 0.5. We then need to choose the heights of the rectangles. We are going to choose

the heights so that these heights are completely dependent of the function f(x) under which we find the

region of interest. That is, we will choose the heights of the rectangles to be f(x)-values for some x. In the

picture below, we choose the height of the first rectangle to be f(0.5) and the height of the second to be

f(1). Though choosing the right side of the rectangles is aesthetically pleasing, the choice of height can be

somewhat arbitrary. For example, we could have chosen f(0.25) and f(0.6) as our heights. You will see this

happening on your homework assignment. You will see soon that our choice of height will not matter, as

long as we choose them in the right interval. Anyway, here is a picture:

We can then estimate the area of the region by adding the areas of the rectangles used (their area is base

by width):

A ≈ AR1+AR2

= 0.5× f(0.5) + 0.5× f(1) = 0.5× 0.52 + 0.5× 12 = 0.625

Five Rectangles: Same deal except that the rectangles should now have width 0.2. We will again choose

the heights of the five rectangles to be the “endpoints”; that is, f(0.2), f(0.4), f(0.6), f(0.8) and f(1). Here

4

is a picture:

We can then estimate the area of the region by adding the areas of the rectangles used:

A ≈ AR1+AR2

+AR3+AR4

+AR5

= 0.2× f(0.2) + 0.2× f(0.4) + 0.2× f(0.6) + 0.2× f(0.8) + 0.2× f(1)

= 0.2× 0.22 + 0.2× 0.42 + 0.2× 0.62 + 0.2× 0.82 + 0.2× 12

= 0.44

Do observe that the estimate using five rectangles is much closer to the true area of1

3than is the estimate

using two rectangles. N

It is clear that using more rectangles in our estimation would produce more accurate estimates of the

true area. In fact, it is not too outlandish a feat to stretch your imagination and imagine using an infinite

number of rectangle would give the true area. This will be more formally discussed next class.

1.2 Riemann Integrals

We have been talking only about functions that are always positive. Consider the following graph detailing

the velocity of a particle on a straight line at a given time t :

5

Negative velocity means that the particle is moving backwards. If one knows how long a particle has been

moving and, at each moment of time, how fast a particle moves, then we can calculate the distance travelled

from its starting place. We call this displacement and we want displacement to be negative if the particle

ends “backwards” of where it started. For example, if I stand in place, take 2 steps forward and 4 steps

backward, I would like to say I have displaced by −2 steps. Calculating displacement at a point in time is

given by multiplying velocity by time. That is, the following region would give me the displacement from

time t = 0 to t = 2 :

From t = 0 to t = 1, the particle has been moving exclusively forwards (its velocity is positive) and thus the

area of this region corresponds to positive displacement; this area is exactly the distance travelled in this

time-frame. From t = 1 to t = 2, however, the particle moves backwards (its velocity is negative now) and

thus its displacement is negative; the area of this region corresponds to the negative distance travelled.

Physically motivated, we are going to define a new pseudo-area in which any areas under the x-axis are

negative. The Riemann integral is defined to be a so-called net area. The Riemann Integral is exactly the

area for regions above the x-axis, the negative area for regions under the x-axis and the difference between

these if both are present. For example the Riemann integral of the curve above from t = 0 to t = 2 is exactly

zero; there is equal amount of positive and negative area and they cancel each other out. Physically, you

can think of a Riemann integral as the “final displacement” of the particle; at the end of the day, how far

has this particle travelled forwards or backwards?

Just like we estimated areas under regions, we can estimate Riemann integrals. In fact, we can still

use rectangles to do so. The only difference is that we are going to consider rctangle under the x-axis to

have “negative heights” (negative height by positive width would give a negative area) given by negative

f(x)-values. These estimates are called Riemann sums and can be calculated using the following (hopefully

familiar from the area example) procedure:

Calculating Riemann Sums

Given: A function f(x) “integrable” (this will be defined next class – don’t worry about it for now)

on an interval [a, b].

Result: A Riemann sum estimating the Riemann integral of f(x) from a to b.

1. Choose a number n of rectangles to use and calculate the width ∆x =b− an

of each rectangle so that

each rectangle is of equal width.

2. Write the interval as a partition (a collection of intervals that do not overlap and comprise the entirety

6

of [a, b]) of n intervals (one for each rectangle) using the following notation:

P : a < a+ ∆x < a+ 2∆x < . . . < a+ n∆x = b

Essentially, start at a and keep increasing by the width of the rectangles until you get to b, the other

end of the interval.

3. Inside each interval of the partition, choose a test point. We use the notation x∗i to mean a test point

chosen in the ith interval. The ith interval is [a+ (i− 1)∆x, a+ i∆x] (try putting in a couple of values

for i to verify this; for example the 2nd interval would be [a+ ∆x, a+ 2∆x] so that x∗2 would be found

here. The choice of test point is irrelevant but must be chosen and stuck with.

4. Each of these test points will give you the height f(x∗i ) of the ith rectangle in your Riemann sum.

Calculate all of the areas and add them up.

The formula for the Riemann sum, then, can be expressed by

R. Sum = AR1 +AR2 + . . .+ARn

= ∆xf(x∗1) + ∆xf(x∗2) + . . .+ ∆xf(x∗n)

=

n∑i=1

∆xf(x∗i )

Example 1.2.1. Consider the function f(x) = −x2 on the interval [0, 2]. Calculate the Riemann sum esti-

mating the Riemann integral of f(x) from 0 to 2 using partition P : 0 < 0.5 < 1 < 1.5 < 2 and intermediate

points x∗1 = 0.25, x∗1 = 0.75, x∗1 = 1.25, and x∗1 = 1.75. Then graph the function and the rectangles used in

this calculation.

Solution: Our first observation should be that our partition contains four equally large intervals, each

having width ∆x = 0.5. This tells us that we are using four rectangles with width 0.5. Notice,, too, that

each of the intermediate points are chosen within these four intervals. The heights of the rectangles will be

given by the graph and will have heights f(0.25), f(0.75), f(1.25) and f(1.75). Here is the picture of this:

7

And here is the calculation of the Riemann sum estimating the Riemann integral:

R. Sum =

n∑i=1

∆xf(x∗i )

=

4∑i=1

0.5f(x∗i )

= 0.5f(x∗1) + 0.5f(x∗2) + 0.5f(x∗3) + 0.5f(x∗4)

= 0.5f(0.25) + 0.5f(0.75) + 0.5f(1.25) + 0.5f(1.75)

= 0.5(−0.252) + 0.5(−0.752) + 0.5(−1.252) + 0.5(−1.752)

= −2.625

A fun fact: for the record (just like areas before, we don’t yet know how to calculate this), the Riemann

integral of f(x) = −x2 from 0 to 2 is 2.67. N

8

Chapter 2

Definite Integrals

2.1 Definite Integrals

Recall that the Riemann sum, an estimate for the Riemann integral (a so-called generalised “area” which

could be negative) under a function f(x) on an interval [a, b], is given by the following formula:

n∑i=1

∆xf(x∗i ).

This is an estimate of the Riemann integral given by adding the areas of n rectangles of width ∆x whose

height are determined by intermediate points x∗i chosen from the intervals comprising a partition

P : a < a+ ∆x < a+ 2∆x < . . . < a+ (n− 1)∆x < b

of [a, b]. Specifically, for each i = 1, 2, . . . , n, x∗i ∈ [a+ (i− 1)∆x, a+ i∆x].

Last class, we observed that as we increase the number of rectangles used, the Riemann sum more

accurately estimates the Riemann integral. In fact, if we can imagine using an infinite number of rectangles

(whose widths must then be infinitely small, or close to zero), then

(i) the choice of each intermediate point x∗i should not matter, as long as they are chosen in the correct

intervals (after all, since the rectangles are infinitely thin and the intervals infinitely small, we would

“eventually” have only one choice of test point) and

(ii) the Riemann sum would give exactly the Riemann integral; that is, an estimate with infinitely small

error is exactly correct.

Because of point (i), we can choose out test points to be those that are both easiest to calculate with and

are notational easy to handle. For this reason, we can choose to always use the right endpoints and define

x∗i = a+ i∆x.

Finally, (ii) motivates the following definition.

Definition 2.1.1. If f is a function that is integrable on the interval [a, b], then the definite Riemann integral

of f from a to b is defined by ∫ b

a

f(x) dx = limn→∞

n∑i=1

∆xf(x∗i ),

where ∆x =b− an

and x∗i = a+ i∆x.

9

Notation. We write∫ baf(x) dx to mean “integrate f with respect to x from x = a to x = b”. This symbol is

formed from an integral sign∫, a lower bound of integration a, and upper bound of integration b, a function

f and a differential operator dx which indicates what is the independent variable. Note that all parts of this

symbol must be present for it to make any sense (perhaps, except, later when we do indefinite integrals).

Every function I give you will usually be integrable, so don’t too much fear the trick question. I said in

class that “integrable” usually means continuous. Formally,

Theorem 2.1.2. Let f be a real-valued function. If either

(i) f is continuous on all of [a, b] or

(ii) f has only a finite number of jump discontinuities in [a, b],

then f is integrable on [a, b] and the definite integral∫ baf(x) dx exists.

Notice that we are talking only of jump discontinuity here. The non-integrable functions that we will be

interested in for this course will be non-integrable because of an asymptotic discontinuity.

Example 2.1.3.

(a) Any polynomial is integrable on any subset of R.

(b) The function f(x) =1

xis integrable on [1, 4].

(c) The function f(x) =1

xis not integrable on the interval [−1, 1], due to x = 0 being a vertical asymptote.

(d) The ceiling function f(x) = dxe contains jump discontinuities only at each integer and thus is integrable

over the interval [0, 3]. N

Remember that a Riemann integral can be though of as some generalised area where those areas under

the x-axis are negative. More specifically, we can think of the Riemann integral as a net area, which is

calculated by subtracting the negative areas from the positive ones.

Example 2.1.4. Area under f(x) = sinx on [0, π] is 2, [π, 2π] is −2 and [0, 2π] is 0. N

Since we can think of integrals of net areas, we can sometimes calculate Riemann integrals if the area

bounded by the function is a geometric shapes whose area formulas are well known.

Example 2.1.5.

(i)∫ 1

0

√1− x2 dx =

π

4

(ii)∫ 3

0(x− 1) dx = 1.5 N

2.2 Properties of Definite Integrals

In this section, assume that every integral written exists on the interval of interest.

When calculating Riemann sums, we implicitly assumed that a < b; that is, that we were calculating

areas from left to right. We can also calculate areas from right to left, or from b to a. All of the calculations

are the same, except with a new rectangle width

∆′x =a− bn

= −b− an

= −∆x.

10

That is, the rectangles each now have negative widths. Taking the limit of this new sum with all of the signs

reversed, we get the “boundary reversal” rule:

(2.1)

∫ b

a

f(x) dx = −∫ a

b

f(x) dx

If a = b, then we are integrating from a to a; we are finishing where we start. Then each of the rectangles in

the Riemann sum would have width ∆x =a− an

= 0 and thus (after all, how much area does a line have?)

(2.2)

∫ a

a

f(x) dx = 0.

The following theorem follows from the linearity of limits and sums.

Theorem 2.2.1 (Properties of Integrals).

(i)∫ bac dx = c(b− a), for any constant c ∈ R

(ii)∫ bacf(x) dx = c

∫ baf(x) dx, for any constant c ∈ R

(iii)∫ ba

[f(x) + g(x)] dx =∫ baf(x) dx+

∫ bag(x) dx

(iv)∫ ba

[f(x)− g(x)] dx =∫ baf(x) dx−

∫ bag(x) dx

(v)∫ caf(x) dx+

∫ bcf(x) dx =

∫ baf(x) dx

Proof.

(i) “Proof” by picture. Draw the rectangle bounded by f(x) = c, f(x) = 0, x = a and x = b and calculate

its area.

(ii) Write the definite integral as an infinite limit of partial sums and remove the constant using the

properties of sums and limits.

(iii) Write the definite integral as an infinite limit of partial sums, use the properties of sums and limits

over sums and then rewrite the resulting limits of partial sums as integrals.

(iv) This follows from (ii) and (iii).

(v) Proof by picture. Integrate from a to c and integrate from c to b. The integral from a to b, then will

be a glueing of the previous two areas.

Example 2.2.2.

(i) Given that∫ π0

sinx dx = 2 and∫ π0

cosx dx = 0, evaluate∫ π0

[5 + 3 sinx− cosx] dx.

(ii) Given that∫ 2

0[4x3 + 3x2] dx = 24 and

∫ 2

1[4x3 + 3x2] dx = 22 evaluate

∫ 1

0[4x3 + 3x2] dx. N

11

Chapter 3

The Fundamental Theorem of

Calculus

3.1 Antiderivatives

We have been exclusively thinking of definite integrals geometrically; we have been considering them as the

net area of the region bounded by a function between two values x = a and x = b. Another way to think

of integration is algebraically: we can think of integration as an operation that is in some way an “inverse

function” for differentiation. This approach is motivated by the following question: “If I differentiate a

function, how can I recover it?”. This section will answer this question for some specific functions. In the

next section, we are going to see that the geometric net area problem is intimately related with this algebraic

approach and this algebraic approach will gives us a method to calculate definite integrals.

Definition 3.1.1. A function F (x) is a called an antiderivative of f(x) on an interval [a, b] if F ′(x) = f(x)

for all x ∈ [a, b].

Proposition 3.1.2. Suppose that F1 and F2 are antiderivatives of a function f on an interval [a, b]. That

is, F ′1 = f = F ′2. Then F1 = F2 + C for some real constant C ∈ R; any two antiderivatives of a function

differ only by some constant.

Proof. Consider the difference function F1−F2. Since F ′1 = F ′2, (F1−F2)′ = F ′1(x)−F ′2(x) = 0 for all x ∈ [a, b].

That is, F1 − F2 is a constant function and thus F1 = F2 = C for some C ∈ R and F1 = F2 + C.

Corollary 3.1.3. If F is an antiderivative of f on an interval [a, b], then the most general antiderivative of

f on [a, b] is

F (x) + C,

where C ∈ R is any constant.

Note. Notice that sinced

dxC = 0 for all constants C ∈ R, any function which has an antiderivative has an

(uncountably) infinite number of them.

Notation. Any antiderivative of a function f is a called an an indefinite integral of f and is denoted∫f(x) dx = F (x) + C,

where F ′(x) = f(x) and C ∈ R is some constant.

12

Note. Pay special attention to the difference between a definite and indefinite integral. A definite integral∫ baf(x) dx is a number representing net area and the indefinite integral

∫f(x) dx is a function, whose

derivative is f.

Example 3.1.4. Find each of the following general indefinite integrals.

(a)

∫sinx dx

Sinced

dx(− cosx) = − d

dx(cosx) = −(− sinx) = sinx, we can write

∫sinx dx = − cosx+ C.

(b)

∫cosx dx = sinx+ C

(c)

∫1

xdx

Sinced

dx(ln |x|) =

1

xso that ∫

1

xdx = ln |x|+ C.

Notice that we need absolute value here because1

xcan be defined on intervals containing negative values

but lnx can not.

(d)

∫xn dx for n 6= −1 (the power rule for integration)

Using the power rule of differentiation, for n 6= 1,

d

dx

(xn+1

n+ 1

)=

(n+ 1)xn

n+ 1= xn

and thus ∫xn dx =

xn+1

n+ 1+ C.

(e)

∫ex dx = ex + C N

In all of the above examples, we should have been more careful about the intervals on which the indefinite

integrals are defined. For example,∫xn dx would be defined on any interval for n > −1 but would be defined

only on intervals not containing zero for n < −1 since the denominator would be zero. From now one, when

we write an indefinite integral or antiderivative, we are going to assume it is valid only on some interval and

that the interval will not be explicitly mentioned (unless it is special in some way).

You are responsible to have the previous indefinite integrals committed to memory.

Following is a collection of indefinite integrals that you should also get to know quite well. All of these

indefinite integrals can be verified by differentiating. These will be provided on a formula sheet for the

midterm and final exam.

13

∫[f(x)± g(x)] dx =

∫f(x) dx±

∫g(x) dx

∫Cf(x) dx = C

∫f(x) dx∫

sec2 x dx = tanx+ C

∫csc2 x dx = − cotx+ C∫

secx tanx dx = ln(secx) + C

∫cscx cotx dx = − cscx+ C∫

tanx dx = ln(secx) + C

∫cotx dx = ln(sinx) + C∫

secx dx = ln(secx+ tanx) + C

∫cscx dx = ln(cscx− cotx) + C∫

1

1 + x2dx = tan−1 x+ C

∫1√

1− x2dx = sin−1 dx+ C

Example 3.1.5. Find each of the following general indefinite integrals.

(a)

∫(10x4 − 2 sec2 x) dx =

10x5

5− 2 tanx+ C = 2x5 − 2 tanx+ C

(b)

∫ √x dx =

∫x1/2 dx =

x3/2

2/3=

3x3/2

2+ C

N

3.2 Fundamental Theorem of Calculus, Part 2

The Fundamental Theorem of Calculus, Part II, gives us a way to easily calculate definite integrals using

indefinite integrals:

Theorem 3.2.1 (FTC, Part II). If f is a function continuous on [a, b], then∫ b

a

f(x) dx = F (b)− F (a),

where F is any antiderivative of f ; that is, F ′ = f.

Note. The choice of antiderivative does not matter. Suppose that G is another antiderivative of f. Then G

and F differ by a constant C and thus

F (b)− F (a) = (G(b) + C)− (G(a) + C) = G(b)−G(a).

Notation. We will write F (x)]ba to denote the difference F (b) − F (a). This can also be written as F (x)|ba

or

∫f(x) dx

]ba

(this one is probably rare in the wild but makes total sense).

The book calls this the Evaluation Theorem. We will prove this next class. In the meantime, here are

some examples to prove to you that this is a useful theorem to us.

Example 3.2.2. Calculate the following definite integrals.

(a)

∫ 3

0

(x3 − 6x) dx = −27

4

(b)

∫ 2

0

(2x3 − 6x+

3

x2 + 1

)dx = 3 tan−1(2)− 4

(c)

∫ 1

−1eu du = e− 1

e

14

(d)

∫ 2

1

(x

2− 2

x

)dx =

3

4− ln(4) N

I will end this lecture with the protypical physics example: the particle on a line.

Example 3.2.3. Suppose that the velocity (m/s) of at a time t of a particle moving along a line has the

following formula:

v(t) = t2 − t− 6

Find the distance travelled and displacement of the particle between the times t = 1 and t = 4. Solution:

Remember that the displacement s of the particle in this time interval is given by the definite integral

s =

∫ 4

1

v(t) dt

=

∫ 4

1

(t2 − t− 6) dt

=

(t3

3− t2

2− 6t

)]41

=

(43

3− 42

2− 6(4)

)−(

13

3− 12

2− 6(1)

)= −4.5;

that is, the particle at t = 4 is 4.5 metres behind where it started at t = 1.

To calculate the total distance travelled, we need to calculate the actual area (where the displacement is

the net area) of the following region:

Notice that v(t) is positive from t = 3 to t = 4 and is negative from t = 1 to t = 3. When the area is negative,

15

we need to take the absolute value in order to calculate distance travelled. That is,

A =

∣∣∣∣∫ 3

1

v(t) dt

∣∣∣∣+

∫ 4

3

v(t) dt

=

∣∣∣∣∫ 3

1

(t2 − t− 6)) dt

∣∣∣∣+

∫ 4

3

(t2 − t− 6) dt

=

∣∣∣∣∣(t3

3− t2

2− 6t

)]31

∣∣∣∣∣+

(t3

3− t2

2− 6t

)]43

=

∣∣∣∣[(33

3− 32

2− 6(3)

)−(

13

3− 12

2− 6(1)

)]∣∣∣∣+

[(43

3− 42

2− 6(4)

)−(

33

3− 32

2− 6(3)

)]=

∣∣∣∣−22

3

∣∣∣∣+17

6

=61

6N

3.3 Fundamental Theorem, Part I

Considering differentiation as a unary operation of sorts on the space of functions, we can think of integration

as a function which is an inverse for differentiation. The Fundamental Theorem of Calculus, Part I, formalises

this relationship. To prove it, we will need the following fact about the upper and lower limits of a definite

integral:

Proposition 3.3.1. If m ≤ f(x) ≤M for a ≤ x ≤ b, then

m(b− a) ≤∫ b

a

f(x) dx ≤M(b− a).

Proof. Proof by picture (stolen from Stewart):

Theorem 3.3.2 (FTC, Part I). If f is a function continuous on the interval [a, b], then the function g

defined by g(x) =∫ xaf(t) dt for a ≤ x ≤ b is an antiderivative of f. That is, g′(x) = f(x) for a < x < b.

16

Proof. Suppose that x and x+ h are in the open interval (a, b). Then

g(x+ h)− g(x) =

∫ x+h

a

f(t) dt−∫ x

a

f(t) dt

=

(∫ x

a

f(t) dt+

∫ x+h

x

f(t) dt

)−∫ x

a

f(t) dt

=

∫ x+h

x

f(t) dt

and so, for h > 0, we can divide both sides of the equation

g(x+ h)− g(x) =

∫ x+h

x

f(t) dt

by h to get that

(3.1)g(x+ h)− g(x)

h=

1

h

∫ x+h

x

f(t) dt.

Since f is continuous on the closed interval [x, x+ h], there is a minimum and maximum f -value m and M,

respectively. That is, for all t ∈ [x, x+ h], m ≤ f(t) ≤M. By the extreme value theorem, there are t-values

that are assigned to these by f. That is, there exist u and v in [x, x+ h] such that f(u) = m and f(v) = M.

By Proposition 3.3.1, we know that the definite integral on the interval [x, x+ h] is bounded by

mh ≤∫ x+h

x

f(t) dt ≤Mh.

That is,

f(u)h ≤∫ x+h

x

f(t) dt ≤ f(v)h.

Dividing all terms in this equality by h > 0, we have

f(u) ≤ 1

h

∫ x+h

x

f(t) dt ≤ f(v)

and thus, by Equation 3.1, we have

f(u) ≤ g(x+ h)− g(x)

h≤ f(v).

We will now take the limit as h → 0 of all parts of this inequality. Notice that this limit in the middle is

exactly the limit definition of the derivative g′ of g. Because u, v ∈ [x, x + h], both u → x and v → x as

h→ 0. That is,

limh→0

f(u) = limu→x

f(x) = f(x)

and

limh→0

f(v) = limv→x

f(x) = f(x)

By the squeeze theorem, then, we get that

f(x) = limh→0

g(x+ h)− g(x)

h= g′(x)

17

and g(x) =∫ xaf(t) dt is an antiderivative for f(x).

Note. We can also write the statement of this theorem using Leibniz notation by

d

dx

∫ x

a

f(t) dt = f(x).

Aside from formalising the inverse relationship between differentiation and integration as functions, the

first part of the Fundamental Theorem can also be used to easily prove the second part of the Fundamental

Theorem.

3.4 Fundamental Theorem, Part II

Last class, we saw that the Fundamental Theorem, Part II, is very useful in that it allows us to easily

calculate indefinite integrals of functions whose antiderivatives are known. We will now prove that this

actually works.

Theorem 3.4.1 (FTC, Part II). If f is a function continuous on [a, b], then∫ b

a

f(x) dx = F (b)− F (a),

where F is any antiderivative of f ; that is, F ′ = f.

Proof. Suppose that F is an antiderivative of f. Consider the function g defined by g(x) =∫ xaf(t) dt for

x ∈ [a, b]. By the first part of the fundamental theorem, we can differentiate this function and g′(x) = f(x).

That is, g is also an antiderivative of f.

Since F and g are both antiderivatives of f, they differ by a constant. That is, we can write F (x) =

g(x) + C for some C ∈ R. Then

F (b)− F (a) = (g(b) + C)− (g(a) + C)

= g(b)− g(a)

=

∫ b

a

f(t) dt−∫ a

a

f(t) dt (definition of g)

=

∫ b

a

f(t) dt (same bounds rule)

=

∫ b

a

f(x) dx.

3.5 Application: Probability Distributions

A (real-valued) random variable is a function X : Ω→ R, where Ω is a set of possible outcomes. Less formally,

we usually just think of a random variable as a collection of outcomes labelled using real numbers. For

example, we can think of Height to be a random variable whose possible values are heights. Let us consider

continuous (or unlistable) random variables only. A probability density function for X is a continuous

function f : X → [0, 1] satisfying the following axioms:

1. For all x, 0 ≤ f(x) ≤ 1, (this comes from the definition of the function),

2.

∫ ∞−∞

f(x) dx = 1

18

For those familiar with probability, you can interpret Axiom 2 as “the probability of something happening

is 1”.

You are probably familiar with the phrase “is Normally distributed”. What this means is that a contin-

uous random variable has the Normal distribution as its probability density function. For example, suppose

that the Height is a (continuous) random variable which is Normally distributed. Then its standardized

(meaning we measure in standard deviations instead of height units) probability density function is given by

f(x) =1√2πe−x

2/2

and has graph

Suppose that we want to calculate the probability of one’s height falling being between −1 standard deviation

(one standard deviation less than average) and 2 standard deviations (above the average height). Recall that

this probability corresponds to the area under the Normal curve; that is, the following region whose area is∫ 1

−2f(x) dx.

In a statistics course, you will be taught that, to calculate this area, you take the left areas to the left of −1

and 2 (from a table giving such areas) and subtract them. This is clear from the following pictures:

19

Believe me that f is continuous on all of (−∞,∞). Define an “area to the left of x” function

g(x) =

∫ x

−∞f(x) dx.

In the above photos, I have actually given regions whose areas are g(−1) and g(2). By the first part of the

Fundamental theorem, g(x) is an antiderivative of f. Then by the second part of the Fundamental Theorem,

we can use these “area to the left of x” functions to calculate definite integrals. Unfortunately, these functions

do not have antiderivatives that can be written as elementary functions (they are written as series – we will

not be doing that in this class) and we will thus use a calculator or table to find them). In particular,∫ 1

−2f(x) dx = g(1)− g(−2)

=

∫ 1

−∞f(x) dx−

∫ −2−∞

f(x) dx

= 0.8413− 0.0228 (from a calculator)

= 0.8185

That is, there is a 0.8185 probability that a randomly chosen person has a height between one standard

deviation below average and two standard deviations above average.

20

Chapter 4

The Substitution Rule

4.1 More Complicated Integrands

Think back to when you first started learning how to differentiate functions. We would have started with

establishing some intuition using geometry: one learns to think of the derivative as the slope of the line

tangent to a point on the curve. Then, an algebraic approach is introduced: you learn that derivatives

can be thought of as a function giving a family of slopes (one for each point) and that they satisfied some

algebraic properties (they can be added, subtracted and multiplied by constants. After playing around

with how derivatives work a little bit, you would have then learned (and perhaps proved using limits) some

elementary derivatives, derivatives like (sinx)′ = cosx or (lnx)′ = 1/x. Almost immediately, though, one

runs into more complicated functions whose derivatives we would like to calculate. You then learned many

rules that allowed you to calculate more complicated derivatives using the known elementary ones. For

example:

• The product rule was formulated to calculate derivatives of products.

• The quotient rule was formulated to calculate derivatives of quotients.

• The chain rule was formulated to calculate the derivatives of composites.

There were also several techniques for differentiating functions of two variables, functions involving trigono-

metric functions and functions involving logarithmic or exponential functions.

Naturally, this course is going to take the same direction. We have already seen the integral geometrically

– as a net area under a curve – and we have shown the integral to be an algebraic operation that is, in a special

sense, inverse to differentiation. We have also seen some elementary integrals. We are now going to spend

some time developing techniques for using these elementary functions to evaluate more complicated integrals.

More specifically, we are going to develop techniques that are inverse to our techniques of differentiation:

how can we undo the product rule? How can we undo the quotient rule? How can we undo the chain rule?

We are going to start this course with the simplest case: we will undo the chain rule. The process, called

substitution, is a change of variables which, in special cases will give us easier integrands to work with. Let

us start with an example of the chain rule to see what happens.

Example 4.1.1. Consider the function f(x) = sin(x2). By the chain rule, f ′(x) = 2x cos(x2) and thus∫2x cos(x2) dx = sin(x2) + C.

Notice that, when using the chain rule, we are multiplying the derivative of the “inside”. Substitution is

going to become helpful in the situation that the derivative of some inside function appears as a factor of the

21

integrand. Pretend that we had just been asked to evaluate

∫2x cos(x2) dx but that nobody had told us

what its antiderivative is. Since we know an indefinite integral for cosine (just sine!), it would be really nice

if that cos(x2) was, say, a cos(u) and we could integrate with respect to u instead of x. Let us try assigning

a new variable by letting u = x2. Then∫2x cos(x2) dx =

∫2x cos(u), dx.

This looks like it could be good, but two problems: one, there are two variables in this expression and two,

we are integrating with respect to x instead of the desired u. We need to get rid of that dx by substituting it

somehow with a du. We can do this by noticing that u is a variable in terms of x, so that we can differentiate:

du

dx=

d

dxx2 = 2x =⇒ du = 2x dx =⇒ dx =

du

2x.

We can now plug this in for dx, simplify and finally evaluate:∫2x cos(u), dx =

∫2x cos(u)

du

2x=

∫cos(u) du = sin(u) + C

Since the original question was asking for the antiderivative in terms of x, we need to change the variable

back from u to x : ∫2x cos(x2) dx = sin(u) + C = sin(x2) + C N

Notice how when we take the derivative in u in this example, the derivative of u appeared as a factor of

the integrand and thus cancelled when we substituted dx for our du. These cases are the most recognisable

candidates for substitution.

4.2 Examples

Only with practice can one become proficient at recognising and evaluating integrals with substitution.

Usually, the crux of a solution lies in recognising what elementary antiderivative is present in the integrand.

In this section, I will present a variety of exercises, all of which are fair game for a homework, midterm or

final. These were done in class and I will only include details for the more tricky ones.

Example 4.2.1. Evaluate

∫(x+ 1)215 dx.

Hint: use u = x+ 1. N


∫ (1− 1

w

)cos(w − lnw) dw.

Hint: use u = w − lnw. N


∫2x

1 + (x2 + 2)2dx.

Hint: notice that

∫1

1 + x2dx = tan−1(x) + C and use u = x2 + 2. N


∫sin(t)[4 cos2(t) + 6 cos2(t)− 8] dt.

Hint: use u = cos(t). N


∫ [x cos(x2 + 1) +

1

x2 + 1

]dx.

Solution: This one is tricky because the most obvious substitution, u = x2 + 1, doesn’t work. The problem

22

is that there is nothing to cancel with in the fraction1

x2 + 1. The trick to evaluating this integral is to first

break it into two separate integrals: ∫x cos(x2 + 1) dx+

∫1

x2 + 1dx

At this point, the substitution u = x2 + 1 will work on the first integral and the second integral is simply

tan−1(x) + C. N


∫sin(x) cos(x) dx.

Hint: use u = sin(x). N


∫(x4 + 4x2 + 5)(x3 + 2x) dx

Hint: use u = x4 + 4x2 + 5 and notice that 4x3 + 8x = 4(x3 + 2x). N


∫x7√

2 + x4dx.

Solution: Let u = 2 + x4. Then du = 4x3 and dx =du

4x3. We substitute these into the integral:

∫x7√u

du

4x3=

∫x4√udu

The problem here is that we still have a remaining x4 in the numerator of the integrand. The trick here is

to realise that the formula for u can be rearranged as x4 = u− 2, so that we can make another substitution

and the integrand becomes possible to evaluate and resubstitute:∫x4√udu =

∫u− 2√u

du

=

∫u√udu−

∫2√udu

=

∫u

12 du− 2

∫u−

12 du

=2

3u

32 − 2(2u

12 ) + C

=2

3(2 + x4)

32 − 4(2 + x4)

12 + C N

4.3 The Substitution Rule

Last class, I did some examples of integration by substitution. This is the formal statement of the Substitution

Rule, along with a proof that justifies why our arithmetic works.

Theorem 4.3.1 (The Substitution Rule). If u = g(x) is a differentiable function with range [a, b] and f is

function continuous on [a, b], then ∫f(g(x))g′(x) dx =

∫f(u) dx.

Proof. Consider and antiderivative F of f ; that is, a function F such that F ′ = f. Consider, then, the

composite F (g(x)). Taking its derivative, we have, by the chain rule, (F (g(x)))′ = F ′(g(x))g′(x). Then

23

F (g(x)) is an antiderivative of F ′(g(x))g′(x); that is,∫f(g(x))g′(x) =

∫F ′(g(x))g′(x)

= F (g(x)) + C

= F (u) + C

=

∫F ′(u) + C

=

∫f(u) + C


Notice that when making a substitution u = g(x), one is effectively changing the variable of integration; one

changes the independent variable from x (or whatever it was) to u. So, when evaluating definite integrals,

we need to be a little more careful about how the bounds are handled. There are two ways, mainly, to deal

with this:

1. We can remove the bounds, solve it as an indefinite integral, change the u’s back into x’s and then use

the Fundamental Theorem of Calculus to evaluate it.

2. We can change the x-valued bounds to u-valued bounds by plugging the bounds into the formula for

the u substitution. This doesn’t require changing back into x-values.


∫ 1

0

e1−x dx.

Solution (Method 1): We can treat this as an indefinite integral by stripping the bounds and solving. We

would do a substitution using u = 1− x. Then dx = −du and∫e1−x dx = −

∫eu du = −eu + C = −e1−x + C.

Now we can use this antiderivative to evaluate using FTC, Part II:∫ 1

0

e1−x dx = −e1−x]10

= −e1−1 − (−e1−0) = −1 + e.

Solution (Method 2): We can solve the definite integral directly by changing the bounds of integration

into u-values. We will still have the same substitution, u = 1 − x, except now that we will be integrating

from u(0) = 1− 0 = 1 to u(1) = 1− 1 = 0. Then∫ 1

0

e1−x dx = −∫ u(1)

u(0)

eu du = −∫ 0

1

eu du = −eu]01 = −e0 − (−e1) = −1 + e. N

It isn’t hard to imagine my relief when we got the same answer using both methods. Because the nature

of substitution is to simplify an integrand into an elementary integral, it is normal that Method 2 (changing

the bounds) is computationally easier. Of course, you are free to choose either method when doing homework

or exams. Following are further examples that were worked out completely in class.

Example 4.4.2. Show that∫ π0

sec2(t/4) dt = 4.

Hint: Let u = t/4 and recall that∫

sec2(x) dx = tan(x) + C. The new bounds: u = 0 to u = π/4. N

24

Example 4.4.3. Show that

∫ 1

0

ez + 1

ez + zdz = ln(e+ 1).

Hint: Let u = ez + z and change the bounds: u = 1 to u = 1 + e. N

Example 4.4.4. Show

∫ 2

1

e1/x

x2dx = e−

√e.

Hint: Let u = 1/x and change bounds: u = 1 to u = 1/2. N

Example 4.4.5. Show

∫ 3

1

[x cos(x2 + 1) +

1

x2 + 1

]dx ≈ −0.26301.

Hint: First break the integral into a sum of two integrals. One of the integrals will be tan−1(x) and the

other can be solved using a substitution of u = x2 + 1. N

Example 4.4.6. Show

∫ 1/2

0

sin−1(x)√1− x2

dx =π2

72.

Hint: Let u = sin−1(x) and change bounds: u = sin−1(0) = 0 to u = sin−1(1/2) = π/6. N

25

Chapter 5

Integration by Parts

Theorem 5.0.1 (Integration by Parts Formula).∫f(x)g′(x) dx = f(x)g(x)−

∫g(x)f ′(x) dx.

Proof. Consider the product f(x)g(x) of f and g. By the product rule of derivatives,

[f(x)g(x)]′ = f(x)g′(x) + f ′(x)g(x).

That is, f(x)g(x) is an antiderivative of f(x)g′(x) + f ′(x)g(x) and thus∫[f(x)g′(x) + f ′(x)g(x)] dx = f(x)g(x).

Note that I am choosing here a specific antiderivative here, f(x)g(x) + C with C = 0. By the linearity of

integration, we can break the left-hand sum up as∫f(x)g′(x) dx+

∫f ′(x)g(x) dx = f(x)g(x)

and finally, we can rearrange for the desired result:∫f(x)g′(x) dx = f(x)g(x)−

∫f ′(x)g(x) dx

To make this formula easier to read, remember and use, we make a few substitutions. Let u = f(x) and

v = g(x). Then dv = g′(x) dx, du = f ′(x) dx and we have the following formula for integration by parts:

(5.1)

∫u dv = uv −

∫v du

To solve a problem using integration by parts, we fill out the following table:

Let u = and dv = .

Then du = and v =∫dv = .

So we choose a u and a dv. Then we differentiate u to calculate du and integrate dv to get v. The formula

can then be applied.

Knowing what to choose for u and dv so that this formula produces requires some practice. However,

it usually involves a balance of choosing u’s that are easy to differentiate and “simplify” in some way (e.g.,

26

eliminates natural logarithms or decreases the degree of a polynomial) and choosing dv’s so that the integral∫dv is possible (and not more difficult than the original problem.

We now do some examples of problems that you are likely to see in the wild.


∫xex dx.

Solution: This is a straightforward problem. Choose the parts:

Let u = x and dv = ex dx.

Then du = dx and v =∫ex dx = ex.

Then ∫xex dx = uv −

∫v du

= xex −∫ex dx

= xex − ex + C. N


∫lnx dx.

Solution: This is a problem where we want to eliminate a natural logarithm. Choose the parts:

Let u = lnx and dv = dx.

Then du =1

xand v =

∫dx = x.

Then ∫lnx dx = uv −

∫v du

= x lnx−∫x

1

xdx

= x lnx−∫

dx

= x lnx− x+ C. N


∫(x2 + 2x) cosx dx.

Solution: This problem requires two applications of integration by parts. Choose the parts:

Let u = x2 + 2x and dv = cosx dx.

Then du = (2x+ 2) dx and v =∫

cosx dx = sinx.

Then ∫(x2 + 2x) cosx dx = uv −

∫v du

= (x2 + 2x) sinx−∫

(2x+ 2) sinx dx.

27

We are now left with a new integrand which is the product of sinx and some simpler polynomial. Polynomials

easily become simpler so this is a good candidate for another application of integration by parts. We can’t

use the same variables twice, so let’s use new ones:

Let u1 = 2x+ 2 and dv1 = sinx dx.

Then du1 = 2 dx and v1 =∫

sinx dx = − cosx.

Then ∫(x2 + 2x) cosx dx = uv −

∫v du

= (x2 + 2x) sinx−∫

(2x+ 2) sinx dx.

= (x2 + 2x) sinx−[u1v1 −

∫v1 du1

].

= (x2 + 2x) sinx−[−(2x+ 2) cosx+ 2

∫cosx dx

].

= (x2 + 2x) sinx+ (2x+ 2) cosx− 2 sinx+ C. N


∫tan−1(4t) dt.

Solution: In this problem, we need to eliminate an inverse trigonometric function and the resulting in-

tegral needs to be evaluated using substitution. Choose the parts:

Let u = tan−1(4t) and dv = dt.

Then du =4

1 + (4t)2and v =

∫dt = t.

Then ∫tan−1(4t) dt = uv −

∫v du

= t tan−1(4t)−∫

4t

1 + 16t2dt.

We can solve the resulting integral now using a substitution. We can’t use the seemingly standard u for our

variable of substitution (it has already been used) and our variable of integration is t, so let’s use x. Let

x = 1 + 16t2. Then dx = 32t dt and∫tan−1(4t) dt = uv −

∫v du

= t tan−1(4t)− 1

8

∫1

xdx.

= t tan−1(4t)− 1

8lnx+ C

= t tan−1(4t)− 1

8ln(1 + 16t2) + C N


∫eθ cos θ dθ.

Solution: This problem involves integrating by parts, where neither parts become simpler when differ-

entiating. This problem contains a common trick for how to deal with these cases. Choose the parts:

28

Let u = cos θ and dv = eθdθ.

Then du = − sin θ dθ and v =∫eθ dθ = eθ.

Then ∫eθ cos θ dθ = uv −

∫v du

= eθ cos θ +

∫eθ sin θ dθ.

Choosing some new variables for a second application of integration by parts:

Let u1 = sin θ and dv1 = eθdθ.

Then du1 = cos θ dθ and v1 =∫eθ dθ = eθ.

Then ∫eθ cos θ dθ = eθ cos θ +

∫eθ sin θ dθ

= eθ cos θ +

[u1v1 −

∫v1 du1

]= eθ cos θ +

[eθ sin θ −

∫eθ cos θ dθ

]= eθ cos θ + eθ sin θ −

∫eθ cos θ dθ

Notice that∫eθ cos θ dθ appears as a term on both sides of the equation. That is, we can collect terms and

2

∫eθ cos θ dθ = eθ cos θ + eθ sin θ.

Divide both sides by 2 and we have successfully evaluated the desired integral:∫eθ cos θ dθ =

1

2

(eθ cos θ + eθ sin θ

). N


∫cos√x dx.

Solution: This problem requires a preliminary substitution before parts can be used. We are going to

avoid using u again, so let t =√x. Then dt =

1

2√xdx and dx = 2

√xdt = 2t dt. The integral we are

evaluating now, then, is ∫cos√x dx = 2

∫t cos t dt.

Choose the parts:

Let u = t and dv = cos t dt.

Then du = dt and v = sin t.

29

Then ∫cos√x dx = 2

∫t cos t dt

= uv −∫v du

= t sin t−∫

sin t dt

= t sin t+ cos t+ C N

5.1 Integration by Parts

Recall the following formula for integration by parts:

(5.2)

∫u dv = uv −

∫v du

To solve a problem using integration by parts, we fill out the following table:

Let u = and dv = .

Then du = and v =∫dv = .

So we choose a u and a dv. Then we differentiate u to calculate du and integrate dv to get v. The formula

can then be applied.


Definite integrals for those integrals requiring integration by parts are relatively straight forward:∫ b

a

u dv = uv]ba −

∫ b

a

v du

Of course, one can always simply compute the indefinite integral (i.e., remove the bounds and generate an

antiderivative) and then use the FTC with that antiderivative.

5.3 Examples

Example 5.3.1. Compute

∫ π

0

e2θ sin 3θ dθ ≈ 123.81. N

Example 5.3.2. Compute

∫ 12

0

cos−1(x) dx ≈ 0.6576. N


∫t2 sin(βt) dt. [This is messy, remember that β is a constant here.] N

Example 5.3.4. Prove the reduction formula∫(lnx)n dx = x(lnx)n − n

∫(lnx)n−1 dx

N

30

Example 5.3.5. Using the previous exercise’s reduction formula, compute

∫ 3

1

(lnx)3 dx ≈ 0.89039. N

31

Chapter 6

Trigonometric Integrals

In this class, we learn how to deal with certain classes of trigonometric integrals; that is, we will study some

integrals whose integrands are combinations of some trigonometric functions. Trigonometric integrals could

be things like ∫sin2 x dx,

∫cosx sinx dx, or

∫secx tanx dx.

Obviously, some of these will have easily calculable (or even elementary antiderivatives). We are going to

study in this class some combinations of sine/cosine and tangent/secant integrands that can be massaged

into integrals that are easily solved by substitution.

6.1 Powers of Sine and Cosine

We look at integrals of the form ∫sinn x cosm x dx.

We consider the three following possibilities for m and n :

(i) n is odd,

(ii) m is odd, or

(iii) both m and n are even.

Here are examples and strategies for each case:

(i) If the power of sine is odd, we can split off a sine, use the identity sin2 x = 1−cos2 x and then make the

substitution u = cosx dx to compute the integral. Note that du = − sinx dx, so that this will cancel

the sine you had originally split off.

32

Example 6.1.1.∫sin7 x cos4 x dx =

∫sin6 x sinx cos4 x dx

= −∫

(1− cos2 x)6 sinx cos4 x dx

= −∫

(1− u2)6u4 dx

= −∫

(1− u2)6u4 dx

= −∫

(u16 − 6u14 + 15u12 − 20u10 + 15u8 − 6u6 + u4) dx

= − 1

17u17 +

6

15u15 − 12

16u16 +

20

11u11 − 15

9u9 +

6

7u7 − 1

5u5 + C N

(ii) If the power of cosine is odd, we do something similar. We split off one of the cosines, use the identity

cos2 x = 1 − sin2 x and then do a substitution with u = sinx. Notice that du = cosx dx so that this

choice, too, results in the split term cancelling.

(iii) If both the power of sine and cosine are even, then you need to be a little more creative. In fact, no

good strategy – aside from practising many problems – exists for this case. You may, however, lean

heavily on these four identities.

Half-angle identities:

sin2 x =1

2(1− cos 2x) cos2 x =

1

2(1 + cos 2x)

Double-angle identities:

sin 2x = 2 sinx cosx cos 2x = cos2 x− sin2 x

Example 6.1.2. ∫sin2 x cos2 x dx =

∫(sinx cosx)2 dx

=

∫ (1

2sin 2x

)2

dx

=1

4

∫sin2 2x dx

=1

4

∫ (1

2(1− cos 2(2x))

)dx

=1

8

∫(1− cos 4x) dx

=1

8

(x− 1

4sin 4x

)=x

8− 1

32sin 4x+ C N

33

Example 6.1.3 (Method 2).∫sin2 x cos2 x dx =

∫ (1

2(1− cos 2x)

)(1

2(1 + cos 2x)

)dx

=1

4

∫(1− cos 2x)(1 + cos 2x) dx

=1

4

∫(1− cos2 2x) dx

=1

4

∫(sin2 2x) dx

=1

8

∫(1− cos 4x) dx

=x

8− 1

32sin 4x+ C N

6.2 Powers of Tangent and Cotangent

We look at integrals of the form where BOTH secant and tangents exist∫tanm x secn x dx.

We consider the three following possibilities for m and n :

(i) m is odd,

(ii) n is even, or

(iii) anything else.

Here are examples and strategies for each case:

(i) When the power of tangent is odd, split off a secx tanx, use the identity tan2 x = sec2x− 1 and then

make the substitution u = secx. Notice, then, that du = secx tanx dx and this this will cancel with

the secx tanx we have split off.

Example 6.2.1. ∫tan5 x sec3 x dx =

∫tan4 x sec2 x secx tanx dx

=

∫(sec2 x− 1)2 sec2 x secx tanx dx

=

∫(u2 − 1)2u2 du

=

∫(u6 − 2u4 + u2) du

=1

7u7 − 2

5u5 +

1

3u3 + C

=1

7sec7 x− 2

5sec5 x+

1

3sec3 x+ C N

(ii) When the power of secant is even, split off a sec2 x, use the identity sec2 x = 1 + tan2 x and then make

the substitution u = tanx. Notice that du = sec2 x dx and this again will cancel with the split term.

34

Example 6.2.2. ∫tan4 x sec6 x dx =

∫tan4 x sec4 x sec2 x dx

=

∫tan4 x(1 + tan2 x)2 sec2 x dx

=

∫u4(1 + u2)2 du

=

∫(u4 + 2u6 + u8) du

=1

5u5 +

2

7u7 +

1

9u9 + C

=1

5tan5 x+

2

7tan7 x+

1

9tan9 x+ C N

(iii) If we are not in Case (i) or (ii) or one of either secant or tangent don’t even appear, there is no clear-cut

way to play and could use some combination of identities, elementary integrals or integration by parts.

You can also use these integrals:∫tanx dx = ln(secx) + C

∫secx dx = ln(secx+ tanx) + C

Example 6.2.3.∫tan5 x dx =

∫tan4 x tanx dx

=

∫(sec2 x− 1)2 tanx dx

=

∫(sec4 x− 2 sec2 x+ 1) tanx dx

=

∫(sec4 x tanx− 2 sec2 x tanx+ tanx) dx

=

∫sec4 x tanx dx− 2

∫sec2 x tanx dx+

∫tanx dx

=

∫sec3 x secx tanx dx− 2

∫secx secx tanx dx+

∫tanx dx

The right-hand integral is given by the formulas above. Now, the two left-hand integrals are a Case

(ii) and can be solved by making the (coincidentally, the same) substitution u = secx. Then du =

secx tanx dx and∫tan5 x dx =

∫sec3 x secx tanx dx− 2

∫secx secx tanx dx+

∫tanx dx

=

∫u3 du− 2

∫u du+

∫tanx dx

=1

4u4 − u2 + ln(secx+ tanx) + C

=1

4sec4 x− sec2 x+ ln(secx+ tanx) + C N

35

Example 6.2.4.

∫sec3 x dx can be solved using integration by parts. See your textbook for this; this

exercise is worked out in detail as an example in the textbook. N

36

Chapter 7

Trigonometric Substitutions

7.1 Inverse Substitution

When we previously integrated functions using substitution, to solve an integral like∫

sin 2x dx, we would

let u = 2x; that is, we would define a new variable u which is written in terms of x. This allows us to

replace “blocks” of the integrand with this new variable. This simplifies the integrand and, if done properly,

generates an integrand whose antiderivative is easy enough to find.

In this lecture, we are going to work backwards. Rather than condensing integrands to make the integral∫f(x) dx easier to compute, we are going to replace x with some expression written in terms of another

variable. This will have the effect of “blowing up” the integrand and making it seemingly more difficult. Of

course, we are only going to do this is the blown-up integrands simplify nicely.

Example 7.1.1. Solve

∫ √4 + x2 dx by substituting x = 2 tan θ (we will talk more about which substitu-

tion to make and when after this example).

Solution. Since x = 2 tan θ, we have dθ = 2 sec2 θ dθ and thus∫ √4 + x2 dx =

∫ √4 + (2 tan θ)2 2 sec2 θ dθ

=

∫ √4 + 4 tan2 θ 2 sec2 θ dθ

=

∫ √4(1 + tan2 θ) 2 sec2 θ dθ

= 4

∫ √1 + tan2 θ sec2 θ dθ

= 4

∫ √sec2 θ sec2 θ dθ

= 4

∫sec3 θ dθ

= 2(sec θ tan θ + ln(sec θ + tan θ)) + C.

That last equality is an example in your textbook (the integral of∫

sec3 θ dθ). We now must change the

new variable θ back into the original variable x. We do this using the ratio-definition of the trigonometric

functions. Since x = 2 tan θ, tan θ = x2 and since tangent is the ratio “opposite over adjacent”, we can use

the following triangle to find the values of any other trigonometric function (notice that I have added the

hypotenuse as the distance formula gives):

37

Reading from this diagram, then,

sec θ =Hypotenuse

Adjacent=

√4 + x2

2and tan θ =

x

2

and therefore ∫ √4 + x2 dx = 2

(√4 + x2

2

x

2+ ln

(√4 + x2

2+x

2

))+ C. N

7.2 Techniques for√a2 + x2,

√a2 − x2 and

√x2 − a2.

In this course, the following three situations are going to crop up. Here are presented the substitution to

make and how to manipulate the new integrands into something manageable. In each of the cases, be sure

to emulate the above example to get a feel for how it works mechanically.

(i) When√a2 + x2 appears (for example, we just did this with a = 2), then try the substitution x = a tan θ

and use the identity 1 + tan2 θ = sec2 θ :√a2 + x2 =

√a2 + (a tan θ)2 =

√a2 + a2 tan θ = a

√1 + tan2 θ = a

√sec2 θ = a sec θ.

Then, when resubstituting the x’s into the θ’s, use the following triangle:

(ii) When√a2 − x2 appears, then try the substitution x = a sin θ and use the identity 1− sin2 θ = cos2 θ :√a2 − x2 =

√a2 − (a sin θ)2 =

√a2 − a2 sin2 θ = a

√1− sin2 θ = a

√cos2 θ = a cos θ.


38


∫x2√

1− x2 dx. N

(iii) When√x2 − a2 appears, then try the substitution x = a sec θ and use the identity sec2 θ− 1 = tan2 θ :√x2 − a2 =

√(a sec θ)2 − a2 =

√a2 sec2 θ − a2 = a

√sec2 θ − 1 = a

√tan2 θ = a tan θ.



∫1

t3√t2 − 1

dt.

Hint: Recall that

∫1

sec2 θdθ =

∫cos2 θ dθ. N

7.3 Examples

Here are two examples demonstrating fractional exponents and non-unit coefficients for the x2 term.


∫dx

(a2 + x2)3/2, a > 0.

Hint: Write (a2 + x2)3/2 =(√a2 + x2

)3. N


∫x2√

9− 25x2dx.

Hint: Let x =3

5sin θ. N

39

Chapter 8

Partial Fractions

In this section, we are going to be evaluating integrals whose integrands are rational functions.

Definition 8.0.1. A function f(x) is said to be a rational function if f is of the form f(x) =p(x)

q(x), where

p and q are polynomials and q 6= 0.


∫3x+ 11

x2 − x− 6dx.

Solution: We first note that the integrand can be rewritten as

−1

x+ 2+

4

x− 3=−x+ 3 + 4x+ 8

(x+ 2)(x− 3)=

3x+ 11

x2 − 6x− 6.

That is, the integrand can be broken up into “smaller” or “easier” rational functions whose antiderivatives

are known: ∫3x+ 11

x2 − x− 6=

∫ [−1

x+ 2+

4

x− 3

]dx = − ln |x+ 2|+ 4 ln |x− 3| N

So, this section deals with the following

Problem: Evaluating integrands are rational functions.

This can be solved with

Solution: Write the integrand as a sum of smaller rational functions whose antiderivatives are known

or are easily calculated.

To do this, we need a

Technique: Partial Fractions.

8.1 Finding Partial Fractions

Suppose that f(x) =p(x)

q(x)is a rational function. To find the partial fractions of f, we need a fewdefinitions

facts from Abstract Algebra (so all proofs will be omitted).

Definition 8.1.1. A polynomial f(x) is said to be irreducible whenever deg f > 0 (i.e., f is not a constant

function) and f can not be written as a product f(x) = p(x)q(x) where deg p > 0 and deg q > 0.

40

Theorem 8.1.2. Any polynomial f over a domain with unique factorisation (for example, integer or real

coefficients) can be written uniquely as a product of irreducible polynomials.

Theorem 8.1.3. If f is a polynomial over R, then f is either linear or quadratic. More specifically, either

f(x) = ax+b for some real a and b or f(x) = ax2+bx+c for some real a, b and c such that b2−4ac < 0.

To write a function in terms of its partial fractions, it is meant that we write f(x) as a sum of rational

functions of the formA

(ax+ b)ior

Ax+B

(ax2 + bx+ c)i.

Notice that, in the motivating example for partial fractions, the denominators of partial fractions were ex-

actly the factors of the denominators of the original integrands. This is the case, too, in general since when

adding fractions, the least common multiple of these factors is exactly the original denominator! The first

step in finding partial fractions, then, should understandably be to factor the denominator q(x). Since we

know that every polynomial can be uniquely written as a product of linear and quadratic polynomials, there

are exactly five cases to consider when calculating partial fractions.

Special Note: This technique for finding partial fractions works ONLY when the degree of the numerator

is strictly less than that of the denominator. That is, if the degree of the numerator is equal to or greater

than the degree of the denominator, then we can not immediately use this technique. We will cover that

situation later in the notes.

Finding Partial Fractions

In each of the following four cases, capital Roman letters will be non-quantified real constants and will need

to be solved for (usually using elimination or some calculator).

(i) If q(x) = (a1x+ b1)(a2x+ b2) · · · (akx+ bk) is product of k distinct linear factors, then write

f(x) =A1

a1x+ b1+

A2

a2x+ b2+ . . .+

Akakx+ bk

and solve for the constants A1, A2, . . . , Ak.


∫x+ 1

x2 − 4dx.

Solution: Write

x+ 1

x2 − 4=

x+ 1

(x+ 2)(x− 2)=

A

x+ 2+

B

x− 2=A(x+ 2) +B(x− 2)

(x+ 2)(x− 2)=A(x+ 2) +B(x− 2)

x2 − 4.

Since the original fraction and the new fraction with the A and B are equal and have the same

denominator, they must have the same numerator and thus

x+ 1 = A(x+ 2) +B(x− 2) = Ax+ 2A+Bx− 2B = (A+B)x+ (2A− 2B).

Since two polynomials are equal if and only if all of their coefficients are, we have the following system

of equations: A + B = 1

2A − 2B = 1

This system has a unique solution A = 3/4 and B = 1/4 and thus∫x+ 1

x2 − 4dx =

∫ [A

x+ 2+

B

x− 2

]dx =

∫ [3

4

1

x+ 2+

1

4

1

x− 2

]dx =

3

4ln |x+ 2|+ 1

4ln |x−2|+C N

41

(ii) If q(x) = (a1x + b1)n1(a2x + b2)n2 · · · (akx + bk)nk is product of k linear factors, each of which are

repeated n1, n2, . . . , nk times (some of these ni could be equal to one which means some of these

factors could be unique), then write

f(x) =A1,1

a1x+ b1+

A1,2

(a1x+ b1)2+ . . .+

A1,n1

(a1x+ b1)n1

+A2,1

a2x+ b2+

A2,2

(a2x+ b2)2+ . . .+

A2,n2

(a2x+ b2)n2

+ . . .+Ak,1

akx+ bk+

Ak,2(akx+ bk)2

+ . . .+Ak,nk

(akx+ bk)nk

and solve for the constants A1, A2, . . . , Ak. This looks complicated but is in practice quite straightfor-

ward; we are simply writing fractions whose denominators are powers of the linear factors of q and we

stop taking powers when we have as many as q does.


∫1

(x2 − 9)2dx.

Solution: Write

1

(x2 − 9)2=

1

(x+ 3)2(x− 3)2=

A

x+ 3+

B

(x+ 3)2+

C

x− 3+

D

(x− 3)2

=A(x+ 3)(x− 3)2 +B(x− 3)2 + C(x− 3)(x+ 3)2 +D(x+ 3)2

(x+ 3)2(x− 3)2.

Then we can equate the denominators and

1 = A(x+ 3)(x− 3)2 +B(x− 3)2 + C(x− 3)(x+ 3)2 +D(x+ 3)2

= x3(A+ C) + x2(−3A+B + 3C +D) + x(−9A− 6B − 9C + 6D) + (27A+ 9B − 27C + 9D)

and we have the following system of equations:A C = 0

−3A B 3C D = 0

−9A −6B −9C 6D = 0

27A 9B −27C 9D = 1

This system has the unique solution A = 1108 , B = 1

360 , C = − 1108 and D = 1

360 and thus∫1

(x2 − 9)2dx =

∫ [1

108

1

x+ 3+

1

360

1

(x+ 3)2− 1

108

1

x− 3+

1

360

1

(x− 3)2

]dx

=1

180ln |x+ 3| − 1

360(x+ 3)− 1

108ln |x− 3| − 1

360(x− 3)+ C N

(iii) If q(x) is product of distinct quadratic factors, then this is almost the same as when q is a product of

distinct linear factors except for the numerators are now of the form Ax+B instead of simply A.

Example 8.1.6.1

(x2 + 1)(x2 + 2)=Ax+B

x2 + 1+Cx+D

x2 + 2. N

(iv) If q(x) is product of quadratic factors, some of which may be repeated, then we play the same game as

in the repeated linear factors case, where we include all powers of the factors until we have the same

power as that which occurs in q.

42

Example 8.1.7.1

(x2 + 4)4=Ax+B

x2 + 4+

Cx+D

(x2 + 4)2+

Ex+ F

(x2 + 4)3+

Gx+H

(x2 + 4)4.

N

We can, of course, combine these:

Example 8.1.8.

1

(x3 − 2x2 + x)(x2 + 2)(x4 + 2x2 + 1)=

1

x(x− 1)2(x2 + 2)(x2 + 1)2

=A

x+

B

x− 1+

C

(x− 1)2+Dx+ E

x2 + 2+Fx+G

x2 + 1+

Hx+ I

(x2 + 1)2N

For quadratic terms, the integrals you need to evaluate may be somewhat difficult, but is usually done by

completing the square of the denominator, making a substitution and then using logarithms and the formula∫1

a2 + x2dx =

1

atan−1

(xa

)+ C.


∫x− 1

2x2 + 8x+ 11dx.

Solution: We first complete the square in the denominator to write

2x2 + 8x+ 11 = 2(x2 + 4x) + 11 = 2(x2 + 4x+ 4− 4) + 11 = 2((x2 + 4x+ 4)−+11 = 2(x+ 2)2 + 3.

Now we can make the substitution u = x+ 2. Then du = dx, x− 1 = u− 3 and thus∫x− 1

2x2 + 8x+ 11dx =

∫u− 3

2u2 + 3du

=

∫u

2u2 + 3du− 3

∫1

2u2 + 3du

=

∫u

2u2 + 3du− 3

2

∫1

u2 + 32

du

=1

2ln |2u2 + 3| − 3

2

√2

3tan−1

(√2

3u

)+ C

=1

2ln |2(x+ 2)2 + 3| − 3

2

√2

3tan−1

(√2

3(x+ 2)

)+ C N

8.2 Larger Degree Numerators

In the case that f(x) =p(x)

q(x)is a rational function such that the degree of the numerator is at least the

degree of the denominator (i.e., deg p ≥ deg q), then we can’t immediately use partial fractions; we must

first use long division to write f as the sum of two smaller polynomials.

Example 8.2.1. If we wanted to the partial fractions of f(x) =x6

(x2 + 1)(x2 + 2), we could not immediately

write this as the sum of two fractions whose denominators are the factors x2 + 1 and x2 + 2 since 6, the

degree of the numerator, is greater than 4, the degree of the denominator. To solve this, we first do long

43

division:

x2 − 3

x4 + 3x2 + 2)

x6

− x6 − 3x4 − 2x2

− 3x4 − 2x2

3x4 + 9x2 + 6

7x2 + 6

Then we can write

f(x) =x6

x4 + 3x2 + 2= x2 − 3 +

7x2 + 6

x4 + 3x2 + 2

and we now have a fraction with which we can use the partial fraction technique. N

44

Chapter 9

Improper Integrals

In this section, we are going to generalise the definition of the definite integral in two ways. These generali-

sations are called improper integrals. Recall:

Definition 9.0.1. If f is a function that is integrable on the interval [a, b], then the definite Riemann integral

of f from a to b is defined by ∫ b

a

f(x) dx = limn→∞

n∑i=1

∆xf(x∗i ),

where ∆x =b− an

and x∗i = a+ i∆x.

There are two key characteristics of the definite integral and these two characteristics will be generalised

in the following ways:

Type 1: Definite integrals are defined on closed and bounded intervals [a, b]. This will be generalised to infinite

intervals of the form

(i) [a,∞),

(ii) (−∞, b], or

(iii) (−∞,∞).

Type 2: Definite integrals are defined only for an integrable function f on an interval [a, b] (i.e., real functions

that are continuous or contain a finite number of jump discontinuities). This will be generalised to

include general discontinuities (vertical asymptotes):

(i) f can be continuous on [a, b) but discontinuous at b,

(ii) f can be continuous on (a, b] but discontinuous at a, or

(iii) f can have a discontinuity at some c between a and b.

We will now formally define these two types of improper integrals.

Definition 9.0.2 (Improper Integral: Type 1). Let f be a real-valued function implicitly defined so that

the assumptions in these cases can be met.

(i) If∫ taf(x) dx exists for each t ≥ a, define∫ ∞

a

f(x) dx = limt→∞

∫ t

a

f(x) dx.

Should this limit exist, the integral∫∞af(x) dx is said to be convergent and converges to its limit.

Otherwise, this integral is said to be divergent.

45

(ii) If∫ btf(x) dx exists for each t ≤ b, define∫ b

−∞f(x) dx = lim

t→−∞

∫ b

t

f(x) dx.

(iii) If both∫∞af(x) dx and

∫ a−∞ f(x) dx are convergent, then define∫ ∞

−∞f(x) dx =

∫ a

−∞f(x) dx+

∫ ∞a

f(x) dx.


∫ 0

−∞ze2z dz.

Solution: First notice that this is an improper integral because we are integrating over the infinite in-

terval (−∞, 0]. We now check that the appropriate limit, and thus this integral, converges. If it converges,

we will get the value of the integral. Otherwise, we will state that it diverges. We first find an antiderivative

for the integrand by integrating by parts. Let u = z and dv = e2z dz. Then du = dz and v =1

2e2z and thus

∫ze2z dz =

∫u dv = uv −

∫v du =

z

2e2z − 1

2

∫e2z dz =

z

2e2z − 1

4e2z =

e2z

2

(z − 1

2

)+ C.

We can now evaluate the integral:∫ 0

−∞ze2z dz = lim

t→−∞

e2z

2

(z − 1

2

)]0t

= limt→−∞

(e2(0)

2

(0− 1

2

)− e2t

2

(t− 1

2

))= −1

4. N


∫ ∞3

1

(x− 2)3/2dx. N


∫ ∞−∞

xe−x2

dx. N

Definition 9.0.6 (Improper Integral: Type 2). Let f be a real-valued function implicitly defined so that

the assumptions in these cases can be met.

(i) If f is continuous on [a, b) but discontinuous at b, define∫ b

a

f(x) dx = limt→b−

∫ t

a

f(x) dx.

(ii) if f is continuous on (a, b] but discontinuous at a, define∫ b

a

f(x) dx = limt→a+

∫ b

t

f(x) dx.

(iii) If f has a discontinuity at some c between a and b, and both∫ caf(X) dx and

∫ bcf(x) dx are convergent,

define ∫ b

a

f(x) dx =

∫ c

a

f(x) dx+

∫ b

c

f(x) dx.

46


∫ 1

−1

ex

ex − 1dx.

Solution: First notice that this is an improper integral because it has a vertical asymptote where ex−1 = 0,

or at x = 0, and that −1 < 0 < 1 is in the interval of integration. The integrand has no further discontinuities

so that ∫ 1

−1

ex

ex − 1dx =

∫ 0

−1

ex

ex − 1dx+

∫ 1

0

ex

ex − 1dx

should, of course, the integrals

∫ 0

−1

ex

ex − 1dx and

∫ 1

0

ex

ex − 1dx be convergent. We will check that these two

integrals are convergent separately, keeping in mind that if either of them diverge, then we are done (and the

larger integral also diverges). We first find an antiderivative forex

ex − 1. We let u = ex − 1. Then du = ex dx

and thus ∫ex

ex − 1dx =

∫1

udu = ln |u| = ln |ex − 1|+ C.

We can now calculate the right-hand integral:∫ 1

0

ex

ex − 1dx = lim

t→0+

∫ 1

t

ex

ex − 1dx = lim

t→0+ln |ex − 1|]1t = lim

t→0+

(ln |e− 1| − ln |et − 1|

)= −∞

Since the right-hand summand diverges, the whole integral does, too. N


∫ 1

0

3

x5dx. N


∫ 1

0

1√1− x2

dx. N

47

Chapter 10

Areas Between Curves

Suppose that f(x) and g(x) are continuous functions such that, for all x ∈ [a, b], we have f(x) ≥ g(x); i.e.,

we consider the case when f is “equal to or larger” than g on the interval [a, b]. We wish to define the “area

captured between f and g between a and b” so that this agrees with our intuitive notion of area. Just like we

defined the area under a curve by partitioning [a, b] into n subintervals of length ∆x = b−an , we will do that

here to define the area between f and g. The only difference now is that height of the subdividing rectangle

will be f(x∗i )− g(x∗i ) for chosen intermediate points x∗i in our partitions. The area captured between f and

g can be approximated, then, by the partial sum of these rectangles

S ≈n∑i=1

∆x[f(x∗i )− g(x∗i )].

The true area captured between these curves, then, could be defined to be the limit of these partial sums

(just as the definite integral is defined):

S = limn→∞

n∑i=1

∆x[f(x∗i )− g(x∗i )]

This is, of course, the definition of the definite integral of the function f − g, which motivates the following

definition.

Definition 10.0.1. Let f(x) and g(x) be continuous functions such that, for all x ∈ [a, b], we have f(x) ≥g(x). We define the area captured between f and g on the interval [a, b] as∫ b

a

[f(x)− g(x)] dx.

Due to the geometric nature of the problem “find the area captured between curves”, it is crucial that

one draws a picture before attempting any calculations.

48

Example 10.0.2. Sketch and find the area captured between f(x) = 12− x2 and g(x) = x2 − 6.

Solution: We want to find the area of the following region S :

Notice that the interval over which S exists is defined by the points at which f and g intersect. These

intersection points occur at x = ±3 and we will therefore be integrating over the interval [−3, 3]. Notice

further that, on this interval, f is always larger than g (or is equal to, at the points of intersection).

Therefore, we can use the definition of area between two curves to calculate the area of S :

A(S) =

∫ 3

−3[f(x)− g(x)] dx

=

∫ 3

−3[(12− x2)− (x2 − 6)] dx

=

∫ 3

−3[18− 2x2] dx

=

(18x− 2

3x3)]3−3

=

(18(3)− 2

3(3)3

)−(

18(−3)− 2

3(−3)3

)= 72 N

49

Example 10.0.3. Sketch and find the area captured between f(x) = cosx and g(x) = sin 2x for x ∈ [0, π2 ].

Solution: We want to find the area of the two regions R1 and R2 :

Why did we have to break the region of interest up into two separate regions? Notice that, in R1, the

function on top is f(x). This means that, when calculating the area of R1, we need to integrate f(x)− g(x).

However, in R2, the function on top is g(x) and we will need to integrate g(x)− f(x). Because the functions

trade places as the top function at the intersection point (x = π6 ), we need to calculate the regions separately.

The area captured between f and g on [0, π2 ], then, is

A(R1) +A(R2) =

∫ π6

0

[f(x)− g(x)] dx+

∫ π2

π6

[g(x)− f(x)] dx

=

∫ π6

0

[cosx− sin 2x] dx+

∫ π2

π6

[sin 2x− cosx] dx N

50

Example 10.0.4. Sketch and find the area captured between f(x) =1

x, g(x) = x and h(x) =

x

4for x > 0.

Solution: We want to find the area of the two regions R1 and R2 :

Notice here that we also have to break this region up into two smaller regions. At their intersection point,

the functions g and f trade roles as the “function on top of h ”. This means that relevant integrand for R1

over [0, 1] will be g(x) − h(x) and for R2 over [1, 2] will be f(x) − h(x). The total area captured between

these three curves, then, is

A(R1) +A(R2) =

∫ 1

0

[g(x)− h(x)] dx+

∫ 2

1

[f(x)− h(x)] dx

=

∫ 1

0

[x− x

4

]dx+

∫ 2

1

[1

x− x

4

]dx N

51

Chapter 11

Volumes

Read the textbook chapter on volumes (the argument relies too much on pictures for me to reproduce it here

very well). The point, though, is that if we can put a solid on the x-axis so that its bottom sits at x = a

and its top is at x = b, then the volume can be calculated using

V =

∫ b

a

A(x) dx,

where A(x) is the area of the cross-section at x.

Example 11.0.1. The textbook example is to find the volume of a sphere of radius r. Check out the textbook

for an image of this. Basically, we centre the sphere at the origin (0, 0) so that the cross-sections are circles

with radius√r2 − x2. The area of each circle – which defines the area function – is A(x) = π

(√r2 − x2

)2=

π(r2 − x2). The volume of a sphere is given, then, by

V =

∫ r

−rA(x) dx =

∫ r

−rπ(r2 − x2) dx =

4

3πr3. N

The volumes that we will be focusing on in this class will be volumes of revolution. These are solids

generated by rotating a function f(x) around the x-axis between x = a and x = b. Draw a picture to

convince yourself that the cross-sections of this solids are circles (since we are rotating) and have radius

f(x) at each x (since this is how “high” the function is at x). The area function for each x will then be

A(x) = π[f(x)]2 and thus the volume of the solid of revolution will be

V =

∫ b

a

A(x) dx = π

∫ b

a

[f(x)]2 dx.

Example 11.0.2. Find volume of the solid obtained by rotating f(x) = lnx about the x-axis between x = 1

and x = e.

Solution: Here is a visual of the solid we are working with:

52

We first find an antiderivative for [f(x)]2 = [lnx]2. We do this using integration by parts. Let u = [lnx]2

and dv = dx. Then du =2 lnx

xand v = x. Then

∫[lnx]2 dx = x[lnx]2 −

∫2 lnx dx

= x[lnx]2 − 2x lnx+ 2x+ C

The volume of the solid of revolution, then, is

V = π

∫ e

1

[lnx]2 dx

= π(x[lnx]2 − 2x lnx+ 2x

)]e1

= π(e[ln e]2 − 2e ln e+ 2e

)− π

([ln 1]2 − 2 ln 1 + 2

)= π(e− 2) N

Example 11.0.3. Let R be the region captured between the functions f(x) = x and g(x) = x3, for x ≥ 0.

Find the volume of revolution obtained by rotating R around the x-axis.

Solution: Here is a visual of the solid we are working with:

53

Notice that f(x) and g(x) intersect at x = 0 and x = 1 and that x3 ≤ x for 0 ≤ x ≤ 1. Therefore, we can

think of this solid as the cone defined by rotating f(x) = x being hollowed out by the cone-like structure

defined by rotating g(x) = x3. To calculate this volume, we can calculate the volume of the larger cone and

subtract the volume of the smaller cone:

V = Vouter − Vinner = π

∫ 1

0

x2 dx− π∫ 1

0

x6 dx =π

3x3]10− π

7x7]10

=10π

21. N

54

Chapter 12

Arc Length

We now consider the problem of finding the “length of a function” on some interval. What is the length of

a function? A naive approach could be to define the length of a function to be its net change. This is useful

for questions such as “how far away from my starting point do I end up?” but doesn’t help much to answer

something like “how much pencil did it take me to draw this curve?”. By arc length, we will want to think

of laying a string on the function, trimming to size and measuring the length of that string.

Think back to when we defined the definite integral as a true area under the curve. How did we define

this? We defined the definite integral to be an infinitely precise estimate of the area and the estimates of

the area were calculated using certain rectangles. That is, we estimated a two-dimensional area with smaller

two-dimensional objects. When we calculated volumes, we played a similar game; we estimated volumes of

three-dimensional solids by subdividing the solid into smaller three-dimensional solids.

It makes sense, then, to estimate the arc length – a one-dimensional structure – by subdividing the curve

into smaller, easily calculable one-dimensional objects: straight lines. This is the intuition of the following

construction.

Consider a curve C defined by the equation y = f(x), where f is continuous on the interval [a, b]. Let’s

approximate the length of the arc by dividing the interval into n subintervals of width ∆x = b−an , and

approximating the curve with a straight line on each interval. By the pythagorean theorem, the length of

the straight line over the ith subinterval is given by√[xi − xi−1]2 + [f(xi)− f(xi−1)]2 =

√(∆x)2 + (∆yi)2.

The length s of the curve over the interval [a, b] is then approximated by the sum of these lengths:

s ≈n∑i=1

√(∆x)2 + (∆yi)2 =

n∑i=1

∆x

√1 +

(∆yi∆x

)2

By the mean value theorem, there exists, in each of the subintervals, an x∗i such that ∆xf ′(x∗i ) = ∆yi. Then

we have

s ≈n∑i=1

∆x

√1 + (f ′(x∗i ))

2

Taking the limit as n→∞, we have

s =

∫ b

a

√1 +

(dy

dx

)2

dx.

55

Definition 12.0.1. If f ′ is continuous on [a, b], then the length of the curve y = f(x), a ≤ x ≤ b, is

L =

∫ b

a

√1 + [f ′(x)]2 dx.

Example 12.0.2. Find the length of the curve f(x) = 1 + 6x3/2 from x = 0 to x = 1.

Solution: We are interested in following the length of the line segment L :

We first compute f ′(x) :

f ′(x) = 9√x

Then we can write 1 + [f ′(x)]2 = 1 + (9√x)2 = 1 + 81x. The length of f on [0, 1], then, is

L =

∫ 1

0

√(1 + [f ′(x)]2) dx =

∫ 1

0

√(1 + 81x) dx.

This integral can be solved using substitution. Let u = 1 + 81x. Then du = 81 dx, u(0) = 1, u(1) = 82 and

L =

∫ 1

0

√(1 + 81x) dx =

1

81

∫ 82

1

√u du =

2

243u

3/2

]821

=2

243

(82√

82− 1)

N

Example 12.0.3. Find the length of the curve f(x) = ln(cosx) from x = 0 to x = π/3. N

Example 12.0.4. Find the length of the curve f(x) = ln(secx) from x = 0 to x = π/4. N

Example 12.0.5. Find the length of the curve f(x) =√x− x2 + sin−1(

√x).

Hint: Notice that sin−1(x) is defined on the interval [−1, 1] (because −1 ≤ sin(x) ≤ 1) and square roots can

only be taken on non-negative interval. That is, the interval on which this function is defined is [0, 1]. N

56

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Math 1114: Introductory Calculus II Lecture Notes Mount ...€¦ · Math 1114: Introductory Calculus II Lecture Notes Mount Saint Vincent University Darien DeWolf [email protected]