integration techniques - university of alaska systemafmaf/classes/math252/lesson/... · 2013. 4....

Chapter 1

Integration Techniques

1.1 Integration Techniques

The Riemann integral is defined to be∫ ba

f(x) dx = lim‖∆x‖→0

n∑i=0

f(c0)(xi+1 − xi)

where the xi partition (a, b). However the Fundamental Theorem of Calculus allows integrals to be calculated by∫ ba

f(x) dx = F (b)− F (a)

where dF (x)dx = f(x). This technique of finding an anti-derivative works well if an anti-derivative exists and can befound. Consider the following problems.

1. Integrate the following by finding an anti-derivative.∫sin(x2)2x dx

2. Why will the technique for finding the anti-derivative above not work on this integral? Note this function isintegrable.

∫xex dx

Because there are no simple rules for finding all anti-derivatives many techniques must be learned.

1.2 Integration by Parts

1.2.1 Derivation

Consider the following use of the derivative product property and be prepared to explain the steps.

f(x) = g(x)h(x).

f ′(x) = g(x)h′(x) + g′(x)h(x).∫f ′(x) dx =

∫g(x)h′(x) + g′(x)h(x) dx.

g(x)h(x) =

∫g(x)h′(x) dx+

∫g′(x)h(x) dx.∫

g(x)h′(x) dx = g(x)h(x)−∫g′(x)h(x) dx.∫

udv = uv −∫v du.

1

2 CHAPTER 1. INTEGRATION TECHNIQUES

1.2.2 Example

Consider the following technique called integration by parts. Remember

Equation 1∫udv = uv −

∫v du.

∫xex dx

∫xex dx =∫xex dx

u = x dv = ex dxdu = dx v = ex

= xex −∫ex dx

= xex − ex + C.

1.2.3 Practice

Use integration by parts to find the following anti-derivatives.

1.

∫x sinx dx

2.

∫x2 sinx dx

3.

∫x lnx dx

4.

∫lnx dx

5. Complete the WebAssign problems.

1.2.4 Example

The practice problems required using integration by parts in different ways. The following examples demonstratesanother use of this technique.

∫ex sinx dx =∫exsinx dx

u = ex dv = sinx dxdu = ex dx v = − cosx

= ex(− cosx)−∫− cosxex dx

= −ex cosx+∫ex cosx dx

u = ex dv = cosx dxdu = ex dx v =sinx

= −ex cosx+ exsinx−∫

sinxex dx

= −ex cosx+ ex sinx−∫ex sinx dx.

1.3. INTEGRATION OF TRIGONOMETRIC FUNCTIONS 3

Compare the remaining integral on the right to the original integral on the left of the equation.

∫ex sinx dx = −ex cosx+ ex sinx−

∫ex sinx dx.

Glob =

∫ex sinx dx.

Glob = −ex cosx+ ex sinx−Glob.2Glob = −ex cosx+ ex sinx.Glob =

1

2(−ex cosx+ ex sinx).∫

ex sinx dx =1

2(−ex cosx+ ex sinx) + C.

1.2.5 Practice

Find the following anti-derivatives. Try using integration by parts and another method.

1.

∫ex sinx dx

2.

∫ex sinhx dx

3. How many types of problems like this exist?


1.3 Integration of Trigonometric Functions

1.3.1 Motivation

• Integrate∫

sin2 y cos y dy

• Integrate∫

cos57 θ sin θ dθ.

• Note that both were easy to integrate using a u-substitution.

• Note that∫

sin2 y cos3 y dy cannot be directly integrated using a u-substitution.

• Identities, however, can be used to convert such products of trigonometric functions into a u-substitution form.

1.3.2 Example

∫sin2 y cos3 y dy =∫

sin2 y cos2 y cos y dy = Look at the u-substitution to see why cos y is left.

Remember: sin2 y + cos2 y = 1. Explain how this is used in the next step.∫sin2 y(1− sin2 y) cos y dy = u = sin y.

du = cos y.∫u2(1− u2) du = Make the substitution.∫u2 − u4 du =


1

3u3 − 1

5u5 + C =

1

3sin3 y − 1

5sin5 y + C. Undo the substitution.

1.3.3 Example ∫sin2 y cos2 y dy =

Remember:sin2 y = 12 [1− cos(2y)],cos2 y = 12 [1 + cos(2y)].

Note nothing is left this time.∫1

2[1− cos(2y)]1

2[1 + cos(2y)] dy = Which identities were used?

1

4

∫1− cos2(2y) dy = Note the identity is needed again.

1

4

∫1− 1

2[1 + cos(4y)] dy =

1

4

∫1

2− 1

2cos(4y) dy =

1

8

∫1− cos(4y) dy =

1

8

[y − 1

4sin(4y)

]+ C.

1.3.4 Motivation

• Integrate∫

tan θ sec2 θ dθ.

• Integrate∫

tan θ sec3 θ dθ.

• Note that both were easy to integrate using a u-substitution.• Note that

∫tan3 θ sec4 θ dθ cannot be directly integrated using a u-substitution.

• Identities, however, can be used to convert such products of trigonometric functions into a u-substitution form.

1.3.5 Example ∫tan3 θ sec4 θ dθ =

Remember: sec2 θ = 1 + tan2 θ.∫tan3 θ sec2 θ sec2 θ dθ = Look at the u-substitution to see why sec2 θ is left.∫

tan3 θ(1 + tan2 θ) sec2 θ dθ =u = tan θ,du = sec2 θ dθ.∫

u3(1 + u2) du = Make the substitution.∫u3 + u5 du =

1

4u4 +

1

6u6 + C =

1

4tan4 θ +

1

6tan6 θ + C. Undo the substitution.

1.4. TRIGONOMETRIC SUBSTITUTION 5

1.3.6 Example

Note the different approach in this example.∫tan2 θ csc θ dθ =∫ (

sin θ

cos θ

)2(1

sin θ

)dθ = What identities were used here?∫

sin θ

cos2 θdθ =∫

1

cos2 θsin θ dθ =

u = cos θ,du = − sin θ dθ.∫

− 1u2

du =

1

u+ C =

1

cos θ+ C =

sec θ + C

1.3.7 Practice

1. Integrate

∫3 tan θ sec6 θ dθ.

2. Integrate

∫tan2 θ sec θ dθ as far as possible.


1.4 Trigonometric Substitution

1.4.1 Motivation

1. Identify the anti-derivative for each of the following. At most one might not be known.∫1√

1− x2dx

∫1

1 + x2dx

∫x

1 + x2dx

∫1

1− x2 dx

2. All of these have what type (group) of function as anti-derivative?

3. What is similar about each of the integrands above?

1.4.2 Example

∫y√

1− y2dy =

Since 1− y2 looks like 1− sin2 θLet 1− y2 = 1− sin2 θ,

or y = sin θ

and dy = cos θ dθ.

6 CHAPTER 1. INTEGRATION TECHNIQUES∫sin θ√

1− sin2 θcos θ dθ y and dy are substituted everywhere.∫y√

1− y2dy =

∫sin θ√

1− sin2 θcos θ dθ

Remember cos2 θ = 1− sin2 θ.=

∫sin θ√cos2 θ

cos θ dθ.

=

∫sin θ

cos θ

cos θdθ.

=

∫sin θ dθ.

= − cos θ + C= −

√1− y2 + C. See Figure 1.1 below.

sin(θ) =y

1

1 y

θ

1 − y2p

Figure 1.1: Reference for trig substitution

1.4.3 Example

The following looks more complicated, but algebra enables the same technique to work.

∫1√

2y2 − 5dy =∫

1√5( 25y

2 − 1)dy = Force the constant to be 1.

1√5

∫1√

25y

2 − 1dy = Factor out the constant.

Since2

5y2 − 1 looks like sec2 θ − 1.

Let2

5y2 − 1 = sec2 θ − 1.

2

5y2 = sec2 θ.√2

5y = sec θ.

y =

√5

2sec θ.

dy =

√5

2sec θ tan θ dθ.

1.5. INTEGRATION USING PARTIAL FRACTION DECOMPOSITION 7

∫1√

2y2 − 5dy =

1√5

∫1√

sec2 θ − 1

√5

2sec θ tan θ dθ.

=1√2

∫1√

sec2 θ − 1sec θ tan θ dθ.

=1√2

∫1√

tan2 θsec θ tan θ dθ.

=1√2

∫1

tan θsec θ tan θ dθ.

=1√2

∫sec θ dθ.

=1√2

ln | sec θ + tan θ|+ C

=1√2

ln

∣∣∣∣∣√

2

5y +

√2y2 − 5

5

∣∣∣∣∣+ C. See Figure 1.2 below.

θ

sec(θ) =2y

5

2y5

2y2 − 5

p

p p

p

p

Figure 1.2: Reference for trig substitution

1.5 Integration Using Partial Fraction Decomposition

1.5.1 Motivation

Partial fraction decomposition is an algebraic technique for breaking up rational expressions into smaller pieces.Read the following examples and answer the questions.

2

5+

4

7=

7

7· 2

5+

5

5· 4

7

=14

35+

20

35

=14 + 20

35

=34

35.

• Where did the 5 and 7 in the denominator’s end up in the final expression?• What is the name of this process (from elementary school)?• What does this tell us about the way 4330 might have resulted from a sum?

The next example repeats this process on a rational expression.

2

x+ 3+

3

x− 5 =x− 5x− 5 ·

2

x+ 3+x+ 3

x+ 3· 3x− 5


=2x− 10

(x+ 3)(x− 5) +3x+ 9

(x+ 3)(x− 5)

=(2x− 10) + (3x+ 9)

(x+ 3)(x− 5)

=5x− 1

x2 − 2x− 15 .

• Where did the x+ 3 and x− 5 in the denominator’s end up in the final expression?

• What is the name of this process (from algebra 1)?

• To break apart 4x−71x2−13x+22 what should be (un)done first?

• What does this tell us about the way 4x−71x2−13x+22 might have resulted from a sum?

1.5.2 Example

Notice how the process of adding fraction is worked backwards in the following example.

4x− 71x2 − 13x+ 22 = First factor the denominator.

4x− 71(x− 2)(x− 11) = Write the atomic fractions.

4x− 71(x− 2)(x− 11) =

A

x− 2 +B

x− 11 . Clear the denominators.

(x− 2)(x− 11) · 4x− 71(x− 2)(x− 11) = (x− 2)(x− 11) ·

(A

x− 2 +B

x− 11

).

4x− 71 = A(x− 11) +B(x− 2).

Finding Coefficients Trick

Note the equation 4x− 71 = A(x− 11) +B(x− 2) has a function (polynomial) on both sides.

1. If these two functions are equal, what will be true about their outputs for x = 2? x = 11? x = 908?

2. If any two functions are equal, at how many inputs will their outputs be the same?

Notice how we use this concept of equal functions to find the coefficients.

4x− 71 = A(x− 11) +B(x− 2).x = 2.

4(2)− 71 = A(2− 11) +B(2− 2).−63 = −9A.A = 7.

x = 11.

4(11)− 71 = A(11− 11) +B(11− 2).−27 = 9B.B = −3.

4x− 71(x− 2)(x− 11) =

7

x− 2 −3

x− 11 .

1.5. INTEGRATION USING PARTIAL FRACTION DECOMPOSITION 9

Finding Coefficients Method

4x− 71 = A(x− 11) +B(x− 2).4x− 71 = Ax− 11A+Bx− 2B.4x− 71 = (A+B)x+ (−11A− 2B).

Note the equation 4x− 71 = (A+B)x+ (−11A− 2B) has a polynomial on both sides.

1. If 4x− 71 = Mx+N (two polynomials are equal), what should M and N be?

2. To prove this try the following. Plug in x = 0 on both sides. What does the result state about N?

3. Differentiate both sides. What does the result state about M?

4. If any two polynomials are equal, at how many inputs will their outputs be the same?

5. If 4x− 71 = (A+B)x+ (−11A− 2B), what will be true about 4 and A+B and −71 and −11A− 2B?

A+B = 4. This is the original equation.

11(A+B) = 11(4). Scale both sides by 11.

11A+ 11B = 44.

−11A− 2B = −71. This is the other original equation.9B = −27. Added the previous two equations.B = −3.

A+ (−3) = 4.A = 7.

1.5.3 Example

Consider the following decomposition of a fraction.

41

28=

3

4+

5

7.

=3

22+

5

7

=1

2+

1

22+

5

7.

Why the 34 is broken into two fractions will be explained in a later course.This example breaks up a complex rational expression into atomic rational expressions using this repeated term

technique.

−2x2 − 9x− 16(x+ 2)2(x− 1) =

A

x+ 2+

B

(x+ 2)2+

C

x− 1 .

(x+ 2)2(x− 1)−2x2 − 9x− 16

(x+ 2)2(x− 1) =

(x+ 2)2(x− 1)(

A

x+ 2+

B

(x+ 2)2+

C

x− 1

).

−2x2 − 9x− 16 = A(x+ 2)(x− 1) +B(x− 1) + C(x+ 2)2.


Use the first trick for finding the coefficients.

−2x2 − 9x− 16 = A(x+ 2)(x− 1) +B(x− 1) + C(x+ 2)2.x = −2.−6 = −3B.B = 2.

x = 1.

−27 = 9C.C = −3.

x = 0.

−16 = A(2)(−1) +B(−1) + C(4).−16 = −2A− 2− 12.A = 1.

−2x2 − 9x− 16(x+ 2)2(x− 1) =

1

x+ 2+

2

(x+ 2)2+−3x− 1 .

1.5.4 Practice

1.∫

7x−2 − 3x−11 dx

2.∫ −2x2−9x−16

(x+2)2(x−1) dx


1.5.5 Motivation

• Before starting PFD what must be done with x2−x−1x2−3x+1?

• What is the clue that this needs to be done?

• What happens with 3−4xx2−4x+13?

• If we use Ax−1 why does Ax+Bx2−4x+13 make sense?

1.5.6 Example

This example demonstrates the use of the quadratic term.

2x2 + 3

(x− 1)(x2 + 4) =A

x− 1 +Bx+ C

x2 + 4.

2x2 + 3 = A(x2 + 4) + (Bx+ C)(x− 1).2x2 + 3 = Ax2 + 4A+Bx2 −Bx+ Cx− C.2x2 + 3 = (A+B)x2 + (−B + C)x+ (4A− C).

2 = A+B. (1.1)

0 = −B + C. (1.2)3 = 4A− C. (1.3)

1.6. IMPROPER INTEGRATION 11

B = C. From 1.2

A+ C = 2. From 1.1

−4A− 4C = −8. Scaling above equation.4A− C = 3. This is 1.3−5C = −5. Adding the above equations.C = 1.

B = 1. From B = C above. (1.4)

A+ 1 = 2. From 1.1

A = 1. (1.5)

(1.6)

2x2 + 3

(x− 1)(x2 + 4) =1

x− 1 +x+ 1

x2 + 4(1.7)

1.5.7 Practice

1.∫

2x2+3(x−1)(x2+4) dx


1.6 Improper Integration

1.6.1 Derivation

Review

For each of the following partitions graph the function, draw rectangles for the associated lower sum, and calculatethe value of the lower sum for f(x) = 1− x2.

• {−1, 0, 1},

• {−1,− 12 , 0, 12 , 1},

• {−1,− 34 ,− 12 ,− 14 , 0, 14 , 12 , 34 , 1},

• {−1,− 78 ,− 34 ,− 58 ,− 12 ,− 38 ,− 14 ,− 18 , 0, 18 , 14 , 38 , 12 , 58 , 34 , 78 , 1}.

1. Describe the difference between the areas represented by your lower sums and the actual area between f(x)and the x-axis.

2. How are the partitions changing? Consider them in order from the first listed to the last listed.

3. How does the area estimate change as the partitions change?

4. How is the exact area obtained from the area estimates (the sums)?

New Application

g(x) =1

x3/2.

1. Begin calculating∫ 1

0g(x) dx.

2. Why can’t this calculation be finished?


3. Evaluate each of the following integrals.∫ 10.1g(x) dx,

∫ 10.01

g(x) dx,∫ 1.001

g(x) dx,∫ 1.0001

g(x) dx.

4. Describe the difference between the areas represented by your integrals and the total area between the curveand the x-axis.

5. How are the integrals changing? Consider them in order from the first listed to the last listed.

6. How does the area estimate change as the limits of integration change?

7. How might you calculate the actual area from the area estimates?

Another New Application

h(x) = e−x.

1. Begin calculating∫∞

0h(x) dx.

2. Why can’t this calculation be finished?

3. Evaluate each of the following integrals.∫ 10e−x dx,

∫ 100e−x dx,

∫ 1000

e−x dx,∫ 1000

0e−x dx.

4. Describe the difference between the areas represented by your integrals and the total area between the curveand the x-axis (to the right of the y-axis).

5. How are the integrals changing? Consider them in order from the first listed to the last listed.

6. How does the area estimate change as the limits of integration change?

7. How might you calculate the actual area from the area estimates?

1.6.2 Example ∫ 10

1√xdx

∫ 10

1√xdx = lim

a→0+

∫ 1a

1√xdx

= lima→0+

2√x∣∣1a

= lima→0+

2√

1− 2√a= 2.

1.6.3 Practice

1.

∫ 10

1

x2dx

2.

∫ ∞1

1

x2dx

3.

∫ 1−1

1

x2dx

4.

∫ 1−1

1

xdx


Chapter 2

Series

2.1 Sequences

The limits and integrals of previous parts of calculus are defined on functions defined on every real number. Thereare also limit and sum problems for which functions are only defined on every integer.

2.1.1 Motivation

Complete the following problems to see examples of sequences.

The Story of Bakbuk the Traveler

Bakbuk the traveler has a bushel of grain. Each night he offers half of the grain he has to a person in exchange fora room for the night.

1. How much grain does he have when he departs on his voyage?

2. How much grain does he have after the 1st night, 2nd, 3rd, and 4th nights?

3. How much grain does he have after the nth night?

4. As he continues to travel what will happen with the amount of grain he carries? Assume he does not buy more.

The Story of the Inheritance of Bigvai

Bigvai is given a bar of gold that is one unit wide. At the end of his life he cuts out the middle third to pay remainingdebts and passes on the remaining gold to his children. His children at the ends of their lives remove the middlethirds of each piece of gold to pay remaining debts and pass on the remaining gold to their children. This is repeatedeach generation.

Bigvai Generation 1 Generation 2 Generation 3

1. How many bars does Bigvai have?

2. How many bars do the first, second, and third generations have?

3. How many bars do the nth generation have?

4. How wide is the Bigvai’s bar?

5. How wide are the first, second, and third generation’s bars?

6. How wide are the nth generation’s bars?

13

14 CHAPTER 2. SERIES

7. How much of a bar of gold does the nth generation possess?

8. As the generations pass what is the number of bars approaching?

9. As the generations pass what is the width of the bars approaching?

10. As the generations pass what is the amount of gold approaching?

2.1.2 Definition

Definition 1 (Sequence) A function is a sequence if and only if its domain is a subset of the integers.

Below are four notations for sequences. The second is most common and will be used primarily in this text.

1. {0, 2, 4, 6, 8, . . .}

2. an = 2n, n = 0, 1, 2, . . .

3. f(n) = 2n, n = 0, 1, 2, . . .

4. {2n}∞n=0

2.1.3 Uses

Because sequences are integer value, the only limit of interest is to infinity.

Definition 2 (Limit of a Sequence) limn→∞

an = L if and only if for all � > 0 there exists an integer N > 0 such

that n > N implies |an − L| < �.

Using notes from Calculus 1 if needed, write the formal definition for the following sequence limit.

limn→∞

an =∞.

Note that the only difference between these definitions and the limit definitions of other functions is that N hasto be an integer. Because the integers are a subset of the reals and due to a theorem presented in analysis, all thelimit techniques learned previously still apply.

Find the limits of each of these sequences.

an 2, 1,12 ,

14 ,

18 ,

116 , . . .

bn21 ,

42 ,

83 ,

164 ,

325 ,

646 , . . .

cn1e ,− 1e2 , 1e3 ,− 1e4 , 1e5 ,− 1e6 , . . .

dn35 ,

115 ,

165 ,

375 ,

795 ,

1635 , . . .

fn12 ,− 34 , 78 ,− 1516 , 3132 ,− 6364 , . . .

Why can limn→k

an (the limit of a sequence as the input approaches any number) not exist?

2.1.4 Examples

Algebra

limn→∞

n2 + 2

n3=

limn→∞

n2

n3+

2

n3=

limn→∞

1

n+

2

n3=

limn→∞

1

n+ limn→∞

2

n3= 0.

2.1. SEQUENCES 15

L’Hôpital’s Rule

limn→∞

√2n + n2

2n=

limn→∞

√1 +

n2

2n= , because radicals are continuous√

limn→∞

1 +n2

2n=√

limn→∞

1 + limn→∞

n2

2n.

Aside

limn→∞

n2

2n=

limn→∞

n2

en ln 2∞/∞

= L’Hôpital’s Rule

limn→∞

2n

(ln 2)en ln 2∞/∞


limn→∞

2

(ln 2)2en ln 2= 0.

limn→∞

√2n + n2

2n=√

limn→∞

1 + limn→∞

n2

2n=

√1 + 0

= 1.

Squeeze

Where is Ω(n) = sin(nπ/4)n going? To begin to understand this sequence try the following.

1. Write the first 8 terms.

2. What appears to be happening?

3. Graph the points (n,Ω(n)).

4. What appears to be happening?

5. Speculate where Ω(n) is going.

−1 ≤ sin(nπ/4) ≤ 1.

− 1n ≤sin(nπ/4)

n≤ 1n .

limn→∞

− 1n≤ lim

n→∞

sin(nπ/4)

n≤ lim

n→∞

1

n.

0 ≤ limn→∞

sin(nπ/4)

n≤ 0.

limn→∞

sin(nπ/4)

n= 0.

Now complete the WebAssign problems.


Monotonic Bounded

This example illustrates a concept about sequences which provides a method for determining if a limit exists. Notethis concept does not tell us what the limit value is. Given a0 =

√2, an =

√2 + an−1. The goal is to find lim

n→∞an.

1. Evaluate the first 8 terms.

2. Approximate each term using a computer or calculator. Record as many decimal places as possible.

3. Looking at these terms what is the relation between a element and the element before it?

4. Guess the limit.

5. Prove that your guess is correct using the following ideas.

Show that an < 2 for all n, that is, an is bounded above.

1. a0 < 2

√2 < 2.

2. a1 < 2

2 +√

2 < 2 + 2√2 +√

2 <√

2 + 2

=√

4

= 2.

a1 < 2.

3. Show a2 and a3 are less than 2.

4. Show an < 2.

Show that an > an−1 for all n, that is, an is strictly increasing.

1. a1 > a0.

2 +√

2 > 2, so√2 +√

2 >√

2.

2. Show a2 > a1

3. Show an > an−1

1. In what direction is an tending?

2. Is there a boundary an does not cross in this direction?

3. Because of this, does an converge?


2.2. SERIES 17

2.2 Series

2.2.1 Definition

Remember, a function is a sequence if and only if its domain is a subset of the integers. A series is a sum of theelements of a sequence. Below are a sequence and its associated series.

Sequence: ak =12k

{12 ,

14 ,

18 ,

116 , . . .

}

Series:

∞∑k=1

1

2k12 +

14 +

18 +

116 + . . .

The following steps illustrate the definition of a series.

an =

(1

2

)nS1 = a1. S2 = a1 + a2. Likewise Sn is the sum of the first n elements of the sequence.

1. Evaluate S1, S2, S3, S4, S5.

2. What is the relationship between S1, S2, S3, S4, S5 and the sum of all (infinite) elements of the sequence.

3. What technique could we use to find the total of the infinite sequence?

4. Find a pattern to the sums you calculated (i.e., find a formula for Sn).

5. Find S∞ using the pattern.

Definition 3 (Series)

∞∑k=1

ak = limn→∞

Sn where Sn =

∞∑k=1

ak.

2.2.2 Example

bn = 2n for n = 0, 1, 2, . . . Calculate

∞∑k=0

bn.

First find calculate a few values for Sn.

n Sn0 1 = 1.1 1 + 2 = 3.2 1 + 2 + 4 = 7.3 1 + 2 + 4 + 8 =154 1 + 2 + 4 + 8 + 16=31

The Sn values are each one less than the next value of bn, i.e., 2− 1, 4− 1, 8− 1, 16− 1, 32− 1, . . . Thus

∞∑k=0

bn =

∞∑k=0

2n =

limn→∞

2n+1 − 1 = ∞.

If the series value exists and is finite, the series is called convergent otherwise it is called divergent.


2.2.3 Warning

an = (−1)n, n = 0, 1, 2, . . .Attempt to find this sum using the same technique as the previous example. The steps are below.

1. Calculate Sn for n = 1, 2, 3, 4, 5, 6.

2. Write a pattern for Sn

3. What kind of sequence is this (describe its action)?

4. Does this limit exist?

5. Does the series value exist?

2.3 Series Techniques

2.3.1 Lower Limit

1. What is

∞∑k=1

1?

2. What is

∞∑k=2

1?

3. What is

∞∑k=3

1?

4. What is

∞∑k=50

1?

5. If we only care whether a series is finite or infinite do we care about the first 2 terms? 5 terms? 500 terms?

2.3.2 Algebra

1. Write out∑3i=1 i and

∑3i=1 i

2. Do not add the terms.

2. Write out∑3i=1(i+ i

2).

3. Compare this to the previous two.

4. What might we do with∑∞i=1

(1i − 1i2

)?

5. What condition would this require?

2.4 nth Term Divergence

2.4.1 Derivation

Suppose an ≥ 2 for all k.

1. What is the

∞∑n=1

2?

2. As a result what is the

∞∑n=1

an?

2.5. GEOMETRIC SERIES 19

We will now discover a test for infinite (divergent) series. Suppose limk→∞

ak = 3.

1. Because the limit is 3, the distance |ak − 3| approach what?

2. Will all the sequence values ak eventually be larger than 2?

3. What is the

∞∑n=1

2?

4. Therefore what must

∞∑n=1

ak be?

2.4.2 nth term divergence test

If limn→∞

an 6= 0, then∞∑k=0

an diverges.

2.4.3 Practice

Use the nth term divergence test to determine if the series is definitely divergent or not.

1. an = 1− 12n .

2. bn =1

1+n .

3. cn =1

2+3n .


2.5 Geometric Series

Definition 4 (Geometric) A series is geometric if it can be written∑∞k=0 ar

k where r is a constant.

The following steps calculate the partial sum Sn.

sn = a+ ar + ar2 + . . .+ arn−1.

rsn = ar + ar2 + ar3 + . . .+ arn.

sn − rsn = (a+ ar + ar2 + . . .+ arn−1)−(ar + ar2 + ar3 + . . .+ arn)

= a− arn.sn(1− r) = a(1− rn).

sn =a(1− rn)

1− r .

Thus the sum of a geometric series is given by the following.

∞∑k=0

ark = limn→∞

n∑k=0

ark

= limn→∞

a(1− rn)1− r

=a

1− r .

However, there are restrictions.


1. The following are all geometric series.

(a)

∞∑n=0

(27

133

)n(b)

∞∑n=0

(27

27

)n(c)

∞∑n=0

(133

27

)n2. Which of the above will converge?

3. Based on the derivation provided what is the sum of the convergent series above?


2.6 Integral Series

2.6.1 Derivation

• (Review) If ak ≤ 5 and ak ≥ ak−1 does the sequence converge?

• Suppose ak = 1/k2 Note ak = 1/k2 ≥ 0 for all k ≥ 1.

– Which is bigger a1 or a1 + a2? a1 + a2 or a1 + a2 + a3?

– Remember Sn is the sum of the first n terms of ak.

– Which is bigger Sn or Sn−1?

– Thus the sequence of Sn is doing what?

• We need one more piece to determine convergence or divergence.

1 2 3 4 5 6 7 8 9 10

1/9

1

1/4

Figure 2.1: Integrals and Series (part 1)

• What in Figure 2.1 represents ∑∞k=2 1k2 ?• What in Figure 2.1 represents

∫∞1

1x2 dx?

• Which is bigger?

• Using this compare ∑∞k=2 1k2 and ∫∞1 1x2 dx.• What is

∫∞1

1x2 dx?

2.7. COMPARISON TESTS 21

1 2 3 4 5 6 7 8 9 10

1

1/2

1/31/4

Figure 2.2: Integrals and Series (part 2)

• Does this series converge or diverge?

• What in Figure 2.2 represents ∑∞k=1 1k?• What in Figure 2.2 represents

∫∞1

1x dx?

• Which is bigger?

• What is∫∞

11x dx?

• Using this compare ∑∞k=1 1k and ∫∞1 1x dx.• Does this series converge or diverge?

2.6.2 Summary∞∑k=1

ak and

∫ ∞1

a(x) dx converge or diverge alike.

2.6.3 Practice

•∫ ∞

1

1

x3/2dx

•∫ ∞

1

1

x1dx

•∫ ∞

1

1

x1/2dx

• Based on your calculations conjecture for what values of p the following series converges and diverges∞∑n=1

1

np.

• Complete the WebAssign assignment.

2.7 Comparison Tests

2.7.1 Motivation

1. Decide which of the series types in Figure 2.3 the series below look like.

(a)

∞∑n=1

7n

3n − 1


(b)

∞∑n=1

1

n2 + 1

(c)

∞∑n=1

7n

3n + 1

(d)

∞∑n=2

1

n3 − 1

2. Do the series they look like converge or diverge?

3. Based on this information conjecture whether these series converge or diverge?

2.7.2 Examples

Example 1

∞∑n=1

1

n2 + 1looks like

∞∑n=1

1

n2. Note n2 + 1 > n2 so 1n2+1 <

1n2 . Thus

∞∑n=1

1

n2 + 1<

∞∑n=1

1

n2. Also note

∞∑n=1

1

n2is a

convergent p-series. Thus this series must converge as well.

Example 2

∞∑n=1

7n

3n − 1 looks like∞∑n=1

(7

3

)n. Note 3n − 1 < 3n so 7n3n−1 > 7

n

3n . Thus

∞∑n=1

7n

3n − 1 >∞∑n=1

(7

3

)n.Also note that∑∞

n=1

(73

)nis a divergent geometric series. Thus this series must diverge as well.

2.8 Limit Comparison

2.8.1 Motivation

1. Attempt to determine if the series∑∞n=2

1n3−1 converges or diverges by comparing it to

∑∞n=2

1n3 .

2. What difficulty do you encounter.

3. The next method addresses this issue.

2.8.2 Derivation

Consider the positive series

∞∑n=1

an and

∞∑n=1

bn. Suppose limn→∞

anbn

= c. According to the definition of a limit∣∣∣anbn − c∣∣∣ < �

for large enough n. We will re-arrange this statement to compare an and bn. Suppose limn→∞

anbn

= c.

−� < anbn − c < �c− � < anbn < c+ �

(c− �)bn < an < (c+ �)bn

Geometric p-series∞∑n=0

arn∞∑n=1

1

np

Figure 2.3: Known Series

2.9. ROOT & RATIO TEST 23

an < (c+ �)bn.∞∑n=1

an <

∞∑n=1

(c+ �)bn.

∞∑n=1

an < (c+ �)

∞∑n=1

bn.

So if∑∞n=1 bn converges, then

∑∞n=1 an also converges.

Again suppose limn→∞

anbn

= c.

−� < anbn − c < �c− � < anbn < c+ �

(c− �)bn < an < (c+ �)bnan > (c− �)bn.

∞∑n=1

an >

∞∑n=1

(c− �)bn.

∞∑n=1

an > (c− �)∞∑n=1

bn.

So if∑∞n=1 bn diverges, then

∑∞n=1 an also diverges.

2.8.3 Example

Compare

∞∑n=2

1

n3 − 1 to∞∑n=2

1

n3.

limn→∞

1n3−1

1n3

=

limn→∞

n3

n3 − 1 =

limn→∞

1n3n

3

1n3 (n

3 − 1) =

limn→∞

1

1− 1n3= 1.

Thus the series

∞∑n=2

1

n3 − 1 converges like∞∑n=2

1

n3.

Now complete the WebAssign assignment.

2.9 Root & Ratio Test

2.9.1 Notation

1. Find the first 10 derivatives of x10. Do not expand the coefficient (you will miss the point).

2. With each succeeding derivative how does the coefficient change (what do you do to it)?

3. The tenth derivative is just a coefficient produced by doing what?

4. Because this is a common operation we call it ‘factorial.’


n! = n(n− 1)(n− 2)(n− 3) . . . (2)(1).1. Write 5!3! using the definition of factorial above.

2. Divide the obvious parts of the expression to simplify.

3. Do the same with 7!5! .

4. What is n!(n−5)!?

2.9.2 Derivation

1. Suppose an ≥ 0 for all n.2. If an+1an < 1 what is the sequence doing?

3. If an+1an > 1 what is the sequence doing?

4. If limn→∞

an+1an

< 1 what is the sequence doing?

5. If limn→∞

an+1an

> 1 what is the sequence doing?

6. Since the sequence is positive, limn→∞

an > (what number)?

7. Because of these facts we can apply what sequence technique to prove convergence?

2.9.3 Ratio Test

If

∞∑n=1

an is a positive series, the series

converges when limn→∞

an+1an

< 1

diverges when limn→∞

an+1an

> 1

confuses us when limn→∞

an+1an

= 1

2.9.4 Example∞∑n=1

n

en

limn→∞

an+1an

=

limn→∞

n+1en+1

nen

=

limn→∞

(n+ 1)en

nen+1=

limn→∞

n+ 1

ne= L’Hôpital’s Rule

limn→∞

1

e=

1

e< 1.

2.10. NON-POSITIVE SERIES 25

Thus the series converges.

2.9.5 Root Test

If

∞∑n=1

an is a positive series, the series

converges when limn→∞

n√an < 1

diverges when limn→∞

n√an > 1

confuses us when limn→∞

n√an = 1

2.9.6 Example

∞∑n=1

(1

n

)n

limn→∞

n

√(1

n

)n=

limn→∞

((1

n

)n)1/n=

limn→∞

1

n= 0

< 1.

Convergent

2.10 Non-Positive Series

2.10.1 Case 1: Review

an = (1/2)n, n = 1, 2, 3, . . .

1

2

3

4

7

8

15

16

0

15 16

7 8

3 4

1 2

31

3231 32

Figure 2.4: Case 1


1. Complete the following table. Note Sn =∑ni=1 ai.

anSn

2. Compare this sequence to cases 2-4. All the terms an are what kind of numbers?

3. Because of this each new sum is what in comparison to the previous one?

4. In order to converge each term must be what size in comparison to previous ones?

5. How do the an values correspond to the arrows above?

6. How do the Sn values correspond to the axis labels above?

2.10.2 Case 2

bn = −1

2n, n = 1, 2, 3, . . .

1. Complete the following table.bnSn

2. All the terms bn are what kind of numbers? Compare to cases 1, 3, and 4.

3. Because of this each new sum is what in comparison to the previous one?

4. How is this different from the previous case?

5. Sketch an arrow diagram for this example.

2.10.3 Case 3

cn =(−1)n+1

n, n = 1, 2, 3, . . .

1. Complete the following table.cnSn

2. What do the terms cn do? Compare to cases 1, 2, and 4.


4. What size is each term in comparison to the previous?

5. How is that seen in the arrow diagram?

6. What is limn→∞

|cn|?

7. Do the sums appear to be converging?

8. How is this seen in the arrow diagram?

2.11. ALTERNATING SERIES 27

2.10.4 Case 4

dn = (−1)n+1(

1 +1

n

), n = 1, 2, 3, . . .

1. Complete the following table.dnSn

2. What do the terms dn do? Compare to cases 1-3.


4. What size is each term in comparison to the previous?

5. What is limn→∞

|dn|?

6. To what do all the downward pointing arrows seem to converge?

7. To what do all the upward pointing arrows seem to converge?

8. What is the difference between these two values?

9. Compare this answer to the limit.

10. Do the sums appear to be converging?

2.11 Alternating Series

Use the examples in the previous section to answer these questions.

1. What was different between the last two series examples (Case 3 and Case 4)?

2. Why did one converge when the other did not?

3. Use these ideas to determine if en =(−1)nlnn converges.

A series

∞∑n=a

bn is alternating if

∞∑n=a

bn =

∞∑n=a

(−1)nan where an > 0 for all n.

The alternating series

∞∑n=a

(−1)nan is convergent if

limn→∞

an = 0. (2.1)

an+1 ≤ an. (2.2)

2.12 Absolute Convergence

2.12.1 Motivation∞∑n=1

cosn

n2

1. Write out the first 5 terms (calculators might be helpful).

2. Is this series positive?

3. Is this series alternating?

4. How do we handle this series?


2.12.2 Derivation

1. Which is bigger 1 + 12 +14 +

18 or 1− 12 + 14 − 18?

2. Why?

3. If 1 + 12 +14 +

18 + . . . converges should 1− 12 + 14 − 18 + . . . converge?

2.12.3 Absolute Convergence Test∞∑n=1

cosn

n2

A series

∞∑n=a

an is absolutely convergent if

∞∑n=a

|an| is convergent.

Consider

∞∑n=1

∣∣∣cosnn2

∣∣∣ =∞∑n=1

| cosn|n2

<

∞∑n=1

1

n2

which is finite.

The series is absolutely convergent

2.12.4 Terminology

• If a series is absolutely convergent, it is convergent.

• A series∞∑n=a

an is absolutely convergent if

∞∑n=a

|an| is convergent.

• A series∞∑n=a

an is conditionally convergent if it is convergent but not absolutely convergent.

Absolute convergence is a series test not a type of series as this terminology may seem to imply.

Chapter 3

Power Series

3.1 Definition

Definition 5 (Power Series) A series is a power series iff it looks like

∞∑n=0

cnxn = c0 + c1x+ c2x

2 + c3x3 + . . . .

The following questions connect the concept and calculations of power series with simple, previous concepts.

1. What are power series?

(a) What type of function is 1− 3x+ 7x2?(b) What type of function is 1− 3x+ 7x2 − 11x3?(c) What type of function is 1− 3x+ 7x2 − 11x3 + 15x4 + 17x7 − 3x10 + 4x11 − 256x13 − 31155 x27?(d) Look at the definition of power series again. What type of function is a power series?

2. What do we calculate?

(a) Test if the following converge or diverge.

i.

∞∑n=0

(3

5

)nii.

∞∑n=0

(7

5

)niii.

∞∑n=0

(x5

)n(b) For what values of x will this series converge?

(c) For what values of x is f(x) =∑∞n=0

(x5

)ndefined?

(d) What do we call this (algebra terminology)?

3.2 Power Series Domains

To determine the values at which a particular power series converges, the following theorem will be needed in additionto the previously learned series tests.

1. Compare(

13

)nand

(23

)n.

2.∑∞n=1

(x3

)nconverges for x = 2. Why?

3. What does the previous line imply (directly) about∑∞n=1

(13

)nconverging or diverging?

29

30 CHAPTER 3. POWER SERIES

4. What does this imply (directly) about∑∞n=1

(2.53

)n5. Note the use of limits in the following proof.

Theorem 1 (Radius of Convergence) If∑cnx

n converges for x = b, then it converges for all |x| < b.

Proof:Suppose

∑cnx

n converges for x = b.limn→∞

cnbn = 0. Which series test tells us this?

By definition of a limit there exists N such that n > N implies |cnbn| < 1. Why?|cnxn| =

∣∣∣ cnbnxnbn ∣∣∣ Next follows some algebra.|cnxn| = |cnbn|

∣∣xb

∣∣n|cnxn| <

∣∣xb

∣∣n .If |x| < b then this is convergent. Which test is this?∑cnx

n <∑cnb

n. Why?Thus

∑cnx

n is convergent. Which test is this?

3.2.1 Example

Using the ratio test

limn→∞

xn+1

(n+ 1)2÷ x

n

n2=

limn→∞

xn+1

(n+ 1)2· n

2

xn=

limn→∞

x · n2

(n+ 1)2∞/∞


limn→∞

x · 2n2(n+ 1)

∞/∞= L’Hôpital’s Rule

limn→∞

x = x.

Thus by the ratio test p(x) converges when |x| < 1, that is the domain contains (radius of convergence) (−1, 1).However, this does not determine whether −1 or 1 are in the domain. These must be checked separately.

p(1) =

∞∑n=1

1n

n2

=

∞∑n=1

1

n2.

p(−1) =∞∑n=1

(−1)nn2

.

The first converges, because it is a p-series. The second converges by the absolute convergence test. Thus the domain(called the interval of convergence) is [−1, 1].

3.3 Generating Power Series

One of the uses of power series is as an alternate representation of known functions. The following sections demon-strate how such a power series can be generated.

3.3. GENERATING POWER SERIES 31

3.3.1 Derivation∫ xc

f ′(t) dt = f(x)− f(c). Note how this definite integral is rearranged.∫ x0

cos t dt = sinx− sin 0.

sinx = sin 0 +

∫ x0

cos t dt. Note how integration by parts is used below to continue this integral.

u = cos t dv = dtdu = − sin t dt v = t− x

= 0 +

(cos t(t− x)|x0 −

∫ x0

− sin t(t− x) dt)

= [cosx(x− x)]− [cos 0(0− x)] +∫ x

0

sin t(t− x) dt

= x+

∫ x0

sin t(t− x) dt Note the specific choice for u and dv made to continue integration by parts.

u = sin t dv = (t− x) dtdu = cos t dt v = 12 (t− x)2

= x+

(sin t

1

2(t− x)2

∣∣∣∣x0

−∫ x

0

cos t1

2(t− x)2 dt

)= x+

(sinx

1

2(x− x)2 − sin 01

2(0− x)2

)−∫ x

0

cos t1

2(t− x)2 dt

= x−∫ x

0

cos t1

2(t− x)2 dt

Complete the following steps to understand how this process works.

1. Continue this integration by parts process for four more steps. You may not guess and skip.

2. How long could this integration by parts problem continue?

3. If this process could be “finished,” what would the result be?

4. What is sin(1) approximately?

3.3.2 Calculation

Definition 6 A function is smooth of order n if its nth derivative exists and is continuous.

1. sinx is infinitely smooth. What other functions are infinitely smooth (at least where they are defined)?

2. The coefficient of each term (power of x) came from what calculation?

For a sufficiently smooth function

f(x) = f(c) + (x− c)f ′(c) + (x− c)2

2!f ′′(c) +

(x− c)33!

f ′′′(c) + . . .

+(x− c)n

n!f (n)(c) + (−1)n

∫ xc

(t− x)nn!

f (n+1)(t)dt.

=

n∑j=0

(x− c)jj!

f (j)(c) + (−1)n∫ xc

(t− x)nn!

f (n+1)(t)dt

= Pn(x) +Rn(x)

Note the following.


• The polynomial portion Pn(x) is called the Taylor polynomial about c of degree n.

• Rn(x) is referred to as the remainder.

• If c = 0 these are also called Maclaurin series.

• If the process is (can be) continued infinitely it is called a Taylor series.

1. Calculate the Taylor series for sinx with c = 0.

2. Calculate the Taylor series for cosx with c = 0.

3. Calculate the Taylor series for ex with c = 0.

4. Calculate the Taylor series for lnx with c = 1. Why is c = 1 instead of c = 0?

5. Calculate the following Taylor series.

(a) x1/2 with c = 1.

(b) x1/3 with c = 1.

(c) (1 + x)−1 with c = 0.

(d) xr with c = 1.

6. Complete the WebAssign assignment.

3.4 Generating Power Series Efficiently

3.4.1 Motivation

The following problem illustrates why directly using the Taylor method to generate power series is not always themost efficient means to do so. Generate the Taylor series (infinite) for f(x) = sin(x2) about a = 0. List at least thefirst two, non-zero terms. Use of a computer is recommended for the steps. You may not let the computer generatethe whole series, however.

3.4.2 Techniques

Memorize these four series for future use.

sinx =x

1!− x

3

3!+x5

5!− x

7

7!+x9

9!− x

11

11!+ . . .

cosx = 1− x2

2!+x4

4!− x

6

6!+x8

8!− x

10

10!+ . . .

ex = 1 +x

1!+x2

2!+x3

3!+x4

4!+x5

5!+ . . .

lnx =(x− 1)

1− (x− 1)

2

2+

(x− 1)33

− (x− 1)4

4+ . . .

3.4. GENERATING POWER SERIES EFFICIENTLY 33

Algebra

1. Evaluate sin(x2) using the power series provided above.

2. Compare this to your previous result.

3. Generate the power series for sin(x3)/x using a similar method.

4. Generate the power series for (1 + x)ex.

5. Generate the power series for ex

1+x .

Derivatives

1. What is the derivative of sinx?

2. Copy the Taylor series for sinx.

3. Take the derivative of this Taylor series.

4. Identify this Taylor series in the provided list.

5. Does the result make sense?

Integrals

1. What is the anti-derivative of ex?

2. Integrate the power series for ex.

3. Use e0 = 1 to calculate +C in the integral.

4. Integrate sin(x2).

5. Write the Taylor series for (1 + x)−1 about c = 0.

6. Write the Taylor series for (1 + x2)−1 about c = 0.

7. Integrate this Taylor series.

8. Write the Taylor series for arctanx about c = 0.

Complex Series

Calculate the following using their Taylor series, then compare.

1. eix

2. cosx

3. i sinx


3.5 Taylor Polynomial Error

When a Taylor polynomial is used instead of a Taylor series, the value produced is not exact. Consider the followingusing T7(x), a 7th degree Taylor polynomial for sinx with c = 0.

f(x) = sinx

T7(x) = x− x3

3! +x5

5! − x7

7!

1. What is T7(1)?

2. How close is this value to sin 1? Note your calculator does not know, so don’t ask it.

3. But wait, your calculator claims to know. What is it actually giving you?

-Π -3 Π

4-

Π

2-

Π

4

Π

4

Π

2

3 Π

4Π

-2

-1

1

2

f(x) = sinx black

T3(x) = x− x3

3! red

T5(x) = x− x3

3! +x5

5! blue

T7(x) = x− x3

3! +x5

5! − x7

7! green

Use the examples above to answer the following.

1. Which Taylor polynomial looks the least like sinx?

2. Which Taylor polynomial looks the most like sinx?

3. Which Taylor approximation would be better than all three of these?

4. Where is the red Taylor polynomial (T3) closest to sinx?

5. Where is the blue Taylor polynomial (T5) closest to sinx?

6. All of these Taylor polynomials were about c = 0.

7. If we wanted to estimate sin(π + 1) what might we do?

3.5.1 Error Estimation

Remember

f(x) =

n∑j=0

(x− c)jj!

f (j)(c)︸︷︷︸Pn(x)

+ (−1)n∫ xc

(t− x)nn!

f (n+1)(t)dt︸︷︷︸Rn(x)

f(x) ≈n∑j=0

(x− c)jj!

f (j)(c)︸︷︷︸Pn(x)

.

3.5. TAYLOR POLYNOMIAL ERROR 35

For a sufficiently smooth function f(x)

f(x) =

n∑j=0

(x− c)jj!

f (j)(c)︸︷︷︸Taylor Polynomial

+f (n+1)(z)

(n+ 1)!(x− c)n+1︸︷︷︸

Remainder Term

for some z between c and x.T4(x) = 1+x+

x2

2! +x3

3! +x4

4! is the 4th degree Taylor polynomial for ex with c = 0. The procedure below estimates

the error for T4(2).According to the remainder theorem for some z between c = 0 and x = 2,

f (n+1)(z)

(n+ 1)!(x− c)n+1 = Rn(x).

f (5)(z)

5!(2− 0)5 = , because n = 4 and x = 2

ez

5!25 ≤ e

2

5!25, because ex is increasing

= 1.97041

Thus the error in our estimate of e2 is less than or equal to 1.97041.

3.5.2 Error control

If error is calculated after the fact, it may be too large to be useful. The following example determines what degreeTaylor polynomial is required to approximate e2 with an error less than 0.1.

Again according to the remainder theorem for some z between c = 0 and x = 2,

f (n+1)(z)

(n+ 1)!(x− c)n+1 < 0.1.

f (n+1)(z)

(n+ 1)!(2− 0)n+1 < 0.1, because x = 2.

ez

(n+ 1)!2n+1 < 0.1,

e2

(n+ 1)!2n+1 < 0.1, because ex is increasing.

• Since we don’t know an exact value for e we will replace it with a nearby integer, 3.

• Note e2(n+1)!2n+1 < 32

(n+1)!2n+1, because e < 3.

• Also if 32(n+1)!2n+1 < 0.1, then e2

(n+1)!2n+1 < 0.1 since it is even smaller.

• Thus we need to find a value for n that satisfies 32(n+1)!2n+1 < 0.1.

• Use a computational device to plug in numbers n until the result is less than 0.1.

Chapter 4

Analytic Geometry

4.1 Parametric Form

4.1.1 Motivation

This example will demonstrate limitations of the expressing curves as functions with scalar inputs and outputs (e.g.,y = f(x).). Consider the flight in the provided animation.

1. What is the shape?

2. What is the perimeter of the shape?

3. How much distance did the plane cover?

4. How far did it get?

5. What is an equation for this shape?

6. Is this a function (remember formal definition)?

7. From the equation can you tell how many times the plane flew around the circle?

8. From the equation can you tell what direction the plane flew?

9. Do we need a better model?

4.1.2 New Notation

Parametric form expresses curves as functions of one variable with vector outputs (e.g., P (t) = (x, y)). Use thefollowing example to practice graphing in parametric form.

x = cos t.y = sin t.t ∈ [0, 2π].

Plot the curve by first completing the table below, then plotting the points ((x, y) coordinates).

t 0 π4π2

3π4 π

5π4

3π2

7π4 2π

xy

37

38 CHAPTER 4. ANALYTIC GEOMETRY

4.1.3 Converting

While not frequently useful, it is possible to convert from parametric form to standard function form in some cases.This is used below to demonstrate the greater flexibility of parametric form. Consider x = t2 + 5, y = t4 + 10t2 + 25,t ∈ [−10, 10]. Suppose we wanted to go back to standard functional notation.

Note

y = t4 + 10t2 + 25

= (t2 + 5)(t2 + 5)

= x2. For,

t ∈ [0, 10],x ∈ [5, 105].

Consider x = cos θ, y = sec θ, θ ∈ (−π/2, π/2). Suppose we wanted to go back to standard functional notation.Note

y = sec θ

=1

cos θ

=1

x. For

θ ∈ (−π/2, 0],x ∈ (0, 1].

4.1.4 Flexibility

Ax = 2t+ 3,y = 10t+ 18,t ∈ (−∞,∞)

Bx = 2t,y = 4t2 + 4,t ∈ (−∞,∞)

Cx = cos(2t),y = cos2(t),t ∈ (−∞,∞)

Dx = 3t2,y = 15t2 + 3,t ∈ (−∞,∞)

Ex = t+ 1,y = t2 + 2t+ 5,t ∈ (−∞,∞)

Fx = 5− t,y = 28− 5t,t ∈ [0,∞)

Gx = sin t,y = sin2 t+ 4,t ∈ (−∞,∞)

1. Graph each function.

2. Determine the direction for each function.

3. Determine the number of times the curve is traversed.

4. Identify the shape.

5. Convert from parametric form to standard function notation.

4.2 Parametric Derivatives

4.2.1 Piecewise Derivatives

For curves expressed as scalar functions the first derivative indicates where the curve is increasing (going up) anddecreasing (going down). Because the x and y portions are separate in parametric the (two) derivatives providedifferent information. Use the following example to discover what information they provide.

x = cos(2t).y = sin(3t).t ∈ [−π, π].

4.2. PARAMETRIC DERIVATIVES 39

1. Graph this curve. Use technology.

2. Calculate dxdt anddydt .

3. Calculate where dxdt = 0.

4. Label these points on the graph.

5. Repeat these two steps with dydt .

6. What do these derivatives indicate about the curve?

4.2.2 Slope in Parametric

Derivation

1. Remember the following mantras about slope

• Slope is rise over run.• Rise is change in height (y).• Run is change in distance (x).• Thus slope is change in y over change in x.

2. What calculus concept expresses “change in”?

3. Using this, how can we calculate “change in x” for x = cos(2t)?

4. Using this, how can we calculate “change in y” for y = sin(3t)?

5. If slope is change in y over change in x how do we use the previous two parts to calculate slope?

Example

x = cos(2t).y = sin(3t).t ∈ [−π, π].

What is the slope at t = π/4?

dy

dt= 3 cos(3t).

dx

dt= −2 sin(2t).

dy

dx=

dy/dt

dx/dt

=3 cos(3t)

−2 sin(2t) .

3 cos(3t)

−2 sin(2t)

∣∣∣∣t=π/4

=−1/√

2

−2

=1

2√

2.


4.2.3 Practice

x = cos(2t).y = sin(3t).t ∈ [−π, π].

1. Calculate the slope at t = −π/6, π/6, and π/3.2. Evaluate dxdt and

dydt at t = −π/2 and π/2.

3. What is the slope of the tangent at these points?

4. What is happening at these points?


4.3 Modeling

Parametric form is convenient for modeling a variety of physical motions. The following is a classic example. Firstview the illustration provided then read and answer the questions. Do not adjust the parameters yet. The model(parametric function) will be completed in class.

1. Why does the light (red dot) move?

2. What is the shape of the wheel?

3. What is the parametric form of this shape? You know this from trigonometry.

4. This parameterization expresses points by giving the x and y distance from what part of the wheel?

5. What is the y value of this part at all times (suppose this is a 27 in diameter wheel)?

6. We want to move this part of the wheel up by this value. How can we do this arithmetically?

7. Why does the hub move forward?

8. How far does the hub move forward if the light has gone from the ground all the way around back to theground?

9. How far does the hub move forward if the light moves from the ground to straight above?

10. In general how far does the hub move as the wheel turns?

11. What is the x value of this part at all times?

12. We want to move this part of the wheel right by this value. How can we do this arithmetically?

13. In this animation where did the light begin?

14. Does our representation need to be modified for this starting point?

15. How can the starting point be adjusted?

4.4 Polar Coordinates

4.4.1 Illustrated

Use the provided illustrations to answer the following.

1. Use the Cartesian illustration to move the skidsteer from (0, 0) to (4, 4).

2. Note how the skidsteer drives like it is on roads.

3. Use the Polar illustration to move the skidsteer from (0, 0) to (5, π/4).

4. Experiment to find more ways to get to the same location.

4.4. POLAR COORDINATES 41

4.4.2 Graphing

Old Problem

y = 1 + sinx. x ∈ [0, 2π].

1. Fill out the table below.

2. Graph the points (Cartesian coordinates).

3. In what direction (order) did you plot the points.

x 0 π/4 π/2 3π/4 π 5π/4 3π/2 7π/4 2πy

New Problem

r = 1 + sin θ.


2. Graph the points (polar coordinates).


θ 0 π/4 π/2 3π/4 π 5π/4 3π/2 7π/4 2πr

Another New Problem

r = sin 2θ.


2. Graph the points (polar coordinates).


θ 0 π/4 π/2 3π/4 π 5π/4 3π/2 7π/4 2πr

4.4.3 Advanced Polar Graphing

r = 1− cos θ.

1. Find the roots.

2. Where is any point of the form (r; θ) = (0, θ)?

3. Find where the function is increasing and decreasing.

4. What does it mean for r to increase?

5. What would you expect to happen between roots in terms of increasing/decreasing?

6. Graph this curve using this information.


4.5 Polar Slopes

Slope is described as rise over run. In Section 4.2 this was calculated using parametric form. In this section it willbe calculated from polar coordinates.

1. Graph the polar point (1;π/4).

2. Sketch the line segment from the origin to this point.

3. Calculate the x and y coordinates of this point.

4. Calculate the x and y coordinates of r = cos(2θ) using the same technique.

5. Calculate dxdθ anddydθ .

6. Calculate dydx .

7. Determine where the slope of the tangents are 0.

4.6 Practice

1. Convert the following points from polar to Cartesian form. (0;π/2), (1;π/2), (2; 3π/4).

2. Convert the following points from Cartesian to polar form. (1, 1), (2, 0), (3, 4).

3. Graph the following curves expressed in polar coordinates.

(a) r = cos(θ).

(b) r = cos(3θ).

(c) r = 1 + sin(θ).


4.7 Polar Area

4.7.1 Derivation

Area in Cartesian

f(x) = 1− x2, x ∈ [−1, 1].

1. Graph the function on this domain.

2. Draw rectangles to approximate the area as done in Calculus I.

3. In Cartesian which variable is the input (domain)?

4. How is this used in calculating the area of the rectangles?

5. Where does this show up in the area integral?

6. In Cartesian which variable is the output (codomain)?

7. How is this used in calculating the area of the rectangles?

8. Where does this show up in the area integral?

4.7. POLAR AREA 43

Area in Polar

r(θ) = 1 + cos θ.

1. Graph the function.

2. What are the domain values in polar?

3. Draw lines to show divisions in the domain.

4. What shape is produced by two of these divisions and the function?

5. What is the area of a whole circle of radius r?

6. What is the area of 1/2, 1/4, 1/8 of a circle of radius r?

7. What is the angle for a 1/2, 1/4, 1/8 circle?

8. Use this to write the area of a circular wedge in terms of the radius and angle.

4.7.2 Example

Find the area inside r = cos θ for θ ∈ [0, π].

∫ π0

1

2cos2 θ dθ =∫ π

0

1

2

(1

2[1 + cos(2θ)]

)dθ =∫ π

0

1

4[1 + cos(2θ)] dθ =

1

4

[θ +

1

2sin(2θ)

]∣∣∣∣π0

=1

4([π + 0]− [0 + 0])

=π

4.

4.7.3 Advanced Polar Area

Find the area inside both r = sin θ and r = cos(2θ).

1. Graph the functions.

2. Identify the region(s) enclosed between the two curves.

3. Determine at what angle these regions begin and end.

4. Setup the integrals.

5. Integrate to find the area.



4.8 Arclength

4.8.1 Derivation

1. Sketch y = cosx for x ∈ [0, π].

2. Sketch a line segment from (0, 1) to (π,−1).

3. Calculate the length of the line segment.

4. What is the line segment (length thereof) with respect to length of the curve cosx?

5. Sketch a line segment from (0, 1) to (π/2, 0) and another line segment from (π/2, 0) to (π,−1).

6. Calculate the total length of these line segments.

7. What is the total length with respect to length of the curve cosx?

8. How can we obtain a more accurate length for cosx?

9. How can we obtain the exact length?

length ≈n∑i=0

√(xi+1 − xi)2 + (yi+1 − yi)2

length = limn→∞

n∑i=0

√(xi+1 − xi)2 + (yi+1 − yi)2

= limn→∞

n∑i=0

√(x′(c))2 + (y′(d))2 by Mean Value Theorem

=

∫ ba

√(x′(t))2 + (y′(t))2 dt.

1. The formula above is written in which form?

2. Can we directly use the above to find the length of r = cos(2θ)?

3. Can we calculate x and y for r = cos(2θ)?

4. Now can we calculate the arclength?

5. How could we handle y = x2 for x ∈ [0, 4]?


4.9 Volumes of Objects of Revolution

If an object has sufficient regularity its volume and surface area can be calculated. Note how the formulas are derivedin the same fashion as was the arclength formula.

1. Pyramids

(a) Long, long ago, in a land far, far away, a ruler built a pyramid in his own honor. It consisted of 8 levels.The bottom level measured 10 × 10 cubits; the height of each level was 5/4 cubit. Each level leaves 5/9cubits of the previous level exposed (size of the ledges). What was the volume of this pyramid? Supposeeach level is solid.

4.9. VOLUMES OF OBJECTS OF REVOLUTION 45

(b) Almost as long ago, in the same land far, far away, the ruler’s son built a pyramid in his own honor. Hecould not afford to make it taller than his father’s pyramid, but he could make one with more levels. Itconsisted of 16 levels. The bottom level measured 10× 10 cubits; the height of each level was 5/8 cubit.Each level leaves 5/17 cubits of the previous level exposed (size of the ledges). What was the volume ofthis pyramid? Suppose each level is solid.

(c) How could these be used to compute the volume of a smooth sided pyramid (think Egypt)?

2. Cones

(a) Guido is making a conical cake. He begins by baking five cake rounds. Each is 2 inches tall. The bottomcake has a radius of 20/3 inches. Each cake is 4/3 inches smaller. What is the total volume of cake?

(b) Guido makes a second conical cake. To waste less cake when it is cut to form a cone, he begins by bakingten cake rounds. Each is 1 inch tall. The bottom cake has a radius of 80/11 inches. Each cake is 8/11inches smaller. What is the total volume of cake?

(c) How could these be used to compute the volume of a cone?

4.9.1 Disc Method

If the object can be produced by continuously rotating a curve (see the provided illustration), then this method ofcalculating volumes produces a simple formula.

Derivation

1. Sketch the curve y = 2x− x2.

2. Draw one of the rectangles that would be used to estimate area.

3. Imagine rotating this curve around the x-axis. (See the provided illustration.)

4. What does the 3D object look like?

5. What shape does the rectangle form?

6. What is the volume of this shape?

7. How can we use this in an integral?

Example

Find the volume of the object formed by revolving y = 2x− x2 around the x-axis.

∫ 20

π(2x− x2)2 dx =

π

∫ 20

4x2 − 4x3 + x4 dx =

π

(4

3x3 − x4 + 1

5x5)∣∣∣∣2

0

=

π

(32

3− 16 + 32

5

).


Practice

1. Calculate the volume of the solid produced by rotating the following region around the y-axis. The region isenclosed by y = x2, y = 2, x = 0.

2. Calculate the volume of the solid produced by rotating the following region around the y-axis. The region isenclosed by y = x2, y = 0, x = 3.

3. Calculate the volume of the solid produced by rotating the following region around the x-axis. The region isenclosed by y = x2, y = 2, x = 0.


4.9.2 Shell Method

Sometimes calculating volumes using discs is overly difficult. Note the problem in the following example.

Derivation

1. Attempt to find the volume of the object generated by rotating the following region about the y-axis. Theregion is enclosed between y = (x− 1)2 and y = 1.

2. What is the difficulty?

3. Sketch the region enclosed between y = (x− 1)2 and y = 1.

4. Sketch one of the rectangles that would estimate area.

5. If this rectangle is rotated about the y-axis what shape would it form? See the provided illustration.

6. If the rectangle had width 0.1 and height 2, what would its volume be?

Example

Find the volume formed by rotating the following region around the y-axis. The region is enclosed between y = (x−1)2and y = 1.

∫ 20

(2πx)[1− (x− 1)2] dx =

2π

∫ 20

2x2 − x3 dx =

2π

(2

3x3 − 1

4x4)∣∣∣∣2

0

=

2π

(16

3− 4).

Practice

1. Calculate the volume of the solid produced by rotating the following region around the x-axis. The region isenclosed by y = x2, y = 2, x = 0.

2. Calculate the volume of the solid produced by rotating the following region around the x = 3. The region isenclosed by y = x2, y = 0, x = 2.



4.9.3 Surface Area

Derivation

1. Sketch f(x) = 2 + sinx for x ∈ [0, 2π].

2. Sketch a line segment between the points (π/2, f(π/2)) and (π, f(π)).

3. Sketch the rotation of the curve and the line segment.

4. Use the provided video to identify the shape the line segment forms.

5. The surface area of a right, circular cone is πr√r2 + h2 where r is the radius of the base and h is the height

(orthogonal from point to base). Note√r2 + h2 is just the length of the hypotenuse.

6. Calculate the surface area for the frustrum (partial) cone with radii R and r in Figure 4.1.

7. Note r and R form bases of similar triangles. Thus lrr =lr+lRR . Clear the denominators and use this to simplify

the surface area formula.

8. As the side length approaches zero, what will happen to r and R?

9. Use this to derive the integral formula for surface area.

r

R

lr

lR

Figure 4.1: Frustrum of a Cone

The surface area of an object of revolution can be calculated by∫ ba

2πr(arclength) dx

Examples

Find the surface area of the object produced by revolving the region enclosed by y = 2x− x2 and the x-axis aboutthe x-axis.


First find the curve segment.

2x− x2 = 0.x(2− x) = 0.

x = 0, 2.

P (t) = (x, 2x− x2).P ′(t) = (1, 2− 2x).

arclength =√

12 + (2− 2x)2.

0.5 1.0 1.5 2.0

-1.0

-0.5

0.5

1.0

Because the curve is rotated around the x-axis, the radius is given by the function. The surface area is∫ 10

2π(2x− x2)√

1 + (2− 2x)2 dx.

Find the surface area of the object produced by revolving y = 2x− x2 about the y-axis.

-2 -1 1 2

0.2

0.4

0.6

0.8

1.0

Because the curve is rotated around the y-axis, the radius is given by the x-value. The surface area is∫ 10

2πx√

1 + (2− 2x)2 dx.

Practice

Use technology if needed.

1. Calculate the surface area of the solid produced by rotating y = x2 for x ∈ [0, 2] about the x-axis.

2. Calculate the surface area of the solid produced by rotating y = x2 for x ∈ [0, 2] about the y-axis.

3. Calculate the surface area of the solid produced by rotating the following region around the x = 3. The regionis enclosed by y = x2, y = 0, x = 2.



4.9.4 Generalizing

1. The first two illustrations in this lesson were constructed by doing what?

2. As the number of levels was increased each object became more like what?

3. How is this similar to the arclength or area (rectangles) calculations?

integration techniques - university of alaska systemafmaf/classes/math252/lesson/... · 2013. 4....

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