contents · 2016-12-09 · w. ethan duckworth, loyola university maryland, 2016 de nitions of...

Contents

5 Integrals reviewed 35.1 Basic facts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35.5 U -substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

6 Integral Applications 116.1 Area between two curves . . . . . . . . . . . . . . . . . . . . . . . 116.2 Volumes by rotation . . . . . . . . . . . . . . . . . . . . . . . . . . 166.3 Volumes by Cylindrical Shells . . . . . . . . . . . . . . . . . . . . . 266.5 Average Value of a Function . . . . . . . . . . . . . . . . . . . . . 32

7 Techniques of Integration 347.1 Integration by Parts . . . . . . . . . . . . . . . . . . . . . . . . . . 347.2 Trigonometric Integrals . . . . . . . . . . . . . . . . . . . . . . . . 397.3 Trigonometric Substitution . . . . . . . . . . . . . . . . . . . . . . 457.4 Integration of Rational Functions . . . . . . . . . . . . . . . . . . . 527.5 Strategy for Integration . . . . . . . . . . . . . . . . . . . . . . . . 617.8 Improper Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

8 Further Integral Applications 748.1 Arc Length . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 748.2 Optional: Area of a Surface of Revolution . . . . . . . . . . . . . . 788.5 Probability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81

10 Parametric and Polar 8710.3 Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . 8710.4 Areas in Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . 9710.1 Curves Defined by Parametric Equations . . . . . . . . . . . . . . 10310.2 Calculus with Parametric Curves . . . . . . . . . . . . . . . . . . . 109

11 Sequences and Series 12111.2 Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12111.6 Absolute Convergence and the Ratio Test . . . . . . . . . . . . . . 13011.8 Power Series. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13511.10 Taylor and Maclaurin Series . . . . . . . . . . . . . . . . . . . . . . 14111.11 Applications of Taylor Series . . . . . . . . . . . . . . . . . . . . . 154

11.11.1 Calculating Important Numbers . . . . . . . . . . . . . . . . 15511.11.2 Calculating “Impossible” Integrals . . . . . . . . . . . . . . . 15811.11.3 Solving “Impossible” Equations . . . . . . . . . . . . . . . . . 159

1

CONTENTS 2

11.11.4 Finding Difficult Limits . . . . . . . . . . . . . . . . . . . . . 16111.11.5 Calculating Transcendental Functions . . . . . . . . . . . . . 162

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 tripartite 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, 2016

Definitions of Integrals

DefinitionIf f(x) is any function, and [a, b] any interval that f is defined

on, we define a number

∫ b

a

f(x) dx as follows:

∫ b

a

f(x) dx = the number that the sum

f(x1)∆x + f(x2)∆x + · · · + f(xn)∆x

approaches as n gets larger and larger (weassume that x1, . . . , xn, are n equallyspaced points in the interval [a, b], and

∆x =b− a

n).

We call the sum f(x1)∆x+ f(x2)∆x+ · · ·+ f(xn)∆x a Rie-

mann Sum, and we call

∫ b

a

f(x) dx the definite integral.

Details

• We divide the interval [a, b] into n equal pieces, each ofwidth ∆x

· · · · · ·a b

= ∆x

• There are some standard ways to pick each xi:

Left Hand Rule:x1, . . . , xn are the x-values on the left of the intervals

x1 x2 x3 xn· · · · · ·a b

Right Hand Rule:x1, . . . , xn are the x-values on the left of the intervals

x1 x2 xn−1 xn· · · · · ·a b

Midpoint Hand Rule:x1, . . . , xn are the x-values on the left of the intervals

x1 x2 xn· · · · · ·

a b

Names and notationWe can abbreviate a Riemann sum using

∑notation

f(x1)∆x + · · · + f(xn)∆x =

n∑

i=1

f(xi)∆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, 2016

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 inExample 1. In this example we break it down into more steps than many readerswill need.

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


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.

This is where we ended on Tuesday, September 6


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 substitution,

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 Wednesday, September 7

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 how the area of all four can be written as a sum of things ofthe form 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.

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

= −(

10.4773

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

)

− −0.4773

3+ 5(−0.477)2 + 5(−0.477)

≈ 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).)

This is where we ended on Friday, September 9

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 · zWe 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

This is where we ended on Monday, September 12

Example 1. I’m about to make the world’s best egg salad, but the recipe calls for30 cubic inches of hard boiled eggs (i.e. 2 cups of hard boiled eggs). Since I wantto make the world’s best egg salad, I need to be incredibly precise, and so I decideto calculate the volume of a single hard boiled egg, shown below:


How do you think I’ll go about this?...

Being a mathematician, I obviously won’t measure this with water displace-ment and a measuring cup. I’ll use a ruler, because that is the foundation of allmeasurement2.

As a first approximation, I’ll just measure the whole egg. When I do this, I findthat it is about 2 in long and 1 + 5/8 in in diameter.

(a) Using these measurements, find an estimate of the volume of one egg.

Well, to make the world’s best egg salad, I need to have the world’s most precisemeasurement of the volume of an egg, and we haven’t done that. To be more precise,I’ll slice the egg, and then measure the slices with a ruler:

3

1

2

1/2

2Historically, we first learned how to measure along one dimension, and then we defined area ofsquares to measure in two dimensions, and then we extended it to three dimensions, etc.


So, I measure the slices of the egg, and this is what I find: (1) I have 11 slices,so each one is 2/11 in = 0.182 in thick. The diameters, and radii, are as follows

diameters radii

1 + 1/4 0.6251 + 3/8 0.6881 + 5/8 0.8131 + 3/4 0.8751 + 3/4 0.8751 + 5/8 0.8131 + 5/8 0.813

1.5 0.751 + 3/8 0.6881 + 1/4 0.625

1 0.5

(b) Using these measurements find a better estimate of the volume of one egg.

Well, that still isn’t precise enough. Suppose instead of measuring slices we use amathematical model. The outline of the egg described above is pretty well modeledby an ellipse

x2 +y2

0.8132= 1

Note three things: (1) the above calculations should have involved adding things of

the form πr2∆x, (2) to make the calculation exact we should use

∫πr2 dx, and (3)

that r = y, when applied to the equation of the ellipse.(c) Use the above observations to find an integral for the volume of an egg.

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 +4 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 + 4 and get

V =

∫ 2

−2π(−x2 + 4)2 dx = π

∫ 2

−2x4 − 8x2 + 16 dx

= π

(1

5x5 − 8

3x3 + 16x

) ∣∣∣∣2

−2

= π

(1

525 − 8

3· 23 + 32−

(− 1

525 +

8

3· 23 − 32

))

= 2π

(1

525 − 8

3· 23 + 32

)

= 2π

(25

5− 64

3+ 32

)

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

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 regionbounded by y = −2x2 +18, and y = 2x2 +2 (shown below) around the line y = −20(you may assume that the graphs intersect at x = ±2).

Solution. The rotated shape looks roughly as follows:


The larger radius is defined by the top curve, y = −2x2 + 18. It is the distancebetween this curve and the horizontal axis, y = −20. Thus,

R = −2x2 + 18− (−20) = −2x2 + 38

The smaller radius, the radius of the hole, is defined by the bottom curve y = 2x2+2.It is the distance between this curve and the horizontal axis y = −20. Thus

r = 2x2 + 2− (−20) = 2x2 + 22.

Now we integrate (skipping some of the messy steps)

V =

∫ 2

−2π(−2x2 + 38

)2 − π(2x2 + 22

)2dx

= π

∫ 2

−24x4 − 152x2 + 1444−

(4x4 + 88x2 + 484

)dx

= π

∫ 2

−2−240x2 + 960 dx

= π(−80x3 + 960x

) ∣∣∣∣2

−2

= 2000π

Extra examples

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

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?


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)

as a function of y gives x =

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

(nor do we know how to integrate

(√sin−1(y)

)2

).

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



6.5 Average Value of a Function

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

K(t) = 22 + 0.8t− 11 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 12 equally spaced times.

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

Solution. Part (a), discrete case. For 12 times we should have ∆t = 24/2 = 2hours. We will use the following t-values:

t = 0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22.

Now we simply plug these into K(t), add them up, and divide by 12:

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

12

≈

22.00 + 18.10 + 15.67 + 15.80 + 18.87 + 24.50+31.60 + 38.70 + 44.33 + 47.40 + 47.53 + 45.10

12

≈ 369.6

12≈ 30.8◦F

Part (b), integral case. The main trick here is to see how the kind of sum wefound above can be turned into an integral. Adding the values of K(t) simply turns

into

∫ 24

0K(t) dt. But what about dividing by 12? Here’s a way to rewrite that

part:

K(0) +K(2) +K(4) + · · ·+K(20) +K(22)

12

=1

12(K(0) +K(2) +K(4) + · · ·+K(20) +K(22))

=∆t

24(K(0) +K(2) +K(4) + · · ·+K(20) +K(22))


=1

24(K(0) +K(2) +K(4) + · · ·+K(20) +K(22)) ∆t

≈ 1

24

∫ 24

0K(t) dt

The formula we want is on the last line:

Average value of K(t) =1

24

∫ 24

0K(t) dt

=1

24

∫ 24

022 + 0.8t− 11 sin(πt/12) dt

=1

24

(22t+ 0.4t2 + 11 cos(πt/12) · 12

π

) ∣∣∣∣∣

24

0

=1

24

(22(24) + 0.4(242) + 11 cos(2π)− (0 + 0 + 11 cos(0))

)

=1

24

(22(24) + 0.4(242)

)

=1

24(528 + 230.4)

= 31.6◦F

Comparing Answers: There is one exact answer. Part (a), 30.8◦F is a goodapproximation, but part (b), 31.6◦F, is the exact average.

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.

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:

34

CHAPTER 7. TECHNIQUES OF INTEGRATION 35

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 ′′′

xxG+ . . .

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√

11x− 5dx.

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

2x3 − 3x+ 10√11x− 2

dx =

∫(2x3 − 3x+ 10) · (11x− 2)−1/2 dx.

We make two columns of functions

F G

2x3 − 3x+ 10 (11x− 5)−1/2 (both original functions)

6x2 − 3(11x− 5)1/2

1/2· 1

11 = 111 · 2(11x− 5)1/2

12x1

11· 2(11x− 5)3/2

3/2· 1

11 =22

3 · 112(11x− 5)3/2

1222

3 · 112

(11x− 5)5/2

5/2· 1

11 =23

3 · 5 · 113(11x− 5)5/2

023

3 · 5 · 113

(11x− 5)7/2

7/2· 1

11 =24

3 · 5 · 7 · 114(11x− 5)5/2

Now we take the terms that are connected by arrows in the above tables (i.e. thefirst function under F connects down to the right, to the second one under G, etc.):

∫2x3 − 3x+ 10√

11x− 5dx =

2

11(2x3−3x+10)(11x−5)1/2− 22

3 · 112(6x2−3)(11x−5)3/2

+23

3 · 5 · 113(12x)(11x− 5)5/2 − (12)

24

3 · 5 · 7 · 114(11x− 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

)


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)|.

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)|

)

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 cases 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

Complete

Table

forTrigonometric

Substitution

Follow

thetable

from

left

torigh

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onerow

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from

x=a

andx

=b

Ind

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nit

ein

tegra

l:R

ewri

teθ

and

other

trig

funct

ions

as

funct

ions

ofx

√A

2−x2

x=A

sin(θ

)

√A

2−A

2si

n2(θ

)N

owth

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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


The following page summarizes the main steps in Trigonometric substitution.

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

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 we ended on Monday, October 3


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

∣∣∣∣

This is where we ended on Tuesday, October 4

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

∣∣∣∣

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 4. (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|

Example 5. [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 6. [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|)

This is where we ended on Wednesday, October 5

Example 7. 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)

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.The degree of a polynomial is the biggest power of x that appears (or would

appear 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 8. 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)


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 9. 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


∫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 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



∫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 we ended on Friday, October 7

7.5 Strategy for Integration


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

)

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. Which technique would you use for each of the following integrals?

(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) Simplify/rewrite, and then trig powers. Specifically, we can rewrite

this as

∫tan3(x) sec3(x) dx.


(b) This is a kind of u-substituition known as a “rationalizing substitution”.

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) Polynomial division followed by partial fractions. Specifically, after you do

polynomial division, the remainder will be of the form∗

x3 − 3x2 − 10x. Factor

this to get∗

x(x− 5)(x+ 2).

(d) U -substitution. Specifically, u = ln(x), du =1

xdx, and the integral becomes

∫1√udu.

(e) This is a special kind of algebraic trick known as “multiplying by the con-jugate”. Specifically, 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) Basic anti-derivative. Specifically, note that e2 is just a constant number, so∫e2 dx = ex + C.

(g) Complete the squareS. Note that we can’t easily factor the bottom and so thisis not partial fractions.

(h) Polynomial division followed by basic anti-derivatives. Note that we have tostart 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 integralfrom Table 7.1.

(i) This is a rationalizing substitution. 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 polynomial division.


(j) Simplify/rewrite. Certainly this looks like trig-powers, and I won’t even saythat’s wrong, but when you simplify, you find out that you have basically no

non-trivial powers left. Specifically you getsin(x)

tan(x)+

sec(x)

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

You know

∫cos and you should look up

∫csc.

(k) Integration by parts (actually probably tabular integration by parts).(l) Completing the square, and then trig substitution. 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) Partial fractions. Note that it is easy to factor the denominator:x− 1

(x− 5)(x+ 1).

(n) Either a couple of fairly predictable u-substitions, or one big u-substitution,or a trig-substitution.

Two fairly predictable u-substitutions:

u = 1− x2

du = −2x dx∫= −1

2

∫1

u+√udu

w =√u (note u = w2)

dw =1

2u−1/2 du

2u1/2 dw = du

2w dw = du∫= −1

2

∫1

w2 + w2w dw

= − 1

w + 1dw

= − ln |w + 1|= . . . (not going to finish it)

You can combine the above double substitution into one substitution:

u =√

1− x2 (note 1− x2 = u2)

du =1

2√

1− x2· (−2x) dx

−√

1− x2 du = x dx

−u du = x dx∫= −

∫1

u2 + uu du

= −∫

1

u+ 1du

= − ln |u+ 1|


= . . . (not going to finish it)

Or you can do a trig-substitition:

x = sin(θ)

dx = cos(θ) dθ√

1− x2 = cos(θ) (note 1− x2 = cos2(θ))∫

=

∫sin(θ)

cos2(θ) + cos(θ)cos(θ) dθ

=

∫sin(θ)

cos(θ) + 1dθ

u = cos(θ)

du = − sin(θ) dθ∫= −

∫1

u+ 1du

= − ln |u|= . . . (not going to finish it)



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 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

af(x) dx =

F (b)− limx→−∞

F (x) if a = −∞limx→∞

F (x)− F (a) if b =∞limx→b−

F (x)− F (a) if x = b is a V.A.

F (b)− limx→a+

F (x) if x = a is a V.A.

This leaves two extra cases, where the above do not apply directly:


1. For

∫ ∞

−∞f(x) dx you MUST split the integreal into two pieces than fit into

the previous cases:

∫ ∞

−∞f(x) dx =

∫ 0

−∞f(x) dx+

∫ ∞

0f(x) dx

Note that 0 can be replaced here by any real number.

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

∫ b

af(x) dx MUST be split

into two pieces that fit into the previous cases:

∫ 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

x→∞F (x)− F (0), where F (x) = −e−x

= −(

limx→∞

e−x − e0)

= −(e−∞ − 1)

= 1

Example 2. 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(

limx→∞

F (x)− F (1)), where F (x) =

−1

x

= π

(limx→∞

−1

x− −1

1

)


= π

(−1

∞ + 1

)

= π(0 + 1)

= π

Here’s one way to think about this example. The shape we get by rotatation is aphysical solid, that is infinitely long, but has a finite volume. Here’s a nice wayto think about this: if you turned this horn vertically, you could fill it with liquid,using a finite amount of liquid.

Example 3. 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

x→∞ln(x)− ln(1)

=∞− 0

=∞

So in this case, the integral is divergent.

(b)

∫ ∞

1

1

x1.5dx =

∫ ∞

1x−1.5 dx

= limx→∞

F (x)− F (1) where F (x) =x−0.5

−0.5

=1

−0.5

(limx→∞

1

x0.5− 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

= limx→∞

F (x)− F (1) where F (x) =x0.3

0.3


=1

0.3

(limx→∞

x0.3 − 1)

=1

0.3(∞− 1)

=∞

So in this case the integral is divergent.


Example 4. 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


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) = ±∞”.

Example 5. 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

= limx→0−

F (x)− F (−2) + F (1)− lima→0+

F (x), where F (x) = −1

x

= −(

limx→0−

1

x− 1

−2

)−(

1

1− limx→0+

1

x

)

= −(−∞+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 6. Figure out which of the following converge.

(a)

∫ 1

0

1

xdx

(b)

∫ 1

0

1

x1.3 dx


(c)

∫ 1

0

1

x0.6 dx

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− lima→0

a−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

x0.6 dx =

∫ 1

0x−0.6 dx

= lima→0

x0.40.4

∣∣∣∣1

a

=1

0.4(

1− lima→0

a0.4)

=1

0.4 (1− 0)

=1

0.4So 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 traveled 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

74

CHAPTER 8. FURTHER INTEGRAL APPLICATIONS 75

(−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

y = 4(x+ 4)3/2, 0 ≤ x ≤ 2, y > 0

Solution. We start by taking the derivative of y:

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

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).


8.2 Optional: Area of a Surface of Revolution

This section is similar in spirit and in some details to section 8.1, so we won’t spendas much time explaining the geometric basics this time.

We call the following kind of shape a band

r

`

To find its surface area imagine cutting it and flattening it out, and straighteningit out.

2πr

`

2πr

`

From this we see thatArea of band A = 2πr`

This leads directly to a formula for surface area of rotation:

Definition. The surface area of rotation generated by rotating the curve y =f(x), from x = a to x = b, around the x-axis equals

Surface Area =

∫ b

a2πr` dx

where ` =

√1 +

(dy

dx

)2

and r = f(x)

Example 1. (Stewart, 6e, 8.2#10) Find the surface area generated by rotating

y =x3

6+

1

2x,

1

2≤ x ≤ 1, around the x-axis.

Solution. We find the derivative, square it, and put it inside of the square root

dy

dx=x2

2− 1

2x2√1 +

(dy

dx

)2

=

√1 +

(x2

2− 1

2x2

)2


=

√1 +

x4

4− 2 · x

2

2· 1

2x2+

1

4x4

=

√1 +

x4

4− 1

2+

1

4x4

=

√x4

4+

1

2+

1

4x4

=

√(x2

2+

1

2x2

)2

=x2

2+

1

2x2

Now we put this in the integral, along with 2πr = 2πf(x).

∫ 1

1/22π

(x3

6+

1

2x

)(x2

2+

1

2x2

)dx = 2π

∫ 1

1/2

x5

12+

x

12+x

4+

1

4x3dx

= 2π

∫ 1

1/2

x5

12+

4x

12+

1

4x3dx

= 2π

∫ 1

1/2

x5

12+x

3+

1

4x3dx

= 2π

(x6

72+x2

6− 1

8x2

) ∣∣11/2

= 2π

(1

72+

1

6− 1

8− (

1/26

72+

1/22

6− 1

8(1/2)2)

)

= 2π

(1

72+

1

6− 1

8− (

1/26

72+

1/22

6− 1

8(1/2)2)

)

= 2π

(1

72+

1

6− 1

8− (

1

4608+

1

24− 1

2)

)

Extra examples

Example 2. Find the surface area obtained by rotating y = ex around the x-axis,from x = 0 to x = 1.

Solution. We start by taking the derivative, squaring this, and putting it inside ofa square root

dy

dx= ex

√1 +

(dy

dx

)2

=√

1 + e2x

∫ 1

02πf(x)

√1 +

(dy

dx

)2

dx = 2π

∫ 1

0ex√

1 + e2x dx

= 2π

∫ √1 + u2 du

= 2π

(u

2

√1 + u2 +

1

2ln(u+

√1 + u2)

)integral formula #21


= π(ex√

1 + e2x + ln(ex +√

1 + e2x)) ∣∣1

0

= π(e√

1 + e2 + ln(e+√

1 + e2)− (1 ·√

1 + 1 + ln(1 +√

1 + 1)))

= π(e√

1 + e2 + ln(e+√

1 + e2)−√

2− ln(1 +√

2))

Example 3. [Gabriel’s Horn, part II] Find the surface area of Gabriel’s Horn (seeExample 2 in Section 7.8).

This is the shape obtained by rotating1

xaround the x-axis, with the range

1 ≤ x <∞.

Solution. We start with the derivative and the square root.

dy

dx= − 1

x2√1 +

(dy

dx

)2

=

√1 +

1

x4

Now we put this into the integral

∫ ∞

12π

1

x

√1 +

1

x4dx

My intuition at this point is to do a u-substitution to turn the square root into

something of the form√

1 + u2. This means we should try u =1

x2

u =1

x2

du = −2x−3 dx

−1

2x3 du = dx

Plugging this into our original integral we get∫

2π1

x

√1 + u2

(−1

2

)x3 du = −π

∫x2√

1 + u2 du

= −π∫

1

u

√1 + u2 du

(using x2 =

1

u

)

= −π(√

1 + u2 − ln

∣∣∣∣∣1 +√

1 + u2

u

∣∣∣∣∣

)integral formula 23

= −π

√

1 +1

x4− ln

∣∣∣∣∣∣

1 +√

1 + 1x4

1/x2

∣∣∣∣∣∣

∣∣∣∣∣

∞

1

= −π(√

1 +1

x4− ln

∣∣∣∣∣x2 + x2

√1 +

1

x4

∣∣∣∣∣

) ∣∣∣∣∣

∞

1

To evaluate this integral, we need to consider the limit as x approaches ∞:

limx→∞

−π(√

1 +1

x4− ln

∣∣∣∣∣x2 + x2

√1 +

1

x4

∣∣∣∣∣

)= −π

(√1 +

1

∞ − ln

∣∣∣∣∣∞+∞√

1 +1

∞

∣∣∣∣∣

)


= −π(√

1 + 0− ln∣∣∞+∞

√1 + 0

∣∣)

= −π(1− ln |∞|)= −π(−∞)

=∞

Thus, this integral is infinite, and it diverges. But we can say a little more thanthis. It is infinitely large, i.e. the surface area really is infinite. If you were going topaint Gabriel’s Horn, you would need to buy an infinite amount of paint! However,this is quite paradoxical since we saw in Example 2, in Section 7.8, that the volumeof this horn is finite! Thus, you could fill the horn with a finite amount of paint,but you could not paint the horn. This kind of paradox shows that infinite thingsdo not obey our usual rules of common sense.

This is where we ended on Friday, October 21

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 = 0.15.

t10

C = 0.15

0.05

p(t)

(a) Show that p(t) satisfies the defining property of a probablity density function,

that

∫ ∞

−∞f(x) dx = 1.

(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) =0.15− 0.05

10t+ 0.05 = 0.01t+ 0.05. Then

∫ ∞

−∞p(t) dt =

∫ 10

00.01t+ 0.05

=0.01

2t2 + 0.05t

∣∣∣∣10

0

=0.01

2(100) + 0.05(10)

=1

2+ 0.5

= 1

(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

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, or Desmos (https://www.desmos.com/calculator/t9krookvci)3. 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.


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 .3In one line I entered f(x) and the formula, in the next line I typed “int” and Desmos turned

this into an integral sign. Then I entered the numbers

https://www.desmos.com/calculator/t9krookvci

https://www.desmos.com/calculator/t9krookvci

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 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


Example 2. 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

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 11 points on the curve correspondingto θ = 0, π/10, 2π/10, . . . , 10π/10.

(b) Calculate the cartesian coordinates of each point from part (a), and, using thecartesian graph paper, plot the 11 points from part (a).


(c) Use your calculator1 or Desmos2, 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

0

0.5

1

1.5

2

2.5

3

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

1To change the graphing mode to polar, go to MODE , then choose POL instead of FUNC2Just enter r and θ. You can change the grid under the wrench/settings.


Solution. (a) We make a table of values

θ r

0 3.5 sin(0) = 0π/10 3.5 sin(π/10) ≈ 1.12π/10 2.13π/10 2.84π/10 3.35π/10 3.56π/10 3.37π/10 2.88π/10 2.19π/10 1.110π/10 0

and put them on the graph paper

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 cos(0) = 0 0 sin(0) = 0π/10 1.1 1.1 cos(π/10) ≈ 1.03 1.1 sin(π/10) ≈ 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 3.56π/10 3.3 −1.03 3.177π/10 2.8 −1.66 2.298π/10 2.1 −1.66 1.219π/10 1.1 −1.1 0.3310π/10 0 0 0

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).

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. 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

We have

x0 = r cos(θ)

∣∣∣∣θ=π/12

= (1− sin(π/12)) cos(π/12) ≈ 0.72

y0 = r sin(θ)

∣∣∣∣θ=π/12

= (1− sin(π/12)) sin(π/12) ≈ 0.19

For the slope, we first need to take some derivatives:

dy

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

= − cos(θ)(2 sin(θ)− 1)

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 slope.

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 (short version: put the calculator in parametric mode and writeboth the cartesian line and the polar equation as parametric equations). For now,we just graph each of them separately, and mentally try to combine the pictures tosee 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, andanalogous 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).

This is where we ended on Tuesday, November 1

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. Find the area contained in one loop for the following shape:

r = 4 cos(3θ)

Solution. The hardest part of a lot of polar graph areas is figuring out the boundsof integration. For this, it usually helps to look at a graph, and then to solve somesort of equation. Here’s the graph:


From the graph, we can see the following: So now we need to figure out the boundsfor θ to describe one loop.

You can also make a very rough guess for these bounds by the graph by imaginingtangent lines that go through the origin and contain one loop:

From this graph, you might guess that the bounds of the integral will by θ = ±π/4,or maybe ±π/5, or ±π/6. To be sure, we have to solve it algebraically.

How do we solve for θ to get the bounds? The simplest way to do this is to setsomething equal to 0 and solve. There are three reasonable choices in this problemfor what to set equal to 0:

set y = 0: y = r sin(θ)⇒ 0 = 4 cos(3θ) sin(θ)

set x = 0: x = r cos(θ)⇒ 0 = 4 cos(3θ) cos(θ)

set r = 0: r = 4 cos(3θ)⇒ 0 = 4 cos(3θ)


Any of these will work to some degree, but the last one works best: it gives onlythe values of θ where both x and y are 0. So, finally, let’s solve this and integrate:

set r = 0: r = 4 cos(3θ)⇒ 0 = 4 cos(3θ)

cos(3θ) = 0

cos(anything) = 0 at “anything” = ±π/2,±3π/2,±5π/2, . . .

3θ = ±π/2,±3π/2,±5π/2, . . .

θ = ±π/6,±3π/6,±5π/6, . . .

Thus, it appears that our bounds of integration are θ = ±π/6. To confirm this, wegraph the function between these bounds

So, finally, we set up the integral and finish:

A =

∫ π/6

−π/6

1

2r2 dθ

=

∫ π/6

−π/6

1

2(4 cos(3θ))2 dθ

=

(1

2

)(16)

∫ π/6

−π/6cos2(3θ) dθ

=

(1

2

)(16)

(1

3

)(1

2

(3x+

1

2sin(6x)

)) ∣∣∣∣π/6

−π/6=

(1

2

)(16)

(1

3

)(2)

(1

2

)(3(π/6) +

1

2sin(6(π/6))

)

=4

3π

This is where we ended on Wednesday, November 2

Example 3. (a) Set up an integral to find the area of that part of the circler = 2 sin(θ) + 5 cos(θ) that is in quadrant II.

(b) Use your calculator/computer/online-math to evaluate the integral from part(a).


Solution. We start by graphing the whole curve

As you can see, only a little part of the circle is in quadrant II. We need to solve forthe θ values that just give us that part of the circle. The part of the circle that isin quadrant II is defined by where this graph crosses the y-axis, which means thatthe x-value is 0. So, we set x equal to 0 and then solve for θ:

x = r cos(θ)

0 = (2 sin(θ) + 5 cos(θ)) cos(θ)

cos(θ) = 0⇒ θ = ±π/2,±3π/2,±5π/2,±7π/2, . . .

or: 2 sin(θ) + 5 cos(θ) = 0

2 sin(θ)− 5 cos(θ)

2sin(θ)

cos(θ)= −5

tan(θ) = −5/2

θ = tan−1(−5/2) + π

So, the integral is∫ tan−1(−5/2)+π

π/2

1

2r2 dθ =

∫ tan−1(−5/2)+π

π/2

1

2(2 sin(theta) + 5 cos(θ))2 dθ

When we enter this in our calculator/computer/online-math we get

1

2

(21

4sin(2θ) +

29

2θ − 10 cos2(θ)

)

=29

8π − 5

2+

29

4tan−1(−5/2)

≈ 0.259

Can you tell if that looks about right?Yes: the part of the graph we want is part of a circle, with height 2 and width

about 0.25. If this were a rectangle, the area would be 0.5, but it’s not the wholerectangle, so it’s reasonable for the area to be 0.259.


Extra Examples

Example 4. A chocolate company is coming out with a new chocolate product. Itis going to make a mold which they can pour chocolate into. They need to knowthe volume of the mold, but basically this means they need to know area of theshape of the mold. The mold has shape given by the following

r = −6(θ − π/2)2 + 4|θ − π/2|+ 2,π

2− 1 ≤ θ ≤ π

2+ 1, units for r = inches.

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

θ =π

2− 1, 0.2, 0.35, π/6, 0.7, 0.9, π/3, 1.2, 1.4 and π/2, 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 4

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. (a) We can put our calculators in polar mode, and enter the formula,and use the table function to get the following:

π

2− 1 0.00

.7 0.93

.9 1.98π

32.45

1.2 2.661.4 2.51π

22.00

and the resulting points like like this


(b) When we graph the whole function with appropriate settings we get this

(c) To find the area we integrate

∫ π2

+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 equations 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 calculator3 We plot this for 0 ≤ t ≤ 5

3First 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 =.


This is where we ended on Friday, November 4

Example 3. A few years ago I made and printed out a decorative award for mykid successfully passing swimming lessons. I want to have their name, the date, thename of the swimming school, etc., but I also want a decorative curlicue (aka “curlycue”) around the outside of the document. Use your calculator to see which of thefollowing will work best:{x = 1.5 cos5/7(t) + 0.1 cos(81t)

y = sin5/7(t) + 0.2 sin(21t)

}or

{x = 1.5 cos5/7(t) + 0.1 cos(81t)

y = sin5/7(t) + 0.1 sin(81t)

}

If you can, see if you can understand why the curves look the way the do.

Solution. Both sets of parametric equations start with the same basic shape, x =1.5 cos5/7(t), y = sin5/7(t). This shape is called a “super-ellipse” because it lookslike an ellipse, but a little bit flattened out, sort of half way between an ellipse anda rectangle:


The other formulas that are added, either 0.1 cos(81t), or 0.2 sin(21t) or 0.1 sin(81t),litterally add a small variation to the basic shape. Imagine moving your hand aroundthe shape of the super-ellipse, but wiggling your pencil in small circular motions asyou move your hand, that’s what the other shapes will look like.

The first one looks like this

And the second looks like this

The second shape is about right for an award certificate, especially if we drop theaxes and put some words in the middle:


Congratulations!Liam you’ve passed your ad-

vanced swimming certifi-cation, September 1, 2010

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 cartesianequations. 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 4. 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 + sin(7π/4) = 7/4−√

2/2 ≈ 1.04

y0 = g(7/4) = 7/4 + cos(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. We will use the easiest possible way to write x as afunction of t:

x = t

y = (t− 1.04) + 2.46

The same approach works with every equation of the form y = f(x): to put it inparametric mode 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 we ended on Monday, November 7

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 calculator 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

graph of 2 cos

(πt

10

)+ 4 cos

(2πt

5

)

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 cycloid4

(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))

4This 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 equation5:

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 parametric 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

5If 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

Extra Examples

Example 6. Graph the following and find the area in the loop (exact values onlyplease): {

x = t3 + t2 − ty = −t3 + t2 + t

}

Solution. We start by looking at the whole graph:

We can see that there is one loop, and it appears to intersect itself at (x, y) = (1, 1).We need to solve for the t values at this intersection

x = 1

t3 + t2 − t = 1


t3 + t2 − t− 1 = 0

Cubics are a bit tricky to solve, but sometimes you get lucky and this is one of thosetimes. The first two terms and the last two terms both have a factor of t+ 1:

t3 + t2 − t− 1 = t2(t+ 1)− (t+ 1)

= (t+ 1)(t2 − 1)

= 0

t+ 1 = 0⇒ t = −1

t2 − 1 = 0⇒ t2 = 1⇒ t = ±1

Thus, we have t = ±1.To make sure we have the right values of t we graph it from t = −1 to t = 1

Now we set up our integral∫ 1

−1g(t)f ′(t) dt =

∫ 1

−1(−t3 + t2 + t)(3t2 + 2t− 1) dt

=

∫ 1

−1−3t5 + t4 + 6t3 + t2 − t dt

= −1

2t6 +

1

5t5 +

3

2t4 +

1

3t3 − 1

2t2∣∣∣∣1

−1

= −1

2+

1

5+

3

2+

1

3− 1

2−(−1

2− 1

5+

3

2− 1

3− 1

2

)

=31

15

Example 7. Graph the following and find the area in one of the small inner loops(exact values only please):

x = cos(t) +1

2cos(3t)

y = sin(t) +1

2sin(3t)


Solution. We start by looking at the whole graph:

We can see that there are two small inner loops, one at the top and one at thebottom. Let’s look at the loop on top, the one that is between y = 0.5 and y = 0.8.To set up our integral we need to solve for t, and to do this we should set x equalto 0

x = 0

cos(t) +1

2cos(3t) = 0

This is a difficult equation to solve, but we need to have a sense of hope and faiththat the values come out to be simple. When is cosine equal to 0? At π/2, 3π/2,etc. Would these values make the equation 0? Yes, in fact t = π/2 makes each termequal to 0:

cos(π/2) +1

2cos(3π/2) = 0 +

1

2· 0 = 0

If you plug t = π/2 into both equations you get the point (x, y) = (0, 1/2), which isthe bottom of the loop. We also need to solve for the top.

There are various ways we could figure out the other solution: if it’s one of ourstandard trig values then we could just guess and check, t = 0, π/6, π/4, π/3, π/2,etc. Maybe we could be a bit more clever: the “1/2” in front of the second cosineterm should probably cancel a 1/2 that would come out of cosine in the first term.So maybe we should guess and check t = π/3, since cos(π/3) = 1/2. It turns outthis works:

cos(π/3) +1

2cos(3π/3) =

1

2+

1

2(−1) = 0

So now we may think we have the numbers we need, t = π/2 and t = π/3, but weshould graph them to make sure we’ve got it right:


Hmm, that’s only half the loop, with t = π/2 showing up on the bottom. We needa different value, one that makes us go all the way around the loop. Maybe we canget another value that’s a bit past t = π/2, and still gives some sort of 1/2 comingout of the first cosine. Maybe we should try t = 2π/3. It turns out this works:

cos(2π/3) +1

2cos(3 · 2π/3) = −1

2+

1

2(1) = 0

Now let’s try graphing again to make sure we got it right.

That’s it. Let’s set up and finish our integral:

∫ 2π/3

π/3g(t)f ′(t) dt =

∫ 2π/3

π/3

(sin(t) +

1

2sin(3t)

)(− sin(t)− 3

2sin(3t)

)dt

=

∫ 2π/3

π/3− sin2(t)− 2 sin(t) sin(3t)− 3

4sin(3t)2 dt


= . . . (note that each term here is something you know how to do)

=√

3/2− 7π/24

≈ −0.05

Something must be wrong here, can you see what it is?The answer can’t be negative. Did we make a big mistake, or do we just need

to reverse something? Note that π/3 ≤ t ≤ π/2 covers the first half of the loop“backwards”: It starts at the top, then goes to the left. Of course, when we integrate

an ordinary cartesian function, y = f(x), we always want

∫ b

af(x) dx where x moves

from x = a to the right to get to x = b. Since our parametric integral is really justa substitution rule for our cartesian integral, we should set it up to move the same

way. Long story short: we should use the integral

∫ π/3

2π/3. . . dt. This will reverse our

numbers and we get7π/24−

√3/2 ≈ 0.0503

Now, does this make sense? If we zoom in on the loop and read off values from thegraph, we could approximate the area with a rectangle

Area rectangle = width×height

= (0.08− (−0.08))× (0.85− 0.5)

= 0.0560

It looks like we got it right!

Chapter 11

Infinite Sequences and Series

11.2 Series

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.

121

CHAPTER 11. SEQUENCES AND SERIES 122

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 2. List the first 5 terms of the sequence, simplify if possible

(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

16=

1

16

a5 = (−1)5 1

52= (−1) · 1

25= − 1

25

(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


Definition. Given a sequence an, the series

k∑

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 3. 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 4. Find10∑

n=1

(9n− (−1)n)

Solution.

10∑

n=1

(9n− (−1)n) =10∑

n=1

9n−10∑

n=1

(−1)n

= 910∑

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 5. 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 traveled 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 distancetraveled is

1− 1

2n

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

limn→∞

1− 1

2n= 1− 0 = 1

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

∞∑

n=1

(1/2)n = 1

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 for all n

Finally, we can simplify this a little:

A series is geometric if it can be written in the following form∞∑

n=0

a0rn



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 canceled 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 for

N∑

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

Solution. The distance traveled 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 7. 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 multipliedby, 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)


Checkboxes• lim

n→∞cn 6= 0 2

=⇒∑

cn Div.

Example 8. 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

=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 10 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.



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.

Example 9. 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 10. [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

+ . . .

≥ 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 11. 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


Checkboxes•

∑|cn| Conv 2

=⇒∑

cn Conv.

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

11.6 Absolute Convergence and the Ratio Test

Comments. BMSC (Bounded Monotonic Series Comments) Suppose a series∑

anhas only positive terms and we want to decide whether it converges or diverges. Ifit diverges, it’s because the limit doesn’t exist. If the limit doesn’t exist, it’s notbecause the sum goes up and down, because we’re only adding positive terms, sothe sum can only go up. So, if the limit doesn’t exist, it’s because the sum goes toinfinity. Therefore, if the sum doesn’t go to infinity, the limit must exist. In otherwords, if the series doesn’t go to infinity, then it converges.

Definition. We say

∞∑

n=0

an is absolutely convergent if the series

∞∑

n=0

|an| con-

verges.

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.

Proof. Suppose∑

an is abolutely convergent. By definition, this means∑|an|

converges. Let L =∑|an|. Then 2L =

∑2|an|. Note that

∑(an+|an|) is a series

with only positive terms. Furthermore, an + |an| ≤ 2|an| and so∑

(an + |an|) ≤∑2|an| = 2L. Thus,

∑(an + |an|) does not sum to infinty, and so it converges

by the BMSC Comments. Note that∑

an =∑

2|an| −∑

(an + |an|). Therefore,∑an equals a combination of two convergent series, and so

∑an is 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 equaling ∞, 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 1. Apply the Ratio Test to∞∑

n=1

(−1)nn10

1.5n.

Solution.

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

Now we claim that

limn→∞

(n+ 1)10

n10= lim

n→∞

n10

n10.

To see this, 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 tobegin with. In any case, we get

limn→∞

(n+ 1)10

n10= lim

n→∞

n10

n10= 1

and so

L = 1 · 1

1.5· 1

=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 2. Apply the Ratio Test to the following series:

∞∑

n=1

(−1)n+1 n

7n2 + n+ 1

Solution.

L = limn→∞

∣∣∣∣an+1

an

∣∣∣∣

= limn→∞

∣∣∣∣∣(−1)n+2 n+1

7(n+1)2+(n+1)+1

(−1)n+1 n7n2+n+1

∣∣∣∣∣


= limn→∞

∣∣∣∣(−1)n+2

(−1)n+1· n+ 1

7(n+1)2+(n+ 1)+1· 7n2 + n+ 1

n

∣∣∣∣

= limn→∞

∣∣∣∣(−1)n+2

(−1)n+1

n+ 1

n· 7n2 + n+ 1

7(n+1)2+(n+1)+1

∣∣∣∣

Note that n+ 1 and n are both polynomials, i.e. whole number powers of n. There-

fore,n+ 1

nis a rational function. The degree of n on top and on the bottom is the

same and so for the limit we take the ratio of leading terms

limn→∞

n+ 1

n= lim

n→∞

n

n= 1

Note that 7n2 + n + 1 and 7(n + 1)2 + (n + 1) + 1 are both polynomials, i.e.

whole number powers of n. Therefore,7n2 + n+ 1

7(n+1)2+(n+1)+1is a rational function.

The degree of n on top and on the bottom is the same and so for the limit we takethe ratio of leading terms

limn→∞

7n2 + n+ 1

7(n+1)2+(n+1)+1= lim

n→∞

7n2

7(n+ 1)2= lim

n→∞

7n2

7n2= 1

Putting these limits into our above calculation for L:

L = limn→∞

∣∣∣∣(−1)n+2

(−1)n+1

n+ 1

n· 7n2 + n+ 1

7(n+1)2+(n+1)+1

∣∣∣∣= 1 · 1 · 1= 1

Since the limit equals 1, the Ratio Test gives us no information about whether ornot this series will converge.


Example 3. Apply the Ratio Test to∞∑

n=1

(−1)n10n

n!.

Solution.

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)!

∣∣∣∣

The main trick here is to simplify the factorials:

2!

3!=

2 · 13 · 2 · 1 =

�2 · �13 · �2 · �1

.


Similarly5!

6!=

�5 · �4 · �3 · �2 · �16 · �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.

Plugging this into the above formula for L we have

L = 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 the following series:

∞∑

n=1

(−1)n−1 1.1n

n10 + n2 + 1

Solution.

L = limn→∞

∣∣∣∣an+1

an

∣∣∣∣

= limn→∞

∣∣∣∣∣∣(−1)n 1.1n+1

(n+1)10+(n+1)2+1

(−1)n−1 1.1n

n10+n2+1

∣∣∣∣∣∣

= limn→∞

∣∣∣∣(−1)n

(−1)n−1

1.1n+1

(n+ 1)10 + (n+ 1)2 + 1· n

10 + n2 + 1

1.1n

∣∣∣∣

= limn→∞

∣∣∣∣(−1)n

(−1)n−1

1.1n+1

1.1n· n10 + n2 + 1

(n+ 1)10 + (n+ 1)2 + 1

∣∣∣∣

= limn→∞

∣∣∣∣(−1)n

(−1)n−1· 1.1 · n10 + n2 + 1

(n+ 1)10 + (n+ 1)2 + 1

∣∣∣∣= 1 · 1.1 · 1= 1.1

Since the limit equals 1.1, and since 1.1 is greater than 1, the Ratio Test tells usthat this series will diverge (unconditionally).

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 analyze 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 · 1

So, this test tells us nothing about this series1.


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 + . . .

1Of 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


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 + . . .

Example 1. Consider the power series

∞∑

n=0

xn

View this series as a function of x and use it to define f(x)

f(x) =∞∑

n=0

xn

(a) Find the domain of f(x). (Hint: just try guessing some values of x until yousee a pattern, then confirm your pattern by using some previous results.)

(b) Find a closed form formula for f(x). Compare the graph of f(x) to the firstfew terms of the power series. Specifically, compare f(x) to 1 + x + x2 + x3

and 1 + x+ x2 + x3 + x4 + x5 + x6.

Solution. (a) Let’s try f(1):

f(1) =∞∑

n=0

1n

= 1 + 1 + 1 + 1 + 1 + . . .

= diverges

Since

∞∑

n=0

1n diverges, we say that f(1) is not defined, and so 1 is not in the domain.

Let’s try f(10):

f(10) =

∞∑

n=0

10n

= 1 + 10 + 100 + 1000 + . . .

= diverges

Since∞∑

n=0

10n diverges, we say that f(10) is not defined, and so 10 is not in the

domain.In fact, it’s pretty clear x = 10 was even worse than x = 1, and similarly that

any number bigger than x = 1 will make the series diverge. Let’s try a smallernumber.

Let’s try f(1/2):

f(1/2) =∞∑

n=0

(1/2)n


= 1 +1

2+

1

4+

1

8+

1

16+ . . .

Do you recognize this series? What kind is it? Does it converge?It’s geometric with r = 1/2. Since 1/2 < 1 the series converges.Let’s try some similar values:

f(1/4) =

∞∑

n=0

(1/4)n geometric r = 1/4 < 1 converges

f(3/4) =∞∑

n=0

(3/4)n geometric r = 3/4 < 1 converges

f(0.1) =∞∑

n=0

(0.1)n geometric r = 0.1 < 1 converges

f(0.9) =

∞∑

n=0

(0.9)n geometric r = 0.9 < 1 converges

It would appear that all x-values satisfying 0 < x < 1 are in the domain. Butthere are more x-values that work.

Let’s try f(−1/2):

f(−1/2) =∞∑

n=0

(−1/2)n

This is geometric with r = −1/2. Since |r| = 1/2 < 1 the series converges.Generalizing, any value of x = r with |r| < 1 will make the series converge.

Thus, the domain is

domain = all x such that −1 < x < 1

= the interval (−1, 1)

(b) In part (a) we saw that f(x) gives a geometric series for any x satisfying

−1 < x < 1. If we think of a specific number, r, then we know that∞∑

n=0

rn =1

1− r .

Replacing r with x we get

f(x) =1

1− x

The graphs below show f(x) =1

1− x , together with 1 + x + x2 + x3 and 1 + x +

x2 + x3 + x4 + x5 + x6,


The graphs show the following behavior. For numbers close to 0, like x = 0.2,the polynomials look almost identicial to f(x). For numbers farther from 0, likex = 0.6, there’s a gap between f(x) and the polynomials. But, the gap is smallerfor 1 +x+ · · ·+x6 than for 1 +x+x2 +x3. In general, if we use more terms in ourpolynomials, the gaps will get smaller.

Theorem 1. Let∞∑

n=0

cn(x − a)n be any power series centered at a. There is an

R ≥ 0 such that one of the following holds:• If R = 0 implies that the series converges only for x = a• If R =∞ implies that the series is absolutely convergent for all x• If 0 < R < ∞ 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

We call R the radius of convergence

Definition. Let∞∑

n=0

cn(x− a)n be any power series centered at a and let R be the


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?”.

radius of convergence.

The open interval of con-vergence

=

(a−R, a+R) if R > 0, R 6=∞(−∞,∞) R =∞a R = 0 (here we deliberatley abuse

the phrase “open interval”)

Example 2. Find the Radius of Convergence, and the open interval of convergence

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

∣∣∣∣

= 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

Open I.O.C. = (-1/2,3/2)

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 . Then do this:

solve something involving x < 1

until you get to this:for |x− a| <?

Then whatever you have for “?” is equal to R, i.e. R = “?”This is where we ended on Tuesday, November 22


One algebraic rule for absolutevalues states that if a > 0,then a|x+ b| = |ax+ ab|.

Example 3. Find the Radius of convergence and the open interval of convergence

∞∑

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

Open I.O.C. = (1,2)

Example 4. Find the Radius of Convergence and the open 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.


Recall the notation “f (n)(x)”,which means the nthderivative.

−4

?

18

?

7

div abs conv div

Open I.O.C. = (-4,18)

Extra Examples

Example 5. Find the radius of convergence and open interval of convergence forthe following power series

∞∑

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

Open I.O.C. = (-8,-6)

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

.

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. This example is about the Maclaurin series for f(x) = ex.(a) Find f(x), f ′(x), f ′′(x), f ′′′(x), and f (4)(x).(b) Find f(0), f ′(0), f ′′(0), f ′′′(0), and f (4)(0).

(c) Findf(0)

0!,f ′(0)

1!,f ′′(0)

2!,f ′′′(0)

3!, and

f (4)(0)

4!.

(d) Write the degree 4 Maclaurin polynomial for f(x) = ex.(e) Compare the graphs of ex and your answer to part (d) on the interval −1 ≤

x ≤ 1. What are your observations?(f) Explore: how many terms do you have to add to the Maclaurin polynomial

to make the graphs identical on the interval −5 ≤ x ≤ 5?

(g) Convert the Maclaurin polynomial to∑

-notation. If you’re using Desmos

(or some other computer graphing program), use the∑

-notation to graph a

Maclaurin polynomial with very large degree, say degree 20, or 50, or 100.(h) Find the radius of convergence of the Maclaurin series. What does this mean

in terms of graphs?

Solution. (a)

f(x) = ex

f ′(x) = ex

f ′′(x) = ex

f ′′′(x) = ex

f (4)(x) = ex

(b)

f(0) = e0 = 1

f ′(0) = e0 = 1

f ′′(0) = e0 = 1


f ′′′(0) = e0 = 1

f (4)(0) = e0 = 1

(c)

f(0)

0!= 1

f ′(0)

1!= 1

f ′′(0)

2!=

1

2!f ′′′(0)

3!=

1

3!

f (4)(0)

4!=

1

4!

Note that you should really avoid multiplying out the factorials. You need to becomecomfortable with expressions like “4!” without multiplying them out. Also, the wayit’s written now, it’s easy to say what the next three numbers would be if we need

them:1

5!,

1

6!, etc.

(d) Since we are writing down a polynomial of degree 4, we know, even beforewe calculate anything for parts (a)–(d), that we will have something of this form:

+ x+ x2 + x3 + x4

This is because all polynomials of degree 4 can be written in the form just given.The only question is what goes in the blanks? Again, for all polynomials, eachblank should have a number. In this case, the numbers come from (c):

1 + x+1

2x2 +

1

3!x3 +

1

4!x4

(e) To enter this polynomial in a calculator or computer, it might be nice tosimplify how we write it first:

1 + x+x2

2+x3

3!+x4

4!

As above, you should really enter this using “!” not multiplying out the numbers2.The result is shown below:

2Here’s how to enter “!” on different devices: In Desmos on a computer, use the keyboarddirectly. In Desmos on a smart phone, go to the “funcs” button, then the “stats” tab, thenchoosing n! near the bottom. On a TI-84 calculator, go to MATH then choose PRB then choose

4:! .


Our observations are this: the graphs of ex and 1 +x+x2

2!+x3

3!+x4

4!look identical

on the interval −1 ≤ x ≤ 1.(f) We don’t have a formula for this, we simply try adding more terms, and

graphing the result on −5 ≤ x ≤ 5 and seeing when the graphs look identical.Below we show the graph of ex, in red, and Maclaurin polynomials of degree 5, 6,7, . . . , 13, all in black.

You can see around x = −5 that some of the black lines are close to ex, and someare pretty far away. The ones that are close have the higher degrees.

Below we show just ex, still in red, and the degree 13 Maclaurin polynomial: it’sa pretty good match.


(g) To convert into∑

notation, look at something like this:

1 +x2

2!+x3

3!+x4

4!+x5

5!+x6

6!+x7

7!+ . . .

x13

13!

and look for the patterns. We want to put these patterns into formulas. Hopefullyyou see that we have:

13∑

n=0

1

n!xn

Using this notation for Desmos, or some other computer system, it’s easy to graph

50∑

n=0

1

n!xn


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.

Note that now the interval where the graphs look identical appears to be −22 ≤x ≤ 22.

(h) 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 =∞.

What this means in terms of the graphs, is that we could, if we wanted to, makethe graphs between ex and a Maclaurin polynomial look identical on as large ofa range of x-values as we want. However, there is a cost. Suppose you wantedthe graphs to be the same for −100 ≤ x ≤ 100, we might need to use 200 terms,

i.e.200∑

n=0

xn

n!. That’s a huge amount of terms so there’s probably (there is) a more

efficient way to calculate ex on such a large range.


Example 2. (a) Find the Taylor series of f(x) = ln(x) with center equal to 1

(b) Write the series using∑

notation

(c) Find the radius of convergence,(d) Compare f(x) = ln(x) to it’s Taylor polynomials of degree 2 and 4 on the

interval [0, 2].

Solution. (a) We follow the pattern of the last example as closely as possible: Westart by 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

(b) 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

(c) 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.(d) Shown below are the original graph of f(x) and the two polynomials


Example 3. (a) Find the Maclaurin series for f(x) = sin(x),


(b) write the series using∑

notation,

(c) find its radius of convergence,(d) Compare the graph of f(x) to the the degree 3 and degree 7 polynomials on

the interval [−5π/4, 5π/4].

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 =∞.

Shown below are the graphs


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 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

22 · 2!x2 +

1 · 323 · 3!

x3 − 1 · 3 · 524 · 4!

x4 + . . .

=∞∑

n=0

(−1)n+1 (2n− 1)!!

2n · n!xn

where (2n−1)!! denotes the “double factorial” of (2n−1) (i.e. start at 2n−1 and godown by 2 at each step till you get to 1); out of convenience, let’s define (−1)!! = 1and (−3)!! = −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, . . . , or, to put it differently, +, +, −, −, +, +, −, −, etc. I’ll show belowthree ways to do this. Why three ways? Because I’m not 100% satisfied with anyof them.

One way to get +, +, −, −, +, +, −, −, etc. is to use sine, with some extraconstants. In fact, if we are going to use sin, we might as well use it to give part ofthe coefficient with

√2 too:

sin(π/4) =1√2, sin(3π/4) =

1√2, sin(5π/4) = − 1√

2,

sin(7π/4) = − 1√2, sin(9π/4) =

1√2, sin(11π/4) =

1√2, etc.

So, here’s one way to write the above series:

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, one alternative is to define a simple function, that allows us to round

numbers. 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.

Here’s a third approach: since + and −’s come in twos, we can use two termsat a time in the series, two positive terms, then two negative terms, etc. The twoterms will come from a single value of n, so we’ll use n to give an even term and anodd term together. Here’s how it looks:

sin(x) =

∞∑

n=0

((−1)n

1

(2n)!√

2(x− π/4)2n + (−1)n

1

(2n+ 1)!√

2(x− π/4)2n+1

)

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.

This is where we ended on Monday, December 5

11.11 Applications of Taylor Series

Comments. At this point, we take the following series as “known”

1

1− x =∞∑

n=0

xn


ex =

∞∑

n=0

xn

n!

ln(x) =∞∑

n=1

(−1)n+1 (x− 1)n

n

sin(x) =∞∑

n=0

(−1)nx2n+1

(2n+1)!

cos(x) =

∞∑

n=0

(−1)nx2n

(2n)!

tan−1(x) =∞∑

n=0

(−1)nx2n+1

2n+ 1

(1 + x)k =

∞∑

n=0

(k

n

)xn (for any real number k)

11.11.1 Calculating Important Numbers

Example 1. Things to think about in this example: What is the number π? Howis it defined? How can we calculate it?

(a) How is π defined?(b) Based on the definition, how could we approximate π?(c) Draw a circle with radius 1. Inscribe an equilateral triangle inside (i.e. draw

an equilateral triangle that touches the circle at the three corners.) Calculatethe perimeter of the equilateral triangle. Use this to approximate π.

(d) Repeat the previous part with the triangle replaced by a square.(e) Repeat the previous part with the square replaced by a hexagon.(f) Repeat the previous part with the hexagon replaced with a dodecagon (12-

sided figure).(g) Start with the Macluarin series for tan−1(x), and x = 1, to find a series that

calculates π.(h) Add enough terms in your series to calculate π accurately to one decimal

place.

(i) Start with the Macluarin series for tan−1(x), and x =1√3

, to find a series

that calculates π.(j) Add enough terms in your series to calculate π accurately to 3 decimal places.(k) Use the series below, discovered by Ramanujan in the early 1900’s

Ramanujan:1

π=

2√

2

9801

∞∑

n=0

(4n)!(1103 + 26390k)

(n!)4 · 3964n

with one term, to find an approxmiation of π.(l) Use the series from the previous part with two terms to find an approximation

of π.


(m) Use the series below, discoverd in 1997 by Bailey, Borwein and Plouffe,

π =∞∑

n=0

120n2 + 151n+ 47

16n(8n+ 1)(2n+ 1)(8n+ 5)(4n+ 3).

with one term, find the approximation of π that this gives.(n) Using the series from the previous part, with five terms, to find the approxi-

mation of π that this gives.

Solution. The number π is usually defined as the ratio of the circumference of acircle to it’s diameter. Given a physical circle, we can measure the circle’s diameterand circumference, but not very accurately. To calculate it more accuratley we cancalculate perimeters of inscribed polygons, e.g. calculate the perimeter of a 12-goninscribed in a unit circle:

r = 1

This dodecagon has perimeter 24 sin(π/12) = 6√

2(√

3−1) ≈ 6.2116. The circle hascircumference of 2π, and so this shows π ≈ 3.1058. (Archimedes used this approachand a regular 96-gon to approximate π as 3.14185.)

(a) A better way to calculate π is to use an infinite series. For example, if westart with the Maclaurin series for tan−1(x) we have

tan−1(x) =

∞∑

n=0

(−1)n1

2n+ 1x2n+1

π

4= tan−1(1)

=∞∑

n=0

(−1)n

2n+ 1

π =

∞∑

n=0

(−1)n4

2n+ 1

=4

1− 4

3+

4

5− 4

7+ · · · − 4

23


≈ 3.1

This series converges very slowly (i.e. you have to add a lot of terms before you geta good approximation of π). For example, you need to add 400 terms before youget the first 2 decimal places correct.

(b) As we’ve seen in the previous section, Taylor series converge more rapidlynear the center of the series. Since we are starting with a Maclaurin series, we couldget better convergence by plugging in a value for x that is closer to 0. If you thinkabout the standard trig values that we know, π/6 is closer to 0 than π/4. So, itwould be better to use tan−1(π/6). Let’s try:

tan−1(x) =∞∑

n=0

(−1)n1

2n+ 1x2n+1

π

6= tan−1

(1√3

)

=

∞∑

n=0

(−1)n(

1√3

)2n+1

2n+ 1

π =∞∑

n=0

(−1)n63(−1/2)(2n+1)

2n+ 1

=∞∑

n=0

(−1)n61

3n+1/2(2n+ 1)

=6

31/2− 6

33/2(3)+

6

35/2(5)− 6

37/2(7)+

6

39/2(9)− 6

311/2(11)

≈ 3.141

This series converges faster than the previous one.

There are series that converge much faster than the series given above. Forinstance, we could use a half angle identity to find that tan(π/12) = 2 −

√3, and

then use x = 2 −√

3 in

∞∑

n=0

(−1)nx2n+1

2n+ 1to find π/12. This converges to 3.14159

using just 5 terms.Srinivasa Ramanujan, 1887–1920, lived in India, except for a 5 year period of

time where he collaborated with mathematicians in England. This series was dis-covered by Ramanujan in the 1910’s. He could do things with infinite series thatno one else could, before or since, so it’s hard to say how he came up with it. Formore information about Ramanujan see en.wikipedia.org/wiki/Ramanujan orwww-groups.dcs.st-and.ac.uk/~history/Mathematicians/Ramanujan.html or readthe book The Man who knew infinity: A Life of the Genius Ramanujan by RobertKanigel. Here’s one of my favorite looking series formulas that he discovered (Ican’t say I understand it well enough to evaluate it’s mathematical merit, but itdoes look nice):

∫ ∞

0e−3πx2 sinh(πx)

sinh(3πx)dx =

1

e2π/3√

3

∞∑

n=0

e−2n(n+1)π

(1 + e−π)2(1 + e−2π)2 · · · (1 + e−(2n+1)π)2

en.wikipedia.org/wiki/Ramanujan

www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Ramanujan.html


The first term alone gives1

π≈ 0.31830987 which turns into π ≈ 3.141592. The

first two terms, i.e. k = 0 and k = 1, gives 1/π ≈ 0.3183098861837906 which turnsinto π ≈ 3.141592653589793.

Ramanujan’s formula was discovered in the early 1900’s. In 1997 an amazingdiscovery of another series was made by David H. Bailey, Peter Borwein and SimonPlouffe. Just finding such a series, and proving that it works, is something to beproud of, i.e. something to publish. Their paper about it is “On the Rapid Com-putation of Various Polylogarithmic Constants,” in Mathematics of Computation,66 (218), p903-913.. This series does not converge more quickly than Ramanujan’s,but it does have the advantage that it can be used to calculate a range of digits inπ without first calculating all the digits that came before. Here it is

π =

∞∑

n=0

120n2 + 151n+ 47

16n(8n+ 1)(2n+ 1)(8n+ 5)(4n+ 3).

For n = 0 this gives 471·1·5·3 which equals 3.1. The sum for n = 0, 1, . . . , 4 gives

3.1415926 which is correct out to the digit 6.

This is where we ended on Tuesday, December 6

This is where we ended on Wednesday, December 7

11.11.2 Calculating “Impossible” Integrals

Example 2. (a) Use Maclaurin Series to find an approximation of the integral

∫ 1

0e−x

2dx

(b) Find the Maclaurin Series for

∫e−x

2dx.

Solution. Before we start, it’s worth repeating that the anti-derivative

∫e−x

2dx is

not possible to solve using our usual methods. To be more specific: it is not possibleto write down a finite formula involving just our basic functions that equals thisanti-derivative. However, it is possible if we are allowed to use infinite combinationsof basic functions, for example, infinite combinations of powers of x, which is whatwe’ll do in part (b).

(a) Here’s an outline of the steps we’ll take: start with the series for ex, substitutein −x2 in place of x, then integrate the series, and evaluate at x = 1.

ex = 1 + x+x2

2+x3

3!+ . . .

e−x2

= 1− x2 +x4

2!− x6

3!+x8

4!− . . .

∫ 1

0e−x

2dx =

∫ 1

01− x2 +

x4

2!− x6

3!+x8

4!− . . . dx

=

(x− x3

3+

x5

5 · 2!− x7

7 · 3!+

x9

9 · 4!− . . .

) ∣∣∣∣1

0


= 1− 1

3+

1

5 · 2 −1

7 · 3!+

1

9 · 4!− 1

11 · 5!. . .

If we add the terms shown here we get 0.747 which is accurate to to the thirddecimal place.

(b) Basically we did this part above, before we plugged in x = 1. So, writing∫e−x

2dx is a series we have this

∫e−x

2dx = x− x3

3+

x5

5 · 2!− x7

7 · 3!+

x9

9 · 4!− . . .

=∞∑

n=0

(−1)n1

(2n+ 1)n!

Using this series, we have now defined a new function. This function may not befamiliar to you, but once upon a time ex, and sin(x), and cos(x), were not familiarto you either, and they were just names. Let’s give this one a name as well: “Erf(x)”(this stands for “Error Function”). Actually, let’s name Erf(x) after a multiple ofthis series:

Erf(x) =2√π

∞∑

n=0

(−1)n1

(2n+ 1)n!

It’s not much of a name, but Erf(x) is a very useful function in probability andstatistics. People know a lot about this function, and if they graph it, and learnsome standard values for it, and calculate it’s derivatives and integrals, before toolong it can become as familiar as ex or sin(x) or cos(x), etc.

11.11.3 Solving “Impossible” Equations

Example 3. Can you solve the equation

cos(x) = x

algebraically? Can you use a Taylor series?

Solution. We cannot solve cos(x) = x algebraically, or using basic functions. Forinstance, if we try to divide by x or try to apply cos−1 we get

cos(x)

x= 1 or x = cos−1(x)

But in both cases we fail to isolate x.On the other hand, if I draw the graphs by hand I get something like the

following:


I can tell that they intersect, and that the intersection is sort of near the origin.Thus, I can approximate a solution using a degree 2 Taylor polynomial for cos(x).Since

cos(x) = 1− x2

2!+x4

4!− . . .

we use 1− x2

2!. Now we have a quadratic equation:

1− x2

2!= x

−1

2x2 − x+ 1 = 0

x2 + 2x− 2 = 0

x =−2±

√4− (−8)

2

=−2±

√12

2

=−2± 2

√3

2

= −1±√

3

≈ −2.732, .732

= 0.732

The approach given above can be summarized: replace a function with it’s Taylorapproximation, and then solve. In fact, the same approach can work even if weuse a lower degree Taylor polynomial. What is a lower degree Taylor polynomial?The degree 1 polynomial is exactly the same thing as the tangent line. If you solvean equation using a tangent line then you are doing exactly one step in Newton’smethod of root finding.


11.11.4 Finding Difficult Limits

Example 4. Compare two solutions to finding the following limit,

limx→0

1− cos(x20)

x40

(a) Using your calculator and (b) using Maclaurin Series.

Solution. (a) This is roughly what the graph on my calculator looks like:

Perhaps you can tell that this graph isn’t right, and maybe you blame the calculatorfor this. Perhaps you think that a fancy computer system could get this graphright? Here’s the same function graphed using Maple (a fancy computer system)under default settings:


Maybe this graph is better, but it’s still fairly inaccurate. From these graphs youshould conclude that the limit equals 0. After all, when you x gets close to 0, say|x| < 0.25, the y-values also get close to 0. But this answer is wrong.

(b) We replace cosine with it’s degree 2 Taylor polynomial

cos(x) ≈ 1− 1

2x2

cos(x20) ≈ 1− 1

2x40

limx→0

1− cos(x20)

x40= lim

x→0

1−(1− 1

2x40)

x40

= limx→0

12x

40

x40

= limx→0

1

2

=1

2

The reason why the calculator and computer graphs are so inaccurate is that thisexample is designed to exagerate differences between numbers in the 20th decimalplace! Let’s look at x = 0.25 and what happens when we try to plug this into ourfunction. According to Maple (with extra precision settings), we have this:

1− cos(0.2520)

0.2540=

4.135× 10−25

8.2718× 10−25

=0.

24 zeros︷︸︸︷000 000 . . . 000 4135

0. 000 000 . . . 000︸︷︷︸24 zeros

82718

You can see that the numbers on the top and bottom of that fraction are waaaaaaaaytoo small for your calculator to use: they are too close to 0. Actually, the problemreally starts just before the calculator gets 4.135× 10−25. The step before this is

1− 0. 999 999 . . . 999︸︷︷︸24 9s

58641

This second number is way to close to 1 for your calculator to see the difference.At this calculation it just gets 0 on top of the fraction.

11.11.5 Calculating Transcendental Functions

Example 5. You are a math consultant hired by an app company that is writingan app for a fitness tracker. The fitness tracker is worn on the wrist, and has a veryprimitive computer in it, which is where you come in. Their app is going to helppeople analyze their throwing motion, for things like throwing baseballs, softballs,frizbees, you name it. One of the things the app will need to do is calculate sine, soit can convert rotational motion (the arm and ball moving around the sholder) tovertical distance. The company has hired you to help with tihs part of the program.

Here are the requriments: write an algorithm (use “pseudo-code”, i.e. pretendcomputer code, but in a very precise, step-by-step format) that calculates sine. Your“program” should obey the following:


(1) Your “program” should be able to calculate sin(x) for a wide range of x-values,such as the following:

sin(0.78), sin(1.47), sin(3.05), sin(25.232 741 23), sin(3, 891, 498), sin(−4, 677, 113)

(2) Your program cannot use any built-in transcendental functions such sine, co-sine, etc.: it can only use +, −, × and ÷ (because the transcendental func-tions are not available on the fitness tracker device. Silly engineers, didn’tthey know that something originally designed to count steps would need tocalculate sine?!)

(3) Your “program” should return values that are accurate to the 8th decimalplace,

(4) Your program should be efficient in the following sense: + and − each cost1 something (it could be 1 cent, or 1 second, or 1 Watt-seconds of electricity,etc.), but × and ÷ “cost” 10 somethings.

(5) Your program can have some logic (if-then statements, conditions, differentcases, etc.) and stored values (maybe a handful of numbers that are part ofthe functions used), but not a huge amount (for some reasonable definition ofhuge).

Solution. Let’s consider one of the values listed above, sin(25.23274123). At firstglance, you should think that you just plug x = 25.23274123 into

sin(x) = x− x3

3!+x5

5!− x7

7!+ . . .

and keep adding terms until you get an accurate enough answer (as needed for part(3) ). However, the formula just given is most accurate for values close to 0, and 25is too big! We would have to add a lot of terms to make this accurate. But thiswould violate requirment (4).

To avoid this problem, we reduce the value of x, by subtracting multiples of 2πfrom it.

sin(25.23274123) = sin(25.23274123− 2π) = sin(18.94955592)

= sin(25.23274123− 4π)

= sin(25.23274123− 6π)

= sin(25.23274123− 8π) = sin(0.1)

Therefore, we can use the formula above with x = 0.1, and since 0.1 is “close” to0, we should only need a few terms of the Maclaurin series. So, now we have thebeginning of our pseudo-code:

Let x = input

Add/subtract 2π from x until result in [0, 2π)

So the first step of our pseudo-program is to reduce values of x as we’ve justdone. By subtracting (or adding) multiples of 2π we can put every x into the rangeof [0, 2π):

sin(0.78) = sin(.78)


sin(1.47) = sin(1.47)

sin(3.05) = sin(3.05)

sin(25.232 741 23) = sin(0.1)

sin(3, 891, 498) = sin(4.462 592 654)

sin(−4, 677, 113) = sin(2.178592654)

Some of these numbers will be easy to calculate with the MacLaurin series for sine,but some won’t. For instance, if we use

4.462 592 654− (4.462 592 654)3

3!+

(4.462 592 654)5

5!

to approximate sin(4.462 592 654) we get 4.399 which is really inaccurate. Again,we could more terms to our approximation, but we can also reduce x more.

Let’s think about how to reduce x. How would you usually find something likesin(330◦) by hand? You would use one of the values of sine in the range 0 to 90.In particular, sin(330) = − sin(30) = −1/2. We can have our pseudo-program dothat too. We reduce every x value to the range [0, π/2], and sometimes make thewhole result negative depending on which quadrant the original angle was in. Inparticular, we have this:

0 ≤ x ≤ 2π −→ 0 ≤ x ≤ π/2

x 7→

π − x if π/2 < x ≤ πx− π if π < x ≤ 3π/2

2π − x if 3π/2 < x < 2π

So, based on the last step we know which quadrant the angle is, and then we subtractmultiples of π/2 until we get something in the first quadrant.

sin(0.78) = sin(.78) = sin(0.78)

sin(1.47) = sin(1.47) = sin(1.47)

sin(3.05) = sin(3.05) = sin(3.05− π/2) = sin(1.4792)

sin(25.232 741 23) = sin(0.1) = sin(0.1)

sin(3, 891, 498) = sin(4.462 592 654) = − sin(4.462 592 654− π) = − sin(1.321)

sin(−4, 677, 113) = sin(2.178 592 654) = sin(2.178 592 654− π/2) = sin(0.607 796 326)

Putting this together with our work above, we have the following start:

Let x = input

Add/subtract 2π from x until result in [0, 2π)If x > π remember to make result negative

Get reference angle in [0, π/2]:replace x with π−x if π/2 <x≤ πreplace x with x −π if π <x≤ 3π/2replace x with 2π−x if 3π/2 <x< 2π

Plug x into x− x3

3!+ · · ·+???

Make result negative if needed


So this reduces the whole problem to calculating sin(x) for 0 ≤ x ≤ π/2. Now wefigure out how many terms we need in the polynomial to make our answer accurateto the 8th decimal place. There are two reasonable ways to determine how big theerror will be. For this example, it makes sense to compare our polynomial to sin(x)on our calculators or computers. After all, the fitness tracker doesn’t have sine, butwhile we are writing our program for the fitness tracker we can use more advancedtechnology. I’ll describe below how to calculate the error using calculus.

If we compare the graph of sine on [0, π/2] to the degree 5 MacLaurin polynomial,this is what we get:

That looks like a great match. But is it accurate to the 8th decimal place? Totell more precisely we should not look at the original graphs, but at the differencebetween them. In other words, instead of graphing both sin(x) and T5(x) (whereT5(x) is the degree 5 polynomial) and then trying to see the gaph between them, weshould simply graph f(x)− T5(x). If f(x) = T5(x) then the graph of f(x)− T5(x)would be 0. If the gap between f(x) and T5(x) has size 0.001, then f(x) − T5(x)will have size 0.001, etc. Here’s the grapd of f(x)− T5(x):


sin(x)− degree 5 poly

This shows that the our approximation (the degree 5 polynomial) is off by about0.004, which is too much.

To figure out how many terms we would need to make our answer more accurate,we can just guess and check. (There are ways to calculate how big this gap is, whichI’ll discuss at the end.) We will keep increasing the degree of Taylor polynomial touse, until we see the graph being contained within ±0.000 000 009 = 9× 10−9





So, that pretty much does it. Now to describe the above process in psuedo-code:

Let x = input

Add/subtract 2π from x until result in [0, 2π)If x > π remember to make result negative


Plug x into x− x3

3!+ · · ·+ x13

13!Make result negative if needed

We are done, but as promised, here’s an alternative way to figure out the rightdegree of Taylor polynomial to use to make the error be less than 0.000 000 009.


The following result shows how big the error can be in a Taylor series:

error between MacLaurinpolynomial of degree n andsin(x) on interval [0, π/2]

≤ (π/2)n+1

(n+ 1)!

We want this error to be less than 0.000 000 01, so at most it can be 0.000 000 009.Thus we solve the following for n:

error between MacLaurinpolynomial of degree n andsin(x) on interval [0, π/2]

≤ (π/2)n+1

(n+ 1)!< 0.000 000 009

There is not a good algebraic way to solve this. We could proceed by trial and error:

n(π/2)n+1

(n+ 1)!degree of MacLaurin poly upper bound for error

1 1.233 700 5503 0.253 669 5075 0.020 863 4807 0.000 919 2609 0.000 025 20211 0.000 000 47113 0.000 000 006

Or we could replace (n+ 1)! with an approximation:

n! ≈√

2πn(ne

)n

which yields(π/2)n+1

√2π(n+ 1)

(n+1e

)n+1 ≈ 9× 10−9

This is not easy to solve, but our computer or calculator can approximate a solution:

n ≈ 12.848

which confirms as above that we should use n = 13, i.e. a degree 13 polynomial.Finally, just for fun, suppose your clients come back to you and say that a degree

13 polynomial is just too expensive: it’s sucking all the battery life out of the fitnesstracker. On the other hand, suppose they say you could use a little more memoryif you want: that turns out to be relatively cheap.

Well, we could be smarter in how we use our polynomials. After all, it looks likeour polynomial is about perfect at x = 0 (this is true) and it’s worst at x = π/2.Maybe we could make the results better if we center the Taylor series at π/4? Thisis true. Maybe if memory is really cheap, we can even use two polynomials, one for[0, π/4] and one for [π/4, π/2]. This is true. Maybe we can tweak the polynomialsso that the error is spread a little more evenly. This is true. If we combine all ofthe above ideas, we can produce the following.


Let x = input

Subtract 2π from x until result in [0, 2π)If x > π remember to make result negative


Subtract π/2 from x until result in [0, π/2]If x < 0.7353981635 plug x into

− 2.6055× 10−9 + (1.000000350 + (−0.0000076606040 + (−0.1666038290

+ (−0.000245781867 + (0.008828111005− 0.0004965572149x)x)x)x)x)xIf x > 0.7353981635 plug x into

− 0.00021837049 + (1.001466317 + (−0.00420144159 + (−0.1599845671

+ (−0.00640798292 + (0.01207764189− 0.001260610577x)x)x)x)x)xMake result negative if needed

Here are some comments about the above pseudo-code. The polynomials thatappear above are combinations of something called Chebyshev polynomials, pro-duced by an iterative algorithm known as the Remez algorithm. The polynomials,as written, are vastly cheaper than the degree 13 polynomial we used above. Firstly,they are just degree 6, which is a lot less than 13. Secondly, they have been writtenso that no divisions are used. The idea behind this is simple: a polynomial likex5

5!can be rewritten as 0.008333333333x5. The version with

1

5!requires an extra

4 multiplications and one division. Thirdly, these polynomials have x factored outof each term: this minimizes the number of products. It is exactly 6 here, whereasif we write a polynomial in the usual way, we calculate products for each power ofx. The intervals are not exactly [0, π/4] and [π/4, π/2: I adjusted them by hand alittle bit (the dividing line is π/4− 0.05) to make the error on both intervals nearlyequal. The net result for the error is shown below: note that it is still less than10−8, as desired (but at a much lower cost).


This example goes beyond elementary calculus, but provides what I think is a niceexample of how much work sometimes goes into engineering someting to performoptimally under very constrained conditions.

This is where we ended on Friday, December 9.

contents · 2016-12-09 · w. ethan duckworth, loyola university maryland, 2016 de nitions of...

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