Math 10b Final Review Sheet – SOLUTIONS

• The Math 10b Final Exam will be Tuesday, May 7th, 1:30 pm – 4:30 p.m.

• Location: Abelson 131

• The final exam is cumulative, but the material covered since the second midterm (fromChapters 6, 7 & 8) will make up at least 60% of the exam. In particular, at least 60% ofthe exam will be on:

– Section 6.2: Volumes (Disks and Washers)

– Section 7.1: Differential Equations

– Section 7.3: Separable Equations

– Section 8.1: Sequences

– Section 8.2: Series

– Section 8.3: The Integral & Comparison Tests

– Section 8.4: Other Convergence Tests

– Section 8.5: Power Series

– Section 8.7: Taylor Series

Note: Some questions may require knowledge from earlier in the semester, from calc I,and/or from precalculus.

• You must bring your student ID to the final exam.

• No calculators will be allowed on the exam.

Our exams are closed book: no notes, books, calculators, cell phones or internet-connecteddevices will be allowed.

• On the exam, you will be asked to show all your work.

We are grading your ability to clearly communicate your understanding of how to solveeach problem much more than your ability to get the correct answer. In fact, a correctanswer with little or no work might not receive any points at all.

Showing your work clearly gives us justification for assigning partial credit.

• The solutions to this review material will be available on LATTE by Fri, May 3.

• On the exam, you will be given the following three summation formulae (so you do *not*need to memorize these):

n∑i=1

i =n(n+ 1)

2

n∑i=1

i2 =n(n+ 1)(2n+ 1)

6

n∑i=1

i3 =

[n(n+ 1)

2

]2• Note: On several problems the instructions say “set up, but do NOT simplify or

evaluate the integral that gives . . . ”. For these problems, your answer should havespecific numbers for the limits of integration and should have a function for the integrand,but you do not need to simplify the function or evaluate the integral. For example, if

your answer was

∫ 5

2

(x− 1)2dx, you would not have to multiply out the (x− 1)2 and you

would not have to actually integrate (i.e., you can leave the answer like that).

OVER for Solutions to Practice Exam 1 →

Practice Exam 1 – SOLUTIONS (Problems 1 - 14)

1. Find the area between the graphs of f(x) = x2 − 5x − 7 and g(x) = x − 12 over theinterval [−2, 5].

Solution: Find points of intersection:

x2 − 5x− 7 = x− 12

x2 − 6x+ 5 = 0

(x− 5)(x− 1) = 0

So the graphs intersect at x = 1 and x = 5. Note that f(x)− g(x) = x2 − 6x+ 5. Doingsign analysis shows that f(x)−g(x) is positive on (−∞, 1)∪(5,∞) and negative on (1, 5),so f(x) > g(x) on (−∞, 1) ∪ (5,∞) and g(x) > f(x) on (1, 5). So:

area =

∫ 1

−2f(x)− g(x) dx+

∫ 5

1

g(x)− f(x) dx

=

∫ 1

−2x2 − 6x+ 5 dx+

∫ 5

1

−x2 + 6x− 5 dx

=x3

3− 3x2 + 5x

∣∣∣∣1−2

+−x3

3+ 3x2 − 5x

∣∣∣∣51

=7

3− (−74)

3+

25

3− (−7)

3

=113

3

2. Find the equation of the tangent line to the graph of g(x) =

∫ sinx

0

√8− 4t dt at the point

x =π

2.

Solution: g′(x) =√

8− 4 sinx cosx, so the slope is m = g′(π/2) =√

8− 4(1)(0) = 0

g(π/2) =∫ sin(π/2)

0

√8− 4t dt =

∫ 1

0

√8− 4t dt = (−1

4)(2

3)(8 − 4t)3/2

∣∣10

= −16(43/2 − 83/2) =

−16(8− (2

√2)3)

so the tangent line is y = −16(8− (2

√2)3)

3. Solve the following integrals:

(a)

∫ex

e2x + 2ex − 3dx

Solution: Let u = ex so du = exdx. Then

∫ex

e2x + 2ex − 3dx =

∫1

u2 + 2u− 3du.

Use Partial Fractions.1

u2 + 2u− 3=

A

u+ 3+

B

u− 1, so we get

1 = A(u− 1) +B(u+ 3).

Letting u = 1, we get B = 14

and letting u = −3, we get A = −14. So:∫

1

u2 + 2u− 3du = −1

4

∫1

u+ 3du+

1

4

∫1

u− 1du

= −1

4ln |u+ 3|+ 1

4ln |u− 1|+ C

= −1

4ln |ex + 3|+ 1

4ln |ex − 1|+ C

(b)

∫(lnx)2dx

Solution: Let w = lnx, so x = ew. Then dw = 1xdx, so dx = ewdw. Then∫

(lnx)2dx =

∫w2ewdw.

Now do Integration by Parts, with u = w2 and dv = ewdw, so du = 2wdw andv = ew. Then:∫w2ewdw = w2ew −

∫2wewdw

Do Integration by Parts again, with u = w and dv = ewdw, so du = dw and v = ew.Then:

w2ew −∫

2wewdw = w2ew − 2(wew −∫ewdw) = w2ew − 2wew + 2ew + C

Finally, we substitute back in to get x(lnx)2 − 2x lnx+ 2x+ C.

4. The base of a solid S is the region in the x-y plane bounded by graphs of x = y2 andx + y = 2. Each cross section taken perpendicular to the y-axis is a square. Set up, butdo NOT simplify or evaluate, the integral(s) that give the volume of S.

Solution: Find the points of intersection by setting y2 = 2 − y ⇒ y2 + y − 2 = 0 ⇒y = −2 or y = 1.

The base of the solid is:

The volume is then

∫ 1

−2(2− y − y2)2dy.

5. Determine whether the following sequences converge or diverge. If a sequence converges,find its limit.

(a)

{n2 − 5n+ 6

3 + 4n− 5n2

}∞n=5

Solution:

limn→∞

n2 − 5n+ 6

3 + 4n− 5n2= lim

n→∞

n2

−5n2= −1

5

(b)

{6− n

2 + 3n2

}∞n=0

Solution:

limn→∞

6− n2 + 3n2

= limn→∞

−n3n2

= 0

(c)

{ln(n+ 3)

ln(2 + 5n)

}∞n=1

Solution: Using l’Hopital’s rule:

limn→∞

ln(n+ 3)

ln(2 + 5n)l′H= lim

n→∞

1n+35

2+5n

= limn→∞

2 + 5n

5(n+ 3)= 1

6. Find the general solution to the differential equationdy

dx= y2 + 1.

Solution:dy

dx= y2 + 1

1

1 + y2dy = dx∫

1

1 + y2dy =

∫dx

tan−1 y = x+ C

y = tan(x+ C)

7. Find the following integrals. If an integral is improper, determine whether or not itconverges and find its value if it does.

(a)

∫ ∞e

dx

x lnx 3√

lnxLet u = lnx so du = 1

xdx and x = e→ u = 1 and as x→∞ we have u→∞, so∫ ∞

e

dx

x lnx 3√

lnx=

∫ ∞1

1

u 3√udu

= limt→∞

∫ t

1

u−4/3du

= limt→∞−3u−1/3

∣∣t1

= limt→∞−3(

13√t− 1)

= 3

(b)

∫ 3

0

dx√9− x2 ∫ 3

0

dx√9− x2

=

∫ 3

0

dx√9− x2

=1

3

∫ 3

0

dx√1− (x

3)2

=

∫ 1

0

du√1− u2

(let u = x/3, so 3du = dx)

= limt→1−

∫ t

0

du√1− u2

= limt→1−

sin−1 u∣∣t0

= limt→1−

sin−1 t− sin−1 0

=π

2

(c)

∫ 1

0

x

x2 − 2dx

Let u = x2 − 2 so du = 2xdx, and x = 0→ u = −2 and x = 1→ u = −1, so∫ 1

0

x

x2 − 2dx =

1

2

∫ −1−2

1

udu

=1

2ln |u|

∣∣−1−2

=1

2(ln | − 1| − ln | − 2|)

= −1

2ln 2

8. Suppose you are trying to install some new software on your computer. The installationprogram displays a progress bar, showing how much of the installation has been completedat any given time – something like this:

After one second, 13

of the bar is colored in. After another second, an additional 19

of thebar (that’s 1

9of the whole bar) gets colored in. After the third second, an additional 1

27

of the bar gets colored in. The bar continues to fill in according to this pattern (i.e., atthe nth second, an additional 1

3nportion of the bar gets colored in.

(a) Write down a sum (in sigma notation) that represents the amount of the bar that iscolored in after n seconds.n∑i=1

1

3i

(b) Assuming the bar continues to fill in according to this pattern, how much of the barwould be filled in if you waited forever for the program to install?∞∑i=1

1

3i=

13

1− 13

=1

3(3

2) =

1

2of the whole bar.

9. Determine whether the following series converge or diverge.

(a)∞∑n=0

n!

5n

L = limn→∞

∣∣∣∣(n+ 1)!

5n+1

5n

n!

∣∣∣∣ = limn→∞

∣∣∣∣(n+ 1)

5

∣∣∣∣→∞So the series

∞∑n=0

n!

5ndiverges by the Ratio Test.

(b)∞∑n=1

n3

2n3 + 5

limn→∞

n3

2n3 + 5=

1

26= 0

So the series∞∑n=1

n3

2n3 + 5diverges by the Divergence Test.

(c)∞∑n=1

2√n 5n

Note that2√n 5n

≤ 2

5nfor all n ≥ 1 and that all terms of both sequences are positive.

We know that∞∑n=1

2

5n= 2

∞∑n=1

1

5nand this series is geometric with |r| = 1

5< 1, so it

converges. Therefore, by the comparison test,∞∑n=1

2√n 5n

also converges.

(d)∞∑n=1

(−3)n

2n+1

∞∑n=1

(−3)n

2n+1=

1

2

∞∑n=1

(−3

2)n

This is geometric with |r| = 32> 1, so the series

∞∑n=1

(−3)n

2n+1diverges.

(e)∞∑n=1

(−1)n+1 n2

n!

L = limn→∞

∣∣∣∣(−1)n+2 (n+ 1)2

(n+ 1)!

n!

(−1)n+1 n2

∣∣∣∣ = limn→∞

∣∣∣∣ (n+ 1)2

(n+ 1)n2

∣∣∣∣ = 0

So the series∞∑n=1

(−1)n+1 n2

n!converges by the Ratio Test.

(f)∞∑n=1

(1

(n+ 5)3+( 2

3

)n−1)

∞∑n=1

(1

(n+ 5)3+( 2

3

)n−1)=

∞∑n=1

1

(n+ 5)3+∞∑n=1

( 2

3

)n−1. For the first series, we

can use the integral test since the function f(x) =1

(x+ 5)3is positive, continuous

and decreasing on [1,+∞). You can show the integral

∫ ∞1

1

(x+ 5)3dx converges,

so∞∑n=1

1

(n+ 5)3also converges. The series

∞∑n=1

( 2

3

)n−1is geometric, with r =

2

3.

Since |r| < 1, this series also converges.

Since both series converge, the series∞∑n=1

(1

(n+ 5)3+( 2

3

)n−1)also converges.

10. Find the Taylor polynomial of degree 4 that approximates the solution to the differentialequation y′′ + 2y′ = 0 subject to the following initial conditions:

y(0) = 1 y′(0) = −1

Solution:y′′ = −2y′ → y′′(0) = 2

y′′′ = −2y′′ → y′′′(0) = −4

y(4) = −2y′′′ → y(4)(0) = 8

So

P4(x) = 1− x+2

2!x2 − 4

3!x3 +

8

4!x4 = 1− x+ x2 − 2

3x3 +

1

3x4

11. Find the radius and interval of convergence of the following power series:∞∑n=1

(−1)nxn

32nn

L = limn→∞

∣∣∣∣(−1)(n+1)x(n+1)

32(n+1)(n+ 1)

32nn

(−1)nxn

∣∣∣∣ = limn→∞

∣∣∣∣ x32

n

(n+ 1)

∣∣∣∣ =|x|9

limn→∞

n

(n+ 1)=|x|9

By the Ratio Test, the series converges when L < 1, i.e., when |x|9< 1, or |x| < 9, so the

radius of convergence is 9.

For the interval of convergence, we know that it converges on (−9, 9), but we need to testthe endpoints.

x = −9: We plug x = −9 into the original series to get:

∞∑n=1

(−1)n(−9)n

32nn=∞∑n=1

9n

9nn=∞∑n=1

1

n

This is the Harmonic Series, which diverges, so our series diverges at x = −9.

x = 9: We plug x = 9 into the original series to get:∞∑n=1

(−1)n9n

32nn=∞∑n=1

(−1)n9n

9nn=∞∑n=1

(−1)n

n

This is the Alternating Harmonic Series, which converges, so our series diverges at x = 9.

Thus the interval of convergence for∞∑n=1

(−1)nxn

32nnis (−9, 9].

12. Consider the function f(x) =1

x− 2.

(a) Find the Taylor series for f(x) about x = 1.

(Write out at least the first four terms of the series.)

Solution: We have

f ′(x) =−1

(x− 2)2, f ′′(x) =

2

(x− 2)3, f ′′′(x) =

−6

(x− 2)4, f (4)(x) =

24

(x− 2)5

and so

f(1) = −1, f ′(1) = −1, f ′′(1) = −2, f ′′′(1) = −6, f (4)(1) = −24

and therefore the Taylor series for f about x = 1 is

−1− (x− 1)− 2

2!(x− 1)2 − 6

3!(x− 1)3 − 24

4!(x− 1)4 + · · ·

= −1− (x− 1)− (x− 1)2 − (x− 1)3 − (x− 1)4 + · · ·

(b) Write the series in sigma notation.

∞∑n=0

−(x− 1)n

(c) What is the interval of convergence of the series you found in part 12a?

This is geometric with r = x− 1, so this converges for |x− 1| < 1, so the radius ofconvergence is 1.

For the interval of convergence, we know the center is 1 and the radius of convergenceis 1. Since it is geometric, we also know it diverges on the endpoints, so the intervalof convergence is (0, 2).

(d) Can you use the series you found in part 12a to approximate the number −25

? Whyor why not?

No, we can’t. While the radius of convergence is 1, the center is also 1, so the intervalon which we know it converges is (0, 2). To approximate the number −2

5, we would

need to take x = −1/2, which is outside this interval (the series will diverge at thispoint).

13. We showed in class that the Taylor series1for ex about x = 0 is∞∑n=0

xn

n!with radius of

convergence R =∞.

(a) Show that

1√e

=∞∑n=0

(−1)n

n!2n.

Solution: Since 1√e

= e−1/2, we can just plug x = −12

into the series for ex to get:

1√e

=∞∑n=0

(−12)n

n!=∞∑n=0

(−1)n

n!2n

1It is also true that the Taylor series for ex is equal to the function ex for all x.

(b) Use the first four terms of the series to find an approximation for1√e

.

1√e≈ 1− 1

2+

1

8− 1

48=

48− 24 + 6− 1

48=

29

48

14. Determine whether the following statement is true or false.2

For a series∞∑n=1

an, if the sequence of partial sums {sn}∞n=1 is convergent, then

the sequence {an}∞n=1 must also be convergent.

If the statement is true, briefly explain why. If the statement is false, give an example(a function or a formula or a graph) that shows that it’s false.

TRUE: If the sequence of partial sums {sn}∞n=1 is convergent, that means the series∞∑n=1

an is convergent (by definition) and the Divergence Theorem tells us that if this series

converges then we must have limn→∞

an = 0, so the sequence {an}∞n=1 does, in fact, converge.

OVER for Solutions to Practice Exam 2 →2When we say a statement is “true” we mean it must always be true. When we say a statement is “false”

we mean it could possibly be false.

Practice Exam2 – SOLUTIONS (Problems 15 - 28)

15. We must find

∫ 6

0

[10 + 10 cos

(π12t)]

dt, which equals

∫ 6

0

10 dt+

∫ 6

0

10 cos(π12t)dt. Note

that

∫ 6

0

10 dt = 10(6−0) = 60. For

∫ 6

0

10 cos(π12t)dt, use substitution with u = π

12t and

du = π12dt. Changing limits of integration: t = 0 ⇒ u = 0 and t = 6 ⇒ u = π

2. So we

get12

π

∫ π2

0

10 cosu du =120

πsinu

]π2

0

=120

π

(1− 0

)=

120

π.

So 60 +120

πgallons of water flow into the reservoir during the first six hours after the

pipe was opened.

16. (a)

∫ ∞−∞

ex

1 + e2xdx =

∫ 0

−∞

ex

1 + e2xdx+

∫ ∞0

ex

1 + e2xdx = lim

t→−∞

∫ 0

t

ex

1 + e2xdx+ lim

t→∞

∫ t

0

ex

1 + e2xdx.

Forex

1 + e2xdx use substitution, with u = ex and du = ex dx, getting

∫du

1 + u2,

which equals tan−1(u). So

∫ex

1 + e2xdx = tan−1(ex). Therefore

limt→−∞

∫ 0

t

ex

1 + e2xdx = lim

t→−∞

(tan−1(ex)

]0t

)= lim

t→−∞

(tan−1(1)− tan−1(et)

)= π

4− tan−1(0) = π

4,

and

limt→∞

∫ t

0

ex

1 + e2xdx = lim

t→∞

(tan−1(ex)

]t0

)= lim

t→∞

(tan−1(et)− tan−1(1)

)= π

2− π

4= π

4.

So

∫ ∞−∞

ex

1 + e2xdx converges to π

4+ π

4= π

2.

(b)

∫x

1 + 5x4dx. First rewrite the integral as∫

x

1 + 5x4dx =

∫x

1 + (√

5x2)2dx

Use substitution with u =√

5x2 and du = 2√

5x dx. Then∫x

1 + 5x4dx =

1

2√

5

∫1

1 + u2du =

1

2√

5tan−1 u+ C =

1

2√

5tan−1(

√5x2) + C.

(c)

∫ex

e2x + 5ex + 4dx. Use substitution, with u = ex and du = exdx, getting

∫du

u2 + 5u+ 4.

Then use partial fractions:1

u2 + 5u+ 4=

1

(u+ 4)(u+ 1)=

A

u+ 4+

B

u+ 1, so

1 = A(u + 1) + B(u + 4). Letting u = −4 gives A = −13

and letting u = −1 givesB = 1

3. So ∫

du

u2 + 5u+ 4= −1

3

∫du

u+ 4+

1

3

∫du

u+ 1.

= − 1

3ln |u+ 4|+ 1

3ln |u+ 1|+ C or

1

3ln

∣∣∣∣∣ u+ 1

u+ 4

∣∣∣∣∣+ C.

Therefore

∫exdx

e2x+ 5ex + 4= −

1

3ln |ex+ 4|+

1

3ln |ez + 1|+C =

1

3ln

∣∣∣∣∣ ex + 1

ex + 4

∣∣∣∣∣+C

(d)

∫ π/2

0

cosx

1− sinxdx. The integrand has a vertical asymptote at x = π

2, so

∫ π/2

0

cosx

1− sinxdx = lim

t→π/2−

∫ t

0

cosx

1− sinxdx.

For

∫cosx

1− sinxdx, use u-substitution with u = 1 − sinx and du = − cosx dx,

getting −∫

1

udu, which equals − ln |u|. Resubstituting gives − ln |1− sinx|. Then

limt→π/2−

∫ t

0

cosx

1− sinxdx = lim

t→π/2−

(− ln |1− sinx|

]t0

)= lim

t→π/2−

(− ln 1+ln |1− sin t|

)= lim

t→π/2−ln |1− sin t|.

Note that as t → π/2−, 1 − sin t → 0+, so ln |1 − sin t| → −∞. Therefore

limt→π/2−

(ln 1− ln |1− sin t|

)= −∞. So

∫ π/2

0

cosx

1− sinxdx diverges.

17. Find the area of the region bounded by the graphs of y = 8x2

, y = 8x, and y = x.

Solution: Find points of intersection:

8

x2= 8x

1 = x3

1 = x

8

x2= x

8 = x3

2 = x

0 1 2 3

2

4

6

8

10

So the graphs intersect at x = 1 and x = 2. We can see from the sketch that:

area =

∫ 1

0

8x− x dx+

∫ 2

1

8

x2− x dx

=

∫ 1

0

7x dx+

∫ 2

1

8

x2− x dx

=7x2

2

∣∣∣∣10

+−8

x− x2

2

∣∣∣∣21

=7

2+

5

2= 6

18. (a)

∫ 3

−3f(x) dx. The subintervals are [−3,−1], [−1, 1] and [1, 3]. Note that ∆x = 2.

(i) R3 = f(−1)∆x+ f(1)∆x+ f(3)∆x = [f(−1) + f(1) + f(3)] · 2= (−1 + 1 + 2) · 2 = 4

(ii) L3 = f(−3)∆x+ f(−1)∆x+ f(1)∆x = [f(−3) + f(−1) + f(1)] · 2= (1 + (−1) + 1) · 2 = 2

(iii) M3 = f(−2)∆x+ f(0)∆x+ f(2)∆x = [f(−2) + f(0) + f(2)] · 2= (0 + 0 + 2) · 2 = 4

(iv) T3 = 12

[f(−3) + f(−1)

]∆x+ 1

2

[f(−1) + f(1)

]∆x+ 1

2

[f(1) + f(3)

]∆x

= 12·∆x

[f(−3) + 2f(−1) + 2f(1) + f(3)

]= 1

2· 2[1 + 2(−1) + 2(1) + 2

]= 3

(b) F (x) =

∫ x

−3f(t) dt.

(i) F ′(−2) = f(−2) = 0

(ii) F (−1) = 12− 1

2= 0 and F (0) = 1

2− 1 = −1

2, so F (−1) > F (0).

(iii) F (x) is concave up on (−1, 2), since F ′(x) = f(x) and f is increasing on (−1, 2).

19. Note that limx→0

∫ x

0

ln(1 + t) dt = 0. So limx→0

∫ x

0

ln(1 + t) dt

x2is an indeterminate form of

type0

0. So apply l’Hopital’s rule, using the Fundamental Theorem of Calculus for the

numerator, getting

limx→0

∫ x

0

ln(1 + t) dt

x2= lim

x→0

ln(1 + x)

2x= lim

x→0

1

1 + x

2=

1

2.

20. (a) Note that the domain of y =√

16− x2 is [−4, 4] and the graph intersects the x-axisat x = −4 and x = 4. The cross-sectional area is A(x) = 1

2

√16− x2 · 3. So the

volume of S is

V =

∫ 4

−4A(x) dx =

∫ 4

−4

(1

2

√16− x2 · 3

)dx =

3

2

∫ 4

−4

√16− x2 dx.

Note that

∫ 4

−4

√16− x2 dx is the area of the upper semi-circle of radius 4, which

12π(42). So the volume of S is

3

2· 1

2π(42) = 12π.

(b) The base of the solid S is shown below:

y x3=

y 2 x!=

y 2=

2 x! y 2 0 x 1.01< <if( )< <

x3 y 2 0 x 8< <if( )< <

Since we are integrating with respect to y, write y = 3√x as a function of y, getting

x = y3. Then, since the cross-sections are squares, A(y) = (y3)(y3) = y6. So thevolume of S is ∫ 2

0

y6 dy =y7

7

]20

=128

7.

21. (a)∞∑n=3

4n

(2n)!. Use the ratio test.

limn→∞

∣∣∣∣∣ an+1

an

∣∣∣∣∣ = limn→∞

∣∣∣∣∣∣∣∣∣4n+1

(2(n+ 1))!

4n

(2n)!

∣∣∣∣∣∣∣∣∣ = limn→∞

4n+1

(2(n+ 1))!·

(2n)!

4n= lim

n→∞4·

(2n)!

(2n+ 2)(2n+ 1)(2n)!

= 4 limn→∞

1

(2n+ 2)(2n+ 1)= 0.

Since limn→∞

∣∣∣∣ an+1

an

∣∣∣∣ < 1, the series converges.

(b)∞∑n=1

1

n√n

We can rewrite this integral as∞∑n=1

1

n32

. This is a p-series with p = 32. Since p > 1

the series converges.

(c)∞∑n=1

sin2 n

n2

Note thatsin2 n

n2≤ 1

n2for all n ≥ 1 and that all the terms of both sequences are

positive. Therefore we can use the comparison test with∞∑n=1

1

n2, which converges

because it is a p-series with p = 2 > 1. Therefore∞∑n=1

sin2 n

n2converges.

(d)∞∑n=0

(−3

2

)n+1. The series is geometric with r = −3

2. Since r ≤ −1, the series diverges.

(e)∞∑n=1

e−n. The series is geometric with a =1

eand r =

1

e. Since |r| < 1, the series

converges to

1

e

1−1

e

=1

e− 1.

(f)∞∑n=5

3√n

n− 4

Note that3√n

n− 4≥

3√n

nsince dividing by a larger denominator (n > n− 4) yields a

smaller number. Since all terms of both sequences are positive for n ≥ 5, we can use

the comparison test. We can see that∞∑n=5

3√n

n=∞∑n=5

1

n2/3, which diverges because it

is a p-series with p = 23< 1. Therefore

∞∑n=5

3√n

n− 4diverges.

22. Both differential equations are separable:

(a)dy

dx= x⇒ dy = xdx⇒

∫dy =

∫xdx⇒ y =

x2

2+ C

Plugging in the initial condition (y(1) = 2) yields:

2 =1

2+ C ⇒ C =

3

2

so the solution is y = x2

2+ 3

2.

(b)dy

dx= y ⇒ 1

ydy = dx⇒

∫1

ydy =

∫dx⇒ ln |y| = x+ C

Plugging in the initial condition (y(1) = 2) yields:

ln 2 = 1 + C ⇒ C = ln(2)− 1

so the solution is y = ex+ln(2)−1 or y = 2ex−1.

23. We are looking for a 4th degree Taylor polynomial that approximates the solution to thedifferential equation y′′ + y′ = x, subject to the following initial conditions: y(0) = 1 andy′(0) = 0.

Given these initial conditions, we’ll center the Taylor polynomial at x = 0. We can rewritethe equation as y′′ = x− y′. Then

• y′′′ = 1− y′′

• y(4) = −y′′′

Now use the initial conditions to evaluate the derivatives at x = 0:

• y(0) = 1

• y′(0) = 0

• y′′(0) = 0− y′(0) = 0− 0 = 0

• y′′′(0) = 1− y′′(0) = 1− 0 = 1

• y(4)(0) = −y′′′(0) = −1

So the Taylor polynomial is

P4(x) = f(0) + f ′(0)x+f ′′(0)

2!x2 +

f ′′′(0)

3!x3 +

f (4)(0)

4!x4

= 1 + (0)x+0

2!x2 +

1

3!x3 −

1

4!x4 = 1 +

x3

3!−x4

4!.

24. First rewrite .99 as an infinite sum:9

10+

9

100+

9

1000+

9

10000+ · · · . This is a

geometric series with a =9

10and r =

1

10. Since |r| < 1, the series converges to

a

1− r =

9

10

1−1

10

=

9

10

9

10

= 1.

25. (a)∞∑n=0

n 6n

(n+ 1)!. Use the ratio test, getting

limn→∞

∣∣∣∣∣ an+1

an

∣∣∣∣∣ = limn→∞

∣∣∣∣∣∣∣∣∣(n+ 1) 6n+1

(n+ 2)!

n 6n

(n+ 1)!

∣∣∣∣∣∣∣∣∣ = limn→∞

((n+ 1) 6n+1

(n+ 2)!·

(n+ 1)!

n 6n

)

= limn→∞

(6 ·

n+ 1

n·

(n+ 1)!

(n+ 2)!

)= lim

n→∞

(6 ·

n+ 1

n·

1

n+ 2

)= 6 · 1 · 0 = 0.

Since limn→∞

∣∣∣∣ an+1

an

∣∣∣∣ < 1, the series converges.

(b)∞∑n=0

(−1)n3n+1

n5 . Use the ratio test:

limn→∞

∣∣∣∣∣ an+1

an

∣∣∣∣∣ = limn→∞

∣∣∣∣∣∣∣∣∣(−1)n+13n+2

(n+ 1)5

(−1)n3n+1

n5

∣∣∣∣∣∣∣∣∣ = limn→∞

3n+2

(n+ 1)5·n5

3n+1 = limn→∞

3·n5

(n+ 1)5= 3·1 = 3.

Since limn→∞

∣∣∣∣ an+1

an

∣∣∣∣ > 1, the series diverges.

26. To find the radius of convergence of the Bessel function J0(x) =∞∑n=0

(−1)nx2n

22n(n!)2, we use

the ratio test:

limn→∞

∣∣∣∣∣ an+1

an

∣∣∣∣∣ = limn→∞

∣∣∣∣∣∣∣∣∣∣(−1)n+1x2(n+1)

22(n+1)((n+ 1)!)2

(−1)nx2n

22n(n!)2

∣∣∣∣∣∣∣∣∣∣= lim

n→∞

∣∣∣∣∣ x2(n+1)

22(n+1)((n+ 1)!)2·

22n(n!)2

x2n

∣∣∣∣∣

limn→∞

∣∣∣∣∣ x2n+2

22n+2(n+ 1)!(n+ 1)!·

22nn!n!

x2n

∣∣∣∣∣ = limn→∞

∣∣∣∣∣ 1

4·

1

(n+ 1)2· x2∣∣∣∣∣ = lim

n→∞

1

4·

1

(n+ 1)2·x2 = 0.

Since this limit is < 1 for all x, the radius of convergence is R =∞. Therefore the intervalof convergence is (−∞,∞).

27. f(x) =

∫ x

1

ln√t dt.

(a) Note that f(1) = 0. The Fundamental Theorem of Calculus tells us that f ′(x) =

ln√x, so f ′(1) = 0. Then f ′′(x) =

1√x· 1

2√x

=1

2x, so f ′′(1) =

1

2. Finally

f ′′′(x) = − 1

2x2, so f ′′′(1) = − 1

2. So the third Taylor polynomial for f(x) about

x = 1 is

P3(x) = 0 + 0(x− 1) +12

2!(x− 1)2 −

12

3!(x− 1)3 =

(x− 1)2

4− (x− 1)3

12.

(b) To find an estimate for

∫ 2

1

ln√t dt, first note that

∫ 2

1

ln√t dt = f(2). So we use

the Taylor polynomial we found in part (a) to approximate f(2). We get

P3(2) =(2− 1)2

4− (2− 1)3

12=

1

4− 1

12=

1

6. So

∫ 2

1

ln√t dt ≈

1

6.

28. (a) False. Consider the sequence { 12, 14, 12, 14, . . . }. Then 0 < an < 1 for every n. But the

sequence { an } does not converge.

(b) True. Suppose that the power series has radius of convergence R. Then the seriesconverges for |x| < R, since the series is centered at x = 0. We can write thisinequality as −R < x < R. If x = 2 satisfies this inequality, x = 1 must satisfy it aswell.

OVER for Solutions to Practice Problems →

Practice Problems – SOLUTIONS (Problems 29 - 53)

29.

∫ b

a

f(x) dx = limn→∞

n∑i=1

f(xi)∆x, where ∆x = b−an

and xi is the right endpoint of the ith

subinterval. So xi = a + i · ∆x. In this integral, a = 1, b = 3 and f(x) = 2x2 + 1. So∆x = 2

nand xi = 1 + i ·∆x = 1 + 2i

n. So

n∑i=1

f(xi)∆x =n∑i=1

(2(1 + 2i

n

)2+ 1)

2n

= 2n

n∑i=1

(2(1 + 2i

n

)2+ 1)

= 4n

n∑i=1

(1 + 2i

n

)2+ 2

n

n∑i=1

1

= 4n

n∑i=1

(1 + 4i

n+ 4i2

n2

)+ 2

n

n∑i=1

1 = 4n

n∑i=1

1 + 4n

n∑i=1

4in

+ 4n

n∑i=1

4i2

n2 + 2n

n∑i=1

1

= 4n

n∑i=1

1 + 16n2

n∑i=1

i+ 16n3

n∑i=1

i2 + 2n

n∑i=1

1 = 4n· n+ 16

n2 · n(n+1)2

+ 16n3 · n(n+1)(2n+1)

6+ 2

n· n

= 4 + 8( n2 + n

n2

)+

8

3

( 2n3 + 3n2 + n

n3

)+ 2.

Therefore limn→∞

n∑i=1

f(xi)∆x = limn→∞

(4 + 8

( n2 + n

n2

)+

8

3

( 2n3 + 3n2 + n

n3

)+ 2)

= 4 + 8 +16

3+ 2 =

58

3.

So

∫ 3

1

(2x2 + 1) dx = 583

.

30. Let n(t) be the number of inmates in country prisons at time t, where t is measured in

years. Then n(t) =

∫300e

t5 dt. Use substitution, with u = e

t5 and du = 1

5e

t5 , getting

5

∫eu du, which equals 5eu. So

∫300e

t5 dt = 300 · 5e

t5 + C = 1500e

t5 + C.

We know that n(0) = 2000, so 2000 = 1500e0 + C → 2000 = 1500 · 1 + C → C = 500.Therefore n(t) = 1500e

t5 + 500.

31. Problem 33a. Use The Midpoint Rule to estimate

∫ 3.2

0

f(x) dx, with n = 4 and ∆x = 0.8,

getting∫ 3.2

0

f(x) dx = 0.8[f(0.4)+f(1.2)+f(2.0)+f(2.8)

]= 0.8

[6.5+6.4+7.6+8.8

]= 0.8(29.3) = 23.44.

32. First, use the substitution u = lnx, du =1

xdx. With this substitution, the limits of

integration change: x = 1 becomes u = 0 and x = e8 becomes u = 8. So the new integral

is

∫ 8

0

u2 du.

Now use the midpoint rule with n = 4: ∆x = 8−04

= 2 and the midpoints are 1, 3, 5, and7. Therefore,∫ 8

0

u2 du = 2(f(1) + f(3) + f(5) + f(7) = 2(

12 + 32 + 52 + 72)

= 2 · 84 = 168.

33. F (x) =

∫ x

π

t2

sin2 t+ 1dt

(a) F ′(x) =x2

sin2 x+ 1by the Fundamental Theorem of Calculus.

(b) The slope of the tangent line at x = π is F ′(π) =π2

sin2(π) + 1=

π2

0 + 1= π2. The

point of tangency is (π, F (π)) = (π,

∫ π

π

t2

sin2 t+ 1) = (π, 0). So the equation of the

tangent line isy − 0 = π2(x− π) ⇒ y = π2x− π3.

34.d

dx

(∫ x2

0

√arctan t dt

)=√

arctan(x2)·2x, using the Fundamental Theorem of Calculus

and the chain rule.

35. (a) R is the region bounded by the graphs of y = x2 lnx and y = x2 over the interval[1, 2e].

Solution: Find the points of intersection:

x2 lnx = x2

x2(lnx− 1) = 0

So x = 0 and x = e are the points of intersection and only x = e is in the interval[1, 2e]. Checking e1/2, we see that e(1

2− 1) < 0 so x2 lnx is the larger function on

the interval [1, e]. Checking e3/2, we see that e3(3 − 1) > 0 so x2 lnx is the smallerfunction on the interval [e, 2e]. Thus the area is:

area of R =

∫ e

1

x2 lnx− x2 dx+

∫ 2e

e

x2 − x2 lnx dx

We need to compute the integral

∫x2 lnx dx. Use Integration by Parts with u = lnx

and dv = x2dx so du = 1xdx and v = x3

3. Then

∫x2 lnx dx =

x3

3lnx −

∫x2

3dx =

x3

3lnx− x3

9. So

area of R =

∫ e

1

x2 lnx− x2 dx+

∫ 2e

e

x2 − x2 lnx dx

=

(x3

3lnx− x3

9− x3

3

) ∣∣∣∣e1

+

(x3

3− x3

3lnx+

x3

9

) ∣∣∣∣2ee

=

(e3

3− e3

9− e3

3

)−(−1

9− 1

3

)+

(8e3

3− 8e3

3ln(2e) +

8e3

9

)−(e3

3− e3

3+e3

9

)=

(−e

3

9

)−(−1

9− 1

3

)+

(−8e3

3ln(2e)

)−(

+e3

9

)= −2e3

9+

4

9− 8e3

3ln(2e)

(b) The region is shown below:

0 1 2

1

2

(i) Integrating with respect to x:

A =

1∫0

(x2 − 0

)dx+

2∫1

((2− x)− 0 ) dx

=x3

3

]10

+

(2x− x2

2

]21

)=

5

6.

(ii) Integrating with respect to y: Rewrite y = x2 as x =√y and x + y = 2 as

x = 2− y. Then∫ 1

0

((2− y)−√y

)dy = 2y −

y2

2− 2

3y3/2

]10

=5

6.

36. (a) The cross-sections of the solid are disks with radius1

4√

1− x2. So the cross-sectional

area is A(x) = π

(1

4√

1− x2

)2

=π√

1− x2. So A(0) =

π√1− 02

= π.

(b) V =

∫ 12

−√32

A(x) dx = π

∫ 12

−√3

2

1√

1− x2dx = π

(sin−1 x

] 12

−√3

2

)= π

(sin−1(1

2)− sin−1(−

√32

))

= π( π

6−(− π

3

))=

π2

2.

37. The cross sections of S are washers. The outer radius is R =1

x2and the inner radius is

r =1

x4. The integral is

∫ ∞1

(π( 1

x2

)2− π

( 1

x4

)2)dx = π

∫ ∞1

(1

x4−

1

x8

)dx. If this

integral converges, its value is the volume of the solid. Integrating gives

π

∫ ∞1

(1

x4−

1

x8

)dx = π lim

t→∞

∫ t

1

(1

x4−

1

x8

)dx = π lim

t→∞

(−

1

3x3+

1

7x7

]t1

)

= π limt→∞

((− 1

3t3+

1

7t7

)−(− 1

3+

1

7

))= π

(0 + 0 +

1

3− 1

7

)=

4π

21.

Since the integral converges to4π

21, the volume of the solid S is

4π

21.

38. (a)

∫ 1

−1(3|x|+ x 3

√x) dx. Split up the integral as follows:∫ 1

−1(3|x|+ x 3

√x) dx = 3

∫ 1

−1|x| dx+

∫ 1

−1x4/3 dx.

Compute first integral by finding the area under y = |x| over [−1, 1] (two triangles,each with base 1 and height 1). Compute the second using the FTC. The result is∫ 1

−1(3|x|+ x 3

√x) dx = 3 · 2(1 · 1 · 1

2) + (3

7x7/3)

∣∣∣1−1

= 3 + (37− (−3

7)) = 3 + 6

7= 27

7.

(b)

∫10

2x3 − 3x2 − 2xdx. Use partial fractions.

10

x(2x+ 1)(x− 2)=

A

x+

B

2x+ 1+

C

x− 2⇒ 10 = A(2x+1)(x−2)+Bx(x−2)+Cx(2x+1).

If we let x = 2, we get C = 1; if we let x = −12, we get B = 8; and if we let x = 0,

we get A = −5. So∫10

2x3 − 3x2 − 2xdx = −

∫5

xdx+

∫8

2x+ 1dx+

∫1

x− 2dx

= −5 ln |x|+ 4 ln |2x+ 1|+ ln |x− 2|+ C.

(c)

∫x2

x2 + 1. Use polynomial division to rewrite

x2

x2 + 1as 1− 1

1 + x2. Then

∫x2

x2 + 1=∫ (

1−1

1 + x2

)dx = x− tan−1 x+ C.

(d)

∫ ∞0

dx3√ex

dx = limt→∞

∫ t

0

dx3√ex

dx.

∫dx3√ex

dx =

∫(ex)−

13 dx =

∫e−

x3 dx = −3e−

x3 .

Then limt→∞

∫ t

0

dx3√ex

dx = limt→∞

(− 3e−

x3

]t0

)= lim

t→∞

(− 3e−

t3 + 3

)= 0 + 3 = 3. So∫ ∞

0

dx3√ex

dx converges to 3.

(e)

∫ 2

1

dx

(1− x)2/3. The integrand has a vertical asymptote at x = 1 so

∫ 2

1

dx

(1− x)2/3= lim

t→1+

∫ 2

t

dx

(1− x)2/3.

Fordx

(1− x)2/3, use u-substitution with u = 1−x and du = −dx, getting

∫−u−2/3 du,

which equals −3u13 . Resubstituting gives −3(1− x)

13 . Then

limt→1+

∫ 2

t

dx

(1− x)2/3= lim

t→1+

(− 3(1− x)

13

]2t

)= lim

t→1+

(3− (−3(1− t)1/3)

)= 3.

So

∫ 2

1

dx

(1− x)2/3converges to 3.

(f)

∫ ln 2

0

ex

(ex − 1)3dx. The integrand has a vertical asymptote at x = 0, so

∫ ln 2

0

ex

(ex − 1)3dx =

limt→0+

∫ ln 2

t

ex

(ex − 1)3dx. For

∫ex

(ex − 1)3dx, use substitution, with u = ex − 1 and

du = ex dx, getting

∫du

u3= −

1

2u2. Resubstituting gives − 1

2(ex − 1)2. Then

limt→0+

∫ ln 2

t

ex

(ex − 1)3dx = lim

t→0+

(−

1

2(ex − 1)2

]ln 2

t

)= −

1

2limt→0+

(1− 1

(et − 1)2

)= +∞.

So

∫ ln 2

0

ex

(ex − 1)3dx diverges.

(g)

∫ 34

− 34

dx√

9− 4x2=

∫ 34

− 34

dx√9(1− 4x2

9)

=

∫ 34

− 34

dx

3√

1− (2x3

)2=

1

3

∫ 34

− 34

dx√1− (2x

3)2

.

Use substitution, with u = 2x3

and du = 23dx. Changing the limits of integration

gives: x = −34→ u = −1

2and x = 3

4→ u = 1

2. So we get 1

2

∫ 12

− 12

dx√

1− u2=

12

sin−1 u] 1

2

− 12

= 12

sin−1(12)− 1

2sin−1(−1

2) = π

12−(− π

12

)= π

6.

(h)

∫ +∞

−∞

x√x2 + 2

dx =

∫ 0

−∞

x√x2 + 2

dx+

∫ ∞0

x√x2 + 2

dx = limt→−∞

∫ 0

t

x√x2 + 2

dx+

limt→∞

∫ t

0

x√x2 + 2

dx. For

∫x

√x2 + 2

dx use substitution with u = x2 + 2 and

du = 2x. The new integral is1

2

∫1√udu, which equals =

√u. Resubstituting

gives√x2 + 2. So

limt→−∞

∫ 0

t

x√x2 + 2

dx = limt→−∞

(√x2 + 2

]0t

)= lim

t→−∞

(√2−√t2 + 2

)= −∞.

So

∫ 0

−∞

x√x2 + 2

dx diverges. Therefore

∫ ∞−∞

x√x2 + 2

dx must diverge.

39. Note that the semicircle y =√r2 − x2 intersects the x-axis at x = −r and x = r. So

to find the volume of the sphere with radius r we rotate between y =√r2 − x2 and the

x-axis over the interval [−r, r] around the x-axis. The volume of the resulting solid is

V =

∫ r

−rπ(√

r2 − x2)2dx = π

∫ r

−r

(r2 − x2) dx = π

(r2x−

x3

3

]r−r

)

π

((r3 −

r3

3

)−(− r3 +

r3

3

))= π

(2r3 −

2r3

3

)= π ·

4r3

3=

4πr3

3.

40. Let h represent the depth of the water in the tank at time t.

(a)dh

dt= k√h (where k is some constant)

This is a separable differential equation:

dh

dt= k√h⇒ 1√

hdh = kdt⇒

∫h−1/2dh =

∫kdt⇒ 2h1/2 = kt+ C

Solving for h gives:

h1/2 =kt

2+C

2⇒ h =

(kt

2+C

2

)2

(b) One of our initial conditions is h(0) = 36, so 36 =(C2

)2so 6 = C

2so C = 12. Then

h(t) =

(kt

2+ 6

)2

The other initial condition is h(2) = 16 so 16 = (k + 6)2 so 4 = k + 6 so k = −2.Then the specific solution is

h(t) = (−t+ 6)2

(c) Half of 36 is 18, so we solve for t:

18 = (−t+ 6)2 ⇒√

18 = −t+ 6⇒ t = 6−√

18 (≈ 1.76)

41. y =√

12− 4x2 so y′ =1

2(12− 4x2)−1/2(−8x) =

−4x√12− 4x2

. Then

yy′ + 4x =√

12− 4x2(

−4x√12− 4x2

)+ 4x = −4x+ 4x = 0

So y is indeed a solution to the differential equation.

42. We have (ln y)dy

dx= xy so

ln y

ydy = xdx. Take the integral of both sides:

∫ln y

ydy =∫

xdx. For the dy integral, use the u-substitution u = ln y so du = 1ydy. The integral

then becomes∫udu = u2

2= 1

2(ln y)2. Then we have∫

ln y

ydy =

∫xdx⇒ 1

2(ln y)2 =

x2

2+C ⇒ (ln y)2 = x2+2C ⇒ ln y =

√x2 + 2C ⇒ y = e

√x2+2C

43. We have dydx

= (x− 1)(y − 2) so

1

y − 2dy = (x−1)dx⇒

∫1

y − 2dy =

∫(x−1)dx⇒ ln |y−2| = x2

2−x+C ⇒ y = ±e

x2

2−x+C+2

Plugging in the initial condition (y(2) = 4) gives

4 = e2−2+C + 2⇒ 2 = eC ⇒ C = ln 2

So the specific solution is y = ex2

2−x+ln 2 + 2 or y = 2e

x2

2−x + 2.

44. y = eax so y′ = aeax and y′′ = a2eax We want y′′ + 4y′ − 12y = 0 so

a2eax + 4aeax − 12eax = 0⇒ eax(a2 + 4a− 12) = 0

Since eax is never equal to 0 for any x, we must have (a2+4a−12) = 0 so (a+6)(a−2) = 0so either a = −6 or a = 2.

45. We are looking for a 3rd degree Taylor polynomial that approximates the solution to thedifferential equation y′ + xy = ex, subject to the initial condition y(0) = 0. We’ll centerthe polynomial at x = 0. We can rewrite the equation as

y′ = ex − xy.

Then

• y′′ = ex − xy′ − y• y′′′ = ex − xy′′ − y′ − y′ = ex − xy′′ − 2y′

Now use the initial condition to evaluate the derivatives at x = 0:

• y(0) = 0

• y′(0) = e0 − 0 = 1

• y′′(0) = e0 − 0− 0 = 1

• y′′′(0) = e0 − 0− 2(1) = −1

So the Taylor polynomial is

P3(x) = f(0) + f ′(0)x+f ′′(0)

2!x2 +

f ′′′(0)

3!x3

= 0 + x+1

2!x2 −

1

3!x3 = x+

x2

2!−x3

3!.

46. (a)∞∑n=0

xn

1 + n2 . Use the ratio test:

limn→∞

∣∣∣∣∣ an+1

an

∣∣∣∣∣ = limn→∞

∣∣∣∣∣∣∣∣∣xn+1

1 + (n+ 1)2

xn

1 + n2

∣∣∣∣∣∣∣∣∣ = limn→∞

∣∣∣∣∣ xn+1

n2 + 2n+ 2·n2 + 1

xn

∣∣∣∣∣ = limn→∞

∣∣∣∣∣ n2 + 1

n2 + 2n+ 2· x

∣∣∣∣∣ .

Since n is positive, we write limn→∞

n2 + 1

n2 + 2n+ 2· |x|. Since lim

n→∞

n2 + 1

n2 + 2n+ 2= 1,

limn→∞

n2 + 1

n2 + 2n+ 2· |x| = 1 · |x| = |x|. According to the ratio test, the series will

converge if this limit is < 1. So the series converges for |x| < 1, that is, for x in theinterval (−1, 1). This tells us that the radius of convergence is R = 1.

To find the precise interval of convergence, we need to check the endpoints.

x = 1: Plug x = 1 into our original series to get

∞∑n=0

1n

1 + n2 =∞∑n=0

1

1 + n2

Since n2 < n2 + 1, we know that 1n2+1

< 1n2 . So the series in question is less than the

series∞∑n=0

1

n2 , which is a p-series with p = 2 > 1, so it converges. By the comparison

test,∞∑n=0

1n

1 + n2 also converges.

x = −1: Plug x = −1 into our original series to get

∞∑n=0

(−1)n

1 + n2

Taking absolute values gives∞∑n=0

1

1 + n2 , which we just showed converges. Therefore

∞∑n=0

(−1)n

1 + n2 converges absolutely.

Thus the interval of convergence is [−1, 1].

(b)∞∑n=0

n!

2nxn. Use the ratio test:

limn→∞

∣∣∣∣∣ an+1

an

∣∣∣∣∣ = limn→∞

∣∣∣∣∣∣∣∣∣(n+ 1)!

2n+1 xn+1

n!

2nxn

∣∣∣∣∣∣∣∣∣ = limn→∞

∣∣∣∣∣ (n+ 1)!xn+1

2n+1 ·2n

n!xn

∣∣∣∣∣ = limn→∞

∣∣∣∣∣ n+ 1

2· x

∣∣∣∣∣= lim

n→∞

n+ 1

2· |x| =∞.

Since this limit is never < 1, the series converges only at its center x = 0. So theradius of convergence is R = 0 and the interval of convergence is { 0 }.

47. (a)∞∑n=1

(−1)n+15n

n!. Use the ratio test:

limn→∞

∣∣∣∣∣ an+1

an

∣∣∣∣∣ = limn→∞

∣∣∣∣∣∣∣∣∣(−1)n+25n+1

(n+ 1)!

(−1)n+15n

n!

∣∣∣∣∣∣∣∣∣ = limn→∞

5n+1

(n+ 1)!·n!

5n= lim

n→∞

5

n+ 1= 0.

Since limn→∞

∣∣∣∣ an+1

an

∣∣∣∣ < 1, the series converges.

(b)∞∑n=1

1

n1/3 + 2n

Note that1

n1/3 + 2n≤ 1

2nfor all n ≥ 1 since n1/3 + 2n ≥ 2n for all n ≥ 1. Since all

terms of both sequences are positive for n ≥ 1, we can use the Comparison Test. We

know that∞∑n=1

1

2nconverges because it is geometric with |r| = 1

2< 1, so the series

∞∑n=1

1

n1/3 + 2nalso converges.

(c)∞∑n=3

lnn

n

We can use the integral test since the function f(x) =lnx

xis positive, con-

tinuous and decreasing on [3,+∞). So we look at

∫ ∞3

lnx

xdx, which equals

limt→∞

∫ t

3

lnx

xdx. For

∫lnx

xdx use substitution with u = ln x and du =

1

xdx.

The new integral is

∫u du, which equals

u2

2. Resubstituting gives

(lnx)2

2. Then

limt→∞

∫ t

3

lnx

xdx = lim

t→∞

((lnx)2

2

]t3

)=

1

2limt→∞

((ln t)2 − (ln 3)2

).

Since limt→∞

(ln t)2 = ∞,1

2limt→∞

((ln t)2 − (ln 3)2

)= ∞. So

∫ ∞3

lnx

xdx diverges.

Therefore∞∑n=3

lnn

ndiverges.

(d)∞∑n=3

(−1)n3n+1

2n= 3

∞∑n=3

(−1)n3n

2n= 3

∞∑n=3

(− 3

2

)n

= 3

(− 3

2

)3

+ 3

(− 3

2

)4

+ 3

(− 3

2

)5

+ · · ·

This is geometric with a = 3

(− 3

2

)3

and r =− 3

2.

Since |r| = 32> 1, the series diverges.

(e)∞∑n=1

1√n3 + 2n− 1

This series is dominated by the√n3 term in the denominator, so we guess that it

will converge like a p-series with p = 32> 1. We’ll use the Comparison Test. Note

that we take the positive square root so the terms are all positive. (You can alsocheck that n3 + 2n− 1 is positive for all n ≥ 1, so all the terms are defined.)

We will compare the series∞∑n=1

1√n3 + 2n− 1

to the series∞∑n=1

1√n3

. Note that n3 ≤

n3 + 2n− 1 for all n ≥ 1 since 1 ≤ 2n for all n ≥ 1. Since all the terms are positive,

we can divide on both sides to get1

n3 + 2n− 1≤ 1

n3and so

1√n3 + 2n− 1

≤ 1√n3

for all n ≥ 1. Since∞∑n=1

1√n3

is a p-series with p = 32> 1, we know this converges

and so, by the Comparison Test, the series∞∑n=1

1√n3 + 2n− 1

converges as well.

(f)∞∑n=1

ne−n2

We’ll use the Integral Test. Let f(x) = xe−x2

=x

ex2. Clearly f(x) is positive and

continuous. We find f ′(x) = e−x2 − 2x2e−x

2= e−x

2(1 − 2x2), which is negative for

x > 1√2

and so is negative for x ≥ 1. Therefore f is decreasing for x ≥ 1. We need

to compute the improper integral

∫ ∞1

x

ex2dx. To compute the indefinite integral∫

x

ex2dx, we use u-substitution with u = x2, so du = 2xdx, so∫

x

ex2dx =

1

2

∫1

eudu = − 1

2eu= − 1

2ex2

and therefore∫ ∞1

x

ex2dx = lim

t→∞

∫ t

1

x

ex2dx = lim

t→∞− 1

2ex2

∣∣∣∣t1

= limt→∞− 1

2et2+

1

2e=

1

2e.

Since

∫ ∞1

x

ex2dx converges, we know by the Integral Test that

∞∑n=1

ne−n2

also con-

verges.

(g)∞∑n=0

(−1)n. Note that limn→∞

(−1)n 6= 0, so the series diverges by the divergence test.

Note: You could also have observed that this is a geometric series with r = −1,which means it diverges. Or you could have noted that sn = { 1, 0, 1, 0, 1, 0, . . . };since lim

n→∞sn doesn’t exist, the series diverges.

(h)∞∑n=1

n!

n3

Since this series has a factorial in it, let’s try the Ratio Test.

limn→∞

∣∣∣∣an+1

an

∣∣∣∣ = limn→∞

∣∣∣∣ (n+ 1)!

(n+ 1)3· n

3

n!

∣∣∣∣ = limn→∞

∣∣∣∣(n+ 1)n3

(n+ 1)3

∣∣∣∣ = limn→∞

∣∣∣∣ n3

(n+ 1)2

∣∣∣∣ =∞

Since L =∞, the series∞∑n=1

n!

n3diverges by the Ratio Test.

48. (a) i. f(x) = sinx, n = 6, a = π2. Note that f(π

2) = 1.

• f ′(x) = cos x, so f ′(π2) = 0

• f ′′(x) = − sinx, so f ′′(π2) = −1

• f ′′′(x) = − cosx, so f ′′′(π2) = 0

• f (4)(x) = sinx, so f (4)(π2) = 1

• f (5)(x) = cos x, so f (5)(π2) = 0

• f (6)(x) = − sinx, so f (6)(π2) = −1

Therefore

P6(x) = 1 + 0(x− π

2)− 1

2!(x− π

2)2 +

0

3!(x− π

2)3 +

1

4!(x− π

2)4 +

0

5!(x− π

2)5 − 1

6!(x− π

2)6

= 1−1

2!(x− π

2)2 +

1

4!(x− π

2)4 −

1

6!(x− π

2)6.

ii. f(x) = ln(2− x), n = 4, a = 1. Note that f(1) = 0.

• f ′(x) = − 1

2− x , so f ′(1) = −1

• f ′′(x) = − 1

(2− x)2, so f ′′(1) = −1

• f ′′′(x) = − 2

(2− x)3, so f ′′′(1) = −2

• f (4)(x) = − 6

(2− x)4, so f (4)(1) = −6

So

P4(x) = −(x− 1)−(x− 1)2

2!−

2(x− 1)3

3!−

6(x− 1)4

4!.

We can write this Taylor polynomial in without factorials:

P4(x) = −(x− 1)−(x− 1)2

2−

(x− 1)3

3−

(x− 1)4

4.

(b) Remember that f(x) = ln(2−x), so f(0) = ln 2. So we can use P4(0) to approximatef(0):

f(0) ≈ P4(0) = −(0− 1)− (0− 1)2

2− (0− 1)3

3− (0− 1)4

4= 1− 1

2+

1

3− 1

4=

7

12.

So ln 2 ≈ 7

12.

49. (a) f(x) =1

xand a = 1. Note that f(1) = 1. Compute the derivatives of f(x) and

evaluate them at x = 1:

• f ′(x) = −x−2, so f ′(1) = −1

• f ′′(x) = 2x−3, so f ′′(1) = 2

• f ′′′(x) = −6x−4, so f ′′′(1) = −6

• f (4)(x) = 24x−5, so f (4)(1) = 24

• f (5)(x) = −120x−6, so f (5)(1) = −120, etc.

So the Taylor series for f(x) =1

xcentered at x = 1 is

1− (x− 1) +2

2!(x− 1)2 − 6

3!(x− 1)3 +

24

4!(x− 1)4 − 120

5!(x− 1)5 + . . .

Since 2! = 2, 3! = 6, 4! = 24, 5! = 120, we can write the series without factorials as

1− (x− 1) + (x− 1)2 − (x− 1)3 + (x− 1)4 − (x− 1)5 + · · · =∞∑n=0

(−1)n(x− 1)n.

(b) f(x) = xex and a = 0. Note that f(0) = 0. Compute the derivatives of f(x) andevaluate them at x = 0:

• f ′(x) = xex + ex, so f ′(0) = 1

• f ′′(x) = xex + 2ex, so f ′′(0) = 2

• f ′′′(x) = xex + 3ex, so f ′′′(0) = 3

• f (4)(x) = xex + 4ex, so f (4)(0) = 4

• f (5)(x) = xex + 5ex, so f (5)(0) = 5, etc.

So the Taylor series for f(x) = xex centered at x = 0 is

x+2

2!x2 +

3

3!x3 +

4

4!x4 +

5

5!x5 + · · ·

= x+ x2 +x3

2!+x4

3!+x5

4!+ · · · =

∞∑n=0

xn+1

n!.

50. We’re given the differential equation y′ − y = 0, with initial condition y(0) = 3.

(a) We center the Taylor series at x = 0. Rewrite the equation as y′ = y and computederivatives:

• y′′ = y′ = y

• y′′′ = y′′ = y

• y(4) = y′′′ = y

• y(5) = y′(4) = y, etc.

Clearly y(n) = y for all n ≥ 0. Therefore y(n)(0) = y(0) = 3, for all n ≥ 0. So theTaylor series is

3 + 3x+3x2

2!+

3x3

3!+

3x4

4!+ · · · =

∞∑n=0

3xn

n!.

(b) To find the radius of convergence use the ratio test:

limn→∞

∣∣∣∣∣ an+1

an

∣∣∣∣∣ = limn→∞

∣∣∣∣∣∣∣∣∣3xn+1

(n+ 1)!

3xn

n!

∣∣∣∣∣∣∣∣∣ = limn→∞

∣∣∣∣∣ 3xn+1

(n+ 1)!·n!

3xn

∣∣∣∣∣ = limn→∞

∣∣∣∣∣ 1

n+ 1· x

∣∣∣∣∣ = 0.

Since this limit is < 1 for all x, the radius of convergence of the series is R =∞ andthe interval of convergence is (−∞,∞).

(c) We can write the series as

3

(1 + x+

x2

2!+x3

3!+x4

4!+ · · ·

)= 3 ·

∞∑n=0

xn

n!.

So our series is just the Taylor series for f(x) = 3ex centered at x = 0.

51. Since f ′′(x) + f(x) = 0, we know that f ′′(x) = −f(x). Therefore

• f ′′′(x) = −f ′(x)

• f (4)(x) = −f ′′(x) = f(x)

• f (5)(x) = f ′(x)

• f (6)(x) = −f ′′(x) = −f(x)

• f (7)(x) = −f ′(x)

• f (8)(x) = −f ′′(x) = f(x), etc.

Evaluating the derivatives at x = 0, using the initial conditions f(0) = 1 and f ′(0) = 0,gives:

• f(0) = 1

• f ′(0) = 0

• f ′′(0) = −1

• f ′′′(0) = 0

• f (4)(0) = 1

• f (5)(0) = 0

• f (6)(0) = −1

• f (7)(0)) = 0

• f (8)(0) = 1, etc.

At this point the pattern is clear, and our Taylor series is

P8(x) = f(0) + f ′(0)x+f ′′(0)

2!x2 +

f ′′′(0)

3!x3 + · · ·+

f (8)(0)

8!x8 + · · ·

= 1 + (0)x−1

2!x2 +

0

3!x3 +

1

4!x4 +

0

5!x5 −

1

6!x6 +

0

7!x7 +

1

8!x8 · · ·

= 1−x2

2!+x4

4!−x6

6!+x8

8!· · · .

We can write this series in sigma notation as follows:

∞∑n=0

(−1)nx2n

(2n)!.

We can recognize this as the Taylor series for f(x) = cos x centered at x = 0. This makes

sense: f(x) = cosx satisfies the given differential equation, since (cosx)′′ = − cosx. Thefunction f(x) = cos x also satisfies the initial conditions, since f(0) = cos(0) = 1 andf ′(0) = − sin(0) = 0.

52. (a) False; an is the nth term of the series, not the nth partial sum. The nth partial sum

of the series is sn = a1 + a2 + a3 + · · ·+ an =n∑i=1

ai.

(b) True. This is a theorem that we learned in Section 8.2 (Theorem 6 on page 570).

(c) False. We have seen that the power series∞∑n=0

xn has radius of convergence R = 1.

We saw in Problem 30a on this review sheet that the series∞∑n=0

xn

1 + n2 also has

radius of convergence R = 1. But the coefficients of these series are not identical.

To see why, write out the first few terms of the series∞∑n=0

xn:

1 + x+ x2 + x3 + x4 + · · · .

Clearly the coefficients are all equal to 1. On the other hand, the first few terms of

the series∞∑n=0

xn

1 + n2 are

1 +x

2+x2

5+x3

10+x4

17+ · · · .

Here the coefficients are not all equal to 1.

53. (Choose which of the following is true.) If the power series∑cn5n converges

(a) then all the cn’s must be equal to 0.

(b) then∑cnr

n converges as long as |r| < 5.

(c) then∑cnr

n converges as long as |r| ≤ 5.

(d) then∑cn6n diverges.

Answer: (b) This follows from Theorem 3 on page 595 of the textbook with r = x− a.Since we know the series converges when x − a = 5, we must have R ≥ 5. The reason(c) doesn’t work is that the series might not converge when we replace 5 by −5 (i.e., wemight have R = 5, in which case we don’t know what’s happening at the other endpointof the interval of convergence).