evergreen.loyola.edu · Contents 5 Integrals reviewed 2 5.1 Basic facts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .2 5.5 U-substitution

Contents

5 Integrals reviewed 25.1 Basic facts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25.5 U -substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

6 Integral Applications 106.1 Area between two curves . . . . . . . . . . . . . . . . . . . . . . . 106.2 Volumes by rotation . . . . . . . . . . . . . . . . . . . . . . . . . . 156.3 Volumes by Cylindrical Shells . . . . . . . . . . . . . . . . . . . . . 296.5 Average Value of a Function . . . . . . . . . . . . . . . . . . . . . 35

7 Techniques of Integration 377.1 Integration by Parts . . . . . . . . . . . . . . . . . . . . . . . . . . 377.2 Trigonometric Integrals . . . . . . . . . . . . . . . . . . . . . . . . 427.3 Trigonometric Substitution . . . . . . . . . . . . . . . . . . . . . . 487.4 Integration of Rational Functions . . . . . . . . . . . . . . . . . . . 557.5 Strategy for Integration . . . . . . . . . . . . . . . . . . . . . . . . 657.8 Improper Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . 68

8 Further Integral Applications 768.1 Arc Length . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 768.5 Probability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80

10 Parametric and Polar 8610.3 Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . 8610.4 Areas in Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . 9510.1 Curves Defined by Parametric Equations . . . . . . . . . . . . . . 9810.2 Calculus with Parametric Curves . . . . . . . . . . . . . . . . . . . 103

11 Sequences and Series 11111.1 Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11111.2 Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11611.4 The Comparison Tests . . . . . . . . . . . . . . . . . . . . . . . . . 12311.6 Absolute Convergence and the Ratio Test . . . . . . . . . . . . . . 13211.8 Power Series. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13711.10 Taylor and Maclaurin Series . . . . . . . . . . . . . . . . . . . . . . 142

A Absolute values 156

1

CONTENTS 2

B Trig functions 157

C Horizontal Asymptotes 160

D Extended Reals 161

E L’ Hospital’s Rule 162

F Cartesian and Polar Graph paper 165

Chapter 5

Integrals reviewed

5.1 Basic facts

Comments. The following page summarizes all the basic information you need toknow about definite integrals. You don’t need to understand all the details wellenough to use them in a problem, but it should all look familiar, and you shouldbe ready to do new material that uses some of the same ideas. Also, note that wehave a tripartate understanding of the definite integral: it is defined as the limit ofa sum of values; it is interpreted as an area, or net area; it is calculated using theFundamental Theorem of Calculus and anti-derivatives.

3

W. Ethan Duckworth, Loyola University Maryland, 2013

Definitions of Integrals

Definition

The combination of symbols “∫ b

a

f(x) dx” is called the defi-

nite integral of f(x), from a to b and is defined as the follow-ing limit∫ b

a

f(x) dx = limn→∞

(f(x∗1)∆x + f(x∗2)∆x + · · · + f(x∗n)∆x

)

with details given below.

Details

• We divide the interval [a, b] into n equal pieces

· · · · · ·a b

We define ∆x and x0, x1, . . . , xn in the picture below

x0 x1 x2 xn−1 xn

= ∆x

· · · · · ·a b

The following formulas are equivalent to the picture:

∆x =b− a

n, x0 = a, x1 = x0 + ∆x,

x2 = x1 + ∆x, . . . , xn = xn−1 + ∆x = b

• Each of x∗1, x∗2,. . . , x∗n is an x-value that can be pickedin different ways:

x∗1 = any x-value in the 1st sub-interval shown above

x∗2 = any x-value in the 2nd sub-interval shown above

. . .

x∗n = any x-value in the nth sub-interval shown above

• There are some standard ways to pick each x∗i :

Left Hand Rule:x∗i = left endpoint of the ith interval = xi−1

Right Hand Rule:x∗i = right endpoint of the ith interval = xi

Midpoint Rule:x∗i = midpoint of the ith interval = 1

2 (xi−1+xi)

Names and notation

• We call the sum f(x∗1)∆x + · · · + f(x∗n)∆x a Riemannsum

• We can abbreviate a Riemann sum using∑

notation

f(x∗1)∆x + · · · + f(x∗n)∆x =n∑

i=1

f(x∗i )∆x.

Interpretations

• If f(x) ≥ 0 then we can makethe following interpretation:

∆x

f(x)f(x)∆x =area

• If f(x) ≥ 0, then we interpret∫ b

a

f(x) dx as the area “under the

curve”: i.e. between f(x) and thex-axis and between x = a andx = b.

f(x)

a b

• In general, we interpret

∫ b

a

f(x) dx as the net area: the

area above the x-axis minus the area below the x-axis(and always between f(x) and the x-axis and betweenx = a and x = b).

Fundamental Theorem of CalculusAlthough the definite integral is defined as a limit of sums,and although we interpret it as a net area, we most oftencalculate it using anti-derivatives and the following result.

Theorem. If f(x) is a continuous function defined on theinterval [a, b], then

∫ b

a

f(x) dx = F (b) − F (a)

where F (x) is any anti-derivative of f(x).

Definition

Given a function f(x), the symbol “∫

f(x) dx” is called the

indefinite integral of f(x) or the anti-derivative of f(x). It is,by definition, a function that satisfies the following property

the derivative of

∫f(x) dx is f(x).

Plus CIf F (x) is one anti-derivative of f(x), then we write

∫f(x) dx = F (x) + C

because this describes all possible anti-derivatives of f(x).

W. Ethan Duckworth, Loyola University Maryland, 2013

CHAPTER 5. INTEGRALS REVIEWED 5

Example 1. Fill the following table in with basic facts and and anti-derivative:∫Cf(x) dx =

∫f(x)± g(x) dx =

∫xn dx =

∫sec2(x) dx =

∫ex dx =

∫1

xdx =

∫cos(x) dx =

∫1

1 + x2dx =

∫sin(x) dx =

Solution.∫Cf(x) dx = C

∫f(x) dx

∫f(x)± g(x) dx =

∫f(x) dx±

∫g(x) dx

∫xn dx =

xn+1

n+ 1+ C if n 6= −1

∫sec2(x) dx = tan(x) + C

∫ex dx = ex + C

∫1

xdx = ln |x|+ C

∫cos(x) dx = sin(x) + C

∫1

1 + x2dx = tan−1(x) + C

∫sin(x) dx = − cos(x) + C

Example 2. Find

∫5ex − 10

√x+

π

x− 13.5

1 + x2dx.

Solution. To find F (x) we use the basic facts about anti-derivatives shown inTable ??. In this example we break it down into more steps than many readers willneed.

Before we start, it’s worth remembering the right way to look at three terms ofthis function:

10√x = 10x1/2

π

x= π · 1

x13.5

1 + x2= 13.5 · 1

1 + x2

Now we can break our integral up into pieces:∫

5ex − 10√x+

π

x− 13.5

1 + x2dx

= 5

∫ex dx− 10

∫x1/2 dx+ π

∫1

xdx− 13.5

∫1

1 + x2dx

= 5ex − 10x3/2

3/2+ π ln |x| − 13.5 tan−1(x) + C

= 5ex − 10 · 2

3x3/2 + π ln |x| − 13.5 tan−1(x) + C

This is where we ended on Monday, January 12


5.5 U-substitution

Here is a brief outline of the technique of U -substitution.0. You are given an integral with respect to x that is too complicated to do

directly.1. Fill in the following

u = something involving x (you get to pick this)

du = (derivative of u) · dx (you don’t get to pick this)

2. Fill in the following

given integraltranslate or cancel all x’s and dx

=. . . . . . . . . . . . . . .=

∫f(u) du.

Make sure that all the x’s (including dx) cancel by the last step; whateveryou’re left with, call it f(u).

3. Find the anti-derivative ∫f(u) du = F (u)

4. Inside of F (u), replace u with the same “something” involving x, that youpicked in step 1.

Comments. When you are practicing u-substitution here are some good bits ofadvice and rules of thumb to keep in mind.

To set u = something you should be willing to take a guess, try setting u equalto something, and then take the derivative, and see if you can get rid of all the x’susing u and du.

Above I wrote “translate or cancel all the x’s and dx”. Different people havedifferent approaches to this step, but they all produce the same result; approach itwhichever way you want. Here’s a brief description of two approaches:

1. just circle those x’s that are part of u and those that are part of du, andreplace those parts with u and du,

2. solve for dx, and replace it with a formula involving du and x’s, and thencancel any remaining x’s.

Whichever approach you take, you should get∫h(x) dx =

∫f(u) du

and should be able to double check your work as follows. If you start with theright hand side, and substitute x’s back in for u and du, you should get the originalintegral, the one on the left.

When using u-substitution, it helps to know what sort of targets you might have

for

∫F (u) du. In other words, you want to have a list of integrals that are known,

and then see if you can turn

∫h(x) dx into one of known integrals.

Example 1. Find

∫tan(x) dx. (Note: this gives us the anti-derivative of one of

our last basic functions.)


Solution. We start by rewriting the function as a fraction:

∫sin(x)

cos(x)dx.

To figure out what u is think about what could possible work for du:

∫sin(x)

cos(x)dx =

∫sin(x)

1

cos(x)dx

︸︷︷︸good du?

=

∫1

cos(x)sin(x) dx︸︷︷︸good du?

There’s no way we can make u equal to something so that it’s derivative is1

cos(x)dx,

so the right choice is u = cos(x):

u = cos(x)

du = − sin(x) dx

− du = sin(x) dx∫

sin(x)

cos(x)dx =

∫1

cos(x)︸︷︷︸u

sin(x) dx︸︷︷︸− du

= −∫

1

udu

= − ln |u|+ C

= − ln | cos(x)|+ C

This is a perfectly good form of the answer, but using the properties of natural log,it’s possible to write it just a bit more compactly:

− ln | cos(x)| = ln

∣∣∣∣1

cos(x)

∣∣∣∣ = ln | sec(x)|∫

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

Example 2. Find

∫cos(ln(x))

xdx.

Solution. In this problem there are two ways to make a good guess for what ushould be. First, since we see cosine with something inside it, u is likely to bewhat’s inside of cosine. Second, we should look at what du could be.:

∫cos(ln(x))

xdx =

∫1

xcos(ln(x)) dx︸︷︷︸

good du?

=

∫cos(ln(x))

1

xdx


You should conclude that u = ln(x):

u = ln(x)

du =1

xdx


∫cos(ln(x))

xdx =

∫cos(ln(x)︸︷︷︸

u

)1

xdx

︸︷︷︸du

=

∫cos(u) du

= sin(u) + C

= sin(ln(x)) + C

Example 3. Find

∫7 sin(x) cos(x)

1 + cos2(x)dx

Solution. Let me do this problem first as if we had divine inspiration: I’ll writedown the correct u and solve it. But afterwards I’ll talk about how to find thecorrect u, what sort of guesses you could make, etc.

u = 1 + cos2(x)

du = −2 sin(x) cos(x) dx

− 1

2du = sin(x) cos(x) dx

∫7 sin(x) cos(x)

1 + cos2(x)dx =

∫7 · 1

1 + cos2(x)︸︷︷︸u

· sin(x) cos(x) dx︸︷︷︸−1

2 du

=

∫7 · 1

u(−1

2) du

= −7

2

∫1

udu

= −7

2ln |u|+ C

= −7

2ln |1 + cos2(x)|+ C

Here are some guesses you might have tried in solving this integral

u = sin(x) u = 1 + cos2(x)

u = cos(x) u = sin(x) cos(x)

u = cos2(x) u = 2 sin(x) cos(x)

You do not need to instantly see which of these choices is best, but to figure it outyou would think about how you can make two things come together. One thingis to start with one of these substitutions and take the derivative to find du. Theother thing is to look at the original integral and see what the possibilities for duare. Here are the results of both approaches:

u = sin(x) ⇒ du = cos(x)

u = cos(x) ⇒ du = − sin(x)

u = cos2(x) ⇒ du = −2 cos(x) sin(x)

u = 1 + cos2(x) ⇒ du = −2 cos(x) sin(x)

u = sin(x) cos(x) ⇒ du = cos2(x)− sin2(x)


u = 2 sin(x) cos(x) ⇒ du = 2(cos2(x)− sin2(x))

∫7 cos(x)

1 + cos2(x)sin(x) dx︸︷︷︸good du?∫

7 sin(x)

1 + cos2(x)cos(x) dx︸︷︷︸good du?∫

7

1 + cos2(x)sin(x) cos(x) dx︸︷︷︸

good du?∫7 sin(x) cos(x)

1

1 + cos2(x)dx


Hopefully you see that u = 1 + cos2(x) works, as we showed above. But youmight also see that u = cos(x) could have worked. This would have turned the

integral into −7

∫u

1 + u2du. Then we would have needed another subsitution,

w = 1 + u2.

The next example shows a technique called “backwards substitution”. It startsthe same as before, identify g(x) as some part of your function, set u = g(x), takedu = g′(x) dx. But the last step, where we replace all x’s with u’s, is harder: thereis an x left over after we do the usual steps. To get rid of this x, we substitutebackwards: instead of having u equal to a bunch of x stuff, we’ll solve this and findx equal to a bunch of u stuff.

Example 4. Find

∫x(3x+ 10)99 dx

Solution.

u = 3x+ 10

du = 3 dx

dx =1

3du

∫x(3x+ 10)99 dx =

∫xu99 1

3du

The problem is, there appears to be no way to cancel that other x. To get rid of itwe need “backwards substitution”. We have an equation u = 3x + 10 and we cansolve this equation for x!

u = 3x+ 10

x =1

3(u− 10)

∫xu99 1

3du =

∫1

3(u− 10)u99 1

3du


=1

9

∫(u− 10)u99 du

=1

9

∫u100 − 10u99 du

=1

9

(u101

101− 10

u100

100

)

=1

9

((3x+ 10)101

101− (3x+ 10)100

10

)

This is where we ended on Tuesday, January 13

Chapter 6

Applications of the definiteintegral

All the applications in this chapter can be understood through the following prin-ciple:

Approximation PrincipleAnything that can be approximated by adding f(x)∆x val-

ues, can be found exactly with

∫f(x) dx.

Examples of what the thing being approximated could be: area between a curveand the x-axis, or area between two curves, or a volume generated by revolutions,or arc-length or surface area, or probability, etc.

6.1 Area between two curves

Example 1. The curves h(x) = −x2 +13

3x− 2 and g(x) =

1

3x+ 1 intersect at 1

and 3. We want to find the area between the curves, and we notice that it can beapproximated by using four rectangles.

(a) Figure out how to find the area of one of the rectangles.

11

CHAPTER 6. INTEGRAL APPLICATIONS 12

(b) Use this to see the area of all four can be written as a sum of things of theform f(xi)∆x. What is f(x)?

(c) Use the Approximation Principle to set up an integral for the exact area, andthen solve this integral.

Solution. (a) Let’s pick the third rectangle to look at. I’ll describe how to find thearea using a top down, or big picture approach: start with general comments, andthen fill in the details.

area = height× width

=(top− bottom

)× 3− 1

4

=(h(?)− g(?)

)× 1

2

To fill in the missing numbers, we look at the x-axis and divide the interval [1, 3]into four equal pieces

1 1.5 2 2.5 3

Then we add the midpoints

1 1.5 2 2.5 3

1.25 1.75 2.25 2.75

Now we can fill in the missing numbers

area =(h(2.25)− g(2.25)

)× 1

2≈ 0.469

(b) The area of all four rectangles can be written using the same formula as inpart (a), but with different numbers plugged in:

(h(1.25)− g(1.25)) · 1

2+ (h(1.75)− g(1.75)) · 1

2

+ (h(2.25)− g(2.25)) · 1

2+ (h(2.75)− g(2.75)) · 1

2= 1.375

Notice that the things we added were of the form (h(x)− g(x))∆x. In other words,the function that we used was f(x) = h(x)− g(x).

(c) Since we have figured out what to sorts of things to add to approximate thearea, we can now apply the Approximation Principle and integrate this instead:

A =

∫ 3

1h(x)− g(x) dx = − 1

3x3 + 2x2 − 3x

]3

1

=4

3

Note that 1.375, the area of four rectangles, is a pretty good approximation of4/3, the exact area between the two curves.


Definition. The area above one curve and below the other is given by

Area =

∫ b

atop curve− bottom curve dx

Definition.

If f(x) ≥ g(x), then area between f(x) and g(x)and between x = a and x = b

=

∫ b

af(x)− g(x) dx

In general, area between f(x) and g(x)and between x = a and x = b

=

∫ b

a

∣∣f(x)− g(x)∣∣ dx

To calculate an integral of the second kind, we need to split the integral up into

two (or more) pieces such as

∫ b

a=

∫ c

a+

∫ b

cwhere in each piece we have only

f(x) ≥ g(x) or only g(x) ≥ f(x).

Comments. There are a few variations on this type of problem. Sometimes youhave to solve for a and b if they have not been given explicitly. Sometimes it changeswhich function is on top and which is on the bottom. In this case, you need to splitthe integral into two pieces, corresponding to the change. Finally, sometimes thearea is defined left-to-right instead of to top-to-bottom.

The Approximation Principle can also work for areas that are to the left of onecurve and to the right of the another curve. In this case, you can picture sideways

rectangles, and think of everything as a function of y. Thus, you’ll have

∫. . . dy.

This is where we ended on Wednesady, January 14

Example 2. The shapes below are defined by the functions x1 = −2y + 15 andx2 = y2 − 12y + 10.

(a) Find the measurements of the rectangle shown below (you may assume thatthe edges line up with y = 1 and y = 3).

(b) Find the y-values for the intersections of the two curves.(c) Find the area shown between the straight line and the parabola.


Solution. (a) Since the rectangle lines up with y-values of 1 and 3, we see that theheight is 2. The width is the difference between the x-values on the line and theparabola. The x-values correspond to y = 3, thus we have

x1(3)− x2(3) = −2(3) + 15− (32 − 12 · 3 + 10) = 26

(b) To find the points of intersection, we solve

−2y + 15 = y2 − 12y + 10

and gety = 5±

√30 ≈ −0.477, 10.477.

(c) To find the area, we imagine adding together the area of rectangles of thesort we had in part 1. This means we take the integral of such rectangles. Thevalue that changes is the width of the rectangle, and this is given by subtractingx1(y)− x2(y), i.e. by integrating −2y + 15− (y2 − 12y + 10).

Now we can use the values from part (b) and finish the problem:

∫ 10.477

−0.477x1(y)− x2(y) dy =

∫ 10.477

−0.477−2y + 15− (y2 − 12y + 10) dy

=

∫ 10.477

−0.477−y2 + 10y + 5

= − y3

3+ 5y2 + 5y

]10.477

−0.477

= −−0.4773

3+ 5(−0.477)2 + 5(−0.477)

−(

10.4773

3+ 5(10.477)2 + 5(10.477)

)

≈ 219.09

Definition. The area to the right of one curve and to the left of another curveequals

Area =

∫ y=b

y=aright curve− left curve dy

Definition.

If f(y) ≥ g(y), thenarea between f(y) and g(y)and between y = a and y = b

=

∫ y=b

y=af(y)− g(y) dy.


Extra examples

Example 3. Find the area between f(x) = x2 and g(x) = −x2 + 2x+ 5 (shown tothe side), and illustrate four rectangles in the Riemann Sum corresponding to theintegral you use.

Solution. To find a and b we intersect f(x) = x2 and g(x) = −x2 + 2x+ 5:

x2 = −x2 + 2x+ 5

2x2 − 2x− 5 = 0

x =2±

√4− 4(2)(−5)

4

x =1±√

11

2x ≈ −1.158, 2.158

Note that g(x) is on top, i.e. g(x) ≥ f(x). Thus, our integral is

∫ 2.158

−1.158−x2 + 2x+ 5− x2 dx.

We simplify and calculate this integral

∫ 2.158

−1.158−2x2 + 2x− 5 dx = − 2

3x3 + x2 − 5x

]2.158

−1.158

≈ 12.16

A four step Riemann Sum is pictured below

∫ b

ag(x)− f(x) dx ≈ (g(x1)− f(x1))∆x+ (g(x2)− f(x2))∆x+ (g(x3)−

f(x3))∆x+ (g(x4)− f(x4))∆x ≈ Area

Comments. Note: the type of justification given at the end of the previous examplemay seem shallow and/or unnecessary, but the point is this: the student should notneed to memorize each integral formula that we learn. Rather, we find each ofthese formulas in exactly the same way. Start by asking, “can we break a quantitydown into pieces of the form f(xi)∆x?” If so, then the whole quantity is found by

integrating,

∫f(x) dx. The whole quantity in the previous example can be breaking

down the area into pieces of the form(g(xi)− f(xi)

)∆x.


Example 4. Find the area that is between f(x) = sin(x) and g(x) = 0.25x and be-tween x = 0 and x = 5, as shown below (use your calculator to find the intersectionin the middle, then split the integral into two parts).

Solution. The endpoints are 0 and 5. We do not always have one function on top,so we need to use absolute values; the area is

∫ 5

0

∣∣ sin(x)− 0.25x∣∣ dx.

To solve this we need to split the integral up: one part will have sin(x) on top andone part will have 0.25x on top. The splitting up occurs at the point of intersection.

To find the point of intersection we can use our calculator1, either using 2cnd ,then CALC , then 5: intersect , or just using zoom and trace. The intersection isat x ≈ 2.475. Thus our integral becomes

∫ 2.475

0

∣∣ sin(x)− 0.25x∣∣ dx+

∫ 5

2.475

∣∣ sin(x)− 0.25x∣∣ dx

=

∫ 2.475

0sin(x)− 0.25x dx+

∫ 5

2.4750.25x− sin(x) dx

=(− cos(x)− 0.125x2

)]2.475

0

+(0.125x2 + cos(x)

)]5

2.475

≈ 4.449

(Note: you can double check your answer using your calculator: enter fnInt(abs(sin(x)- 0.25x),x,0,5).)

6.2 Volumes by rotation

The most basic shape for volumes is rectangular:

1On the Macintosh computer, you can use the Grapher application find the intersection. Go tothe /Applications folder, then the /Utilities folder. Open Grapher, enter one of the functions inthe first “y =” field, then go to the menu for Equation → New Equation to enter the second one.Select both the equations, then go to Equation → Find intersection.


x

y z

Volume = x · y · z

We can generalize this: any shape that has constant cross section (i.e. the sameshape in for all “slices”) with area A, and length `, has an easy volume formula:

Cross section:A

`

Volume = A · `

volume of a shape with constant cross-section of A is A · l where l is the length ofthe shape.

The most important case, for us, of this volume formula is for a cylinder:

h

r

Volume = πr2h

And, the most important application of the cylinder is where we have thickness ∆x

r

∆x

Volume = πr2∆x

On Friday, we got through solving part (c) of the next example

Example 1. Imagine we want to find the volume of a hard boiled egg and that wewant to do this my measuring the egg with a ruler. We could do this by cutting theegg into slices, and then measuring the volume of each slice as a cylinder.

(a) Use the ellipsex2

1.1552+

y2

0.82= 1

to model the outline of an egg with length 2.31 in and diameter 1.6 in. Drawthe ellipse and the shape it makes when rotated around the x-axis.

(b) Imagine cutting the shape from (a) into 8 slices, and draw the result.(c) Imagine replacing each slice from (b) with a cylinder and draw the result.(d) Figure out how to find the volume of each cylinder-slice in (c) and write down

a formula for the sum of the volumes.


(e) Translate the formula from (d) into an integral, and solve the integral.

Solution. (a)

1.155−1.155

0.8

−0.8

We imagine the egg being the three dimensional solid defined by rotating thisellipse around the x-axis:

(b) Show how we can slice this volume into pieces, and approximate the volumeof each piece using a disk, and then turn the the sum of volumes of disks into anintegral.

We start by drawing a picture of the egg after it’s cut into a bunch of slices(notice that cutting a shape into pieces doesn’t change the total volume):

Vol

+ + + + + + +

(c) Now each slice has a slight curve on the top and bottom. This makes it hardto find the exact volume of the slice, but we can approximate the volume by usinga cylinder of the same size:

Vol

+ + + + + + +

(d) Now, we can give a formula for the volume of all these disk-shaped slices

above stuff = πr21∆x+ πr2

2∆x+ πr23∆x+ . . .

To find r1, r2, etc., we go back to how this shape is defined: the radius is thedistance of the curve from the x-axis. This distance is what we usually call the


y-value, i.e. r = y. Now we find the y-values from the ellipsex2

1.1552+

y2

0.82= 1 by

solving for y:

y2

0.82= 1− x2

1.1552

y2 = 0.82

(1− x2

1.1552

)

y =

√0.82

(1− x2

1.1552

)

y = 0.8

√1− x2

1.1552

Combining this formula with the above sum of volumes we get

above stuff = π

(0.8√

1− x21/1.1552

)2

∆x+ π

(0.8√

1− x22/1.1552

)2

∆x

+ π

(0.8√

1− x23/1.1552

)2

∆x+ . . .

Now, suppose we want to actually get some numbers here to add up. We look atthe original range of x-values, divide it into 8 equal pieces,

−1.155 −0.866 −0.578 −0.289 0.000 0.289 0.578 0.866 1.155

and then pick an x-value in each interval (I’ll use the midpoint rule)

x∗1 = −1.011, x∗2 = −0.722, x∗3 = −0.433, x∗4 = −0.144,

x∗5 = 0.144, x∗6 = 0.433, x∗7 = 0.722, x∗8 = 1.011.

Now we plug these into the formula we had above(e) Now we apply the Approximation Principle: anything that we can approxi-

mate with a sum of function values, we can make exact by integrating:

V =

∫ 1.155

−1.155π(

0.8√

1− x2/1.1552)2

dx

This last line is the most important conclusion of this example: translating a volumeby rotation into an integral. Of course, it’s also nice to find the integral:

∫ 1.155

−1.155π(

0.8√

1− x2/1.1552)2

dx = π

∫ 1.155

−1.1550.82

(1− x2/1.1552

)dx

= π(0.8)2

∫ 1.155

−1.1551− x2/1.1552 dx

= π(0.8)2

(x− x3

3(1.155)2

)]1.1155

−1.155

= 3.1 in3


This is where we ended on Friday, January 16

Rule. Let V be the volume generated by rotating f(x) around the x-axis, betweenx = a and x = b. Then V is given by the following formula:

V =

∫ b

aπr2 dx, where r = f(x).

Example 2. Find the volume generated by rotating y = −x2 +2 around the x-axis,between x = −

√2 and x =

√2.

Solution. The original shape, and the volume generated by it are pictured below.

•

r = y = −x2 + 2

∆x

We apply

∫ b

aπ(f(x))2 dx with a = −

√2, b =

√2, f(x) = −x2 + 2 and get

V =

∫ √2

−√

2π(−x2 + 2)2 dx = π

∫ √2

−√

2x4 − 4x2 + 4 dx

= π

(1

5x5 − 4

3x3 + 4x

)]√2

−√

2

= π

(1

525/2 − 4

3· 23/2 + 4

√2−

(− 1

525/2 +

4

3· 23/2 − 4

√2

))

= 2π

(1

525/2 − 4

3· 23/2 + 4

√2

)

= 2π

(22√

2

5− 4 · 2

√2

3+ 4√

2

)

= 2 · 4 ·√

2π

(1

5− 2

3+ 1

)

= 8√

2π

(3

15− 10

15+

15

15

)

=64

15

√2π

Example 3. Find the volume generated by rotating y = x2 around the y-axis,between y = 0 and y = 2.

Solution. This is a sideways problem because the way we can cut it into disks isto make cuts that go up and down along the y-axis.


One slice of this volume gives a disk of thickness ∆y and and radius given by thex-value. To get the x-value we need to write x(y), i.e. x as a function of y. Wesolve y = x2 for x =

√y. We also need to know the bounds of integration; this is

from the smallest y-value to the largest one. The y-values are y = 0 and y = 2.∫ 2

0π(√y)2 dy = π

∫ 2

0y dy

= πy2

2

]2

0

= π22

2= 2π

Rule (Generalized from previous example). The volume generated by rotating f(y)around the y-axis, between y = a and y = b is

Volume of rotation =

∫ b

aπr2 dy, where r = f(y)

Example 4. Set up an integral, and use your calculator to find it, for the volume

of the napkin ring made by rotating the region bounded by y =√

1− x2 + 2, andy = 2 around the x-axis.

Solution. We picture the region defined below, along with one “slice” of volume:

The volume of the slice is given by the area of the face, times ∆x. The main workis to figure out the formula for the area. We start with the two dimensional shape,involving r and R


R

r

Area = πR2 − πr2

Volume = (πR2 − πr2)∆x

So, to finish, we just need to figure out formulas for R and r in the shape that wehave.

Rr

Thus, R = y =√

1− x2 + 2 and r = 2. Putting all of this together, we have theintegral that we want.

V =

∫ 1

−1π(√

1− x2 + 2)2 − π22 dx

Even if we’re going to enter this in our calculators, it’s useful to simplify a littlefirst (this makes it less likely that we make a mistake in our calculator):

V = π

∫ 1

−11− x2 + 4

√1− x2 dx.

If we integrate this in our calculator we get

V ≈ π7.62 ≈ 23.93 .

It is possible to do this problem algebraically too.

V =

∫ 1

−1π(√

1− x2 + 2)2 − π22 dx

=

∫ 1

−1π(1− x2 + 4

√1− x2 + 4− 4) dx


= π

∫ 1

−1−x2 + 4

√1− x2 + 1 dx

= π

(4

3+ 4

∫ 1

−1

√1− x2 dx

)

To finish this, we need to figure out what

∫ 1

−1

√1− x2 dx is. This is probably not

something that you know how to find the anti-derivative of. But, you can figureout the integral without knowing the anti-derivative. If you think of what area is

represented by

∫ 1

−1

√1− x2 dx, you should be able to see that it’s the area of the

top half of a unit circle. This has area1

2π. Thus,

V = π

(4

3+ 4 · 1

2π

)=

4

3π + 2π2

This is where we ended on Tuesday, January 21

Comments. We have seen at least three variations on rotation so far: one functionaround the x-axis, one function around the y-axis, and two functions around thex-axis. There are a few more variations, but rather than introducing each oneseparately, it might be nice to see all the variations all at once:

Rule (Volumes of rotation by disk and washer).

Disks and washers around horizontal linesdisks washers

V =

∫πr2 dx V =

∫πR2 − πr2 dx

around x-axis r = f(x) R = f(x), r = g(x)around y = c r = f(x)−c or c−f(x) R = f(x)−c or c−f(x), r = g(x)−c or c−g(x)

(choose the formulas for R and r that are positive, and with R > r)

Disks and washers around vertical linesdisks washers

V =

∫πr2 dy V =

∫πR2 − πr2 dy

around y-axis r = f(y) R = f(y), r = g(y)around x = c r = f(y)−c or c−f(y) R = f(y)−c or c−f(y), r = g(y)−c or c−g(y)

(choose the formulas for R and r that are positive, and with R > r)

Example 5. Find the volume of the napkin ring made by rotating the region

bounded by y = − 4

9x2 + 2, and y =

1

4x2 +

7

16(shown below) around the line

y = −1.


Solution. The rotated shape looks roughly as follows:

The larger radius is defined by the top curve, y = − 4

9x2 + 2. It is the distance

between this curve and the horizontal axis, y = −1. Thus,

R = − 4

9x2 + 2− (−1) = − 4

9x2 + 3.

The smaller radius, the radius of the hole, is defined by the bottom curve y =1

4x2 +

7

16. It is the distance between this curve and the horizontal axis y = −1.

Thus

r =1

4x2 +

7

16− (−1) =

1

4x2 +

23

16.


Now we integrate (skipping some of the messy steps)

V =

∫ 3/2

−3/2π

(− 4

9x2 + 3

)2

− π(

1

4x2 +

23

16

)2

dx

(messy steps of foiling skipped)

= π

∫ 3/2

−3/2

175

1296x4 − 325

96x2 +

1775

256dx

=25π

16

∫ 3/2

−3/2

7

81x4 − 13

6x2 +

71

16dx

=25π

16

(7

405x5 − 13

18x3 +

71

16x

)]3/2

−3/2

=25π

16· 87

10

=435

32π

Example 6. Find the volume of the region between the curves y = g(x) = x3 andy = f(x) =

√x rotated around the line x = 1 (shown below).


Solution. We solve this with washers, that are stacked up and down along thevertical line x = 1. Thus, we will have

∫ b

aπR2 − πr2 dy

where we need R and r to be formulas for the radiuses, written as functions of y.The larger radius, R, is the horizontal distance between x = 1 and the curve f(x).We need the formula in terms of y, so we solve y =

√x for x = y2,

R = 1− y2

The smaller radius, r, is the horizontal distance between x = 1 and the curve g(x).Again we rewrite the equation first y = x3 ⇒ x = y1/3

r = 1− y1/3

Putting it all together we get

V =

∫ 1

0π(1− y2)2 − π(1− y1/3)2 dy

= π

∫ 1

01− 2y2 + y4 − (1− 2y1/3 + y2/3) dy

= π

∫ 1

0−2y2 + y4 + 2y1/3 − y2/3 dy

= π

(− 2

3y3 +

y5

5+ 2

y4/3

4/3− y5/3

5/3

)]1

0

= π

(− 2

3+

1

5+ 2 · 3

4− 3

5

)

=13

30π

Extra examples

Example 7. Derive the formula for the volume of a sphere of radius r.

Solution. We start with the formula for the top half of a circle of radius r:

y =√r2 − x2

and then plug this into our volume by rotation formula

V =

∫ r

−rπ(√r2 − x2)2 dx

=

∫ r

−rπ(r2 − x2) dx

= π

(r2x− x3

3

)]r

−r


= π

(r2r − r3

3−(−r2r +

r3

3

))

= 2π

(r3 − r3

3

)

= 2π2

3r3

=4

3πr3

Challenge. • Can you figure out how to find the volume of a shape rotatedaround the line y = x? What about other lines?• Can you figure out how to apply washers when f(x) and g(x) switch places

with respect to which one is farther from c? What about if they switch placesalso with respect to which side of c they fall on?

This is where we ended on Wednesday, January 21

Pictures of areas and volumes

f(x)

g(x)

f(x)− g(x) = h

∆xArea of this rectangle = (f(x)− g(x))∆x

Area betweentwo curves

=

∫f(x)− g(x) dx


f(x) g(x)

︸︷︷︸g(x)− f(x)

}∆y

Area of this rectangle = (g(y)− f(y))∆y

Area between twosideways curves

=

∫g(y)− f(y) dy

f(x)

•

•

r = f(x)

∆x

Volume of this disk = π(f(x))2∆x

Volume of revolution be-tween one function andthe x-axis, using disks

=

∫π(f(x))2 dx


g(x)

f(x)

Rr

∆x

x-axis

Volume of this washer = (πR2 − πr2)∆x

=(π(f(x))2 − π(g(x))2

)∆x

Volume of rotation between twofunctions, rotated around the x-axis,using washers

=

∫π(f(x))2 − π(g(x))2 dx


line y = c

x-axis

g(x)

f(x)

R

r

∆x

line y = c

Volume of this washer = (πR2 − πr2)∆x

=(π(f(x)− c)2 − π(g(x)− c)2

)∆x

Volume of rotation between twofunctions, rotated around the liney = c, using washers

=

∫π(f(x)− c)2 − π(g(x)− c)2 dx

6.3 Volumes by Cylindrical Shells

For some functions it’s easier to slice the volume a different way than in the previoussection. In particular, we’ll add one more basic volume shape to our repertoire:


cylindrical shells. One is pictured below

h

r

∆r

To figure out this volume you can imagine cutting the shell and unrolling it andflattening it out, as shown below in four steps.

−→

−→ −→h

2πr ∆r

The result is a rectangular solid with measurements ∆r, h and 2πr, and so thevolume is

V (cylindrical shell) = 2πrh∆r

Definition. Let V be the volume generated by the region bounded by f(x), thex-axis, x = a and x = b, rotated around the y-axis. Then V is given by

V =

∫ b

a2πrh dx, wherer = x and h = f(x).

Example 1. Find the volume obtained by rotating the region between y = sin(x2)and y = 0, from x = 0 to x =

√π around the y-axis.

Solution. We show the original area and the volume of rotation it generates below.

We can picture this volume as being cut into cylindrical shells. We show belowhow this looks, both showing first all the shells at the same time, and then one shellat a time.


Note, there is no way to do this problem (easily) with disks or washers. If weused them we’d need to do it sideways using functions of y. But solving y = sin(x2)

for x gives x =

√sin−1(x) and we don’t know how to integrate this.

Thus, the total volume can be found by adding volumes of the form 2πr∆xh.The integral form is

V =

∫2πrh dr

We replace r, ∆x and h with r = x, ∆r = ∆x, and h = f(x). Thus, our integralbecomes

V =

∫ √π

02πx sin(x2) dx

= π

∫ √π

02x sin(x2) dx

Let u = x2, du = 2x dx.

V = π

∫ x=π

x=0sin(u) du

= −π cos(u)

]x=π

x=0

= −π cos(x2)

]x=π

x=0


= −π(

cos(π)− cos(0))

= −π(−1− 1)

= 2π

Rule (Volumes of rotation by cylindrical shells).

Cylindrical shells around vertical linesOne function Two functions∫ b

a2πrh dx

∫ b

a2πrh dx

around y-axis r = x, h = f(x) r = x, h = f(x)− g(x)around x = c r = x−c or c−x, h = f(x) r = x−c or c−x, h = f(x)− g(x)

(choose the formula for r that is positive; make sure f(x) is on top of g(x))

Example 2. Sketch the region and one typical cylindrical shell for the volume

generated by the region between y = sin(πx) and y = x2 − x +1

4, and from

a ≈ 0.0626 and b ≈ 0.938, rotated around the line x = 1.5. Then set up (but donot solve) an integral for this volume.

Solution. Here is a picture of the original region, and a shell in the rotated volume.


As usual, we wish to integrate volumes of the form 2πrh∆x. As stated in ourrule, h is found by subtracting one curve from the other h = f(x) − g(x) with

f(x) = sin(πx) and g(x) = x2 − x +1

4. Also, we find r by subtracting. It is the

distance between a point on the curve and the line x = 1.5. This distance is foundas 1.5− x. Thus, we have the following integral

V =

∫ 0.938

0.06262π(1.5− x)

(sin(πx)− (x2 − x+

1

4)

)dx

Extra Examples

Example 3. Find the volume generated by rotating f(x) = −x2 + 2, betweenx = −

√2 and x =

√2, and rotating it around the line x = 2.

Solution. This is similar to the previous example. The function f(x) is different,but the main change is with the radius. Instead of r = x we have r = 2− x.

V =

∫ √2

−√

22π(2− x)(−x2 + 2) dx

= 2π

∫ √2

−√

2x3 − 2x2 − 2x+ 4 dx

= 2π(x3 − 2x2 − 2x+ 4

]√2

−√

2

)

= 2π16√

2

3


Example 4. [Stewart 6.3#6] Find the volume generated by rotating the regionbounded by the given curves around the y-axis. Sketch the region and a typicalshell.

y = 3 + 2x− x2, x+ y = 3

Solution.

We still want to have an integral of the form

∫2πrh dx where 2πrh∆x is the

volume of a cylindrical shell, and r and h need to be replaced with suitable functionsof x. Since we are rotating around the y-axis, the radius is x. The height of thecylindrical shell is the distance between f(x) = −x2 + 2x + 3 and g(x) = 3 − x.Thus, our integral is

V =

∫ 3

02πx(−x2 + 2x+ 3− (3− x)) dx

=

∫ 3

02πx(−x2 + 3x) dx

= 2π

∫ 3

0−x3 + 3x2 dx

= 2π(−x4

4+ x3)

]3

0


= 2π(−x4

4+ x3)

]3

0

= 2π · 27

4

=27π

2

This is where we ended on Friday, January 25

6.5 Average Value of a Function

Example 1. Suppose that on one day the temperature is given by the followingfunction

K(t) = 0.4t− 45 sin(πt/12)

where t is in hours (with t = 0 corresponding to midnight), and K(t) is the tem-perature in Fahrenheit.

(a) Find a formula for the average value of the temperature for the whole day byusing 4 equally spaced times.

(b) Find a formula for the average value of the temperature for the whole day byusing 12 equally spaced times.

(c) Set up and solve an integral for the average value of the whole day.

Solution. (a) We will use t = 0, 6, 12, 18. Then the average is

A =K(0) +K(6) +K(12) +K(18)

4

Note that the formula is what we are most interested in here, not the actual calcu-lation. However, it might be nice to see the results

0− 42.6 + 4.8 + 52.2

4≈ 3.6◦F

(b) We will use t = 0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22. Then the average is

A =K(0) +K(2) +K(4) + · · ·+K(16) +K(18) +K(20) +K(22)

12(6.1)

As above, we are most interested in the formula, but also include the actual calcu-lation:

0− 21.7− 37.4− 42.6− 35.8− 18.5 + 4.8 + 28.1 + 45.4 + 52.2 + 47 + 31.3

12≈ 4.4◦F

(c) The key to turning the above calculations into an integral are to see them assums of things of the form K(ti)∆t. It’s pretty clear how to see K(ti) in the abovesums, but we need to rewrite things a little to see where ∆t shows up.

In part (b) we divided by 12, the number of steps we were asked to use. Usingn = 12 steps is equivalent to having ∆t = 2; in terms of an equation this means

1

12=

2

24


1

n=

∆t

24

Thus, our calculation in Equation (6.1) is equivalent to

∆t

24(K(0) +K(2) + · · ·+K(22)) =

1

24(K(0)∆t+K(2)∆t+ · · ·+K(22)∆t)

If we use larger values of n, and eventually let n →∞, the 24 doesn’t change, butthe sum on the right changes into an integral. Thus, the exact average is given bythe following integral:

A =1

24

∫ 24

0K(t) dt.

Now we find this integral

A =1

24

∫ 24

00.4t− 45 sin(πt/12) dt

=1

24

(0.2t2 + 45 cos(πt/12) · 12

π

)]24

0

=1

24

(0.2t2 + 45 cos(πt/12) · 12

π

)]24

0

=1

24· 115.2

= 4.8

Definition (Generalized from previous example). The average value of a functionf on an interval [a, b] is

favg =1

b− a

∫ b

af(x) dx

Comments. If f(x) ≥ 0 then we can understand this definition another way:

geometrically. If we think of

∫ b

af(x) dx in terms of area, and b − a in terms of

width, then we can rewrite this definition:

area = average height× width

and so favg =

∫ ba f(x) dx

b− a becomes

average height =area

width

Example 2. Find the average value of the function f(x) = x− 3x2 on the interval[−1, 1].

Solution.

favg =1

2

∫ 1

−1x− 3x2 dx

=1

2

(x2

2− x3

)]1

−1

= −1

Note,when you use thisformula, you have been given

an integral like

∫xex dx.

There are two functions in thisintegral, however neither ofthem looks like f ′, that is tosay, neither looks like aderivative. It’s up to us to“pretend” that one of them isthe derivative of something.

Chapter 7

Techniques of Integration

7.1 Integration by Parts

Comments. The most basic rules for finding anti-derivatives come from doing

derivative rules backwards:

∫sec2(x) dx = tan(x) comes from finding the derivative

of tan(x) and u-substitution comes from doing the chain rule backwards. In the sameway, integration by parts comes from doing the product rule backwards.

This is where we ended on Monday, January 26

Here’s how we turn the product rule into integration by parts

(f · g)′ = f ′ · g + f · g′ product rule∫(f · g)′ =

∫(f ′ · g + f · g′) taking

∫of both sides

f · g =

∫f ′ · g +

∫f · g′ simplifying

Now we solve the last equation for one of the remaining integrals to get the formulawe want:

Rule.

Integration by parts∫f ′ · g = f · g −

∫f · g′

Comments.

Comments. When I forget this formula, I usually write down one or two lines fromthe above calculations starting with the product rule, and then get the result again.

Comments. The book writes this a different way. Let v = f(x) and u = g(x) sodv = f ′(x) dx and du = g′(x) dx. Then we have the following rule.

Rule.

Integration by parts(alternative notation)

∫u dv = u · v −

∫v du

This formula has a mnemonic:

38

CHAPTER 7. TECHNIQUES OF INTEGRATION 39

You devil, ultra-violet voodoo:

∫uyou

dvdevil

= uultra

vviolet

−∫

vvoo

dudoo

Comments. I urge students to try a little bit to look past the notation of theintegration by parts. In other words, it’s OK to use f ’s and g’s or u’s and v’s, andit’s OK to really prefer one over the other. But, also look past what all these lettersare and think about what the formula is saying in words.

Rule.

Integration by parts, verbally∫two functions = original · anti-deriv−

∫deriv · anti-deriv

Comments. When you use the verbal description of Integration by Parts, justremember that you use the same function for both anti-derivatives, and use theother function for the original and derivative spots.

Comments. Usually you are given something to integrate that looks like a product.You have to choose which thing to call f ′ (or du) and which to call g (or v). Thepoint is that the second integral should usually be easier for some reason than theoriginal integral.

Example 1. Find

∫xe3x dx.

Solution. This time we will use the formula∫u dv = uv −

∫v du.

Our original integral has x and e3x. We need to choose which of these functions

to call u and which to call dv. We make this choice so that the integral

∫v du is

simpler; in particular, we should choose u so that du will be simpler. To do this,we choose u = x and therefore dv = e3x dx. Thus, we fill in the following

u = x du = 1

dv = e3x dx v = 13e

3x

(choose this) (fill in this)

and we put the results together using the IBP formula,∫xe3x dx = uv −

∫v du =

1

3xe3x −

∫1

3e3x dx,

and finish by taking the last anti-derivative,

1

3xe3x −

∫1

3e3x dx =

x

3e3x − 1

9e3x.


Heuristic

Comments. One question students often have is which function in the integral tocall f and which to call g′, or, equivalently, which to call u and which to call dv, or,equivalently, which to take the derivative of, and which to take the anti-derivativeof.

Rule. One rule of thumb is to take the derivative of the function (i.e. set g or uequal to the function) that shows up first on the following list:

LIATE: Logarithmic, Inverse trig, Algebraic, Trig, Exponential.

Example 2. Find

∫t2 sin(5t) dt.

Solution. Using the formula

∫f ′g = fg −

∫fg′ we make our choices and fill in

as shown:g = t2 g′ = 2t

f ′ = sin(5t) f = −1

5cos(5t)


Now we plug the results into our IBP formula:∫t2 sin(5t) dt = t2

(− 1

5cos(5t)

)−∫

2t

(− 1

5cos(5t)

)dt.

Now we rewrite the last integral a little as +2

5

∫t cos(5t) dt and apply IBP again.

We let g equal the algebraic function:

g = t g′ = 1

f ′ = cos(5t) f =1

5sin(5t)


and plug the IBP formula into the integrals above

above stuff = − 1

5t2 cos(5t) +

2

5

(1

5t sin(5t)−

∫1

5sin(5t) dt

)

= − 1

5t2 cos(5t) +

2

5

(1

5t sin(5t) +

1

25cos(5t)

)

Example 3. Find

∫ex sin(x) dx. (Hint: do integration by parts twice, and solve

for the equation for the missing integral.)

Solution. Let f ′ = ex and g = sin(x). Then f = ex and g′ = cos(x).∫ex sin(x) dx = ex sin(x)−

∫ex cos(x) dx

Now we apply integration by parts again. Let f ′ = ex and g = cos(x). Then f = ex

and g′ = − sin(x).∫ex sin(x) dx = ex sin(x)−

∫ex cos(x) dx


= ex sin(x)−(ex cos(x)−

∫ex(− sin(x)) dx

)

The equation is now

∫ex sin(x) dx = ex sin(x)− ex cos(x)−

∫ex sin(x) dx

We solve this for the unknown integral

2

∫ex sin(x) dx = ex sin(x)− ex cos(x)

∫ex sin(x) dx =

1

2ex(sin(x)− cos(x))

Example 4. Find

∫ln(x) dx (the last of our basic anti-derivative).

Solution. The trick here is that to use integration by parts, we really, really needto view ln(x) as a product. That way we can choose to view it as f ′ · g. Can youthink of how to make ln(x) into a product?

We write the original integral as

∫1 · ln(x) dx. Then we choose f ′ = 1 and

g = ln(x). From this we get f = x and g′ =1

x.

∫ln(x) dx =

∫1 · ln(x) dx = x ln(x)−

∫x · 1

xdx

= x ln(x)−∫

1 dx

= x ln(x)− x∫

ln(x) dx = x ln(x)− x

Example 5. Find

∫tan−1(x) dx.

Solution. This problem is like the previous one; we apply integration by partsusing one factor equal to 1.

∫tan−1(x) dx =

∫1 · tan−1(x) dx

Now we pick f ′ and g, and calculate f and g′

f ′ = 1 f = x

g = tan−1(x) g′ =1

1 + x2

Now we apply these formulas using integration by parts


∫tan−1(x) dx =

∫1 · tan−1(x) dx

= x tan−1(x)−∫x · 1 + x2

dx

= x tan−1(x)−∫

x

1 + x2dx

To figure out this last integral you can use u-substitution.∫

x

1 + x2dx =

1

2

∫1

udu =

1

2ln |1 + x2|

u = 1 + x2

du = 2x dx

Now we can finish the integral

∫tan−1(x) dx = x tan−1(x)− 1

2ln |1 + x2|

Tabular integration by parts

Comments. An integral of the form

∫x2f(x) dx or

∫x3f(x), etc., can involve

repeated integration by parts. Whenever an integration by parts is repeated, youmight get the sense that a pattern is emerging as you do the steps. However, itcan be hard to see the pattern after you’ve written all those integral symbols, andf ’s and g’s, etc. Tabular integration by parts reveals this pattern. We simplify thenotation a little, by writing this formula using f and g (as opposed to f ′ and g):

Rule (Tabular Integration by Parts). Given an integral of two functions, F and G,take the derivative of one function and the anti-derivative of the other, as shownbelow

F G (both original functions)

F ′wG

F ′′x

G

F ′′′y

G

. . . . . .

+

−

+

−

Multiply the functions that are connected by arrows, and then add or subtract themas indicated. The final result should look like this:

wFG = F

wG− F ′

xG+ F ′′

yG− F ′′′

x xG+ . . .

This is where section 1 ended on Tuesday, January 27

Example 6. Find

∫x3 sin(πx) dx


Solution. We make two columns of functions

F G

x3 sin(πx)

3x2 − 1π cos(πx)

6x − 1π2 sin(πx)

6 1π3 cos(πx)

0 1π4 sin(πx)

Now we take the terms that are connected by arrows in the above tables.∫x3 sin(πx) dx = (+)x3

(− 1π cos(πx)

)− 3x2

(− 1π2 sin(πx)

)+ 6x 1

π3 cos(πx)− 6 1π4 sin(πx)

= −x3

πcos(πx) +

3x2

π2sin(πx) +

6x

π3cos(πx)− 6

π4sin(πx)

Example 7. Find

∫2x3 − 3x+ 10√

2x− 5dx.

Solution. We start by viewing the integral as the product of two functions

∫2x3 − 3x+ 10√

2x− 5dx =

∫(2x3 − 3x+ 10) · (2x− 5)−1/2 dx.

We make two columns of functions

F G

2x3 − 3x+ 10 (2x− 5)−1/2 (both original functions)

6x2 − 3(2x− 5)1/2

1/2· 1

2 = (2x− 5)1/2

12x(2x− 5)3/2

3/2· 1

2 = 13(2x− 5)3/2

12 13

(2x− 5)5/2

5/2· 1

2 = 13 · 1

5(2x− 5)5/2

0 13 · 1

5

(2x− 5)7/2

7/2· 1

2 = 13 · 1

5 · 17(2x− 5)5/2

Now we take the terms that are connected by arrows in the above tables.

∫2x3 − 3x+ 10√

2x− 5dx = (2x3 − 3x+ 10)(2x− 5)1/2 − 1

3(6x2 − 3)(2x− 5)3/2

+ (12x)13 · 1

5(2x− 5)5/2 − (12)13 · 1

5 · 17(2x− 5)7/2

7.2 Trigonometric Integrals

Comments. This section gives tricks for solving integrals of the form

∫sinn(x) cosm(x) dx

and

∫tann(x) secm(x) dx.


Example 1. Find

∫sin(x) cos18(x) dx.

Solution. Let u = cos(x), du = − sin(x) dx and the integral becomes

−∫u18 du

Then the anti-derivative is

−u19

19= −cos19

19

Comments. The same basic approach that we used in the previous example willwork with any power of cosine. We could have had cos2(x), cos−5(x), or even√

cos(x). In fact, we can even extend the same idea to any integral with an oddnumber of sine functions, provided we use a trig identity to get rid of the otherpowers. A similar trick will work if we have an odd number of cosine functions. Wegive an outline of this procedure first, and then do another example.

Rule. To integrate

∫sinn(x) cosm(x) dx:

Case 1 If the power of sine is odd, then let u = cos(x), du = − sin(x) dx, and getrid of all but one power of sin(x) using

sin2(x) = 1− cos2(x)

sin4(x) = (1− cos2(x))2

sin6(x) = (1− cos2(x))3

etc

and then complete your u-substitution and integrate.Case 2 If the power of cosine is odd, then let u = sin(x), du = cos(x) dx, and get

rid of all but one power of cos(x) using

cos2(x) = 1− sin2(x)

cos4(x) = (1− sin2(x))2

cos6(x) = (1− sin2(x))3

etc

and then complete your u-substitution and integrate.Case 3 If both sine and cosine have even powers, then use the identities

sin2(θ) =1

2(1− cos(2θ))

cos2(θ) =1

2(1 + cos(2θ))

then multiply everything out. You now have only powers of cos(2x). Splitthe integral up: odd powers bigger than 1, go to Case 2; even powers, repeatCase 3. In this way you are eventually left with only single powers of cos(2x),cos(4x), cos(8x), . . . , which you can finish immediately.

Example 2. Find

∫sin7(x)

√cos(x) dx.


Solution. Let u = cos(x), du = sin(x) dx. We get rid of sin6(x) by rewriting it as(1− cos2(x))3. Then we have:

∫sin7(x)

√cos(x) dx =

∫sin(x)(1− cos2(x))3(cos(x))1/2 dx

= −∫

(1− u2)3u1/2 du

= −∫

(1− 3u2 + 3u4 − u6)u1/2 du

= −∫u1/2 − 3u5/2 + 3u9/2 − u13/2 du

= −(

2

3u3/2 − 6

7u7/2 +

6

11u11/2 − 2

15u15/2

)

= −(

2

3(cos(x))3/2 − 6

7(cos(x))7/2 +

6

11(cos(x))11/2 − 2

15(cos(x))15/2

)

This is where section 1 ended on Wednesday, January 28

Example 3. Find

∫sin2(x) cos2(x) dx

Solution. We apply the identities mentioned in Case 3 above to get

∫1

2(1− cos(2x))

1

2(1 + cos(2x)) dx =

1

4

∫1− cos2(2x) dx

=1

4

∫1− 1

2(1 + cos(4x)) dx

=1

4

∫1

2− 1

2cos(4x) dx

=1

8

∫1− cos(4x) dx

=1

8

(x− sin(4x)

4

)

Comments. Now we apply the exact same ideas to integrals of the form

∫tann(x) secm(x) dx.

This time, we start with an outline of the approach.

Rule. To integrate

∫tann(x) secm(x) dx

Case 1 If the power of tangent is odd get then let u = sec(x), du = sec(x) tan(x) dx,and get rid of all but one power of tan(x) using

tan2(x) = sec2(x)− 1

tan4(x) = (sec2(x)− 1)2

tan6(x) = (sec2(x)− 1)3

etc

and then complete your u-substitution and integrate.


Case 2 If the power of secant is even then let u = tan(x), du = sec2(x) dx, andget rid of all but two powers of sec(x) using

sec2(x) = tan2(x) + 1

sec4(x) = (tan2(x) + 1)2

sec6(x) = (tan2(x) + 1)3

etc

and then complete your u-substitution and integrate.Case 3 If tangent has an even power and secant an odd power, then get rid of all

the powers of tan(x) using tan2(x) = sec2(x)− 1 as above. Now we have onlypowers of sec(x). Use a little luck, integration by parts, the secant-tangentidentity, and the following:

∫sec(x) dx = ln | sec(x) + tan(x)|.

This is where section 1 ended on Friday, January 30

Example 4. Find

∫tan3(x) sec3(x) dx.

Solution. Note that the power of tangent is odd, so we use

u = sec(x)

du = sec(x) tan(x) dx

If we apply this we can translate part of the integral already

∫tan2(x) tan(x) sec(x)︸︷︷︸

du

sec2(x)︸︷︷︸u2

()dx︸︷︷︸du

=

∫tan2(x)u2 du

Now to finish we need to get rid of the last functions of u. We do this with theidentity

∫tan2(x)u2 du =

∫(sec2(x)− 1)u2 du

=

∫(u2 − 1)u2 du

=

∫u4 − u2 du

=u5

5− u3

3

=sec5(x)

5− sec3(x)

3

Example 5. Find

∫sec3(x) dx.


Solution. We have an even number of tangents (zero) and an odd number ofsecants. The integral is already written in terms of secants, so we apply integrationby parts. To do this, we need to write sec3(x) as a product, namely as sec(x) sec2(x).

∫sec(x) sec2(x) dx = sec(x)

anti-derivative︷︸︸︷tan(x)−

∫ derivative︷︸︸︷sec(x) tan(x)

anti-derivative︷︸︸︷tan(x) dx

= sec(x) tan(x)−∫

sec(x) tan2(x) dx


sec(x)(sec2(x)− 1) dx


sec3(x) +

∫sec(x) dx

∫sec3(x) dx = sec(x) tan(x)−

∫sec3(x) dx+ ln | sec(x) + tan(x)|

2

∫sec3(x) dx = sec(x) tan(x) + ln | sec(x) + tan(x)|

∫sec3(x) dx =

1

2

(sec(x) tan(x) + ln | sec(x) + tan(x)|

)

This is where section 1 ended on Monday, February 2

Example 6. Find

∫tan2(x) sec(x) dx.

Solution. Since we have an even number of tangents, and an odd number of secants,we start by getting rid of all powers of tangent

∫tan2(x) sec(x) dx =

∫(sec2(x)− 1) sec(x) dx =

∫sec3(x)− sec(x) dx.

The integrals are now known:

∫sec(x) dx was given above, and the previous ex-

ample solved

∫sec3(x) dx. We put this in, simplify, and we are done:

1

2

(sec(x) tan(x) + ln | sec(x) + tan(x)|

)− ln | sec(x) + tan(x)|

=1

2

(sec(x) tan(x)− ln | sec(x) + tan(x)|

)

Extra Material: a hard example

Example 7. Find

∫sin4(x) cos2(x) dx

Solution. We start by applying the half angle identities, multiplying everythingout, and then gather like terms:

∫sin4(x) cos2(x) dx =

∫ (1

2(1− cos(2x))

)2 1

2(1 + cos(2x)) dx

=

∫1

22(1− 2 cos(2x) + cos2(2x))

1

2(1 + cos(2x)) dx


=1

23

∫1 + cos(2x)− 2 cos(2x)− 2 cos2(2x) + cos2(2x) + cos3(2x) dx

=1

23

∫1− cos(2x)− cos2(2x) + cos3(2x) dx

Now we break this up so that each part is either already known, or we fits into oneof our casess for even and odd powers:

1

23

∫

1− cos(2x) dx

︸︷︷︸easy

−∫

cos2(2x) dx

︸︷︷︸case 3: even-even

+

∫cos3(2x) dx

︸︷︷︸case 2: odd cosine

Now we solve the last two integrals, and later we’ll plug this back into the bigformula we just had.

We do

∫cos2(2x) dx in two slightly different ways:

Starting from scratch:

∫cos2(2x) dx =

∫1

2(1 + cos(4x)) dx

=1

2

(x+

1

4sin(4x)

)

Using Integral formula 64

∫cos2(u) du =

1

2u+

1

4sin(2u)

︸︷︷︸F (u)∫

cos2(2x) dx =1

2F (2x) =

1

2

(1

2(2x) +

1

4sin(2(2x))

)

Now we do

∫cos3(2x) dx following the Case 2, odd cosine rules:

u = sin(2x), du = 2 cos(2x) dx,1

2du = cos(2x) dx, cos2(2x) = 1− sin2(2x)

∫cos3(2x) dx =

∫cos2(2x) cos(2x) dx

=

∫(1− sin2(2x)) cos(2x) dx

=1

2

∫(1− u2)du

=1

2(u− 1

3u3)

=1

2

(sin(2x)− 1

3sin3(2x)

)

Now we combine all our calculations:

1

23

(∫1− cos(2x) dx−

∫cos2(2x) dx+

∫cos3(2x) dx

)

=1

23

(x− 1

2sin(2x)− 1

2

(x+

1

4sin(4x)

)+

1

2

(sin(2x)− 1

3sin3(2x)

))

=1

23

(1

2x− 1

8sin(4x)− 1

6sin3(2x))

)


Extra Material: deriving the anti-derivative of secant

Example 8. Find the anti-derivative of secant using only a u-substitution, and alot of stubbornness.

Solution.∫

sec(x) dx =

∫1

cos(x)dx def’n of sec(x)

u = sin(x), du = cos(x) dx, dx =1

cos(x)du a desperate u-subst

=

∫1

cos(x)

1

cos(x)du applying u-subst

=

∫1

cos2(x)du

=

∫1

1− sin2(x)du Pythagorean identity

=

∫1

1− u2du applying u-subst

= 12 ln

∣∣∣∣1 + u

1− u

∣∣∣∣ using results from 7.4

= 12 ln

∣∣∣∣1 + sin(x)

1− sin (x)

∣∣∣∣ putting x back in u

We have found the anti-derivative at this point, and it wasn’t so hard. But, wehave a bit more work to do to get our answer to look like the book’s. The work isnot Calculus, it’s pre-calculus. Here it is:

= ln

√1 + sin(x)

1− sin (x)properties of ln(x)

= ln

√1 + sin(x)

1− sin (x)· 1 + sin(x)

1 + sin(x)rewriting fraction

= ln

√(1 + sin(x))2

1− sin2 (x)rewriting fraction

= ln

√(1 + sin(x))2

cos2(x)Pythagorean identity

= ln

(1 + sin(x)

cos(x)

)simplifying square root

= ln

(1

cos(x)+

sin(x)

cos(x)

)splitting fraction up

= ln(sec(x) + tan(x)) def’n of secant and tangent

7.3 Trigonometric Substitution

The basic idea here is that we reverse the usual role of u-substitution. Usually,


we set u equal to some function of x because this “covers up” some complicatedfunction. But here, we’re going to set x equal to a more complicated function (ofθ) because of the special properties of trig functions.

Example 1. Find the area under under y =√

1− x2 from x = 0 to x = 1/2.

Solution. This is the area under part of a circle, as shown,

But this area is not one quarter of the half circle. There is no elementary way tofind this area.

The area we want is given by an integral

∫ 1/2

0

√1− x2 dx.

We do a kind of backwards substitution: instead of letting u = g(x), we will letx = g(θ) where g is some trig function. We will soon give the rules of thumbfor which trig function to use, but for now, as an act of faith, we do the correctsubstitution.

Let

x = sin(θ)

dx = cos(θ) dθ

Plugging these into the above integral we obtain

∫ x=1/2

x=0

√1− sin2(θ) cos(θ) dθ.

We need to do two things to this integral: change the endpoints from x-values toθ-values, and simplify the formula on the inside. Here are the calculations,

x = 0⇒ sin(θ) = 0⇒ θ = 0

x = 1/2⇒ sin(θ) = 1/2⇒ θ = π/6√

1− sin2(θ) =√

cos2(θ) = cos(θ) ∗

Note where we put “∗”: this last equation is true as long as cos(θ) is positive. Ifcos(θ) is negative, then

√cos2(θ) = | cos(θ)|. The integral now becomes

∫ π/6

0cos(θ) cos(θ) dθ =

∫ π/6

0cos2(θ) dθ.

We look up this integral from section 7.2 and get

∫ π/6

0cos2(θ) dθ =

∫ π/6

0

12(1 + cos(2θ)) dθ


=1

2θ +

1

4sin(2θ)

]π/6

0

= π12 + 1

4 sin(π/3)− (0 + 14 sin(0))

= π12 +

√3

8

The following page summarizes the main steps in Trigonometric substitution.This is where section 1 ended on Tuesday, February 3

Example 2. Find

∫ √4− x2 dx.

Solution. This problem is very similar to the previous one: the differences are thatwe have “4” instead of “1” under the square root, and we need to find a generalanti-derivative, not get a definite integral and plug in numbers.

x = 2 sin(θ)

dx = 2 cos(θ) dθ

Plugging these into the above integral we obtain

∫ √4− (2 sin(θ))2 2 cos(θ) dθ =

∫ √4− 4 sin2(θ) 2 cos(θ) dθ

=

∫2

√1− sin2(θ) 2 cos(θ) dθ

= 4

∫ √cos2(θ) cos(θ) dθ

= 4

∫cos(θ) cos(θ) dθ

= 4

∫cos2(θ) dθ

= 4

∫12(1 + cos(2θ)) dθ

= 4 · 12(θ + 1

2 sin(2θ))

= 2(θ + 12 sin(2θ))

To finish, we just need to translate from θ back to x. This is easier if we first applythe double angle identity, sin(2θ) = 2 sin(θ) cos(θ), to get

2(θ + 12 sin(2θ)) = 2(θ + sin(θ) cos(θ))

By the original substitution, x = 2 sin(θ). This allows us to get rid of sin(θ) directlyand to get rid of θ by solving for θ = sin−1(x/2). Plugging this in we get

2(

sin−1(x/2) +x

2cos(θ)

)(partial translation back to x)

To find cos(θ) in terms of x you can draw a right triangle, label an angle as θ, theopposite side as x, the hypotenuse as 2 (this is because sin(θ) = x/2) and solve for

the missing side. You should find that cos(θ) =

√4− x2

2(by the way, it always

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ght.

θ=

sin−1(a/A

)θ

=si

n−1(x/A

)

dx

=A

cos(θ)dθ

=A

√1−

sin2(θ

)θ

=si

n−1(b/A

)co

s(θ)

=

√A

2−x2

A

=A√

cos2

(θ)

=A

cos(θ)

tan(θ

)=

x√A

2−x2

√A

2+x2

x=A

tan(θ

)√A

2+A

2ta

n2(θ

)θ

=ta

n−1(a/A

)θ

=ta

n−1(x/A

)

dx

=A

sec2

(θ)dθ

=A√

1+

tan2(θ

)θ

=ta

n−1(b/A

)si

n(θ

)=

x√A

2+x2

=A√

sec2

(θ)

=A

sec(θ)

cos(θ)

=A

√A

2+x2

√x2−A

2x

=A

sec(θ)

√A

2se

c2(θ

)−A

2θ

=se

c−1(a/A

)θ

=se

c−1(x/A

)

dx

=A

sec(θ)

tan(θ

)dθ

=A√

sec2

(θ)−

1θ

=se

c−1(b/A

)si

n(θ

)=

√x2−A

2

x

=A√

tan2(θ

)=

Ata

n(θ

)ta

n(θ

)=

√x2−A

2

A


works out that the missing side is the√

that you started with in the integral).Thus,

∫ √4− x2 dx = 2(θ + 1

2 sin(2θ)) = 2

(sin−1(x/2) +

x

2·√

4− x2

2

)

Example 3. Find

∫x√

1− x2 dx.

Solution. We could try x = sin(θ) as above, and in all the other problems in thissection, that is the right kind of thing to do. But not here. This problem is au-substitution. Always look for an easy u-substitution first!

u = 1− x2, du = −2x dx, −1

2du = x dx

∫x√

1− x2 dx = −1

2

∫ √u du

= −1

2

u3/2

3/2

= −1

2· 2

3(1− x2)3/2

= −1

3(1− x2)3/2

Example 4. Find

∫x3√

9 + x2 dx

Solution. We follow the substitution on the chart

x = 3 tan(θ)

dx = 3 sec2(θ)√

9 + x2 = 3 sec(θ)

We apply all of these substitutions to our integral

∫(3 tan(θ))33 sec(θ)3 sec2(θ) dθ = 35

∫tan3(θ) sec3(θ) dθ

Now we follow our rules of thumb for powers of trig functions.

u = sec(θ)

du = sec(θ) tan(θ) dθ

tan2(θ) = sec2(θ)− 1

And so our integral becomes this

∫(3 tan(θ))33 sec(θ)3 sec2(θ) dθ = 35

∫tan3(θ) sec3(θ) dθ

= 35

∫(sec2(θ)− 1) sec2(θ) tan(θ) sec(θ) dθ


= 35

∫(u2 − 1)u2 du

= 35

∫u4 − u2 du

= 35

(u5

5− u3

3

)

= 35

(1

5sec5(θ)− 1

3sec3(θ)

)

= 35

1

5

(√9 + x2

3

)5

− 1

3

(√9 + x2

3

)3

= 35

(1

5· 1

35(9 + x2)5/2 − 1

3· 1

33(9 + x2)3/2

)

=1

5(9 + x2)5/2 − 3(9 + x2)3/2

This is where section 1 ended on Wednesday, February 4

Example 5. Find

∫1√

x2 − 2dx

Solution. This integral involves√x2 −A2 where A =

√2. In other words, we can

write it like this ∫1√

x2 − (√

2)2

dx

Now we apply the appropriate substitutions

x =√

2 sec(θ)

dx =√

2 sec(θ) tan(θ) dθ√x2 − (

√2)2 =

√2 tan(θ)

and so our integral becomes

∫1√

2 tan(θ)

√2 sec(θ) tan(θ) dθ =

∫sec(θ) dθ

= ln | sec(θ) + tan(θ)|

= ln

∣∣∣∣∣x√2

+

√x2 − 2√

2

∣∣∣∣∣

= ln

∣∣∣∣∣x+√x2 − 2√2

∣∣∣∣∣

= ln |x+√x2 − 2| − ln |

√2|

= ln |x+√x2 − 2|

where, in the last line, we absorbed ln |√

2| into our “+C”.


Extra examples

Example 6. Find

∫ 2/3

√2/3

1√9x2 − 1

dx.

Solution. The first thing we have to do to this integral is rewrite it in such a way

that we can see what A is for√x2 −A2. We do this by factoring out the 9 from

the square root √9x2 − 1 =

√9(x2 − 1

9) = 3√x2 − 1

9

So now we make our substitution

x =1

3sec(θ)

dx =1

3sec(θ) tan(θ)

and we plug these into our original integral∫

1

3√x2 − 1

9

dx =1

3

∫1√

(13 sec(θ))2 − 1

9

· 1

3sec(θ) tan(θ) dθ

=1

3

∫1√

19 sec2(θ)− 1

9

· 1


=1

3

∫1

13

√sec2(θ)− 1

· 1


=1

3

∫1

13

√tan2(θ)

· 1


=1

3

∫1

��13�

��tan(θ)·��1

3sec(θ)��tan(θ) dθ

=1

3

∫sec(θ) dθ

=1

3ln | sec(θ) + tan(θ)|

To finish the problem, we need either need to change the original endpoints,

∫ 2/3

√2/3

into the correct values for θ, or convert back to x’s and then plug in the numbers.We do the former:

x =2

3⇒ 2

3=

1

3sec(θ)⇒ sec(θ) = 2⇒ cos(θ) = 1/2⇒ θ = π/3

x =

√2

3⇒√

2

3=

1

3sec(θ)⇒ sec(θ) =

√2⇒ cos(θ) = 1/

√2⇒ θ = π/4

Now we finish the integral

1

3ln | sec(θ) + tan(θ)|

]π/3

π/4

=1

3(ln | sec(π/3) + tan(π/3)| − ln | sec(π/4) + tan(π/4)|)

=1

3

(ln |2 +

√3| − ln |

√2 + 1|

)


7.4 Integration of Rational Functions

A rational function is one of the form

polynomial

polynomial

In this section we will cover three techniques to integrate them: polynomial division,partial fractions, and completing the square.

Here are the integrals of the simplest rational functions. The first four we knowalready, the last two we’ll find in examples.

∫1

xdx = ln |x|

∫1

ax± b dx =1

aln |ax± b| (u-subst)

∫1

x2 + 1= tan−1(x)

∫x

x2 ± a dx =1

2ln |x2 ± a| (u-subst)

∫1

x2 + a2dx =

1

atan−1

(xa

)(justified below)

∫1

x2 − a2dx =

1

2aln

∣∣∣∣x− ax+ a

∣∣∣∣ (justified below)

Example 1. Find

∫1

x2 + a2dx.

Solution. We rewrite the fraction so that it looks like1

u2 + 1.

∫1

x2 + a2dx =

∫1

a2(x2

a2+ 1) dx

=1

a2

∫1

(xa

)2+ 1

dx

=1

a2

∫1

u2 + 1a du u =

x

a, du =

1

adx

=a

a2

∫1

u2 + 1du

=1

atan−1(u)

=1

atan−1(x/a)

Example 2. Verify the fact that1

x2 − a2=

1/(2a)

x− a −1/(2a)

x+ aand use this to find

∫1

x2 − a2dx.

Solution. To verify the first statement, we take the fractions on the right, simplifythem, and combine them with a common denominator

1/(2a)

x− a −1/(2a)

x+ a=

1

2a

(1

x− a −1

x+ a

)

=1

2a

(x+ a

(x− a)(x+ a)− x− a

(x+ a)(x− a)

)


=1

2a

((x+ a)− (x− a)

x2 − a2

)

=1

2a

(2a

x2 − a2

)

=1

x2 − a2

Now that we have verified the first statement, we can use it to find the given integral.∫

1

x2 − a2dx =

∫1/(2a)

x− a −1/(2a)

x+ adx

=1

2a

∫1

x− a −1

x+ adx

=1

2a(ln |x− a| − ln |x+ a|)

=1

2aln

∣∣∣∣x− ax+ a

∣∣∣∣


Strategy for integrating rational functions:

Given an integral of the form

∫polynomial

polynomial.

• If you can split the fraction up at + and/or − in the numerator and get abasic anti-derivative, then do this.• If the degree on top is ≥ the degree on the bottom, then do polynomial

division.• If you can factor the bottom then do partial fractions.• If you cannot factor the bottom, and the bottom is a quadratic, then complete

the square.

Example 3. Find ∫2x+ 3

x2 − 9dx

Solution.∫

2x+ 3

x2 − 9dx = 2

∫x

x2 − 9dx− 3

∫1

x2 − 9dx

= 2 · 1

2ln |x2 − 9| − 3 · 1

6ln

∣∣∣∣x− 3

x+ 3

∣∣∣∣

The degree of a polynomial is the biggest power of x that appears (or wouldappear if you multiplied it all out). For rational functions, the top and bottom ofthe fractions are polynomials, and so the top and bottom have degrees.

For instance

• 4

x2 − x+ 1, degree of top = 0, degree of bottom = 2.

• x3 − x+ 1

x2(x+ 1)3, degree of top =3, degree of bottom = 5 (if you multiply everything

out).


We start by recalling long division of numbers.

123

9→

9 ) 1 2 3→

1

9 ) 1 2 39

→

1 3

9 ) 1 2 39

3 32 7

6

→ 123

9= 13 +

6

9.

At each step of long division, we put on number on top. We choose the top so thatwhen we multiply it by the number on the side, we can subtract as much as possibleaway from the numbers inside.

Now we do the same thing with polynomials.

2x2 + 17x+ 2

2x+ 1→

2x+ 1 ) 2x2 + 17x+ 2→

2x

2x+ 1 ) 2x2 + 17x+ 2

2x2 + x

→

2x+ 8

2x+ 1 ) 2x2 + 17x+ 2

−(2x2 + x)

16x+ 2−(16x+ 8)

−6

→ 2x2 + 17x+ 2

2x+ 1= 2x+ 8 +

−6

2x+ 1

At each step of long division, we put a monomial on top. We choose the top sothat when we multiply it by the polynomial on the side, we can subtract as muchas possible away from the polynomial inside.

Example 4. Use polynomial division to find

∫x4 + x3 − 2x2 + 17x+ 2

x2 + 1dx.

Solution. We will first put a x2 on top, because multiplying this by x2 + 1 on theside will allow us to kill the x4 underneath (note, we need to keep track of the x3

column, so either leave that column black, or write “0x3” in it):

x2

x2 + 1 ) x4 + x3− 2x2 + 17x+ 2

−(x4 + x2)

x3− 3x2

Next, we put x on top because when we multiply this by x2 + 1 we can kill off thex3. Then we put −3 on top (but, you can’t see that −3 is the correct thing untilyou’ve finished the step involving the x on top though)

x2 + x − 3

x2 + 1 ) x4 + x3 − 2x2 + 17x+ 2

−(x4 + x2)

x3 − 3x2 + 17x

−(x3 + x)

− 3x2 + 16x+ 2

− (− 3x2 − 3)

16x+ 5


Continuing in this way, we find that the quotient is x2 + x − 3 and the remainderis 16x+ 5. In other words

x4 + x3 − 2x2 + 17x+ 2

x2 + 1= x2 + x− 3 +

16x+ 5

x2 + 1.

Now we can find the integral

∫x4 + x3 − 2x2 + 17x+ 2

x2 + 1dx =

∫x2 + x− 3 +

16x+ 5

x2 + 1dx

=

∫x2 + x− 3 dx+ 16

∫x

x2 + 1dx+ 5

∫1

x2 + 1dx

=x3

3+x2

2− 3x+ 8 ln |x2 + 1|+ 5 tan−1(x)

Here’s a fairly general description of partial fractions.

1. Make sure that you havepoly

polywith the degree top < degree bottom.

2. Factor the bottom into linear and quadratic factors.3. Setting up the fractions: this needs a case-by-case description.

Distinct linear factors (each factor on the left appears as a term on theright: the numbers “1” and “2” could be any numbers, including 0)

∗(x+ 1)(x+ 2) · · ·︸︷︷︸

could be more here

=A

x+ 1+

B

x+ 2+ . . .︸︷︷︸

could be more here

.

Repeated linear factors (with 4 as an example of how many times (x+ 1)is repeated)

∗(x+ 1)4

︸︷︷︸the power here, 4, equals . . . . . . the number of terms here

(x+ 2) . . .=

A

x+ 1+

B

(x+ 1)2+

C

(x+ 1)3+

D

(x+ 1)4

︸︷︷︸+

E

x+ 2+. . .

Distinct quadratic factors

∗(x2 + 2x+ 3)(x2 + 4x+ 5) · · ·︸︷︷︸

could be more here

=Ax+B

x2 + 2x+ 3+

Cx+D

x2 + 4x+ 5+ . . .︸︷︷︸

could be more here

.

Repeated quadratic factors (with 3 as the number of times (x2 + 2x+ 3)is repeated)

∗(x2 + 2x+ 3)3

︸︷︷︸the power here, 3, equals . . .

(x2 + 4x+ 5)

=Ax+B

x2 + 2x+ 3+

Cx+D

(x2 + 2x+ 3)2+

Ex+ F

(x2 + 2x+ 3)3

︸︷︷︸. . . the number of terms here

+Gx+H

x2 + 4x+ 5


Mixed linear and quadratic factors

∗(x+ 1)(x+ 2)2(x2 + 3x+ 4)(x2 + 5x+ 6)2

=A

x+ 1+

B

x+ 2+

C

(x+ 2)2+

Dx+ E

x2 + 3x+ 4+

Fx+G

x2 + 5x+ 6+

Hx+ I

(x2 + 5x+ 6)2

4. After you get the above equation set up, you multiply both sides by thedenominator from the left, and cancel all denominators. To finish there aretwo possible steps:(a) You can plug in x-values that make one of the terms on the right equal

to 0. This might allow you to solve for the other constants A, B, . . . , butit might not (it won’t if there are repeated factors or quadratics withoutroots).

(b) If the previous step doesn’t finish the problem, then you multiply ev-erything out on the right, gather together all the x-terms on the right,gather the x2-terms, the x3-terms etc. Set up a new system of equationsas follows:

constant from left = constant from right

x-coeff from left = x-coeff from right

x2-coeff from left = x2-coeff from right

This gives linear equations in A, B, C, . . . . Solve these equations in theusual way (i.e. solve one equation for one of the letters A, B, C, . . . ,substitute this into the other equations, and repeat: or use matrices andlinear algebra).

Example 5. (a) Use partial fractions to split the fraction−3x− 7

x2 + 7x+ 12into two

fractions.(b) Verify that your partial fraction solution works, by combining your answers

with a common denominator.

(c) Find

∫− 3x+ 7

x2 + 7x+ 12dx.

Solution. (a) We factor x2 + 7x+ 12 as (x+ 3)(x+ 4) and so our set up is

−3x− 7

(x+ 3)(x+ 4)=

A

x+ 3+

B

x+ 4

We multiply both sides of this equation by (x + 3)(x + 4), this cancels all thedenominators

−3x− 7

(x+ 3)(x+ 4)· (x+ 3)(x+ 4)

1=

A

x+ 3· (x+ 3)(x+ 4)

1+

B

x+ 4· (x+ 3)(x+ 4)

1

−3x− 7 = A(x+ 4) +B(x+ 3)

Now we plug in values of x that make one of the terms equal to 0

x = −4 ⇒ 12− 7 = A(0) +B(−1)⇒ 5 = −B ⇒ B = −5


x = −3 ⇒ 9− 7 = A(1) +B(0)⇒ A = 2

Thus, we know−3x− 7

(x+ 3)(x+ 4)=

2

x+ 3− 5

x+ 4

(b) Just this once, we verify that the partial fractions we obtained in the previouspart were correct.

2

x+ 3− 5

x+ 4=

2

x+ 3· (x+ 4)

(x+ 4)− 5

x+ 4

(x+ 3)

(x+ 3)

=2x+ 8− (5x+ 15)

(x+ 3)(x+ 4)

=−3x− 7

x2 + 7x+ 12

(c) We use the partial fractions to finish the integral

∫− 3x+ 7

x2 + 7x+ 12dx =

∫ −3x− 7

x2 + 7x+ 12dx

=

∫2

x+ 3− 5

x+ 4dx

= 2 ln |x+ 3| − 5 ln |x+ 4|

This is where section 1 ended on Tuesady, February 10

Example 6. Find

∫x2 + 3x+ 1

x3 + xdx.

Solution. We start by factoring the bottom, and then setting up the partial frac-tions using the factors of the bottom

x2 + 3x+ 1

x(x2 + 1)=A

x+Bx+ C

x2 + 1.

Multiplying both sides by x(x2 + 1) we get:

x2 + 3x+ 1 = A(x2 + 1) + (Bx+ C)x

It is true that we could plug in x = 0 and solve for A, but we can’t do this againand solve for B and C, so we’ll combine both approaches.

If we plug in x = 0 we gen

x = 0⇒ 1 = A(1) + 0 =⇒ A = 1

Now we combine this with the equation involving B and C

x2 + 3x+ 1 = 1(x2 + 1) + (Bx+ C)x

We start by multiplying everything out, and then gathering terms on the right handside.

x2 + 3x+ 1 = 1(x2 + 1) + (Bx+ C)x


= x2 + 1 +Bx2 + Cx

x2 + 3x+ 1 = (1 +B)x2 + Cx+ 1

Now we set up new equations by making the coefficients from one side equal to thecoefficients from the other side.

x2 coeffs : 1 = 1 +Bx coeffs : 3 = C

constants : 1 = 1.

This gives us A = 1, B = 0 and C = 3. Now we can integrate

∫x2 + 3x+ 1

x(x2 + 1)dx =

∫1

x+

3

x2 + 1dx

= ln |x|+ 3 tan−1(x)

Completing the square is a trick to turn something of the form x2 + ax+ b into(x + c)2 + d. The following pattern works only when the coefficient of x2 is 1 (ifit’s not 1, then factor this coefficient out of the quadratic and complete the squareinside the parentheses).

x2+8x+5 = x2+ 8x + 5÷2 ↓

4∧ 2−−−→16

= x2+ 8x +16 − 16+5÷2 ↓ ↖↗

4∧ 2−−−→ 16

= (x+4)2−11

In short: Take half of the x-coefficient, square this, add and subtract the result intothe formula. Then the first three terms (the x2 term, the x term and the part thatwe added) equal (x+ c)2.

Example 7. Use completing the square to find

∫1

x2 + 6x+ 7dx.

Solution.x2+ 6x +7÷2 ↓

3∧ 2−−−→9

→x2+ 6x +9 − 9+7÷2 ↓ ↖↗

3∧ 2−−−→ 9

Note that x2 + 6x+ 9 = (x+ 3)2, and simplify −9 + 7 to −2 to get

x2 + 6x+ 7 = (x+ 3)2 − 2

Now we can finish the integral∫

1

x2 + 6x+ 7dx =

∫1

(x+ 3)2 − 2dx

=

∫1

u2 − (√

2)2du (u = x+ 3)

=1

2√

2ln

∣∣∣∣∣u−√

2

u+√

2

∣∣∣∣∣

=1

2√

2ln

∣∣∣∣∣x+ 3−

√2

x− 3 +√

2

∣∣∣∣∣


Extra examples

Example 8. [Stewart, 6e, #4b] Set up the partial fraction expansion for

2x+ 1

(x+ 1)3(x2 + 4)2

Solution.

2x+ 1

(x+ 1)3(x2 + 4)2=

A

x+ 1+

B

(x+ 1)2+

C

(x+ 1)3+Dx+ E

x2 + 4+

Ex+ F

(x2 + 4)2

Example 9. [Stewart, 6e, #17] Find

∫ 2

1

4y2 − 7y − 12

y(y + 2)(y − 3)dy

Solution.

4y2 − 7y − 12

y(y + 2)(y − 3)=A

y+

B

y + 2+

C

y − 3

4y2 − 7y − 12 = A(y + 2)(y − 3) +By(y − 3) + Cy(y + 2)

y = 0 =⇒ −12 = A(2)(−3) +B(0) + C(0) =⇒ −12 = −6A =⇒ A = 2

y = −2 =⇒ 4(4) + 14− 12 = A(0) +B(−2)(−5) + C(−2)0 =⇒ 18 = 10B =⇒ B = 9/5

y = 3 =⇒ 4(9)− 21− 12 = A(0) +B(0) + C(3)(5) =⇒ 3 = 15C =⇒ C = 1/5∫ 2

1

4y2 − 7y − 12

y(y + 2)(y − 3)dy =

∫ 2

1

2

y+

9/5

y + 2+

1/5

y − 3dy

= 2 ln |y|+ 9

5ln |y + 2|+ 1

5ln |y − 3|

]2

1

= 2 ln |2|+ 9

5ln |4|+ 1

5ln | − 1| − (2 ln |1|+ 9

5ln |3|+ 1

5ln | − 2|)

= 2 ln |2|+ 9

5ln |4| − 9

5ln |3| − 1

5ln |2|)

Example 10. Find

∫2x3 − 3x+ 7

x+ 1dx

Solution. We start with polynomial division. Note how we write “0x2” in one ofthe columns to help keep things lined up.

x+ 12x2− 2x − 1)2x3+ 0x2−3x+7

−(2x3+ 2x2)

−2x2−3x+7

−(− 2x2−2x)− x+7

−(− x−1)8


Now we can finish the integral

∫2x3 − 3x+ 7

x+ 1dx =

∫2x2 − 2x− 1 +

8

x+ 1dx

=2

3x3 − x2 − x+ 8 ln |x+ 1|

Example 11. Find

∫x

x2 + 4x+ 10dx

Solution. We start by completing the square

x2 + 4x+ 10 = x2 + 4x+ 4− 4 + 10 = (x+ 2)2 + 6

Now our integral becomes∫

x

x2 + 4x+ 10dx =

∫x

(x+ 2)2 + 6dx

To finish we let u = x+ 2, but there’s a problem here∫

x

u2 + 6du

We still need to get rid of that last x on top. We solve u = x + 2 for x to getx = u− 2 and so our integral becomes

∫u− 2

u2 + 6du =

∫u

u2 + 6− 2

u2 + (√

6)2du

=1

2ln |u2 + 6| − 2 · 1

2√

6tan−1

(u√6

)

=1

2ln |(x+ 2)2 + 6| − 1√

6tan−1

(x+ 2√

6

)

Example 12. Apply partial fractions to split upx

(x2 + 2x+ 2)(x2 − x+ 3)

Solution. We set up the partial fraction and clear denominators

x

(x2 + 2x+ 2)(x2 − x+ 3)=

Ax+B

x2 + 2x+ 2+

Cx+D

x2 − x+ 3

x = (Ax+B)(x2 − x+ 3) + (Cx+D)(x2 + 2x+ 2)

In the homework we would stop here, but I’ll show you how to finish it.

x = Ax3 −Ax2 + 3Ax+Bx2 −Bx+ 3B + Cx3 + 2Cx2 + 2Cx+Dx2 + 2Dx+ 2D

x = (A+ C)x3 + (−A+B + 2C +D)x2 + (3A−B + 2C + 2D)x+ (3B + 2D)

Now we equate coefficients

Left hand side Right hand side:

constants terms 0 = 3B + 2D

x-coefficients 1 = 3A−B + 2C + 2D


x2-coefficients 0 = −A+B + 2C +D

x3-coefficients 0 = A+ C

From the first equation we see that 2D = −3B and so D = −32B. From the last

equation we see that C = −A. Plugging these into the other equations (i.e. thesecond and third equations) we get

1 = 3A−B + 2(−A) + 2(−32B)

0 = −A+B + 2(−A) + (−32B)

Simplifying we get

1 = A− 4B

0 = −3A− 12B

From the second equation get B = −6A. Plugging this into the first equation weget 1 = A− 4(−6A), 1 = 25A, A = 1/25.

Now that we know A = 1/25, we get B = −6/25, C = −1/25 and D = −32−625 =

925 .

Thus, our partial fraction solution is

x

(x2 + 2x+ 2)(x2 − x+ 3)=

125x− 6

25

x2 + 2x+ 2+− 1

25x+ 925

x2 − x+ 3

Example 13. Set up (but do not solve) the following as a partial fraction:

x4 + 3x2 − 17x+ 11

x(x+ 1)2(x2 + 2x− 7)

Solution. Applying the above rules we see that we will have fractions with denom-inators of x, x+ 1, (x+ 1)2, and x2 + 2x− 7. This gives

x4 + 3x2 − 17x+ 11

x(x+ 1)2(x2 + 2x− 7)=A

x+

B

x+ 1+

C

(x+ 1)2+

Dx+ E

x2 + 2x− 7

Example 14. Set up (but do not solve) the following as a partial fraction:

x7 − 1324x2 + x− 10

(x2 + 1)(x2 + 10)4

Solution. Applying the above rules we see that we will have fractions with denom-inators of x2 + 1, x2 + 10, (x2 + 10)2, (x2 + 10)3, (x2 + 10)4. This gives

x7 − 1324x2 + x− 10

(x2 + 1)(x2 + 10)4=Ax+B

x2 + 1+Cx+D

x2 + 10+

Dx+ E

(x2 + 10)2+

Fx+G

(x2 + 10)3+

Hx+ I

(x2 + 10)4

This is where section 1 ended on Friday, February 13


Figure 7.1: List of basic anti-derivatives

∫xn dx =

xn+1

n+ 1, if n 6= −1

∫ex dx = ex

∫sin(x) dx = − cos(x)

∫cos(x) dx = sin(x)

∫sec2(x) dx = tan(x)

∫1

1 + x2dx = tan−1(x)

∫1

xdx = ln |x|

∫sec(x) tan(x) dx = sec(x)

∫tan(x) dx = ln | sec(x)|

∫sec(x) dx = ln | sec(x) + tan(x)|

∫1√

1− x2dx = sin−1(x)

∫ln(x) dx = x ln |x| − x

∫1

x+ adx = ln |x+ a|

∫x

x2 + adx =

1

2ln |x2 + a|

∫1

x2 − a2dx =

1

2aln

∣∣∣∣x− ax+ a

∣∣∣∣∫1

x2 + a2dx =

1

atan−1

(xa

)

7.5 Strategy for Integration

Overview• First: Check if you can use any formula in Figure 7.1, maybe after rewriting.• Second: Check, and double check for u-substitution. Maybe u-substitution

will turn the integral into one on the list in Figure 7.1. Look for an insidefunction whose derivative uses all the other x’s on the outside.• The rest are in no particular order: classify the integral by type:

◦ Integration by parts: you have an integral of the product of two functions

(or are in a special case like

∫ln,

∫tan−1, etc). The integral should

become easier if you take the derivative of one function and the anti-derivative of the other.◦ Trigonometric powers: you have an integral with powers of sin and cos,

or of tan and sec (or can rewrite it that way).

◦ Trigonometric substitution: you have an integral containing√±x2 ±A2

(unless u-substitution will work).

◦ Rational functions: you have

∫polynomial

polynomial. Check splitting it up, check

degree, poly division, partial fractions, completing the square.• Try again. Some common tricks: rationalizing substitutions, like u =

√x,

u =√ax+ b, completing the square within a square root, multiplying by the

conjugate, more trig identities.test

Practicing: When you’re practicing this strategy, it’s probably a good idea notto do the whole integral. If you do the whole integral, then you’ll spend a lot oftime practicing partial fractions, for example, instead of practicing how to see thatpartial fractions will be needed.

Example 1. [Integrals mostly from Stewart] For each of the following integrals,


indicate what technique should be used, at least as a starting point. Your answershould be one or more of the following: BA, AS/R, US, etc. Perhaps some workwill need to be done to figure out the strategy, but in all cases stop before doingany integral of any kind. Also stop before doing any partial fraction or polynomialdivision.

BA Basic anti-derivativesAS/R Algebraic simplification or

rewriting.US u-substitution.TP Trigonometric powers.IBP Integration by parts.PF Partial fractions.

PD Polynomial division.CS Completing the square.TS Trigonometric substitution.RS Rationalizing substitution.MC Multiplying by Conjugate.NOA None of the above.

(a)

∫tan3(x)

cos3(x)dx

(b)

∫e√xdx

(c)

∫x5 + 1

x3 − 3x2 − 10xdx

(d)

∫dx

x√

ln(x)

(e)

∫ √1− x1 + x

dx

(f)

∫e2 dx

(g)

∫x− 1

x2 − 4x+ 5dx

(h)

∫x2

1− x2dx

(i)

∫ √2x− 1

2x+ 3dx

(j)

∫sin(x) + sec(x)

tan(x)dx

(k)

∫x8 sin(x) dx

(l)

∫ √3− 2x− x2 dx

(m)

∫x− 1

x2 − 4x− 5dx

(n)

∫x

1− x2 +√

1− x2dx

Solution. (a) AS/R and TP. Specifically, we can rewrite this as

∫tan3(x) sec3(x) dx.

(b) US and/or RS. Specifically, u =√x, so du =

1

2√xdx, dx = 2

√x du. This

would give

∫eu2√x du. Replace the last

√x with u to get

∫eu2u du.

(c) PD followed by PF. Specifically, after you do polynomial division, the remain-

der will be of the form∗

x3 − 3x2 − 10x. Factor this to get

∗x(x− 5)(x+ 2)

.

(d) US. Specifically, u = ln(x), du =1

xdx, and the integral becomes

∫1√udu.

(e) AS/R followed by BA and US (I think it’s OK if you thought this was AS/Rfollowed by TS, read on to see why it’s really BA or US and not TS). Specifi-cally, multiply the top and bottom by

√1− x (this isn’t as random as it may

seem: it’s a common trick with square roots and fractions to multiply by the

conjugate). This gives us

√1− x√1 + x

·√

1− x√1− x =

1− x√1− x2

. To finish this, split


the fraction in half: ∫1√

1− x2dx−

∫x√

1− x2

Note that the first integral is sin−1(x) and that the second is a simple u-substitution, u = 1− x2.

(f) BA. Specifically, note that e2 is just a constant number.(g) CS. Note that we can’t easily factor the bottom.(h) PD followed by BA. Note that we have to start with this since the degree of

the top equals the degree of the bottom. Note that once we do polynomial

division, the remainder becomes∗

1− x2. If “∗” is a constant, or an x or a

constant plus x, we can still get the integral from Table ??.(i) RS, followed by PD. Here are some details:

u =√

2x− 1

du =1

2(2x− 1)−1/2 · 2 dx =

1√2x− 1

dx

dx =√

2x− 1 du

Our integral becomes

∫ √2x− 1

2x+ 3

√2x− 1 du =

∫u2

2x+ 3du. To get rid of

all the x’s solve for backwards substitution: u =√

2x− 1 ⇒ x =1

2(u2 + 1).

Thus we get

∫u2

(u2 + 1) + 3du =

∫u2

u2 + 4du.

Now apply PD.

(j) AS/R followed by BA. Specificallysin(x)

tan(x)+

sec(x)

tan(x)= cos(x) + csc(x). You

know

∫cos and you should look up

∫csc.

(k) IBP (actually probably tabular integration by parts).(l) CS and then TS. Factor out the negative from in front of x2 like so

√−(x2 + 2x− 3)

then complete the square inside the parentheses√−(x2 + 2x+ 1− 1− 3) =√

−((x+ 1)2 − 4

)=√−(x+ 1)2 + 4. Let u = x + 1 and this becomes∫ √

4− u2 du.

(m) PF. Factorx− 1

(x− 5)(x+ 1).

(n) AS/R and US. There are different ways you can do this one, but they allamount to the same thing: u-substitution with some algebra and cancelling.Here’s one approach.

u =√

1− x2

du =1

2√

1− x2· (−2x) dx

−√

1− x2

xdu = dx


Plug this in for dx in the original integral (don’t bother rewriting the integralthis time)

∫x

1− x2 + udx = −

∫x

1− x2 + u

u

xdu

= −∫

u

1− x2 + udu

= −∫

u

u2 + udu

= −∫

1

u+ 1du



7.8 Improper Integrals

The word “improper” here just means that

∫ b

af(x) dx has one (or more) of the

following:• a = −∞, or• b =∞, or• f(x) has a vertical asymptote (VA) in the interval [a, b] (i.e. we have y-values

approaching ±∞).Perhaps it’s best to start with a quick review of vertical and horizontal asymp-

totes.

limx→∞

f(x) = L means that f(x) has a right side horizontalasymptote of y = L

limx→−∞

f(x) = L means that f(x) has a left side horizontalasymptote of y = L

limx→a+

f(x) = ±∞ means that f(x) has a vertical asymptote onthe right of x = a

limx→a−

f(x) = ±∞ means that f(x) has a vertical asymptote onthe left of x = a

Pretty much all the basic functions that you know that have vertical asymptotesare shown below:

• limx→0+

1

x=∞, i.e.

1

xhas a V.A. at x = 0. Notational shortcut: “

1

0+=∞”.

• limx→c

f(x)

g(x)= ±∞ if g(c) = 0 and the top does not equal 0. I.e.

f(x)

g(x)has

a V.A. if we have ÷ by 0 on bottom but not on top. Notational shortcut:

“#(6= 0)

0= ±∞”.

• limx→0+

ln(x) =∞, i.e. ln(x) has a V.A. at x = 0. Notational shortcut: “ln(0) =

∞”.


• limx→π/2

tan(x) = ∞ and limx→−π/2

tan(x) = −∞. I.e. tan(x) has a V.A. at

x = ±π/2 (this is actually ÷ by 0 since cos(±π/2) = 0). Notational shortcut“tan(±π/2) = ±∞”.

Pretty much all the basic functions that you know that have horizontal asymp-totes are shown below (as well as three functions that don’t have them).

• limx→±∞

1

xp= 0 for any real number p > 0. Notational shortcut: “

1

∞ = 0”.

• limx→−∞

ex = 0. Notational shortcut: “e−∞ = 0”.

• limx→∞

tan−1(x) = π/2, limx→−∞

tan−1(x) = −π/2. Notational shortcut: “tan−1(±∞) =

±π/2”.• limx→∞

ln(x) =∞. Notational shortcut: “ln(∞) =∞”.

• limx→∞

ex =∞. Notational shortcut: “e∞ =∞”.

• limx→∞

√x =∞. Notational shortcut: “

√∞ =∞”.

Definition. Suppose the integral

∫ b

af(x) dx involves one or more of the following:

a = −∞, b = ∞, a vertical asymptote at c with a ≤ c ≤ b, and suppose that F (x)

is an anti-derivative of f(x). Then we define

∫ b

af(x) dx as follows:

•∫ b

−∞f(x) dx = F (b)− lim

x→−∞F (x).

•∫ ∞

af(x), dx = lim

x→∞F (x)− F (a).

•∫ ∞

−∞f(x) dx =

∫ c

−∞f(x) dx+

∫ ∞

cf(x) dx where c satisfies −∞ < c <∞

• If x = b is a VA then

∫ b

af(x) dx = lim

x→bF (x)− F (a).

• If x = a is a VA then

∫ b

af(x) dx = F (b)− lim

x→aF (x).

• If x = c is a VA and c and a < c < b, then then

∫ b

af(x) dx =

∫ c

af(x) dx +

∫ b

cf(x) dx.

If, in any of the previous definitions, a limit does not exist (including the case wherethe limit is ±∞), we say the integral is divergent. If each limit exists (and is finite),we say that the integral is convergent. For those cases where one integral is splitinto two integrals, the first integral is convergent if and only if both of of the newintegrals are convergent.

Example 1. Find

∫ ∞

0e−x dx.

Solution.

∫ ∞

0e−x dx = lim

b→∞−e−x

]b

0

= −(

limb→∞

e−x − e0

)


= −(e−∞ − 1)

= 1

Example 2. Find

∫ 1

−2

1

x2dx.

Solution. We do this problem twice: once the WRONG way, and once the correctway.

WRONG WAY

∫ 1

−2

1

x2dx = −1

x

]1

−2

= −(

1

1− 1

−2

)

= −(1 + 1/2)

= −3/2

Can you see how we can tell that this answer must be wrong? Here is the graph ofthis function:

This graph is always positive, so the area under the curve has to be positive, andso there’s no way the integral should be negative. If this integral is defined, it hasto be positive, but we’ll see in a moment that it’s not defined. Here’s the right wayto do this.

CORRECT WAY

∫ 1

−2

1

x2dx =

∫ 0

−2

1

x2dx+

∫ 1

0

1

x2dx

= limb→0−

−1

x

]b

−2

+ lima→0+

−1

x

]1

0

= −(

limb→0−

1

b− 1

−2

)−(

1

1− lima→0+

1

a

)

= −(−∞+1

2)− (1−∞))


=∞− 1

2− 1 +∞

=∞

Since the integral comes out to be infinite, we say that the integral diverges.


Challenge. Can you come up with an example, like the previous one but wherethe integral, when done correctly, converges?

Example 3. Find the volume of the shape known as Gabriel’s Horn. This is

defined by rotating f(x) =1

xaround the x-axis, from x = 1 to x =∞.

Solution. We start by picturing the original region, and the volume generated byrotating it.


I hope the pictures stimulate you to think “volume by disks”. We slice the volumeinto disks, moving along the x-axis. Thus, the volume is given by

V =

∫ ∞

1π

(1

x

)2

dx.

Using the techniques for improper integrals that we have learned, this integral isfairly straight forward,

∫ ∞

1π

(1

x

)2

dx = π

∫ ∞

1

1

x2dx

= π limb→∞

−1

x

]b

1

= π( limb→∞

−1

b− −1

1)

= π(−1

∞ + 1)

= π(0 + 1)

= π

This example shows a physical solid, that is infinitely long, but has a finitevolume. Here’s a nice way to think about this: if you turned this horn vertically,you could fill it with liquid, using a finite amount of liquid.

Example 4. Figure out which of the following converge:

(a)

∫ ∞

1

1

xdx

(b)

∫ ∞

1

1

x1.5dx

(c)

∫ ∞

1

1

x0.7dx

Solution. (a)

∫ ∞

1

1

xdx = lim

b→∞ln(x)

]b

1

= limb→∞

ln(b)− ln(1)

=∞− 0

=∞

So in this case, the integral is divergent.

(b)

∫ ∞

1

1

x1.5dx =

∫ ∞

1x−1.5 dx

= limb→∞

x−0.5

−0.5

]b

1


=1

−0.5

(limb→∞

b−0.5 − 1

)

=1

−0.5

(limb→∞

1

b0.5− 1

)

=1

−0.5

(1

∞ − 1

)

=1

−0.5(0− 1)

=1

0.5

So in this case, the integral is convergent.

(c)

∫ ∞

1

1

x0.7dx =

∫ ∞

1x−0.7 dx

= limb→∞

x0.3

0.3

]b

1

=1

0.3

(limb→∞

b0.3 − 1

)

=1

0.3(∞− 1)

=∞

So in this case the integral is divergent.

Example 5. For which values of p does

∫ ∞

1

1

xpdx converge?

Solution. We saw in the previous example that the integral diverges for p = 1.For p > 1 we look at case (b) and see the same calculations would work in general:

the anti-derivative will come out to be1

xp−1, with p− 1 > 0, and this function has

a horizontal asymptote. So the limit will exist, and the integral converges.For p < 1 we look at case (c) and see that the same calculations will work

in general: the anti-derivative will come out to be x1−p with 1 − p > 0, and thisfunction does not have an horizontal asymptote. So the limit will not exist, and theintegral diverges. In conclusion:

∫ ∞

1

1

xpdx =

{convergent if p > 1

divergent if p ≤ 1

Example 6. Figure out which of the following converge.

(a)

∫ 1

0

1

xdx

(b)

∫ 1

0

1

x1.3dx

(c)

∫ 1

0

1

x.6dx


Solution. (a)

∫ 1

0

1

xdx = lim

a→0ln(x)

]1

a

= ln(1)− lima→0

ln(a)

= 0− (−∞)

=∞


(b)

∫ 1

0

1

x1.3dx =

∫ 1

0x−1.3 dx

= lima→0

x−0.3

−0.3

]1

a

=1

−0.3

(1− lim

a→0a−0.3

)

=1

−0.3

(1− lim

a→0

1

a0.3

)

=1

−0.3

(1− 1

0

)

=1

−0.3(1−∞)

= ±∞


(c)

∫ 1

0

1

x0.6dx =

∫ 1

0x−0.6 dx

= lima→0

x0.4

0.4

]1

a

=1

0.4

(1− lim

a→0a0.4)

=1

0.4(1− 0)

=1

0.4

So in this case the integral is convergent.

Example 7. Figure out which values of p make the following integral converge.

∫ 1

0

1

xpdx


Solution. We saw in the previous example that

∫ 1

0

1

xdx is divergent.

For p > 1 we look at case (b) and see the same calculations would work in

general: the anti-derivative will come out to be1

xp−1, with p − 1 > 0, and this

function has a vertical asymptote. So the limit will not exist, and the integraldiverges.

For p < 1 we look at case (c) and see that the same calculations will work ingeneral: the anti-derivative will come out to be x1−p with 1−p > 0, and this functioncan be evaluated at x = 0. So the limit will exist and the integral converges.

In conclusion: ∫ ∞

1

1

xpdx =

{convergent if p > 1

divergent if p ≥ 1

Example 8. Find the following, or show that it diverges

∫ 3

1(x− 2)−1/3 dx

Solution. Note: you have to spot what the problem is with this example, it’s notmade explicit. You have to remember that the negative power means that you have

1

(x− 2)1/3, and then you have to see that this could give division by 0 when you

plug in x = 2. This means that there is a vertical asymptote at x = 2, and so wehave to split the integral up there:

∫ 3

1

1

(x− 2)1/3dx =

∫ 2

1

1

(x− 2)1/3dx+

∫ 3

2

1

(x− 2)1/3dx

Now we do a very simple u-substitution, u = x− 2 and get

=(x− 2)2/3

2/3

]2

1

+(x− 2)2/3

2/3

]3

2

=3

2

(limx→2

(x− 2)2/3 − (1− 2)2/3 + (3− 2)2/3 − limx→2

(x− 2)2/3)

=3

2(0− 1 + 1− 0)

= 0

Note, this integral comes out to be zero since (x − 2)−1/3 is odd, and the area onthe left, below the x-axis cancels the area on the right, above the x-axis.


Chapter 8

Further Applications ofIntegration

This chapter has a bunch of applications of integration. Unfortunately we are onlygoing to learn two of them: arc-length and probability. The first of these feels to melike a lot of other applications we’ve done: area, volume, now length. But the secondis quite different, and forms the basis of the subject known as Statistics. Statisticsin turn is used in economics, medical research, psychology, biology, agriculture, etc.(actually, every empirical subject).

8.1 Arc Length

Example 1. Suppose a ball is thrown from the ground and has a path given byy = −x2 + 4.

(a) Using three straight line segments, find an approximation for how far the ballhas travelled along its path.

(b) Translate your approximation into an integral, and calculate the integral.

Solution. (a) We’ll approximate this in three steps, using the same ∆x each time.Thus ∆x = 4/3 and our x-values are

x0 = −2, x1 = −2 +4

3= −2/3, x2 =

−2

3+

4

3= 2/3, x3 =

2

3+

4

3= 2.

The y-values corresponding to these x-values are found by plugging in the x-valuesto the formula for the parabola. Thus our y-values are

y0 = 0, y1 = −(−2/3)2 + 4 = 32/9, y2 = −(2/3)2 + 4 = 32/9, y3 = 0

Now we connect these points with line segments

77

CHAPTER 8. FURTHER INTEGRAL APPLICATIONS 78

(−2, 0)

(−2/3, 32/9) (2/3, 32/9)

(2, 0)

We calculate the distance along these straight lines. The distance ` at each stepwill be given by the distance formula

∆x

∆y`

=⇒ ` =√

∆x2 + ∆y2

So we have:

Arc-length ≈√

(4/3)2 + (32/9)2 +√

(4/3)2 + 02 +√

(4/3)2 + (32/9)2

= 8.928

(b) To get an exact answer we need to figure out how to replace each of thosesquare roots by something of the form ∗ ·∆x. If we can do this, then we will replace

the sum we had before with an integral,

∫ 2

−2∗ dx. Here’s the trick: to rewrite each

of those square roots as ∗ ·∆x, factor the ∆x2 out of the square root

√∆x2 + ∆y2 =

√1 +

∆y2

∆x2·∆x.

Now, take lim∆x→0

in the above expression, and we get

√1 +

(dy

dx

)2

dx

In other words, we should integrate the above, from x = −2 to x = 2.

In our example we havedy

dx= −2x. Thus, the exact answer should be

Arc-length =

∫ 2

−2

√1 + (−2x)2 dx

Note that the function is even, so we can integrate from 0 to 2 and multiply theresult by 2. Also, (−2x)2 equals (2x)2, so we can find

2

∫ 2

0

√1 + (2x)2 dx

Now, substitute u = 2x to get

1

2· 2∫ 4

0

√1 + u2 du


We look up this integral in the back of our book (because we’ve already doneintegrals like this in chapter 7) to get

u

2

√1 + u2 +

1

2ln(u+

√1 + u2)

]4

0

= 2√

17 +1

2ln(4 +

√17)− (0 + ln(1 + 0))

= 9.29

Definition. The arc-length of a curve y = y(x), from x = a to x = b equals

Arc-length =

∫ b

a` dx

where ` =

√1 +

(dydx

)2

Advice:• The problems in this section can be long, but it’s mostly algebra. DON’T

PANIC. Go slow and double check every step.• If the stuff the square root is a rational function try the following: (a) Look

for perfect squares, (b) foil, cancel, factor, (c) get common denominators, (d)if it’s a quadratic, complete the square.

Example 2. [Stewart, 6e, 8.1#8] Find the arc-length of the following curve

y2 = 4(x+ 4)3, 0 ≤ x ≤ 2, y > 0

Solution. We solve for y and then take the derivative

y =√

4(x+ 4)3

y = 2(x+ 4)3/2

dy

dx= 2 · 3

2(x+ 4)1/2

So now our integral is∫ 2

0

√1 + (3(x+ 4)1/2)2 dx =

∫ 2

0

√1 + 9(x+ 4) dx

=

∫ 2

0

√9x+ 37 dx

=1

9(9x+ 37)3/2 · 2

3

]2

0

=2

27

(553/2 − 373/2

)

Definition. If we are given a function x = g(y), with c ≤ y ≤ d, then arc-length isgiven by the formula

Arc-length =

∫ d

c` dy

where ` =

√1 +

(dxdy

)2


The following type of algebraic manipulation shows up a lot in these problems(not because it has anything directly to do with arc-length, but because it’s a goodtrick for making up functions that we can put inside of a square root and stillintegrate).

a2 ± 2ab+ b2 = (a± b)2 the prototype perfect square

(xn ± x−n)2 = x2n ± 2 + x−2n

(x− 1

4x

)2

= x2 − 12 +

1

16x2

(x+

1

4x

)2

= x2 + 12 +

1

16x2

1 +

(x− 1

4x

)2

= 1 + x2 − 12 +

1

16x2= x2 + 1

2 +1

16x2=

(x+

1

4x

)2

Generalizing (if the right stuff is the boxes)

1 + (2−2)2

︸︷︷︸perfect square

= 1 + 22 − 12 + 22 = 22 + 1

2 + 22 = (2 + 2)2

︸︷︷︸perfect square

Example 3. Find the arc-length defined by the curve x = 43y

3/2− 14y

1/2 from y = 1to y = 4.

Solution. We finddx

dy:

dx

dy=

d

dy43y

3/2 − 14y

1/2 = 43 · 3

2y1/2 − 1

4 · 12y−1/2 = 2y1/2 − 1

8y−1/2

Now we put this inside of the square root, and simplify, and integrate,

A.L. =

∫ 4

1

√1 +

(2y1/2 − 1

8y−1/2

)2dy

=

∫ 4

1

√1 +

(4y − 1

2 + 164y−1)dy

=

∫ 4

1

√4y + 1

2 + 164y−1 dx

=

∫ 4

1

√(2y1/2 + 1

8y−1/2

)2dy

=

∫ 4

12y1/2 + 1

8y−1/2 dy

= 43y

3/2 + 14y

1/2

]4

1

= 43 · 43/2 + 1

4 · 41/2 − (43 + 1

4)

=115

12


Challenge. Find the arc-length of the curve below

y = ln

(ex + 1

ex − 1

), 2 ≤ x ≤ 3.

(the problem is kind of long).

This is where section 1 ended on Tuesday, February 24

8.5 Probability

Comments. This section is about an application of the definite integral that isvery different from our previous ones: Calculating continuous probabilities (Theprevious applications were mostly geometric: volumes, areas, lengths, etc.). Mostof us have a little understanding finite, discrete probabilities. For instance, mostpeople understand the following questions, and can probably answer them: If youflip a coin, what is the probability that it lands on heads? If you role a single die,what is the probability that it lands on 5 or 6?

But these sorts of questions are harder to answer when the random quantity isinfinite and/or continuous. For instance, consider the following questions: Whatis the probability that a randomly chosen person has height of 6 feet? What isthe probability that a randomly chosen light bulb will burn out in 6 months? Twoanswer these questions we use probability density functions and integrals.

Definition. Let x be some characteristic of some population (e.g. x could be theheight of people, or the age of light bulbs). A probability density function f(x)is any function that satisfies the following conditions

1. f(x) ≥ 0 for all x

2.

∫ ∞

−∞f(x) dx = 1

3.The percentage of the popula-tion satisfying a ≤ x ≤ b =

∫ b

af(x) dx

Comments. Often we have a function f(x) that equals 0 except on some range ofnumbers [c, d]. In this case, condition (2) is equivalent to the following

2′.

∫ d

cf(x) = 1 (if we know that f(x) = 0 for any x outside of [c, d]).

Important points about the phrase “The percentage of the population satisfyinga ≤ x ≤ b”:

1. “percentage” here is given as a fraction of 1. For example, instead of “30%”we’ll have 0.3.

2. The percentage described equals the probability that a randomly chosen in-dividual satisfies a ≤ x ≤ b.

3. We abbreviate this phrase with the following notation: P (a ≤ x ≤ b).4. As described above, we can (and will) replace “percentage” with “probability”.

We will also replace “population” with “set of possible events”. Thus, aprobability density function describes the probability that any set of possibleevents will occur.


Comments. In this section (and in this course) we will study three kinds of prob-ability density functions:

1. The exponential density function.2. The normal distribution3. Various random, made-up ones.

Example 1. A typical jet engine lasts at most 10 years (after that it is replacedautomatically). Let p(t) be the probability density function describing t, the totallife span of a typical engine. The graph of p(t) is given below, where C is unknown.

t10

C

0.05

p(t)

(a) What is the value of C?(b) What is the probability that the jet engine breaks in its first year?(c) What is the probability that the jet engine breaks in its 10th year?(d) What is the probability that the jet engine lasts at least 5 years?

Solution. (a) We have p(t) =C − 0.05

10t+0.05. We solve for C to make

∫ 10

0p(t) dt =

1.

∫ 10

0

C − 0.05

10t+ 0.05 dt =

C − 0.05

20t2 + 0.05t

]10

0

=C − 0.05

20(100) + 0.05(10)

= (C − 0.05)5 + 0.5

(C − 0.05)5 + 0.5 = 1

(C − 0.05)5 = 0.5

(C − 0.05) =0.5

5= 0.1

C = 0.1 + 0.05 = 0.15

For the other parts of the problem, we use C = 0.15 and so p(t) becomes

p(t) = 0.01t+ 0.05

(b) This is

∫ 1

0p(t) dt.

∫ 1

00.01t+ 0.05 dt = 0.005t2 + 0.05t

]1

0

= 0.005 + 0.05

= 0.055


(c) This is

∫ 10

9p(t) dt (note that the 10th year goes from t = 9 to t = 10, in just

the same way that the 1st goes from t = 0 to t = 1)

∫ 10

90.01t+ 0.05 dt = 0.005t2 + 0.05t

]10

9

= 0.005(100) + 0.05(10)− (0.005(9) + 0.05(9))

= 1− 0.855

= 0.145

(d) This is

∫ 10

5p(t) dt (note that the phrase “lasts at least 5 years means the

same thing as its total life span is somewhere between 5 and 10 years).

∫ 10

50.01t+ 0.05 dt = 0.005t2 + 0.05t

]10

5

= 0.005(100) + 0.05(10)− (0.005(5) + 0.05(5))

= 1− 0.375

= 0.625

This is where section 1 ended on Wednesday, February 25

Comments. Now we learn about Exponential Density functions.

Definition. An exponential density function is given by

f(t) =

0 if t < 01

µe−t/µ if t ≥ 0

The number µ is called the mean of f(t).


These density functions describe things like• Waiting time in a restaurant• Waiting time on a phone• Time to the breakage of cheap manufactured items

Example 2. [Stewart, 6e, 8.5#10a] A type of light bulb is labeled as having anaverage lifetime of 1000 hours. It’s reasonable to model the probability of failureof these bulbs by an exponential density function with mean µ = 1000. Use thismodel to find the probability that a bulb

(a) fails within the first 200 hours(b) burns for more than 800 hours

Solution. (a)

∫ 200

0f(t) dt =

∫ 200

0

1

1000e−t/1000 dt

=1

1000(−1000)e−t/1000

]200

0

= −(e−200/1000 − e0

)

= 1− e−.2≈ 0.18 = 18%

(b)

∫ ∞

800f(t) dt =

∫ ∞

800

1

1000e−t/1000 dt

=1

1000(−1000)e−t/1000

]∞

800

= −(

limt→∞

e−t/1000 − e−800/1000)

= −(0− e−.8

)(note: e−∞ =

1

e∞=

1

∞ = 0)

= e−0.8

≈ 0.45 = 45%

Definition. Given any PDF f(x), we define

• the mean of f : the number µ =

∫ ∞

−∞xf(x) dx.

• the median of f : the number m that satisfies

∫ ∞

mf(x) =

1

2(you have to

solve an equation for m).

• the standard deviation of f : the number σ =

√∫ ∞

−∞(x− µ)2f(x) dx .

Example 3. Find the median of the light bulbs described in Example 2.

Solution. We solve for m to make∫ ∞

m

1

1000e−t/1000 dt =

1

2


We start by finding the integral

∫ ∞

m

1

1000e−t/1000 dt =

1

1000(−1000)e−t/1000

]∞

m

= −(

limt→∞

e−t/1000 − e−m/1000)

= −(0− e−m/1000

)

= e−m/1000

Now we set this equal to1

2

e−m/1000 =1

2

ln(e−m/1000

)= ln

(1

2

)

−m1000

= ln

(1

2

)

m = −1000 ln

(1

2

)

m = 693 hours


Definition. An normal density function (or normal distribution) is given by

f(x) =1

σ√

2πe−(x−µ)2/(2σ2)

1

σ√

2π

f(x)

µ

σ

Example 4. [Stewart, 6e, 8.5#12] According to the National Health Survey, theheights of adult males in the United States are normally distributed with mean 69.0inches and standard deviation 2.8 inches.

(a) What is the probability that an adult male chosen at random is between 65inches and 73 inches tall?

(b) What percentage of the adult male population is more than 6 feet tall?

Solution. The heights are described by

f(x) =1

2.8√

2πe−(x−69.0)2/(2(2.8)2)


(a) The problem can be interpreted as “find P (65 ≤ x ≤ 73)”. This means we

should integrate

∫ 73

65f(x) dx. We need to use our calculators to do this: there is no

basic formula for the anti-derivative. We can use the general calculator integrator1orthe calculator’s built-in normal distribution function2. In either case, we should get

∫ 73

65f(x) dx ≈ 0.847 = 84.7%

(b) We integrate

∫ ∞

72f(x) dx. Since your calculator can’t do “∞” we just use a

large value, such as 100, or 1000. You should get

∫ 100

72f(x) ≈ 0.142 = 14.2%

If you repeat the calculation with 200, you should get

∫ 200

72f(x) ≈ 0.142 = 14.2%

This shows that either 100 or 200 are reasonably close approximations of∞ for thisproblem.

This is where section 1 ended on Monday, March 9

1Enter y1 = 1/(2.8√

(2π))e∧(-(X-69)2/(2*2.82

)). Set the graphing window to something

like xmin= 60, xmax= 78, ymin= 0, ymax= 0.15. Go to 2ND , CALC , 7:∫f(x)dx , enter the

lower and upper limits.2There are two built in functions. You can enter y1 = normalpdf(X,69,2.8) in the usual

graphing window (to get “normalpdf” go to DISTR , i.e. 2ND - VARS ). Then you can finish using

7:∫f(x)dx as above. You can also calculate this directly. Enter normalcdf(65,73,69,2.8) ,

where you get “normalcdf” from DISTR .

Chapter 10

Parametric Equations and PolarCoordinates

10.3 Polar Coordinates

Comments. Polar coordinates are different way of labelling points in the plane.These labels can be described in terms of formulas, and the formulas can be used tograph functions and to do the usual Calculus things: find derivatives and integrals.

Definition. For any point in the plane, with Cartesian coordinates (x, y), we definepolar coordinates of this point as follows:

(x, y)cartesian = (r, θ)polar where r and θ satisfy

x = r cos(θ) and

y = r sin(θ)

Note that r and θ are not unique.

We can picture (r, θ). Shown below is an example where (x, y) is in QuadrantII and r and θ are chosen in a natural fashion:

(x, y)

r

θ

Fact. To convert from Cartesian to Polar coordinates, the following formulas canbe used:

r =√x2 + y2

θ =

tan−1(y/x) if (x, y) is in QI or QIV

tan−1(y/x) + π if (x, y) is in QII or QIII

±π/2 as appropriate, if (x, y) is on the y-axis

87

CHAPTER 10. PARAMETRIC AND POLAR 88

Example 1. Translate the following Cartesian coordinates into polar coordinates.(a) (−1,

√3) (Illustrate the point, the Cartesian coordinates, and the polar coor-

dinates graphically. Also, find four pairs of polar coordinates)(b) (1, 2) (Use decimals as needed)

Solution. (a) We calculate

r =

√(−1)2 + (

√3)2 =

√4 = 2

θ = tan−1(√

3) + π = 2π/3

Thus, we have

(−1,√

3)Cartesian = (2, 2π/3)polar

and we picture this information below:

√3

−1

2 2π/3

a point, no coordinates same point, Cartesian coordinates same point, polar coordinates

To get a second pair of polar coordinates we could add 2π to θ, once, twice, orany number of times:

(−1,√

3)Cartesian = (2, 8π/3)polar = (2, 14π/3)polar = . . .

We will illustrate another way of giving polar coordinates to the same point: usinga negative r. The angle θ determines a direction, and a negative value of r meansmoving in the opposite direction:

(−1,√

3)Cartesian = (−2,−π/3)polar−2

−π/3

To double check that this is correct, we simply verify the original definition of polarcoordinates: that x = r cos(θ) an y = r sin(θ). In this case, that means

−1 = −2 cos(−π/3)√

3 = −2 sin(−π/3).

(b) We calculate

r =√

12 + 22 =√

5 ≈ 2.236


θ = tan−1(2/1) = tan−1(2) ≈ 1.107 (radians)

Combining these we get

polar coordinates = (√

5, tan−1(2))

(1, 2)1

22.236

1.107

a point, no coordinates same point, Cartesian coordinates same point, polar coordinates

Example 2. Translate the following polar coordinates into Cartesian coordinatesand illustrate the results graphically.

(a) (1.5, π/3)(b) (17.1, 4.9) (Use decimals as needed.)

Solution. (a) x = 1.5 cos(π/3) = 0.75, y = 1.5 sin(π/3) = 0.75√

3,

(1.5, π/3)polar = (0.75, 0.75√

3)Cartesian

0.75

0.75√

31.5

π/3

(b) x = 17.1 cos(4.9) ≈ 3.19, y = 17.1 sin(4.9) ≈ −16.8

(17.1, 4.9)polar = (3.19,−16.8)Cartesian

3.19

−16.817.1

4.9

This is where section 1 ended on Tuesday, March 10

Definition. A polar equation is an equation relating r and θ, where (r, θ) isinterpreted in terms of polar coordinates. Usually a polar equation is given in theform r = f(θ). A polar curve is the set of all points that satisfy a polar equation.

Example 3. Consider the following polar equation r = 3.5 sin(θ). Use the graphpaper below to help in the following problems.

(a) Using the polar graph paper, plot the 6 points on the curve corresponding toθ = 0, π/10, 2π/10, 3π/10, 4π/10, 5π/10.


(b) Calculate the Cartesian coordinates of each point from part (a), and, usingthe Cartesian graph paper, plot the 6 points from part (a).

(c) Use your calculator1 to plot the curve for 0 ≤ θ ≤ π.

1π/10

2π/10

3π/10

4π/105π/10

6π/10

7π/10

8π/10

9π/10

10π/10

11π/10

12π/10

13π/10

14π/1015π/10

16π/10

17π/10

18π/10

19π/10

0.5

0.5

1

1

1.5

1.5

2

2

2.5

2.5

3

3

3.5

3.5

−3.5

−3.25

−3

−2.75

−2.5

−2.25

−2

−1.75

−1.5

−1.25

−1

−0.75

−0.5

−0.25

0

0.25

0.5

0.75

1

1.25

1.5

1.75

2

2.25

2.5

2.75

3

3.25

3.5

−3.5 −3 −2.5 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 2.5 3 3.5

Solution. (a) We make a table of values

θ r

0 0π/10 1.12π/10 2.13π/10 2.84π/10 3.35π/10 3.5

and put them on the graph paper

1To change the graphing mode to polar, go to MODE , then choose POL instead of FUNC


1π/10

2π/10

3π/10

4π/105π/10

6π/10

7π/10

8π/10

9π/10

10π/10

11π/10

12π/10

13π/10

14π/1015π/10

16π/10

17π/10

18π/10

19π/10

0.5

0.5

1

1

1.5

1.5

2

2

2.5

2.5

3

3

3.5

3.5

(b) We add two more columns to our table of values

θ r x y

0 0 0 0π/10 1.1 1.03 0.332π/10 2.1 1.66 1.213π/10 2.8 1.66 2.294π/10 3.3 1.03 3.175π/10 3.5 0 3.5

and put them on the graph paper

0

0.25

0.5

0.75

1

1.25

1.5

1.75

2

2.25

2.5

2.75

3

3.25

3.5

−2 −1.5 −1 −0.5 0 0.5 1 1.5 2

(c)


Fact. Given a polar equation r = f(θ), the Cartesian derivative equals

dy

dx=

dydθdxdθ

=f ′(θ) sin(θ) + f(θ) cos(θ)

f ′(θ) cos(θ)− f(θ) sin(θ)

To understand this, recall that

dy

dx≈ ∆y

∆x=

∆y∆θ∆x∆θ

and that

x = r cos(θ) = f(θ) cos(θ) and y = r sin(θ) = f(θ) sin(θ)

Comments. The above equation is probably too complicated to memorize, butshould be relatively easy to derive: just take the derivative of x = f(θ) cos(θ) andy = f(θ) sin(θ).

This is where section 1 ended on Wednesday, March 11

Example 4. Find the cartesian equation of the tangent line of the curve r =1− sin(θ), at θ = π/12, and graph the result to verify.

Solution. We have

y = r sin(θ)

= (1− sin(θ)) sin(θ)

dy

dθ= − cos(θ) sin(θ) + (1− sin(θ)) cos(θ)

= − cos(θ)(2 sin(θ)− 1)

x = r cos(θ)

= (1− sin(θ)) cos(θ)


dx

dθ= − cos(θ) cos(θ)− (1− sin(θ)) sin(θ)

= − cos2(θ)− (1− sin(θ)) sin(θ)

Now we plug θ = π/12 in to find the x and y coordinates, and the slope.

x(π/12) = (1− sin(π/12)) cos(π/12) ≈ 0.72

y(π/12) = (1− sin(π/12)) sin(π/12) ≈ 0.19

dy

dx

]

θ=π/12

=− cos(θ)(2 sin(θ)− 1)

− cos2(θ)− (1− sin(θ)) sin(θ)

]

θ=π/12

≈ −0.414

So the equation of the tangent line is given by

y = −0.414(x− 0.72) + 0.19

Later, we will show how to graph this equation and the polar equation in onecalculator window. For now, we just graph each of them separately, and mentallytry to combine the pictures to see that the tangent line looks right:

If you combine these pictures you get the following:


Extra Examples

We show below some familiar examples of functions in cartesian coordinates, andanalagous ones in polar coordinates:

Cartesian Polar

y = x r = θ

y = mx+ b r = mθ + b

y = x2 r = θ2

y =√x r =

√θ

y = ex r = eθ

y = sin(x) r = sin(θ)

y = 1 + cos(x) r = 1 + cos(θ)

The first five of the polar functions will produce spirals: as θ increases so does r.The other two will produce shapes that are closed: after θ goes around far enough,the shape will close and start to repeat.

In addition to functions like the above we have studied a few kinds of equations

Cartesian Polar

y = c r = c

x = c θ = c

x2 + y2 = c2 θ2 + r2 = c2

Example 5. Let r = 2θ + 1.(a) Calculate and plot 11 points from θ = 0 to θ = 2π for this function.(b) Plot this function on your calculator.


(c) Find the equation of the tangent line at 5π/8 (decimals ok).

Solution. (a) Here is the table of numbers

θ = 0 r = 1

θ = π/5 ≈ 0.63 r = 2π/5 + 1 ≈ 2.3

θ = 2π/5 ≈ 1.3 r = 4π/5 + 1 ≈ 3.5

θ = 3π/5 ≈ 1.9 r = 6π/5 + 1 ≈ 4.8

θ = 4π/5 ≈ 2.5 r = 8π/5 + 1 ≈ 6

θ = π ≈ 3.1 r = 2π + 1 ≈ 7.3

θ = 6π/5 ≈ 3.8 r = 12π/5 + 1 ≈ 8.5

θ = 7π/5 ≈ 4.4 r = 14π/5 + 1 ≈ 9.8

θ = 8π/5 ≈ 5 r = 16π/5 + 1 ≈ 11

θ = 9π/5 ≈ 5.7 r = 18π/5 + 1 ≈ 12

θ = 2π ≈ 6.3 r = 4π + 1 ≈ 14

Here is a plot of the points we just calculated.

(b) Here is what the graph looks like


(c)

y = m(x− x0) + y0

x0 = (2(5π/8) + 1) cos(5π/8) ≈ −1.89

y0 = (2(5π/8) + 1) sin(5π/8) ≈ 4.55

dy

dθ= 2 sin(θ) + (2θ + 1) cos(θ)

dx

dθ= 2 cos(θ)− (2θ + 1) sin(θ)

m =dydθdxdθ

]

θ=5π/8

=2 sin(5π/8) + (25π/8 + 1) cos(5π/8)

2 cos(5π/8)− (25π/8 + 1) sin(5π/8)

≈ 0.007

Thus, the tangent line has equation

y = 0.007(x+ 1.89) + 4.55.

If you look at the graph of r(θ) you can see that the slope looks very close to 0 atθ = 5π/8 (note that 5π/8 ≈ 1.96 which just past 3π/5 ≈ 1.88).

10.4 Areas in Polar Coordinates

Comments. In Section 10.3 we learned about polar coordinates, how to graphequations in polar coordinates, and how to take derivatives in polar coordinates. Inthis section we learn about finding integrals in polar coordinates.

Comments. Recall that the area of a circle is πr2. The area of part of a circle,


defined by θ is given below

r

θA = fraction of circle · πr2 =

θ

2ππr2 =

1

2r2θ

Definition. The area of a region defined via a polar curve r = f(θ), between θ = aand θ = b is given by

A =

∫ b

a

12r

2 dθ =

∫ b

a

12(f(θ))2 dθ

Example 1. Find the area of the circle defined by r = 3.5 sin(θ) (see Example 3).

Solution.

A =

∫ π

0

12(3.5 sin(θ))2 dθ

=3.52

2

(12θ − 1

2 sin(θ) cos(θ)) ]π

0

=3.52

22π = (3.5/2)2π (=πr2 where r = 3.5/2)


Example 2. A chocolate company is coming out with a new chocolate product. Itis going to make a mold, like a cookie cutter, given by the following shape, and fillit with chocolate. They need to know how much chocolate will go in each mold,which amounts to knowing the area of the shape.

(a) Use your calculator to calculate some points on the curve

r = −3(θ−π/2)2 +2|θ−π/2|+1,π

2−1 ≤ θ ≤ π

2+1, units for r = inches.

and then plot them on the polar graph paper in Figure 10.1.(b) Use your calculator to graph the polar function.(c) Find the area of the shape.


Figure 10.1: Polar graph paper for Example 2

0.35

0.7

π/3

1.41.75

2π/3

2.4

2.8

π

3.5

3.8

4π/3

4.5 4.9

5π/3

5.6

5.9

0.2

π/6

0.9

1.2

π/2

1.9

2.3

5π/6

3.0

3.3

7π/6

4.0

4.4

3π/2

5.1

5.4

11π/6

6.1

0.5

0.5

1

1

1.5

1.5

2

2

2.5

2.5

3

3

3.5

3.5

Solution. We put our calculators in polar mode, and enter the formula, and graphit, to get this:


To find the area we integrate∫ π

2+1

π2−1

12

(−3(θ − π/2)2 + 2|θ − π/2|+ 1

)2dθ

We perform a u-substitution to simplify this

u = θ − π/2 θ = π2 − 1⇒ u = −1

du = dθ θ = π2 + 1⇒ u = +1

Area = 12

∫ 1

−1(−3u2 + 2|u|+ 1)2 du

=

∫ 1

0(−3u2 + 2u+ 1)2 du (the function is even)

=17

15(skipping steps)


10.1 Curves Defined by Parametric Equations

Comments. Our next topic is parametric equations. These can be viewed asan extension of polar equations. In polar equations, we write both x and y asfunctions of θ using sine and cosine. In parametric equations, we write both x andy as functions of t, possibly using sine and cosine, but possibly not.

Definition. Let (x, y) be a point on a curve. Suppose that x and y are each givenas a function of a third variable,

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


then we call these equations parametric equations and call t the parameter.

Comments. The main point of parametric equations is that we can describe curvesand shapes that are not functions of x, that is to say they don’t pass the verticalline test. When we graph parametric equations we can make a list of points for xand y by hand, or rely upon our calculators. In any case, we graph only x and y,not the parameter t. We can imagine t in the graph: if you imagine tracing thegraph with a pen, the parameter t corresponds to the time the pen is at differentplaces on the graph. Thus, where you start the graph could be t = 0, and t = 0.5 iswhere your pen is half a second later, t = 1 is where your pen is after 1 second, etc.

Example 1. Let x and y be given parametric equations below:

x(t) = t cos(t)

y(t) = t sin(t)

(a) Calculate x and y for 11 values of t: t = 0, π/5, 2π/5, . . . , 10π/5.(b) Plot the x and y values just calculated, and connect the points with a smooth

curve.(c) Explain how the functions and the curve just found can be viewed as a polar

function.

Solution. (a)

t = 0 x = 0 y = 0

t = π/5 x = π/5 cos(π/5) ≈ 0.51 y = π/5 sin(π/5) ≈ 0.37

t = 2π/5 x = 2π/5 cos(2π/5) ≈ 0.39 y = 2π/5 sin(2π/5) ≈ 1.2

t = 3π/5 x = 3π/5 cos(3π/5) ≈ −0.58 y = 3π/5 sin(3π/5) ≈ 1.8

t = 4π/5 x = 4π/5 cos(4π/5) ≈ −2 y = 4π/5 sin(4π/5) ≈ 1.5

t = 5π/5 x = 5π/5 cos(5π/5) ≈ −3.1 y = 5π/5 sin(5π/5) = 0

t = 6π/5 x = 6π/5 cos(6π/5) ≈ −3 y = 6π/5 sin(6π/5) ≈ −2.2

t = 7π/5 x = 7π/5 cos(7π/5) ≈ −1.4 y = 7π/5 sin(7π/5) ≈ −4.2

t = 8π/5 x = 8π/5 cos(8π/5) ≈ 1.6 y = 8π/5 sin(8π/5) ≈ −4.8

t = 9π/5 x = 9π/5 cos(9π/5) ≈ 4.6 y = 9π/5 sin(9π/5) ≈ −3.3

t = 10π/5 x = 10π/5 cos(2π) ≈ 6.3 y = 10π/5 sin(10π/5) = 0

(b) We show below the graph of just the points, and then the graph with asmooth curve through them


(c) To view these quations as a polar function, simply change notation:

parametric polar

x(t) = t cos(t) x = r cos(θ) (t = θ, r = θ)

y(t) = t sin(t) y = r sin(θ)

Example 2. Let x and y be defined by the following parametric equations

x = t+ sin(πt)

y = t+ cos(πt)

(a) Plot by hand 7 values of x and y corresponding to t = 0, 0.25, 0.5, 0.75, 1,1.25 and 1.5.

(b) Use your graphing calculator to plot x and y for 0 ≤ t ≤ 5.

Solution. (a) We start in the same way you probably learned to graph cartesianequations: calculating a table of points and plotting them by hand. Here are thevalues (rounded to the first decimal place).

t x y

0 0 10.25 1 10.5 1.5 0.50.75 1.5 01 1.0 01.25 0.5 0.51.5 0.5 1.5

Now we plot these points, by hand, as shown,


and attempt to draw a line connecting them, just like dot-to-dot, going in the orderthat the points were defined.

(b) We enter the equations in our calculator2 We plot this for 0 ≤ t ≤ 5

2First we need to change the graphing mode to parametric, go to MODE , then choose PAR

instead of FUNC . Now when you go to the Y= menu you’ll see two things to enter for each graph:Y 1T = and X1T =.


Fact. It is always easy, almost trivial, to translate a Cartesian function into para-metric functions: simply change notation and replace x with t:

Cartesian Equation Parametric Equation

You are given y = f(x). Let x = t and y = f(t).

It is sometimes possible to translate from parametric equations into Caretesianequations. It should make sense that this change would be harder than the reversedirection, since many parametric functions produce very complicated graphs thatare certainly not functions of x (for instance the two examples we’ve already donein this section). Here is how such a change can be done sometimes:

Parametric Equation Cartesian Equation

You are given two equations: x = f(t) andy = g(t). Solve one of these for t, and substi-tute this into whichever one you didn’t solve,to eliminate t.

Example 3. Return to Example 4 in the previous section. We had r = 1− sin(θ)and found the tangent line at θ = π/12 to be given by y = −0.414(x− 0.72) + 0.19.Graph both the original polar function and the tangent line in the same calculatorwindow.

Solution. We make both of these functions parametric. For the polar graph, lett = θ, and then use x = r cos(θ) and y = r sin(θ):

x(t) = (1− sin(t)) cos(t)

y(t) = (1− sin(t)) sin(t)

For the cartesian line, simply let t = x and then re-write y as a function of t:

x = t

y = −0.414(t− 0.72) + 0.19


Now we can enter both of these at the same time in the calculator. The resultshould like like the following:


10.2 Calculus with Parametric Curves

Fact. Given parametric equations x = f(t) and y = g(t), the Cartesian derivativeequals

dy

dx=

dydtdxdt

=g′(t)

f ′(t).

To understand this, recall that

dy

dx≈ ∆y

∆x=

∆y∆t∆x∆t

Example 1. Return to the parametric equations in Example 2 from the previoussection:

x = t+ sin(πt)

y = t+ cos(πt)

(a) Find the Cartesian equation of the tangent line at t = 7/4 (decimals ok).(b) Graph the original curve and the tangent line on your calculator.

Solution. (a) As always, for any equation of a tangent line, our goal is to fill in

y = m(x− x0) + y0

with

x0 = x-value of tangent point


y0 = y-value of tangent point

m =dy

dx

In this case, we calculate each of these as a function of t.

x0 = f(7/4) = 7/4 + cos(7π/4) = 7/4−√

2/2 ≈ 1.04

y0 = g(7/4) = 7/4 + sin(7π/4) = 7/4 +√

2/2 ≈ 2.46

dy

dx=g′(7/4)

f ′(7/4)=

1− π sin(πt)

1 + π cos(πt)

]

t=7/4

=1− π sin(7π/4)

1 + π cos(7π/4)

=1− π

√2/2

1− π√

2/2

= 1

Combining the above steps we see that the tangent line is

y = (x− 1.04) + 2.46

(b) To graph both the original curve, and the equation of the tangent line at thesame time, we use parametric mode for both. To put the tangent line in parametricmode we need to make x and y both functions of t. Actually, y is already a functionof x, so as soon as we figure out how to make x a function of t, we will implicitlyalready be done for y too. In other words, if x = f(t) the y = (f(t)− 1.04) + 2.46would give us y. In fact then, we could use just about any formula we want tocalculate x, and then y is take care of. So, we could use something weird, likex = et or x = ln(t), or x = tan(t), but we will use the easiest possible function.

x = t

y = (t− 1.04) + 2.46

The same approach works with every equation y = f(x): to put it in parametricmode simply let x = t and y = f(t).

Now we enter the following in our calculator:

Y 1T = T + cos(πT )

X1T = T + sin(πT )

Y 2T = (T − 1.04) + 2.4

X2T = T

The result is shown below


This is where section 1 ended on Friday, March 20

Example 2. [Based on Dwyer Gruenwald] A double Ferris wheel consists of twosmaller Ferris wheels, attached to a single larger rotating body, somewhat like pic-tured below:

The position of a person riding on a double Ferris wheel is given

x(t) = 20 sin

(πt

10

)+ 10 sin

(2πt

5

)

y(t) = 35− 20 cos

(πt

10

)− 10 cos

(2πt

5

)

where t is in seconds, x and y are in feet.(a) Graph the position of a rider for 0 ≤ t ≤ 20.(b) Take the derivative algebraically, but then use your calcualtor to find when

the slope is 0. What does this mean in terms of the movement of the rider?

Solution. (a) We enter the x- and y-equations into the calculator and get thefollowing graph


(b)

dy

dx=

dydtdxdt

=2 cos

(πt10

)+ 4 cos

(2πt5

)

2 cos(πt10

)+ 4 cos

(2πt5

)

0 =2 cos

(πt10

)+ 4 cos

(2πt5

)

2 cos(πt10

)+ 4 cos

(2πt5

)

0 = 2 cos

(πt

10

)+ 4 cos

(2πt

5

)

To solve this we graph 2 cos

(πt

10

)+4 cos

(2πt

5

)and see when it crosses the x-axis.

The graph looks like this

By zooming in we can solve for t:

t = 0, 2.8, 4.6, 7.8, 10, 12.4, 15.4, 17.2, 20.

These are the positions where the rider of the Ferris wheel reaches a local max ormin in terms of height.


As stated above, all cartesian functions can be turned into parametric functions.We have also seen how to find derivatives with parametric equations. Now we willlook at integrals.

The net area bounded by thecurve x = f(t) and y = g(t),between α ≤ t ≤ β

=

∫ β

αg(t)f ′(t) dt

This formula should be somewhat easy to understand (and therefore remember).

Think about starting with

∫ b

ay(x) dx, and then applying a substitution x = f(t).

Then we have y written as a function of t, i.e. y = g(t) and dx = f ′(t) dt. This

turns the integral

∫ b

ay(x) dx into the one above.

Example 3. The shape defined by

x = 5(t− sin(t)), y = 5(1− cos(t))

is called the cycloid3

(a) Use your calculator to sketch this shape for −2π ≤ t ≤ 4π.(b) Find the equation of the tangent line at t = π/4.(c) Find the area under the “first arch’ (you’ll understand what this is after you

graph it).

Solution. (a) We enter this in our calculators and get the following:

(b) The equation of a line is still given by

y = m(x− x0) + y0

where m is the usual slope, and (x0, y0) is a point on the curve. What’s new in thisproblem is how we find the slope and the coordinates (x0, y0). For the coordinates,we just plug in t = π/4:

x0 = 5(π/4− sin(π/4)) = 5(π/4− 1/√

2) ≈ 0.39

y0 = 5(1− cos(π/4)) = 5(1− 1/√

2) ≈ 1.46

For the slope, we take the derivative as described above:

dy

dx=

dydtdxdt

=5 sin(t)

5(1− cos(t))

3This shape was originally described in the 1600s this way: the path traced out by a fixed pointon the outside of a wheel as the wheel rolls along the ground.


m =dy

dx

]t = π/4 =

5 sin(π/4)

5(1− cos(π/4))

=5/√

2

5(1− 1/√

2)=

1√2− 1

≈ 2.414

Putting it all together, we get the following equation4:

y =5

5√

2− 5(x− 5(π/4− 1/

√2)) + 5(1− 1/

√2) ≈ 2.414(x− 0.39) + 1.46

(c) We integrate

∫ β

αg(t)f ′(t) dt =

∫ 0

−2π5(1− cos(t)) · 5(1− cos(t)) dt

= 25

∫ 0

−2π1− 2 cos(t) + cos2(t) dt

= 25

(3

2t− 2 sin(t) +

1

2sin(t) cos(t)

)]0

−2π

= 75π


Given parametrec equations x = x(t) and y = y(t), we can find the arc-lengthfrom t = a to t = b as follows:

Arc-length =

∫ b

a

√(dx

dt

)2

+

(dy

dt

)2

dt

This should be easy to remember since the length of a diagonal is given by√

∆x2 + ∆y2.

Example 4. Find the arc-length, L, along the top part of a circle of radius 3,between θ = π/3 and θ = π/2, as shown,

3

L

4If you want to graph both this and the cycloid on the same window, put everything in para-metric mode. Try −π ≤ t ≤ π, −1.5 ≤ x ≤ 2.5, −1 ≤ y ≤ 5


Solution. Before we do this using parametric equations, we pause to consider howdifficult it would be using Cartesian equations. We would have to find the x-value

corresponding to π/3, we would have to take the derivative of y =√

32 − x2, squarethe derivative, add 1, put this in a square root, and hope for the best. We will findthat parametric equations are much easier.

We use sine and cosine for our parametric equations. In a circle of radius 3 we

have sin(θ) =y

3and cos(θ) =

x

3. Solving these for x and y, and using t instead of

θ givesx = 3 cos(t), y = 3 sin(t)

Then the arc-length is given by

dy

dt= 3 cos(t),

dx

dt= −3 sin(t)

L =

∫ π/2

π/3

√(−3 sin(t))2 + (3 cos(t))2 dt

=

∫ π/2

π/3

√9 sin2(t) + 9 cos2(t) dt

=

∫ π/2

π/33

√sin2(t) + cos2(t) dt

=

∫ π/2

π/33√

1 dt

= 3t

]π/2

π/3

= 3(π/2− π/3)

= 3π/6 = π/2

Example 5. Find the arc-length of one full arc of the following Archimedean spiral:

x = cos(t) + t sin(t)

y = sin(t)− t cos(t)

0 ≤ t ≤ 2π

Solution.

dy

dt= t sin(t)

dx

dt= t cos(t)


AL =

∫ 2π

0

√(t cos(t))2 + (t sin(t))2 dt

=

∫ 2π

0

√t2 cos2(t) + t2 sin2(t) dt

=

∫ 2π

0

√t2(cos2(t) + sin2(t)) dt

=

∫ 2π

0

√t2 dt

=

∫ 2π

0t dt

=1

2t2]2π

0

=1

24π2

= 2π2

Chapter 11

Infinite Sequences and Series

11.1 Sequences

Definition. A sequence is an infinite ordered list of numbers. The usual notationfor a sequence is something like a1, a2, a3, . . . , where a1 is the first number, a2 isthe second number, etc. We abbreviate this as the sequence an, or {an}.

In many cases our sequences will be given by familiar functions. In other wordswe will have an = f(n) where f(x) is some combination of functions that you mighthave seen in Calculus I. This happens in the next example.

Example 1. Let an =n√n+ 1

.

(a) Find the first 5 terms of the sequence.(b) Find a formula for an+1.(c) Find a formula for a2n.

Solution. (a)

a1 =1

2

a2 =2√3

a3 =3√4

a4 =4√5

a5 =5√6

(b) We simply plug n+ 1 into the formulan√n+ 1

everywhere that has an n:

an+1 =n+ 1√n+ 1 + 1

=n+ 1√n+ 2

.

(c) This is like part (b), except that we plug 2n inton√n+ 1

:

a2n =2n√

2n+ 1.

112

CHAPTER 11. SEQUENCES AND SERIES 113

For us, the most important thing about sequences is that they allow us todescribe limits.

Definition. Let an be a sequence. We write

limn→∞

an = L

if the values of an become infinitely close to L as n approaches ∞. If there is nosuch L, i.e. the limit does not exist, then we say that the sequence is divergent. Ifsuch an L exists then we say that the sequence is convergent.

If an = f(x) where f(x) is a familiar function, then limn→∞

an = L means the same

thing as limx→∞

f(x) = L. This, in turn, means the same thing as saying that f(x)

has a horizontal asymptote, on the right, of y = L.

Example 2. For each part, find limn→∞

an, if it exists.

(a) an = 2n.(b) an = 0.8n.(c) an = (−0.2)n.(d) an = (−5)n.

Solution. For each part, we will write down a few values of an, and then plot them,and then make a guess based on this information.

(a)

n 1 2 3 4

an 2 4 8 16

The graph has no horizontal asymptote on the right, and the sequence has the samelimit, so

limn→∞

2n =∞ = DNE.

(b)


n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15an 0.8 0.64 0.512 0.41 0.328 0.262 0.21 0.168 0.134 0.107 0.086 0.069 0.055 0.044 0.035

The graph has a horizontal asymptote on the right of 0, the sequence has the samelimit, and so

limn→∞

0.8n = 0.

(c) We can’t graph (−0.2)x, because the base is negative. For instance, if x =1/2, this would mean

√−0.2, which isn’t defined. But the sequence an is defined

since we can find (−0.2)1, (−0.2)2, (−0.2)3, . . . . In fact, if we calculate thesenumbers, we see that they fit on two graphs: 0.2x and −(0.2x). We show below thefirst 4 points

Thus, the sequence of numbers we are interested appear on two graphs. Both graphs


have a horizontal asymptote of 0, and so

limn→∞

(−0.2)n = 0.

(d) This part is just like (c), but now the sequence appears on two diverginggraphs

Thus,limn→∞

(−5)n = DNE.


The following fact generalizes what we found in the previous example.

Fact. If a = rn for some number r, then we can say when limn→∞

an converges.

limn→∞

an = limn→∞

rn =

0 if |r| < 1

1 if r = 1

DNE if r = any other number

Comments. The way to see this fact is to imagine something like1

2nas n ap-

proaches infinity. Then you get1

∞ , which is 0 (just like limx→∞

1

x= 0).

Here are two types of formulas that are used very often in sequences which arenot familiar functions from Calculus I

1. n! . This formula is called “n factorial”. Here’s one way to define it

2! = 2 · 13! = 3 · 2 · 14! = 4 · 3 · 2 · 1. . . . . .


n! = n(n− 1)(n− 2) · · · 3 · 2 · 1

In formulas that we encounter later it will be convenient also to extend thisdefinition to 1 and 0 like this

1! = 1 and 0! = 1.

Note that for n ≥ 2 we have n! = n · (n− 1)! .2. (−1)n . This formula just alternates between 1 and −1. (Note: this formula

works if n is an integer, but (−1)x does not work for most real values of xbecause things like (−1)1/2 are not defined.)

Example 3. List the first 5 terms of the sequence, and simplify. Make a conjectureas to whether or not the sequence converges.

(a) an = (−1)n1

n2

(b) an =n!

2n

Solution. (a)

a1 = (−1)1 1

12= −1

a2 = (−1)2 1

22= (1) · 1

4=

1

4

a3 = (−1)3 1

32= (−1) · 1

9= −1

9

a4 = (−1)4 1

42= (1) · 1

24=

1

25

a5 = (−1)5 1

52= (−1) · 1

25= − 1

25

Based on the above evidence, it appears that the sequence is converging to ± 1

∞ .

The “±” doesn’t affect the outcome much, since1

∞ = 0. Thus, the limit appears

to be 0.(b)

a1 =1!

21=

1

2

a2 =2!

22=

2

4=

1

2

a3 =3!

23=

3 · 28

=3

4

a4 =4!

24=

4 · 3 · 216

=3

2

a5 =5!

25=

5 · 4 · 3 · 232

=15

4

It is probably harder to see what happens to this sequence. At first it doesn’tchange, but then it starts to get bigger. Does this trend continue? If so, maybe itcould keep getting a little bigger, but still level out. Or does it keep getting biggerwithout leveling out? If so, then the limit is infinite.


Maybe you can see that each time we move to the next term, we add one factoron top of the fraction, and one factor on the bottom. The factor on the bottom is2 every time. In other words, we multiply by one more 2. But on top, the factorkeeps getting bigger. Moving from a4 to a5, we multiplied the top by 5. At the nextstep we would multiply by 6, then 7 etc. Perhaps this convinces you, correctly, thatthe sequence does not level out. In fact, we can guess now, and will prove later,that

limn→∞

n!

2n=∞.

Example 4. Find limn→∞

an where an =3n2 − 100n+ 5

2n2 + n+ 1.

Solution. It helps in this problem to think in terms of the graph of a function ofx. Thus,

limn→∞

an = limx→∞

3x2 − 100x+ 5

2x2 + x+ 1= H.A. of

3x2 − 100x+ 5

2x2 + x+ 1

Now, I hope you recall the short-cut rules you learned in Pre-Calculus for horizontalasymptotes. Compare the biggest degree on to the top and bottom, if these degreesare equal, as in this case, then take the ratio of the leading coefficients:

H.A. of3x2 − 100x+ 5

2x2 + x+ 1=

3

2.


11.2 Series

Definition. Given a sequence an, the seriesk∑

n=j

an equals the sum aj + aj+1 +

· · ·+ ak.

Comments. • The easiest way to understand this definition the first time yousee it is with the following verbal version:

k∑

n=j

an = the sum of the an starting with aj and ending with ak

• In almost all cases we will take j equal to 0 or 1, i.e. we will start at the first(or zeroth) term. In fact, for convenience we will often start with n = 0, evenif a0 hasn’t been defined. In this case we will assume that a0 = 0.

Example 1. Let an =1

n2. Write down (just write them, don’t calculate them) the

following sums (or at least enough of them to see the pattern):

(a)5∑

n=1

an

(b)

25∑

n=21

an


(c)

∞∑

n=21

an

Solution. (a)1

12+

1

22+

1

32+

1

42+

1

52

(b)1

212+

1

222+

1

232+

1

242+

1

252

(c)1

212+

1

222+

1

232+ . . .

Fact. Let∑

can be any kind of sum (i.e. finite, or infinite, which we define below).

The following property holds, and can be compared to properties of integrals andof numbers:

Property of ordinary numbers 7 · 32 + 7 · 37 + 7 · 42 + 7 · 47 = 7(32 + 37 + 42 + 47)

Property of integrals

∫cf(x) dx = c

∫f(x) dx

Property of series∑

can = c∑

an

Let∑

(an + bn) be any kind of sum. The following property holds, and can be

compared to properties of integrals and of numbers:

Property of ordinary numbers (7 + 32) + (72 + 37) + (73 + 42) + (74 + 47)

= (7 + 72 + 73 + 74) + (32 + 37 + 42 + 47)

Property of integrals

∫f(x) + g(x) dx =

∫f(x) dx+

∫g(x) dx

Property of series∑

(an + bn) =∑

an +∑

bn

Example 2. Find10∑

n=1

(9n− (−1)n)

Solution.

10∑

n=1

(9n− (−1)n) =

10∑

n=1

9n−10∑

n=1

(−1)n

= 9

10∑

n=1

n−10∑

n=1

(−1)n

= 9(1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10)− (−1 + 1− 1 + 1− 1 + 1− 1 + 1− 1 + 1)

= 9(55)− 0

Comments. We will mostly be studying infinite series. We motivate their definitionby looking at functions

an is like f(n) where f(x) is a function

limn→∞

an is like limx→∞

f(x)

Therefore

∞∑

n=1

an is like

∫ ∞

1f(x) dx

So, do you remember how we defined improper integrals like

∫ ∞

1f(x) dx?


Definition. Given a sequence an we define

∞∑

n=1

an as

∞∑

n=1

an = limN→∞

N∑

n=1

an.

• If this limit exists we say that∞∑

n=1

an converges.

• If this limit does not exist then we say that

∞∑

n=1

an diverges.

Example 3. Zeno’s paradox is a problem posed by the ancient Greek philosopherZeno, who was arguing against the position that space and time are infinitely divis-ible. Here’s one version of it.

Suppose we want to walk one city block. We must first walk half the distance,and then half the remaining distance, and half the remaining distance, etc. Sincewe must go through an infinite number of half-way points, we can never finish thetask: we can never walk one block.

Translate this problem into a series and “solve” it.

Solution. The distance travelled after n steps is

1

2+

1

4+

1

8+ · · ·+ 1

2n.

We can write this using∑

notation as

n∑

i=1

1

2i

We can start by giving a formula for how far we have walked after n half-waypoints:

n = 2 :1

2+

1

4=

3

4or 1− 1

4

n = 3 :1

2+

1

4+

1

8=

7

8or 1− 1

8

n = 4 :1

2+

1

4+

1

8+

1

16=

15

16or 1− 1

16. . .

general n1

2+

1

4+ · · ·+ 1

2n=

2n − 1

2nor 1− 1

2n

Assuming that this formula is correct, then after n half-way steps, the total distancetravelled is

1− 1

2n

Taking the limit as n→∞ gives a total distance of

limn→∞

1− 1

2n= 1− 0 = 1


Checkboxes• cn = arn 2

•

{|r| < 1 2 ⇒

∑cn Conv

|r| ≥ 1 2 ⇒∑

cn Div

Thus, we get to the end of the block after all! Another way to summarize this:

∞∑

n=1

(1/2)n = 1

This is where section 1 ended on Friday, March 27

Definition. A series is geometric if

each term equals the previous term times r

In symbols this means that∑

an is geometric if

an = an−1r

Finally, we can simplify this a little:

A geometric series is

∞∑

n=0

arn

Theorem 1 (Geometric Series Theorem).

1. If |r| < 1 then∞∑

n=0

arn =a

1− r (in particular, the series converges).

2. If |r| ≥ 1 then∞∑

n=0

arn diverges.

Proof. We prove just the first claim, and therefore assume |r| < 1. Now we calculate:

(1− r)(1 + r + r2 + · · ·+ rN ) = 1 + r + r2 + · · ·+ rN (here we distributed the 1)

− r − r2 − · · · − rN − rN+1 (here we distributed the − r)(1− r)(1 + r + r2 + · · ·+ rN ) = 1− rN+1 (all terms cancelled except

the first and the last)

1 + r + r2 + · · ·+ rN =1− rN+1

1− r (divide by 1− r)N∑

n=0

rn =1− rN+1

1− r

limN→∞

N∑

n=0

rn = limN→∞

1− rN+1

1− r∞∑

n=0

rn =1− limN→∞ r

N+1

1− r (this is the only spot

∞∑

n=0

rn =1− 0

1− r where we used |r| < 1)

a

∞∑

n=0

rn =a

1− r


Comments. 1. The formula for

∞∑

n=0

rn is proven in an ad hoc manner, meaning

the proof is made up just for this series and does not follow a general strategy.In particular, we will not be able to do anything like this for most other series.

2. The proof gave more than the formula for∞∑

n=0

rn: it gave a formula forN∑

n=0

rn.

Fact.N∑

n=0

rn =1− rN+1

1− r , for any number r. (11.1)

Fact.

∞∑geom =

first #

1− common ratio(if |r| < 1). (11.2)

∑

finite

geom =first #− (last #) · r1− common ratio

(for all r). (11.3)

Example 4. Re-calculate the example about Zeno’s paradox, using our formula.

Solution. The distance travelled in Zeno’s paradox is

∞∑

n=1

(1/2)n.

Thus, the total distance is

first #

1− common ratio=

1/2

1− 1/2=

1/2

1/2= 1.


Example 5. Find the exact formula for each of the following:

(a)∞∑

n=5

πn−2

4n+1

(b)30∑

n=15

2n

5

Solution. (a) It probably helps to write out the first few terms of this series, tomake it easier to see how it’s a geometric series.

∞∑

n=5

πn−2

4n+1=π5−2

45+1+π6−2

46+1+π7−2

47+1+π8−2

48+1+ . . .

=π3

46+π4

47+π5

48+π6

49+ . . .

It does not matter that this series is not written explicitly in the form arn. The mainwork is to identify what r is. Whatever each term successive term gets multiplied


by, that’s r. Look at how each term changes when we go to t. Both the power ofπ and the power of 4 increase. How much to they increase by? Each one increases

by 1. Thus, each term equals the previous one, timesπ

4. In other words, we can

rewrite the given series this way

∞∑

n=5

πn−2

4n+1=π3

46+π3

46· π

4+π3

46·(π

4

)2+π3

46·(π

4

)3+π3

46·(π

4

)4+ . . . = arnX

r =π

4< 1X

Now we can see exactly what a and r equal, we can write down the sum instantly,

∞∑

n=5

πn−2

4n+1=

π3/46

1− π/4

(b) As before, it probably helps to write out a few terms of this series.

30∑

n=15

2n =215

5+

216

5+

217

5+

218

5+ · · ·+ 229

5+

230

5.

Now we apply Equation (11.3) to get

215

5 − 230

5 · 21− 2

=215

5(216 − 1)

Example 6. Write the following number as a fraction of integers,

3.141514151415 . . .

Solution. The definition of decimal notation means that this number equals

3 +1

10+

4

100+

1

1000+

5

10000+ . . .

where as soon as we get to “. . . ” the numerators repeat. If we get a commondenominator, we can write this as follows

3 +1415

10000+ . . .

We can write this more compactly, and see what the pattern is, if we use powers of10

3 +1415

104+

1415

108+ . . .

︸︷︷︸geometric

Now we find a formula for the infinite repeating part

∑geometric =

first #

1− common ratio

=1415/104

1− 1/104


Checkboxes• lim

n→∞cn 6= 0 2

=⇒∑

cn Div.

The harmonic series is a reallyimportant example. It showsthat you can have terms thatare approaching 0, and yetadding them together doesn’tgive a finite answer.

=1415

104 − 1

=1415

9999

Now we add this to 3:

3 +1415

9999=

3 ∗ 9999 + 1415

9999=

31412

9999

Theorem 2.

1. If

∞∑

n=1

an converges then limn→∞

an = 0.

2. (The Divergence test) If limn→∞

an 6= 0 then∞∑

n=1

an diverges.

3. If limn→∞

an = 0 then you can make no conclusions about∞∑

n=1

an.

Comments. Part (3) of the previous result is really subtle. In Example 8 we show

that

∞∑

n=1

1

ndiverges. And yet, lim

n→∞

1

nequals 0. This means that if the terms are

not approaching zero, then the sum will definitely diverge; but if the terms areapproaching zero, well, this fact tells you nothing: the series might go either way.

Example 7. Show that the series∞∑

n=1

n+ 1

10000n+ 109diverges.

Solution. The terms in this series look small, because of the factors of 10000 and109 on the bottom, but, they do not keep getting smaller, and smaller, approaching0. To be more precise:

limn→∞

n+ 1

10000n+ 109=

1

100006= 0X

Since this limit is not 0, the series diverges, by the Divergence Test (just imagine

adding1

10000to itself an infinite number of times: eventually this will get pretty

big!).

Example 8. [Harmonic Series]

Show that

∞∑

n=1

1

ndiverges.

Solution. I’ll give a direct proof now, but later we give a better proof using theIntegral test.

We add groups of terms in this series and show that no matter how far out wego in the terms, we can always group enough of them together to add up to 1/2.

Let’s write out the first 30 terms (or at least some of them) and group the first1 term, the next 2, the next 4, the next 8, the next 16, etc.

1≥ 1

2

+1

2+

1

3︸︷︷︸≥ 1

4+ 1

4

+1

4+

1

5+

1

6+

1

7︸︷︷︸≥ 1

8+ 1

8+ 1

8+ 1

8

+1

8+

1

9+ · · ·+ 1

15︸︷︷︸≥ 1

16+···+ 1

16

+1

16+

1

17+ · · ·+ 1

31︸︷︷︸≥ 1

32+···+ 1

32

+ . . .


Checkboxes• cn ≥ 0 2• c1 + c2 + · · ·+ cn < somefixed number 2=⇒

∑cn Conv.

≥ 1

2+ 2 · 1

4+ 4 · 1

8+ 8 · 1

16+ 16 · 1

32+ . . .

=1

2+

1

2+

1

2+

1

2+ . . .

=

∞∑

n=1

1

2=∞.

No matter how big n gets, we can always group together enough terms to get another

sum ≥ 1

2. Thus, as n goes to ∞, we get an infinite number of

1

2’s. Well, that’s

infinite!


Example 9. Find the sum

∞∑

n=1

1

n2 + nby using partial fractions.

Solution. First we factor1

n2 + n=

1

n(n+ 1).

Then we split the fraction up using partial fractions

1

n(n+ 1)=A

n+

B

n+ 1

1 = A(n+ 1) +Bn

n = −1⇒ 1 = A · 0−B ⇒ B = −1

n = 0⇒ 1 = A · 1 +B · 0⇒ A = 1

Now we can look at the original series and find a lot of cancellation:

∞∑

n=1

1

n2 + n=∞∑

n=1

(1

n− 1

n+ 1

)

= limn→∞

(1

1− 1

2

)+

(1

2− 1

3

)+

(1

3− 1

4

)+ · · ·+

(1

n− 1

n+ 1

)

= limn→∞

(1

1−

��1

2

)+

(

��1

2−

��1

3

)+

(

��1

3−

��1

4

)+ · · ·+

(

��1

n− 1

n+ 1

)

= limn→∞

1

1− 1

n+ 1

= 1− 0

= 1

11.4 The Comparison Tests

Theorem 1 (The Bounded Monotonic Series Theorem (aka You Can’t Walk Through

a Wall Theorem)). Let∑

an be a series. If for all n we have1. an ≥ 02. a1 + a2 + · · ·+ an ≤ some fixed number


then∑

an converges.

Comments. This result is proven by using the Bounded Monotonic Sequence The-orem. In turn, that result depends on a deep, subtle, and defining property of thereal numbers: that every subset of real numbers which is bounded above has a leastupper bound in the real numbers.

Example 1. Define a series∞∑

n=0

an10n

as follows (recursively):

a0 = 1

a1 = int(

(√

2− a0) · 10)

a2 = int

((√2− a0 −

a1

10

)· 102

)

a3 = int

((√2− a0 −

a1

10− a2

102

)· 103

)

a4 = int

((√2− a0 −

a1

10− a2

102− a3

103

)· 104

)

. . .

an = int

((√2− a0 −

a1

10− a2

102− · · · − an−1

10n−1

)· 10n

)

(Note that “int” stands for “integer part” of a number, i.e. the part before thedecimal place. It is a built in function for the TI-84, under MATH , NUM , 5: int( .)

(a) Calculate a0, a1, . . . , a5.

(b) Sketch the proof that∞∑

n=0

an10n

converges, and figure out what it converges to.

Solution. (a)

a0 = 1

a1 = int(

(√

2− 1) · 10)

= 4

a2 = int

((√2− 1− 4

10

)· 102

)= 1

a3 = int

((√2− 1− 4

10− 1

102

)· 103

)= 4

a4 = int

((√2− 1− 4

10− 1

102− 4

103

)· 104

)= 2

a5 = int

((√2− 1− 4

10− 1

102− 4

103− 2

104

)· 105

)= 1


(b) Writing out the first 6 terms of the series we have

5∑

n=0

an10n

= 1 +4

10+

1

102+

4

103+

2

104+

1

105

= 1.41421

Note that5∑

n=0

an10n

is the first 6 digits of√

2. In fact, for all N we have1

N∑

n=0

an10n≤√

2

Thus,

∞∑

n=0

an10n

is bounded above, and must converge, by the Bounded Monotonic

Series Theorem. It converges to√

2.

Example 2. Use the Bounded Monotonic Series Theorem to show that

∞∑

n=1

1

n3.1

converges.

Solution. To get a feel for the series we’ll write out the first few terms:

6∑

n=1

1

n3.1=

1

13.1+

1

23.1+

1

33.1+

1

43.1+

1

53.1+

1

63.1+ . . .

Through insight or inspiration we might think of comparing this to

∫ ∞

1

1

x3.1dx.

There is a connection between the series and the integral: the series is a Riemannsum of the integral, with ∆x = 1:

1Probably this is easy to believe, but here’s the nitty gritty detail. By definition

an = int

((√2− a0 −

a110− a2

102− · · · − an−1

10n−1

)· 10n

).

This means that an is the largest integer d such that

d ≤(√

2− a0 −a110− a2

102− · · · − an−1

10n−1

)· 10n.

This means that d is the largest integer such that

a0 +a110

+a2102

+ · · ·+ an−1

10n−1+

d

10n≤√

2.


First 5 terms in Riemann Sum= f(2)∆x+ f(3)∆x+ f(4)∆x+ f(5)∆x+ f(6)∆x

=1

23.1(1) +

1

33.1(1) +

1

43.1(1) +

1

53.1(1) +

1

63.1(1)

=6∑

n=2

1

n3.1

There is something very important about this particular Riemann Sum: since the

function f(x) =1

x3.1is decreasing, the Right Hand Riemann Sum is below the true

value of the area. Thus6∑

n=2

1

n3.1≤∫ ∞

1

1

x3.1dx

In fact, we have the same result if we replace 6 with any number:

N∑

n=2

1

n3.1≤∫ ∞

1

1

x3.1dx

We can calculate the right hand side

N∑

n=2

1

n3.1≤ x−3.1+1

−3.1 + 1

] limx→∞

1

= 0− 1−3.1+1

−3.1 + 1

N∑

n=2

1

n3.1≤ 1

2.1

Since the above bound is true for all N we can apply the BMST and conclude that

∞∑

n=2

1

n3.1converges

Note that since

∞∑

n=2

1

n3.1converges, so does

∞∑

n=1

1

n3.1.


Theorem 2 (p-series test).

1. If p > 1 then∞∑

n=1

1

npconverges.

2. If p ≤ 1 then

∞∑

n=1

1

npdiverges.

This is where section 1 ended on Wednesday, April 1

Example 3. Show that the following series converges

∞∑

n=1

cos2(n) · (1/2)n

Solution. We will show that cos2(1) · 12

+cos2(2) · 14

+cos2(3) · 18

+ · · ·+cos2(n) · 1

2nis ≤ some fixed number, no matter how big n is. This will imply that that the seriesconverges by the Bounded Monotonic Series Theorem.

Recall that cos2(n) is ≤ 1 for all n, so we have

cos2(1) · 1

2+ cos2(2) · 1

4+ cos2(3) · 1

8+ · · ·+ cos2(n) · 1

2n

≤ 1

2+

1

4+

1

8+ · · ·+ 1

2n

Note that the series on the right is the same one as in Zeno’s Paradox. Thus,

cos2(1) · 1

2+ cos2(2) · 1

4+ cos2(3) · 1

8+ · · ·+ cos2(n) · 1

2n

≤ 1

2+

1

4+

1

8+ · · ·+ 1

2n

≤ 1

Therefore, by the BMST, we have that

∞∑

n=1

cos2(n) · (1/2)n

converges.

Comments. Take a close look at the last example and see if you can figure outwhat makes it work, so that we can apply it to other examples. Here’s how I woulddescribe it: we showed that one series converged, by showing that it was ≤ anotherseries that converged. The main idea in this section is to make a formal test out ofwhat we did in that example.

Comments. We start with a tortured metaphor for the approach we will take inthis section.

Using the Comparison Test is like matchmaking in human relationships. Youstart with someone you know, and think about who you could match them up withthat would make a good partner. To do this, you probably start by thinking aboutwhat the first person is like: Do they have a sense of humor? Do they have anystrong interests? Try to find someone who shares these traits.


Checkboxes: Conv• cn ≥ 0 2• cn ≤ an 2•

∑an Conv 2

=⇒∑

cn Conv.

Checkboxes: Div• cn ≥ 0 2• an ≥ 0 2• cn ≥ an 2•

∑an Div 2

=⇒∑

cn Div.

So, with series, you start with∑

cn. I use the letter “cn” to stand for “com-

plicated” and for “Clyde”, just to give the series a name. Clyde is a lonely serieslooking for a wife. He is complicated, or perhaps that just means confused, andwants a simpler partner who can offer him direction in life. Clyde knows that hisfate is pre-determined: he will either converge or diverge, but he does not knowwhich path is his. His partner will help him find his way. His wife will share hisfate, either they both will remain grounded and converge to a finite real number,or both will rise together on an infinite path to ever greater heights.

More specifically, if it is Clyde’s destiny to converge, then he needs to find apartner who will stay above him, and also converge, thereby keeping him groundedand bounded and under control. On the other hand, if it is Clyde’s destiny todiverge, then he needs a partner who will stay below him, and also diverge, therebycarrying him, lifting him up, to infinity.

Of course, it’s never easy to find the right life partner. Can you help him?

Theorem 3 (Inequality Comparison Test). Given an infinite series∑

cn, find

another infinite series∑

an, and do the following:Case 1, convergence:

(a) Check cn ≥ 0 for almost all n,(b) Check an ≥ cn for almost all n,

(c) Check∑

an converges,

(d) State your conclusion: “therefore, by the Comparison Test,∑

cn converges.”

Case 2, divergence:(a) Check cn ≥ 0 for almost all n,(b) Check an ≥ 0 for almost all n,(c) Check an ≤ cn for almost all n,

(d) Check∑

an diverges,

(e) State your conclusion: “therefore, by the Comparison Test,∑

cn diverges.”

Comments. Logically parts (a) and (b) are equivalent: if the bigger series con-verges (is finite), then so does the smaller one. If the smaller series diverges (goesto infinity), then so does the bigger one. To prove part (a) note that all the finite

sums

N∑

n=1

cn are bounded above by

∞∑

n=1

an, then apply the BMST (aka the Can’t

Walk Through the Wall Theorem).

Example 4. Determine whether or not the following series converge/diverge.

(a)∞∑

n=2

n3

n4 − 100n

(b)

∞∑

n=1

n− 17.5

n2√n

Solution. (a) We start by figuring out what’s going on, but you don’t need to show

this part in your work. Clyde equalsn3

n4 − 100n. Ignoring the “100n”, because it’s


a smaller part, we have a simplified function

n3

n4 − 100n

compare to−−−−−−−−→ n3

n4=

1

n

So, we’re going to compare Clyde to1

n. Since

∞∑

n=1

1

ndiverges, our guess is that

we will have Clyde diverge too, and therefore, we need to show that Clyde is bigger,

i.e. that Clyde is “lifted up” by1

n.

Here are the steps we need to show in our work

CT(a)n3

n4 − 100n≥ 0X (top≥ 0, bottom ≥ 0 for large enough n)

CT(b)1

n≥ 0X

CT(c)n3

n4 − 100n

?≥ 1

n

n4?≥ n4 − 100n cross multiply

0?≥ −100nX

CT(d)∑ 1

ndiv. b/c harmonic seriesX

CT(e) ∴, by C.T.∑ n3

n4 − 100diverges

(b) Again we start with figuring out what’s going, and then show how to write

up our work. Clyde equalsn− 17.5

n2√n

. Ignoring the “17.5”, because it’s a smaller

part, we have a simplified function

n− 17.5

n2√n

compare to−−−−−−−−→ n

n2√n

=1

n3/2

So, we’re going to compare Clyde to1

n3/2. Since

∞∑

n=1

1

n3/2converges, our guess is

that we will have Clyde converge too, and therefore, we need to show that Clyde is

smaller, i.e. that Clyde is “grounded” by1

n3/2.

Here are the steps we need to show in our work:

CT(a)n− 17.5

n2√n≥ 0X (top≥ 0 for large enough n, bottom ≥ 0)

CT(b)n− 17.5

n2√n

?≤ 1

n3/2

(n− 17.5)n3/2?≤ n5/2 cross multiply

n5/2 − 17.5n3/2?≤ n5/2

−17.5n3/2?≤ 0X


CT(c)∑ 1

n3/2conv. p-series, p = 3/2X

CT(d) ∴, by C.T. ,∑ n− 17.5

n2√n

converges

This is where section 1 ended on Tuesday, April 7

Comments. How to wed Clyde and his partner0. You’ve been given a series, Clyde, by your teacher or book.1. Find a simpler partner for Clyde by eliminating the less important parts and

concentrating on the dominant parts (see below for what “dominant” means.You might also use your experience, your knowledge of other series, and stereo-typing.

2. Identify what the fate of Clyde’s partner is, i.e. whether she converges ordiverges. (To do this, at present, Clyde’s partner needs to be a GeometricSeries, or a p-series, or something else that you can apply the Integral Testto. But we will get more possibilities as we go through the remaining sectionsof Chapter 11).

3. If Clyde’s partner converges, you will need to show partner ≥ Clyde.If Clyde’s partner diverges, you will need to show partner ≤ Clyde.

4. Now you get to simply say “By the comparison theorem, Clyde converges/diverges.”.

Comments. How to size up ClydeIn order to look for Clyde’s bride, you have to have some guess as to whether ornot Clyde converges, and then you look for a simple partner with the same fateas Clyde. Here are some rules of thumb for dominance, i.e. which functions are“bigger” than others.

Definition. Let an and bn be two sequences. We will describe when one sequencedominates the other (we’ll show at the end what this means and how it’s impor-tant).

1. If a < b then nbdominatesna. For example, we can arrange n2,√n, n.01, n8,

n, this way

n8 dominates n2 dominates n dominates√n dominates n0.1.

2. If 0 < a < b then an << bn. For example we can arrange 2n, 10n, 3n, en, thisway

10n dominates 3n dominates en dominates 2n.

3. Now combine these with other types of functions, such as

n!, an (if b > 0), np (if p > 0), nn, ln(n),

as follows:

nn dominates n! dominates an dominates np dominates ln(n)

4. So, how to find a partner for Clyde? Concentrate on his dominant parts. Morespecifically, if an dominates bn then

limn→∞

an + bnanything

= limn→∞

ananything


and

limn→∞

anything

an + bn= lim

n→∞

anything

an

Example 5. Determine whether or not the following series converges.

∑ 1 + sin(n)

10n.

Solution. We start by showing how we figure things out. Our first guess won’tquite work, but it’s really useful to see the attempt anyway so you can see how totry things and how to make changes.

Finding a partner:1 + sin(n)

10n≈ 1

10n, so we try matching

∑ 1 + sin(n)

10nwith

∑ 1

10n. Since

∑ 1

10nconverges we’ll try to show that

∑ 1 + sin(n)

10nconverges.

Here is the work that we would need (again, this won’t quite be correct).

CT(a)1 + sin(n)

10n?≥ 0

⇐⇒ 1 + sin(n)?≥ 0

⇐⇒ sin(n)?≥ −1X

CT(b)1 + sin(n)

10n?≤ 1

10n

⇐⇒ 10n(1 + sin(n))?≤ 10n

⇐⇒ 1 + sin(n)?≤ 1

⇐⇒ sin(n)?≤ 0 not true!

This almost works. The problem is that we tried for too much, getting rid of sin(n)entirely. We should have gotten rid of sin(n) with 1. In other words, in 1 + sin(n)

we should replace “sin(n)” with “1”. The correct partner should be∑ 2

10n.

CT(a)1 + sin(n)

10n?≥ 0

⇐⇒ 1 + sin(n)?≥ 0

⇐⇒ sin(n)?≥ −1X

CT(b)1 + sin(n)

10n?≤ 2

10n

⇐⇒ 10n(1 + sin(n))?≤ 2 · 10n

⇐⇒ 1 + sin(n)?≤ 2

⇐⇒ sin(n)?≤ 1X

CT(c)∑ 2

10nconverges, G.S. r =

1

10< 1


CT(d) ∴, by C.T.∑ 1 + sin(n)

10nconverges


Example 6. Determine whether or not the following series converges.

∑ n!

nn.

(Hint: an oracle tells you that n! ≤ nn−2 for all n ≥ 5. Use this.)

Solution. Finding a partner: Using the information given us by the oracle thatn! ≤ nn−2 we have

n!

nn≤ nn−2

nn=

1

n2.

Checkboxes:

CT(a)n!

nn≥ 0

⇐⇒ n! ≥ 0X

CT(b)n!

nn≤ 1

n2

⇐⇒ n2n! ≤ nn

⇐⇒ n! ≤ nn

n2

⇐⇒ n! ≤ nn−2 given by oracle

CT(c)∑ 1

n2converges, p-series, p = 2X

CT(d) ∴, by C.T.∑ n!

nnconverges

11.6 Absolute Convergence and the Ratio Test

Most of the tests we’ve learned so far have restrictions on whether the terms canbe positive or negative (the integral test, p-series test, comparison test and limitcomparison test all require terms that are ≥ 0; the alternating series test requiresterms that alternate perfectly). What about series that have some positive andsome negative terms, but without any real pattern?

Example 1. Can we apply any of the tests we’ve learned so far to the series∞∑

n=1

cos(n)

n2?

Solution. We do not have limn→∞

cn = 0, so we cannot apply the divergence test.

Just by looking at it we can tell that this series is not a geometric series (a geometric

series needs rn, like 10n or1

3n). It’s not a p-series (a p-series needs to have

1

n2and

only1

n2). If we write out the first five terms we get

cos(1)

12+

cos(2)

22+

cos(3)

32+

cos(4)

42+

cos(5)

52+ . . .

+ − − − + . . .


Checkboxes•

∑|cn| Conv 2

=⇒∑

cn Conv.

Checkboxes• L = lim

n→∞

|cn+1||cn|

exists 2

•

{L < 1 2 ⇒

∑cn AbsCon

L > 1 2 ⇒∑

cn Div

where we have indicated the sign of each term with a + or − below the term. Sincesome of the terms are negative, we cannot apply the Comparison Tests. So wecannot apply any of the tests we know.

Definition. We say

∞∑

n=0

an is absolutely convergent if the

∞∑

n=0

|an| converges.

Theorem 1 (Absolute Convergence Implies Convergence).

If∞∑

n=0

|an| converges then∞∑

n=0

an converges too. In other words, absolutely conver-

gent implies convergent.

Comments. Note: in general, it is easier for a series to converge if some of theterms are negative.

This is where section 1 ended on Monday, April 13

Example 2. [Continued] Show that∞∑

n=1

cos(n)

n2is absolutely convergent.

Solution. We take absolute values of each term

∞∑

n=1

∣∣∣∣cos(n)

n2

∣∣∣∣ =∞∑

n=1

| cos(n)|n2

.

This series is better than the original one because now we can apply the comparisontest.

| cos(n)|n2

≥ 0X

1

n2≥ 0X

| cos(n)|n2

≤ 1

n2X

∑ 1

n2converges, p-series test, p = 2X

Therefore, by the Comparison Test,∑∣∣∣∣

cos(n)

n2

∣∣∣∣ converges.

Therefore, by Theorem 1,∑ cos(n)

n2converges too. In other words,

∑ cos(n)

n2

is absolutely convergent.

Theorem 2 (Ratio Test). Let

∞∑

n=1

an be any series. Suppose limn→∞

∣∣∣∣an+1

an

∣∣∣∣ exists or

equals ∞. Call this limit L.

• If L < 1 then∑|an| converges (therefore

∑an converges too, i.e. we have

Abs. Conv.)

• If L > 1 then∑

an diverges (therefore∑|an| diverges too, i.e. we have

Div.).• If L = 1 then the test tells you nothing.


Comments (and Question). • Note that if limn→∞

∣∣∣∣an+1

an

∣∣∣∣ fails to exist for any

reason besides equalling ∞, then the previous test does not apply at all.• Note, the test is easy to remember because for convergence we need the terms

to be decreasing (on average). So we need |an| > |an+1|. This implies|an+1||an|

<

1.• Note, the theorem involves the limit (as n → ∞) of a fraction. We have

learned how to find limn→∞

of many fractions. This could involve horizontal

asymptotes of rational functions, or L’Hospital’s Rule.• Question: If a series is conditionally convergent, which case will it show up in

for the previous theorem?

Proof. (1) Suppose L < 1. By the definition of “limit”, we know that the values of|an+1||an|

become infinitely close to L, as n becomes infinitely large. Let M =L+ 1

2,

so that M is half way between L and 1. Since L < M , and since|an+1||an|

becomes

infinitely close to L, we know that for some value of N that is large enough, we have|aN+1||aN |

< M , and that the same inequality holds for all larger values of N . Thus

|aN+1||aN |

< M ⇒ |aN+1| < M |aN |

|aN+2||aN+1|

< M ⇒ |aN+2| < M |aN+1| < M2|aN |

|aN+3||aN+2|

< M ⇒ |aN+3| < M |aN+2| < M3|aN |

|aN+4||aN+3|

< M ⇒ |aN+4| < M |aN+3| < M4|aN |

etc

Now we compare

∞∑

n=N

|an| to

∞∑

n=0

Mn|aN |. The latter is a geometric series, with

|r| = M < 1, and the inequalities above show that we can apply the Inequality

Comparison Theorem. Therefore,∞∑

n=1

|an| converges.

(2) A similar proof holds in this part as in part (1): mostly we change all theinequalities from “<” to “>”.

Example 3. Apply the Ratio Test to

∞∑

n=1

(−1)n10n

n!.

Solution. The main trick here is to simplify the factorials: 2

2Here’s how you simplify factorials:

2!

3!=

2 · 13 · 2 · 1 = �2 · �1

3 · �2 · �1.


L = limn→∞

∣∣∣∣∣∣(−1)n+1 10n+1

(n+1)!

(−1)n 10n

n!

∣∣∣∣∣∣

= limn→∞

∣∣∣∣(−1)n+1

(−1)n· 10n+1

(n+ 1)!· n!

10n

∣∣∣∣

= limn→∞

∣∣∣∣(−1)10n+1

10n· n!

(n+ 1)!

∣∣∣∣

= limn→∞

1 · 101 · 1

n+ 1

= 1 · 101 · 0= 0

Since L = 0, and 0 < 1, the Ratio Test implies that∞∑

n=1

(−1)n10n

n!is Abs Conv.

Example 4. Apply the Ratio Test to

∞∑

n=1

(−1)nn10

1.5n.

Solution. 3

L = limn→∞

∣∣∣∣∣(−1)n+1 (n+1)10

1.5n+1

(−1)n n10

1.5n

∣∣∣∣∣

= limn→∞

∣∣∣∣(−1)n+1

(−1)n(n+ 1)10

1.5n+1· 1.5n

n10

∣∣∣∣

= limn→∞

∣∣∣∣(−1) · 1.5n

1.5n+1· (n+ 1)10

n10

∣∣∣∣

= limn→∞

1 · 1

1.5· (n+ 1)10

n10

= 1 · 1

1.5· 1

Similarly5!

6!= �5 · �4 · �3 · �2 · �1

6 · �5 · �4 · �3 · �2 · �1=

1

6.

and son!

(n+ 1)!= �n ·��(n− 1) · · ·�2 · �1

(n+ 1) · �n ·��(n− 1) · · ·�2 · �1

=1

n+ 1.

3There are two ways to see that

limn→∞

(n+ 1)10

n10= limn→∞

n10

n10

You can apply the Binomial Theorem to get

(n+ 1)10 = n10 + 10n9 + . . .

and then “erase” the lower degree terms, or just “erase” the “+1” in (n+ 1)10 to begin with.


=1

1.5

Since L =1

1.5, and

1

1.5< 1, the Ratio Test implies that

∞∑

n=1

(−1)nn10

1.5nis Abs. Conv.


Example 5. What happens when you apply the Ratio test to the following series:∞∑

n=1

ln(n)

n2

Solution.

L = limn→∞

∣∣∣∣∣∣

ln(n+1)(n+1)2

ln(n)n2

∣∣∣∣∣∣

= limn→∞

∣∣∣∣ln(n+ 1)

(n+ 1)2· n2

ln(n)

∣∣∣∣

= limn→∞

∣∣∣∣ln(n+ 1)

ln(n)· n2

(n+ 1)2

∣∣∣∣We analyse each the limit of each factor here by itself, starting with the simpler

one,n2

(n+ 1)2,

limn→∞

n2

(n+ 1)2= lim

n→∞

n2

n2= 1

Now we look at the limit ofln(n+ 1)

ln(n)

limn→∞

ln(n+ 1)

ln(n)=

ln(∞)

ln(∞)

=∞∞ use L’Hospital

limn→∞

ln(n+ 1)

ln(n)

LH= lim

n→∞

1n+1

1n

= limn→∞

1

n+ 1

n

1

= limn→∞

n

n+ 1

= limn→∞

n

n= 1

Combining the limits we’ve found with our formula for L above, we have

L = 1 · 1So, this test tells us nothing about this series4.

4Of course, the curious reader will want to know: what does this series do? Does it converge or

diverge? The right way to size it up is to look for the dominant parts:1

n2. Since the series

∑ 1

n2

converges, so will the series in this example. The easiest way to prove this is using the integraltest. You can also use the comparison test, but perhaps surprisingly, you should compare it to

something like1

n1.5



11.8 Power Series.

Definition. A power series centered at the number a is a series of the form

∞∑

n=0

cn(x− a)n = c0 + c1(x− a) + c2(x− a)2 + . . .

where x is a variable. We view such a series as a function f(x). If a = 0 then wewrite the series as

∞∑

n=0

cnxn = c0 + c1x+ c2x

2 + . . .

Theorem 1. Let∞∑

n=0

cn(x − a)n be any power series centered at a. There is a

number R > 0, or R = 0, or R =∞ that• R = 0 implies that the series converges only for x = a• R =∞ implies that the series is absolutely convergent for all x• If R 6= 0,∞ then we can picture those x that make the series converge as

shown below:

a a+Ra−R

??

x = a + R mightgo either way

x = a − R mightgo either way

series is div forall x in here

series is div forall x in here

series is abs conv forall x in here

Example 1. (a) Find the radius of convergence and interval of convergence ofthe series

∞∑

n=0

xn

(b) Let f(x) be the finite formula for the above series. Do you remember whatfunction f(x) is? I mean, do you remember the usual formula for f(x)?Compare what you get by plugging 0.5 into the usual formula, to the sum ofthe first few terms of this series evaluated at x = 0.5. Do the same thing withx = 0.75.

(c) Compare the graph of f(x) (i.e. the graph of the usual formula) and the first

few terms of∞∑

n=0

xn.

Solution. (a) We find R using the Ratio Test.

L = limn→∞

|xn+1||xn|

= limn→∞

|x|


= |x|

So L = |x|. We need L = |x| < 1, so this means R = 1.

−1

?

1

?

0

div abs conv div

Now, we need to test the points x = ±R = ±1 directly.

x = 1: f(1) =∞∑

n=0

1n, diverges.

x = −1: f(−1) =

∞∑

n=0

(−1)n, diverges.

Thus, the interval of convergence is

IC = (−1, 1)

(b) The function is f(x) =1

1− x , because for any real number r with |r| < 1

we have

∞∑

n=0

rn =1

1− r .

Now we can compare f(0.5) to adding the first few terms of the series.

f(0.5) =1

1− 0.5=

1

1/2= 2 1 +

1

2︸︷︷︸1.5

+1

4

︸︷︷︸1.75

+1

8

︸︷︷︸1.875

+ . . .

The conclusion should be that 2 ≈ 1.5, 2 ≈ 1.75, 2 ≈ 1.875 and 2 = 1 + 12 + 1

4 + 18 +

dots.Now we can do the same thing with x = 0.75.

f(0.75) =1

1− 0.75=

1

1/4= 4

1 + 0.75 + (0.75)2 + (0.75)3

︸︷︷︸2.73

+(0.75)4 + (0.75)5 + (0.75)6

︸︷︷︸3.47

+ . . .

The conclusion should be that 4 ≈ 2.73, 4 ≈ 3.47, and 4 = 1 + 34 + 9

16 + 33

43+ . . . .

(c) The graphs below show f(x) =1

1− x , together with T3(x) = 1 +x+x2 +x3

and T6(x) = 1 + x+ x2 + x3 + x4 + x5 + x6,


The graphs show exactly the same kind of thing we saw with plugging in individualnumbers. For numbers closer to 0 (like 0.5 as compared to 0.75), the convergenceof 1 + x+ x2 + x3 + . . . to f(x) happens with fewer terms.

How to find R.• Apply the Ratio Test, and solve for L.• Break into cases depending on L:

* L = # and # < 1, =⇒ R =∞.* L = # and # > 1, =⇒ R = 0.* L = something involving x . Set up and solve the following:

solve something involving x < 1

for |x− a| <?.

Then R =?.This is where section 1 ended on Friday, April 17

Example 2. Find the Radius of Convergence, and draw the open interval

f(x) =

∞∑

n=1

(−1)n(x− 1/2)n

n

Solution.

L = limn→∞

∣∣∣∣∣(−1)n+1 (x−1/2)n+1

n+1

(−1)n (x−1/2)n

n

∣∣∣∣∣

= limn→∞

∣∣∣∣(−1)n+1

(−1)n· (x− 1/2)n+1

n+ 1· n

(x− 1/2)n

∣∣∣∣


The best way to understand|x− 1/2| < 1 is in terms ofdistances. This inequality isequivalent to the question:“which x’s have a distance < 1from 1/2?”.

One algebraic rule for absolutevalues states that if a > 0,then a|x+ b| = |ax+ ab|.

= limn→∞

∣∣(−1)1∣∣ ·∣∣∣∣

n

n+ 1

∣∣∣∣ ·∣∣∣∣(x− 1/2)n+1

(x− 1/2)n

∣∣∣∣

= limn→∞

1 ·∣∣∣∣

n

n+ 1

∣∣∣∣ ·∣∣(x− 1/2)1

∣∣

= 1 · 1 · |x− 1/2|= |x− 1/2|

So, L = |x − 1/2|. We need |x − 1/2| < 1, and so we set R = 1. This means thatthat the interval goes from −1/2 to 3/2.

−1/2

?

3/2

?

1/2

div abs conv div

Example 3. Find the Radius of convergence and draw the open interval

∞∑

n=0

(2x− 3)n

Solution.

L = limn→∞

∣∣∣∣(2x− 3)n+1

(2x− 3)n

∣∣∣∣= lim

n→∞

∣∣(2x− 3)1∣∣

= |2x− 3|

We need L < 1, so we have|2x− 3| < 1

We need to solve this for |x−?| <?, so we multiply by 12

12 |2x− 3| < 1

2 · 1|x− 3/2| < 1/2

From this we see that R = 1/2.1

?

2

?

3/2

div abs conv div

Example 4. Find the Radius of Convergence and the Interval of Convergence

∞∑

n=0

(x− 7)n

11n

Solution.

L = limn→∞

∣∣∣∣∣(x−7)n+1

11n+1

(x−7)n

11n

∣∣∣∣∣

= limn→∞

∣∣∣∣(x− 7)n+1

11n+1· 11n

(x− 7)n

∣∣∣∣


= limn→∞

∣∣∣∣(x− 7)n+1

(x− 7)n· 11n

11n+1

∣∣∣∣

= limn→∞

|x− 7| · 1

11

=1

11|x− 7|

So L =1

11|x− 7|, we need this to be < 1, so

1

11|x− 7| < 1

|x− 7| < 11

From this we see that R = 11.

−4

?

18

?

7

div abs conv div

Extra Examples

Example 5. Find the radius and interval of convergence for the following powerseries

∞∑

n=1

(x+ 7)n

n2 + n+ 1

Solution.

L = limn→∞

∣∣∣∣∣∣∣∣

(x+ 7)n+1

(n+ 1)2 + (n+ 1) + 1

(x+ 7)n

n2 + n+ 1

∣∣∣∣∣∣∣∣

= limn→∞

∣∣∣∣(x+ 7)n+1

(n+ 1)2 + (n+ 1) + 1· n

2 + n+ 1

(x+ 7)n

∣∣∣∣

= limn→∞

∣∣∣∣(x+ 7)n+1

(x+ 7)n· n2 + n+ 1

(n+ 1)2 + (n+ 1) + 1

∣∣∣∣

= limn→∞

∣∣∣∣(x+ 7) · n2n2 + n+ 1

(nn + 1)22 + (n+ 1) + 1

∣∣∣∣= |(x+ 7) · 1|

solve |x+ 7| < 1

center = −7, radius = 1

−8

?

−6

?

−7

div abs conv div

test first endpoint x = −8 :

∞∑

n=1

(−8 + 7)n

n2 + n+ 1


Recall the notation “f (n)(x)”,which means the nthderivative.

∞∑

n=1

(−1)n

n2 + n+ 1?

test second endpoint x = −6 :∞∑

n=1

(−6 + 7)n

n2 + n+ 1

∞∑

n=1

(1)n

n2 + n+ 1

∞∑

n=1

1

n2 + n+ 1

converges (compare to∑

1/n2)

go back to first endpoint1

n2 + n+ 1=

∣∣∣∣(−1)n

n2 + n+ 1

∣∣∣∣∑|cn| converges ⇒

∑cn converges ⇒

∑ (−1)n

n2 + n+ 1converges

Final answerI.O.C = [−8,−6].

This is where section 1 ended on Monday, April 20

11.10 Taylor and Maclaurin Series

Theorem 1. Let f(x) be any function which is well-behaved at x = a. Then f(x)equals the following power series

f(x) =

∞∑

n=0

cn(x− a)n

where cn =f (n)(a)

n!, i.e.

c0 =f(a)

0!= f(a), c1 =

f ′(a)

1!= f ′(a), c2 =

f ′′(a)

2!, c3 =

f (3)(a)

3!, . . . .

We call the infinite series the Taylor series at the number a. If a = 0 we call thisthe Maclaurin series.

Corollary. If

∞∑

n=0

cn(x− a)n is the Taylor series for f(x) then

1. c0 + c1(x− a) is the tangent line at a2. c0 + c1(x− 1) + c2(x− a)2 is the tangent parabola at a3. f(x) can be approximated by a degree n polynomial

f(x) ≈ c0 + c1(x− a) + · · ·+ cn(x− a)n︸︷︷︸Taylor/Maclaurin polynomial

.


Of course, it’s silly taking thederivative a bunch of times forex, but this will be a usefulpattern for future examples.

Comments. Note that all the work in finding a Taylor series is in finding (andsometimes simplifying) the coefficients. We do not, ever, foil out the terms (x−a)2,(x−a)3, etc. In turn, all the work in finding coefficients is in taking derivatives andsimplifying.

When you are given a problem, you are told what f(x) is, and told what thenumber a is. You calculate the derivatives, the coefficients, and then put everythingtogether in a series.

Example 1. (a) Find the Maclaurin series for f(x) = ex.

(b) Write the series using∑

notation.

(c) Find the radius of convergence.(d) Use the series to approximate

√e to the 4th decimal place.

Solution. (a) We start by taking the derivative a few times:

f(x) = ex

f ′(x) = ex

f ′′(x) = ex

f ′′′(x) = ex

f (4)(x) = ex

. . .

f (n)(x) = ex

Now we plug in x = 0 to each of the derivatives:

f(x) = ex f(0) = e0 = 1

f ′(x) = ex f ′(0) = e0 = 1

f ′′(x) = ex f ′′(0) = e0 = 1

f ′′′(x) = ex f ′′′(0) = e0 = 1

f (4)(x) = ex f (4)(0) = e0 = 1

. . .

f (n)(x) = ex f (n)(0) = e0 = 1

Now that we have the derivatives, with x = 0 plugged in, we get the coefficients bydividing by the appropriate factorial:

f(x) = ex f(0) = e0 = 1 c0 = 1

f ′(x) = ex f ′(0) = e0 = 1 c1 = 1

f ′′(x) = ex f ′′(0) = e0 = 1 c2 =1

2

f ′′′(x) = ex f ′′′(0) = e0 = 1 c3 =1

3!

f (4)(x) = ex f (4)(0) = e0 = 1 c4 =1

4!. . .


Note: this time, and almostevery time, the best way towrite the derivative is to writedown the things that aremultiplied together, but not toactually calculate theirproduct. This way, you can seethe pattern. The patternusually involves a factorialand/or a geometric factor.

f (n)(x) = ex f (n)(0) = e0 = 1 cn =1

n!

Now that we have the coefficients we can write down the full series

ex = 1 + x+1

2x2 +

1

3!x3 +

1

4!x4 + . . .

(b) Now we turn the pattern we just wrote into∑

notation.

ex =∞∑

n=0

1

n!xn =

∞∑

n=0

xn

n!.

(c) Now we find the radius of convergence. We use the Ratio Test, and multiply

by the reciprocal ofxn

n!:

L = limn→∞

∣∣∣∣xn+1

(n+ 1)!· n!

xn

∣∣∣∣ = limn→∞

∣∣∣∣x ·1

n+ 1

∣∣∣∣ = 0.

Since the limit is < 1, and this doesn’t depend on x, the series converges for all x.In other words, R =∞.

(d) We plug x = 1/2 into both sides of our first version of the Maclaurin seriesfor ex:

e1/2 = 1 + (1/2) +(1/2)2

2!+

(1/2)3

3!+

(1/2)4

4!+

(1/2)5

5!+

(1/2)6

6!+ . . .

Now we add a few of these terms together and see how many it takes for the answerto settle down in the fourth decimal place.

1 + (1/2) +(1/2)2

2!+

(1/2)3

3!︸︷︷︸1.6458

+(1/2)4

4!+

(1/2)5

5!+

(1/2)6

6!

︸︷︷︸1.64872

+ . . . .

We note that adding more terms would continue to increase the accuracy of ouranswer. (For comparison, my calculator says that e1/2 ≈ 1.6487 so we were able tocalculate the same thing using only the +, × and ÷ signs on our calculator.)

Example 2. Find the Taylor series of f(x) = ln(x) with center equal to 1, write the

series using∑

notation, and find its radius of convergence, and see if you recognize

what you get using this series to calculate ln(2).

Solution. We follow the pattern of the last example as closely as possible: We startby taking the derivative a few times:

f(x) = ln(x)

f ′(x) =1

x= x−1

f ′′(x) = −x−2

f ′′′(x) = 2x−3


f (4)(x) = −3 · 2x−4 = −3!x−4

f (5)(x) = 4! x−5

Now we plug in x = 1 to each of the derivatives:

f(x) = ln(x) f(1) = ln(1) = 0

f ′(x) = x−1 f ′(1) = 1

f ′′(x) = −x−2 f ′′(1) = −1−2 = −1

f ′′′(x) = 2x−3 f ′′′(1) = 2(1)−3 = 2

f (4)(x) = −3! x−4 f (4)(1) = −3!

f (5)(x) = 4! x−5 f (5)(1) = 4!

Now that we have the derivatives, with x = 1 plugged in, we get the coefficients bydividing by the appropriate factorial:

f(x) = ln(x) f(1) = ln(1) = 0 c0 = 0

f ′(x) = x−1 f ′(1) = c1 = 1

f ′′(x) = −x−2 f ′′(1) = −1−2 = −1 c2 = −1

2

f ′′′(x) = 2x−3 f ′′′(1) = 2(1)−3 = 2 c3 =2

3!=

1

3

f (4)(x) = −3! x−4 f (4)(1) = −3! c4 =−3!

4!= −1

4

f (5)(x) = 4! x−5 f (5)(1) = 4! c5 =4!

5!=

1

5

Now that we have the coefficients we can write down the full series

ln(x) = (x− 1)− 1

2(x− 1)2 +

1

3(x− 1)3 − 1

4!(x− 1)4 + . . .

=

∞∑

n=1

(−1)n+1 (x− 1)n

n

Now we find the radius of convergence. We start with the Ratio Test, and multiply

by the reciprocal of(x− 1)n

n:

L = limn→∞

∣∣∣∣(−1)n+2

(−1)n+1

(x− 1)n+1

n+ 1

n

(x− 1)n

∣∣∣∣ = limn→∞

∣∣∣∣(−1)|x− 1|n+ 1

n

∣∣∣∣ = |x− 1|

We set L = |x− 1| < 1 and find R = 1.As an extra, we can use this series to calculate ln(2):

ln(2) =∞∑

n=1

(−1)n+1

n(2− 1)n =

∞∑

n=1

(−1)n+1

nA.H.T.

We’ve seen this series before: it’s the Alternating Harmonic Series!!



Example 3. Find the Maclaurin series for f(x) = sin(x), write the series using∑

notation, and find its radius of convergence.

Solution. We start by taking the derivative a few times:

f(x) = sin(x)

f ′(x) = cos(x)

f ′′(x) = − sin(x)

f ′′′(x) = − cos(x)

f (4) = sin(x)

(After this it repeats)

Now we plug in x = 0.

f(x) = sin(x) f(0) = 0

f ′(x) = cos(x) f ′(0) = 1

f ′′(x) = − sin(x) f ′′(0) = 0

f ′′′(x) = − cos(x) f ′′′(0) = −1

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


Now we can find the coefficients

f(x) = sin(x) f(0) = 0 c0 = 0

f ′(x) = cos(x) f ′(0) = 1 c1 = 1

f ′′(x) = − sin(x) f ′′(0) = 0 c2 =0

2= 0

f ′′′(x) = − cos(x) f ′′′(0) = −1 c3 =−1

3!

f (4) = sin(x) f (4) = 0 c4 =0

4!= 0

1 c5 =1

5!

0 c6 =0

6!

− 1 c7 =−1

7!

Thus we have

sin(x) = x− 1

3!x3 +

1

5!x5 − 1

7!x7 + . . .


We have to work harder to write this in∑

notation, mostly because we need a

formula that just gives the odd powers of x, and the odd factorials. In other words,we need to figure out a formula for the following outputs

value of n in∞∑

n=0

: 0 1 2 3 4 5

resulting power of x and factorial : 1 3 5 7 9 11

A little thought shows that we can use the formula 2n+ 1 for the powers of x andthe factorials. Thus,

sin(x) =

∞∑

n=0

(−1)nx2n+1

(2n+ 1)!.

Now we find the radius of convergence. We start with the Ratio Test, and

multiply by the reciprocal ofx2n+1

(2n+ 1)!. Make sure you see exactly where to put

n+ 1:

L = limn→∞

∣∣∣∣∣(−1)n+1

(−1)n· x2(n+1)+1

(2(n+1) + 1)!· (2n+ 1)!

x2n+1

∣∣∣∣∣ = limn→∞

∣∣∣∣(−1) · x2n+3

x2n+1

(2n+ 1)!

(2n+ 3)!

∣∣∣∣ .

To see how to simplify the factorials, think about what sorts of numbers we have:2n+ 1 is an odd number, and 2n+ 3 is the next odd number. So we have a fractionwith factorials on top and bottom, starting with an odd number on top and thenext odd number on the bottom. For example, the fraction could be something like5!

7!. If you write out all the factors in this fraction, they all cancel except for the 6

and 7 on the bottom. The same thing holds in general, so we get:

L = limn→∞

∣∣∣∣(−1) · x2n+3

x2n+1

1

(2n+ 2)(2n+ 3)

∣∣∣∣ = 1 · |x| · 0 = 0.

Since the limit is 0, and since 0 < 1, we see that the series converges for all x, i.e.R =∞.

Example 4. [The Binomial Series] Let k be any real number. Find the Maclaurin

Series for f(x) = (1 + x)k, write your answer in∑

notation.


f(x) = (1 + x)k

f ′(x) = k(1 + x)k−1

f ′′(x) = k(k − 1)(1 + x)k−2

f ′′′(x) = k(k − 1)(k − 2)(1 + x)k−3

f (4) = k(k − 1)(k − 2)(k − 3)(1 + x)k−4

Now we plug in x = 0.

f(x) = (1 + x)k f(0) = 1


f ′(x) = k(1 + x)k−1 f ′(0) = k

f ′′(x) = k(k − 1)(1 + x)k−2 f ′′(0) = k(k − 1)

f ′′′(x) = k(k − 1)(k − 2)(1 + x)k−3 f ′′′(0) = k(k − 1)(k − 2)

f (4) = k(k − 1)(k − 2)(k − 3)(1 + x)k−4 f (4)(0) = k(k − 1)(k − 2)(k − 3)


f(x) = (1 + x)k f(0) = 1 c0 = 1

f ′(x) = k(1 + x)k−1 f ′(0) = k c1 = k

f ′′(x) = k(k − 1)(1 + x)k−2 f ′′(0) = k(k − 1) c2 =k(k − 1)

2!

f ′′′(x) = k(k − 1)(k − 2)(1 + x)k−3 f ′′′(0) = k(k − 1)(k − 2) c3 =k(k − 1)(k − 2)

3!

f (4) = k(k − 1)(k − 2)(k − 3)(1 + x)k−4 f (4)(0) = k(k − 1)(k − 2)(k − 3) c4 =k(k − 1)(k − 2)(k − 3)

4!

Now we write the power series out without∑

notation:

(1 + x)k = 1 + kx+k(k − 1)

2!x2 +

k(k − 1)(k − 2)

3!x3 + . . .

Now come up with a more compact way to write this. Recall, or define, thebinomial coefficient,

(k

n

)=

n factors︷︸︸︷k(k − 1)(k − 2) · · · (k − n+ 1)

n!

(we set

(k

0

)= 1 and

(k

1

)= k

).

Note that cn =

(k

n

). Thus, we can write Then we can write the

∑notation:

(1 + x)k =

∞∑

n=0

(k

n

)xn

Now we illustrate what this series looks like for k = 3 and k = 1/2.

E.g. if k = 3 then

(3

0

)= 1,

(3

1

)= 3,

(3

2

)=

3 · 22!

= 3,

(3

3

)=

3 · 2 · 13!

= 1.

What about

(3

4

)? We get

3 · 2 · 1 · 04!

= 0. Similarly,

(3

5

)= 0 etc. Thus,

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

which is familiar from elementary algebra.E.g. if k = 1/2 then

(1 + x)k =√

1 + x


= 1 +1

2x+

12 · −1

2

2!x2 +

12 · −1

2 · −32

3!x3 +

12 · −1

2 · −32 · −5

2

4!x4 + . . .

= 1 +1

2x− 1

2 · 2!x2 +

1 · 323 · 3!

x3 − 1 · 3 · 524 · 4!

x4 + . . .

=∞∑

n=0

(−1)n+1 (2n− 3)!!

2n · n!xn

where (2n − 3)!! denotes the “double factorial” of (2n − 3) (i.e. start at 2n − 3and go down by 2 at each step till you get to 1); out of convenience, let’s define(−3)!! = −1 and (−1)!! = 1.

Extra Examples

Example 5. Find the Taylor series of f(x) = sin(x) centered at a = π/4. You do

not have to write this using∑

notation, but you can try if you like.


f(x) = sin(x)

f ′(x) = cos(x)

f ′′(x) = − sin(x)

f ′′′(x) = − cos(x)

f (4) = sin(x)


Now we plug in x = π/4.

f(x) = sin(x) f(0) = 1/√

2

f ′(x) = cos(x) f ′(0) = 1/√

2

f ′′(x) = − sin(x) f ′′(0) = −1/√

2

f ′′′(x) = − cos(x) f ′′′(0) = −1/√

2

f (4) = sin(x) f (4) = 1/√

2



f(x) = sin(x) f(0) = 1/√

2 c0 = 1/√

2

f ′(x) = cos(x) f ′(0) = 1/√

2 c1 = 1/√

2

f ′′(x) = − sin(x) f ′′(0) = −1/√

2 c2 =−1

2!√

2


f ′′′(x) = − cos(x) f ′′′(0) = −1/√

2 c3 =−1

3!√

2

f (4) = sin(x) f (4) = 1/√

2 c4 =1

4!√

2

c5 =1

5!√

2

c6 =−1

6!√

2

c7 =−1

7!√

2

Thus we have

sin(x) =1√2

+1√2x− 1

2!√

2x2 − 1

3!√

2x3 +

1

4!√

2x4 +

1

5!√

2x5 − 1

6!√

2x6 − . . .

The previous equation solves the example, but we can try to put this into∑

notation. It’s easy to find a formula that gives most of each coefficient, namely the

part with1

n!√

2. The only hard part is to find a formula that gives 1, 1, −1, −1, 1,

1, . . . . We could do this using some combinations of sin. In fact, if we are goingto use sin, we might as well use it to give part of the coefficient with

√2 too:

cn =sin((2n+ 1)π/4)

n!sin(x) =

∞∑

n=0

sin((2n+ 1)π/4)

n!(x− π/4)n.

But maybe that looks weird, having a formula with sin on both sides.Well, the alternative is that we need to use (and possibly learn), yet another

function. Let int(x) be the integer part function: it returns the integer part of x(i.e. the part before the decimal point). Then int(0) = 0, int(0.5) = 0, int(1) = 1,int(1.5) = 1, etc. Therefore

∞∑

n=0

(−1)int(n/2) = (−1)0 + (−1)0 + (−1)1 + (−1)1 + (−1)2 + (−1)2 + (−1)3 + (−1)3 + . . .

= 1 + 1− 1− 1 + 1 + 1− 1− 1 + . . .

Combining this with the rest of the formula for cn we have

sin(x) =∞∑

n=0

(−1)int(n/2)

n!√

2(x− π/4)n.

Example 6. Find the Taylor series of f(x) = x−2 centered at a = 2, write the

series using∑

notation, and find its radius of convergence.

Solution. We start by taking the derivative enough times to see what the patternis:

f(x) = x−2

f ′(x) = (−2)x−3


f ′′(x) = 2 · 3x−4

f ′′′(x) = −2 · 3 · 4x−5

f (4) = 2 · 3 · 4 · 5x−6

. . .

f (n) = (−1)n(n+ 1)!x−(n+2)

Now we plug in x = 2

f(x) = x−2 f(2) =1

22

f ′(x) = (−2)x−3 f ′(2) =−2

23

f ′′(x) = 2 · 3x−4 f ′′(2) =3 · 224

f ′′′(x) = −2 · 3 · 4x−5 f ′′′(2) =−4 · 3 · 2

25

f (4) = 2 · 3 · 4 · 5x−6 f (4)(2) =5 · 4 · 3 · 2

26

. . .

f (n) = (−1)n(n+ 1)!x−(n+2) f (n)(2) =(−1)n(n+ 1)!

2n+2

Now we divide by n! to find the coefficients:

f(x) = x−2 f(2) =1

22c0 =

1

22

f ′(x) = (−2)x−3 f ′(2) =−2

23c1 =

−2

23

f ′′(x) = 2 · 3x−4 f ′′(2) =3 · 224

c2 =3 · 2

2! · 24=

3

24

f ′′′(x) = −2 · 3 · 4x−5 f ′′′(2) =−4 · 3 · 2

25c3 =

−4 · 3 · 23! · 25

=−4

25

f (4) = 2 · 3 · 4 · 5x−6 f (4)(2) =5 · 4 · 3 · 2

26c4 =

5 · 4 · 3 · 24! · 26

=5

26

. . .

f (n) = (−1)n(n+ 1)!x−(n+2) f (n)(2) =(−1)n(n+ 1)!

2n+2c5 =

(−1)n(n+ 1)!

n! 2n+2=

(−1)n(n+ 1)

2n+2

We write out a few terms of the series without∑

notation:

1

22− 2

23(x−2)+

3

24(x−2)2− 4

25(x−2)3+

5

26(x−2)4−· · ·+ (−1)n(n+ 1)

2n+2(x−2)n+. . .

and now try to figure out the∑

notation:

∞∑

n=0

(−1)n(n+ 1)

2n+2(x− 2)n


Now we find the radius of convergence. We start with the ratio test, and multi-

plying by the reciprocal of(n+ 1)(x− 2)n

2n+2:

L = limn→∞

∣∣∣∣(−1)n+2

(−1)n+1· (n+ 2)(x− 2)n+1

2n+3· 2n+2

(n+ 1)(x− 2)n

∣∣∣∣ = limn→∞

∣∣∣∣(−1) · n+ 2

n+ 1· (x− 2) · 1

2

∣∣∣∣ =1

2|x−2|

We set this limit to be < 1 and solve for |x− 2|:

1

2|x− 2| < 1⇒ |x− 2| < 2

So R = 2.

Example 7. Recall that the Maclaurin series for ex is as shown:

ex =

∞∑

n=0

xn

n!

(a) Let fn(x) be the Maclaurin polynomial of degree n. Compare the graphs ofex with fn(x) for n = 0, 1, 2, 3, 4 and 5.

(b) For each formula below, what is the smallest degree Maclaurin polynomialyou can use to approximate the value, to within roughly 0.1 accuracy? (Thisaccuracy is basically the smallest gap you can see between the graphs. In otherwords, which graph would you have to use so that the gap and the indicatedx-value is either invisible, or as close to invisible as possible.)

e0.1, e−0.6, e1, e1.8, e−2.

(c) Finish the approximations from the previous part, and compare them withwhat you get using your calculator’s built-in funcition.

Solution. (a) We show 6 graphs

—– ex, − − − 1 —– ex, − − − 1 + x


—– ex, − − − 1 +x2

2!—– ex, − − − 1 + x+

x2

2!+x3

3!

—– ex, − − − 1 + x+ · · ·+ x4

4!—– ex, − − − 1 + x+ · · ·+ x5

5!

(b) For each value we look at the pictures, at the indicated x-value, and gofrom one graph to the next until we find the graph with no gap, or a barely visiblegap. For instance, for x = 0.1, we see that the very first picture has almost no gapbetween the two functions being graphed. Thus, we would use the first picture, andf0(x) = 1. We summarize all the values below

e0.1 ≈ f0(0.1), first picture e1.8 ≈ f4(1.8), fifth picture

e−0.6 ≈ f1(−0.6), second picture e−2 ≈ f5(−2), sixth picture

e1 ≈ f3(1), fourth picture

(c) Now we calculate using the Maclaurin polynomials identified from above:

e0.1 ≈ f0(0.1) = 1 e1.8 ≈ f4(1.8) = 1+1.8+1.82

2!+

1.83

3!+

1.84

4!= 5.8294

e−0.6 ≈ f1(−0.6) = 1 + (−0.6) = 0.4 e−2 ≈ f5(−2) = 1−2 +(−2)2

2!+ . . .+

(−2)5

5!= 0.06

e1 ≈ f3(1) = 1 + (1) +12

2!+

13

3!= 2.6

And here are the results using the built-in function on our calculators:

e0.1 = 1.105170918 e1.8 = 6.049647464


e−0.6 = 0.5488116361 e−2 = 0.1353352832

e1 = 2.718281828

Example 8. Recall that the Taylor series for ln(x), centered at 1, is as shown:

ln(x) =∞∑

n=1

(−1)n+1 (x− 1)n

n

(a) Let fn(x) be the Taylor polynomial of degree n. Compare the graphs of ln(x)with fn(x) for n = 1, 2, 4, 6, 8 and 10.

(b) For each formula below, what is the smallest degree Maclaurin polynomialyou can use to approximate the value, to within roughly 0.1 accuracy? (Thisaccuracy is basically the smallest gap you can see between the graphs. In otherwords, which graph would you have to use so that the gap and the indicatedx-value is either invisible, or as close to invisible as possible.)

ln(1.1), ln(0.4), ln(1.75), ln(2), ln(0.2).

(c) Finish the approximations from the previous part, and compare them withwhat you get using your calculator’s built-in funcition.

Solution. (a) We show 6 graphs

—– ln(x), − − − (x− 1) —– ln(x), − − − (x− 1)− (x−1)2

2!

− − − (x− 1)− (x−1)2

2! + · · · − (x−1)4

4! − − − (x− 1)− (x−1)2

2! + · · · − (x−1)6

6!


− − − (x− 1)− (x−1)2

2! + · · · − (x−1)8

8! − − − (x− 1)− (x−1)2

2! + · · · − (x−1)10

10!

(b) For each value we look at the pictures, at the indicated x-value, and gofrom one graph to the next until we find the graph with no gap, or a barely visiblegap. For instance, for x = 0.1, we see that the very first picture has almost no gapbetween the two functions being graphed. Thus, we would use the first picture, andf1(x) = (x− 1). We summarize all the values below

ln(1.1) ≈ f1(1.1), first picture ln(2) ≈ f8(2), fifth picture

ln(0.4) ≈ f2(0.4), second picture ln(0.2) ≈ f10(0.2), sixth picture

ln(1.75) ≈ f4(1.75), third picture

(c) Now we calculate using the Taylor polynomials identified from above:

ln(1.1) ≈ (1.1− 1) = 0.1

ln(0.4) ≈ (0.4− 1)− (0.4− 1)2

2!= −0.78

ln(1.75) ≈ (0.75− 1)− (0.75− 1)2

2!+

(0.75− 1)3

3!− (0.75− 1)4

4!= 0.5302734375

ln(2) ≈ (2− 1)− (2− 1)2

2!+ · · · − (2− 1)8

8!= 0.6345238095

ln(0.2) ≈ (0.2− 1)− (0.2− 1)2

2!+ · · · − (0.2− 1)10

10!= −1.578874667

And here are the results using the built-in function on our calculators:

ln(1.1) = 0.09531017980 ln(2) = 0.6931471806

ln(0.4) = −.9162907319 ln(0.2) = −1.609437912

ln(1.75) = 0.5596157879

Appendix A

Absolute values

Here is a list of the basic properties of absolute values.1. Absolute value of x is defined as

|x| ={x if x ≥ 0

−x if x < 0

2. The easy way to think of |x|: just make x positive!3. The graph of the absolute value is

4. Absolute value of x− y equals

|x− y| ={x− y if x ≥ yy − x if x < y

5. The easy way to think of |x− y|: the distance between x and y.6. Algebraic properties of absolute value:

a|x− y| = |ax− ay| if a > 0

7. Solving |x− y| < a:

|x− y| < a is equivalent to − a < x− y < a

is equivalent to y − a < x < y + a

157

Appendix B

Trig functions

1. In Calculus all angles will be measured in radians! The reason for this isthat the derivative rules that we will learn, only work in radians.

2. π radians = 180◦

90◦ = π/2, 60◦ = π/3, 45◦ = π/4, 30◦ = π/6.︸︷︷︸the “standard” angles

3. In a right triangle we have

sin =opp

hyp, cos =

adj

hyp, tan =

opp

adj.

4. In a 30-60-90 triangle the sides have lengths (or the ratio of lengths) of 1, 2and√

3. In a 45-45-90 triangle the sides have length (or the ratio of lengths)of 1, 1 and

√2. Therefore,

0 π/6 π/4 π/3 π/2

sin(θ) 0 1/2 1/√

2√

3/2 1

cos(θ) 1√

3/2 1/√

2 1/2 0

note this pattern is easy to remember if youwrite the entries with square roots of 0,1,2,3,4on top:

√0/2,

√1/2,

√3/2,

√4/2

5. To define our functions for an angle θ that is beyond 90◦, we place θ in a circleof radius r and let α be the reference angle, i.e. the smallest angle between θand the x-axis, and let (x, y) be the coordinates on the circle determined byθ.

158

APPENDIX B. TRIG FUNCTIONS 159

θ

(x, y)

α r

Then we have

sin(θ) =y

r= ± sin(α)

cos(θ) =x

r= ± cos(α)

tan(θ) =y

x= ± tan(α)

To determine ± in each case, you can (1) look at whether x or y is + or −, oruse a mnemonic to remember which function(s) are positive in each quadrant:

Positive trig functions in each quadrant

A (All)S (Sine)

T (Tangent) C (Cosine)

The “standard” values for sine and cosine are summarized on the next page.

0 π/6 π/4 π/3 π/2 2π/3 3π/4 5π/6 π 7π/6 5π/4 4π/3 3π/2 5π/3 7π/4 11π/6

sin 0 1/2 1/√2√3/2 1

√3/2 1/

√2 1/2 0 −1/2 −1/

√2 −

√3/2 −1 −

√3/2 −1/

√2 −1/2

cos 1√3/2 1/

√2 1/2 0 −1/2 −1/

√2 −

√3/2 −1 −

√3/2 −1/

√2 −1/2 0 1/2 1/

√2

√3/2

sin

cos

1/2

1/√2

√3/2

−1/2−1/√2

−√3/2

1

−1

π6

π4

π3

2π3

3π4

5π6

5π3

7π4

11π6

7π6

5π4

4π3

π2

π3π2

0

π/6

π/4

π/3π/2

2π/3

3π/2

5π/6

π

7π/6

5π/4

4π/33π/2

5π/3

7π/4

11π/6

sin = 0cos = 1

sin = 1/2

cos =√3/2

sin = 1/√2

cos = 1/√2

sin =√3/2

cos = 1/2

sin = 1cos = 0

sin =√3/2

cos = −1/2sin = 1/

√2

cos = −1/√2

sin = 1/2

cos = −√3/2

sin = 0cos = −1

sin = −1/2cos = −

√3/2

sin = −1/√2

cos = −1/√2

sin = −√3/2

cos = −1/2sin = −1cos = 0

sin = −√3/2

cos = 1/2

sin = −1/√2

cos = 1/√2

sin = −1/2cos =

√3/2

Appendix C

Horizontal Asymptotes

Definition. Let f(x) be any function. We write

limx→∞

f(x) = L

if y = L is a horizontal asymptote (on the right) for f(x). In other words, they-values of f(x) become infinitely close to y = L as the x-values become infinitelylarge.

1. limn→∞

a

bxn= 0

2. limx→∞

axn

b=∞, i.e. there is no horizontal asymptote.

3. Rational functions:

limn→∞

axn + bxn−1 + . . .

rxm + sxm−1 + . . .= lim

n→∞

axn

rxm

To finish, simplify this and apply one of the previous rules.4. lim

x→∞e−x = 0

5. limx→∞

tan−1(x) = π/2

6. limx→∞

ln(x) =∞, i.e. there is no horizontal asymptote.

161

Appendix D

Extended Reals

Note: we can summarize some of the rules for horizontal asymptotes using functionnotation with ∞:

1

∞ = 0

e−∞ = 0

ln(∞) =∞e∞ =∞

tan−1(∞) = π/2

There are two ways to look at the above notation. (1) It is just an abbreviationof the “real” statement which involves limits. Thus, writing “e−∞ = 0” is just ashort hand way of saying lim

x→∞e−x = 0. (2) We can define the extended reals.

These are the usual real numbers, together with ±∞. Then we define the abovecalculations on the extended real numbers.

It’s convenient to define the following combinations of extended reals. Let a beany real number:

a

∞ = 0 a∞ =∞ if a > 0

∞a

=∞ if a > 0 ∞+ a =∞a

0=∞ if a > 0 ∞− a =∞

a∞ =∞ if a > 1 ∞ ·∞ =∞a∞ = 0 if 0 < a < 1

162

Appendix E

L’ Hospital’s Rule

Theorem 2 (L’ Hospital’s Rule).

limf(x)

g(x)=

0

0,∞∞ =⇒ lim

f(x)

g(x)= lim

f ′(x)

g′(x)

where “lim” is any kind of limit (i.e. limx→a

, limx→a+

, limx→a−

, limx→∞

, limx→−∞

).

Comments. Warning: (1) This rule is false if you don’t have0

0or∞∞ : you need

to check for this every time you use this rule. (2) This rule only applies if limf ′(x)

g′(x)exists, or is infinite.

Example 9.

limx→0

sin(x)

x=

0

0X

limx→0

sin(x)

x

LH= lim

x→0

cos(x)

1=

cos(0)

1= 1

Example 10.

limx→∞

ex/2

x2=e∞/2

∞2=∞∞X

limx→∞

ex/2

x2

LH= lim

x→∞

12ex/2

2x=

12e∞/2

2∞ =∞∞X (use L.H. again)

limx→∞

12ex/2

2x

LH= lim

x→∞

14ex/2

2=

14e∞/2

2=∞2

=∞

A similar calculation should convince you that limx→∞

ex

xn=∞ no matter what power

of x we have on the bottom.

Corollary (L’Hospital’s Rule, Variation I, products).

lim f(x) · lim g(x) = 0 · ∞ =⇒ lim f(x) · g(x) = limf(x)

1/g(x)= lim

g(x)

1/f(x).

Apply the previous L’ Hospital’s rule to one of these fractions.

163

APPENDIX E. L’HOSPITAL’S RULE 164

Example 11.

limx→0+

x ln(x) = 0 ln(0+) = 0(−∞)X

limx→0+

x ln(x) = limx→0+

ln(x)

1/x

LH= lim

x→0+

1/x

−1/x2= lim

x→0

−x=

0

Notes: (1) We chose the fractionln(x)

1/xbecause it seemed that taking the derivative

of ln(x) and 1/x were simpler than taking the derivative of x and 1/ ln(x). (2) Wehad to simplify the results after taking the derivative, before we could finish takingthe limit.

Corollary (L’Hospital’s Rule, Variation II, differences).

lim(f(x)− g(x)

)=∞−∞ =⇒ lim

(f(x)− g(x)

)= lim some fraction.

How to rewrite f(x)−g(x) as a fraction will depend on what f(x) and g(x) include,and make take a little creativity or hard work. After finding the fraction, you shouldcheck again and make sure that you can use L’Hospital.

Example 12.

limx→π/2−

(sec(x)− tan(x)

)= sec(π/2−)− tan(π/2−) =∞−∞X

limx→π/2−

(sec(x)− tan(x)

)= lim

x→π/2−

(1

cos(x)− sin(x)

cos(x)

)= lim

x→π/2−1− sin(x)

cos(x)=

1− 1

0X

limx→π/2−

1− sin(x)

cos(x)

LH= lim

x→π/2−0− cos(x)

− sin(x)=

cos(π/2)

sin(π/2)= 0/1 = 0

Corollary (L’Hospital’s Rule, Variation III, powers).

lim f(x)g(x)

= 00, 1∞, ∞0 =⇒ lim f(x)g(x)

= elim g(x) ln(f(x))

.

The limit lim g(x) ln(f(x)) is handled as described above for L’Hospital VariationII, products. Note that the use of e and ln cancel each other out, leaving somethingequal to the original limit.

Example 13.

limx→∞

(1 +

1

x

)x=

(1 +

1

∞

)∞= (1 + 0)∞ = 1∞X

limx→∞

(1 +

1

x

)x= e

limx ln

(1 +

1

x

)

limx→∞

x ln

(1 +

1

x

)= lim

x→∞

ln(1 + 1/x)

1/x

LH= lim

x→∞

11+1/x · (0− 1/x2)

−1/x2= lim

x→∞

1

1 + 1/x=

1

1 + 0= 1

limx→∞

(1 +

1

x

)x= e1 = e

Notes: (1) again we chose to rewrite the product, x ln(1 + 1/x), as a fraction byleaving ln on top. (2) We simplified the fraction after we applied L’Hospital to it.

APPENDIX E. L’HOSPITAL’S RULE 165

Summary1. The cases you use L’Hospital for are

0

0,∞∞ , 0 · ∞, ∞−∞, 00, 1∞, ∞0.

2. In every case, the only part that is really L’Hospital is taking the derivative of

the top and bottom of a fraction (marked in the above examples withLH= ). All

the rest is algebra. You never take the derivative of a product or a differenceor powers of functions directly in L’Hospital problems.

Appendix F

Cartesian and Polar Graphpaper

Example 14. Cartesian graph paper, from (−3.5,−3.5) to (3.5, 3.5)

166

APPENDIX F. CARTESIAN AND POLAR GRAPH PAPER 167

−3.5

−3.25

−3

−2.75

−2.5

−2.25

−2

−1.75

−1.5

−1.25

−1

−0.75

−0.5

−0.25

0

0.25

0.5

0.75

1

1.25

1.5

1.75

2

2.25

2.5

2.75

3

3.25

3.5

−3.5 −3 −2.5 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 2.5 3 3.5

Example 15. Polar graph paper, from (0, 0) to (3.5, 2π)

APPENDIX F. CARTESIAN AND POLAR GRAPH PAPER 168

0.35

0.7

π/3

1.41.75

2π/3

2.4

2.8

π

3.5

3.8

4π/3

4.5 4.9

5π/3

5.6

5.9

0.2

π/6

0.9

1.2

π/2

1.9

2.3

5π/6

3.0

3.3

7π/6

4.0

4.4

3π/2

5.1

5.4

11π/6

6.1

0.5

0.5

1

1

1.5

1.5

2

2

2.5

2.5

3

3

3.5

3.5

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evergreen.loyola.edu · Contents 5 Integrals reviewed 2 5.1 Basic facts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .2 5.5 U-substitution