Math 172-507 Practice Final #1 Solutions
Warning. This has not been proofread. Use at your own peril. Feel free to inform me of any mistakes
you find. (I suspect that if there any mistakes they should be easy to spot and not out-weigh the benefit
of using this as a resource for studying.)
1. True or false?
P1 3 n
D4
a)
nD1 4
Solution. This is almost a geometric series. We just need the summation to start at n D 0.
We fix this by “pulling out” a “3=4” term by the distributive property and changing indices
(letting k D n 1):
1 n
1 1 X
3X 3 n 1 3X 3 k
3
D
D
:
4
4 nD1 4
4
4
nD1
kD0
Now the geometric series applies to say
1 1
3X 3 k
3
3
D
D
D 3 ¤ 4;
4
4
4 1 3=4
4 3
kD0
==
so the claim is false.
b) If limn!1 an ¤ 0, then
P1
nD1
an diverges.
Solution. This claim is true, since it is exactly the statement of The Divergence Test. ==
P
P1
c) If 1
nD1 an diverges so does
nD1 jan j.
Solution. Recall the statement of absolute convergence:
P
P1
“If 1
nD1 jan j converges, then
nD1 an converges.”
It is a logically equivalent statement to say
P
P1
“If 1
nD1 an diverges, then
nD1 jan j diverges.”
Thus the claim is true. (Some of you may be familiar with this. This is known as the
contrapositive statement to the original: It is a fact that if P and Q are statements and we
know that P implies Q, then it is also the case that not Q implies not P . For example,
the statement, “if it is a dog, then it is a mammal,” is true. The statement “if it is not a
mammal, than it is not a dog,” is also true. It’s worth noting that the statement, “if it is a
mammal, then it is a dog,” is not true.)
==
P
1
nC1 n
d) The series 1
2 converges by the Alternating Series Test.
nD1 . 1/
1
Solution. Note that limn!1 2 n D 20 D 1 ¤ 0, so the series fails to converge by the
Alternating Series Test, and the claim is false.
==
1
e) If f .x/ D
P1
nD1
. 1/nC1 x n
,
n.nC1/
1
.
6
then f 00 .0/ D
Solution. Since the derivative of the sum is equal to the sum of the derivatives,
1
X
. 1/nC1 x n
f .x/ D
nC1
nD1
1
0
:
Differentiating once more gives
00
f .x/ D
1
X
. 1/
nC1 n 2
x
nD1
D
1 x
C
3
2
From the expansion, it’s clear that f 00 .0/ D
P1 . 1/n
p
f)
e.
nD0 2n nŠ D
n 1
nC1
3x 2
C :
5
1=3 ¤
1=6 so the claim is false.
Solution. Recall that the power series expansion for e x is
1
1
X
X
. 1=2/n
. 1/n
D
De
2n nŠ
nŠ
nD0
nD0
1=2
xn
nD0 nŠ .
P1
==
So
p
1
Dp ¤ e
e
and the claim is false.
P
P1
n
g) If 1
nD1 cn 3 converges, so does
nD1 jcn j.
==
Solution. True. Think of it like this: limn!1 cn 3n D 0, so cn goes to zero at a rate that’s
faster than the rate at which 1=3n goes to zero. This still holds for jcn j. So eventually we
will have
1
jcn j n
3
P1
P
1 n
for every n. Therefore nD1 jcn j converges by the comparison test with 1
nD1 3 ,
which converges because it is a geometric series and j1=3j < 1.
==
R cos x p
2. If f .x/ D 1
1 C t 3 dt find f 0 .x/.
Solution. Let u D cos x. Then
u
Z
f .x/ D
p
1 C t 3 dt
1
By The Fundamental Theorem of Calculus (part 1) and the chain rule
p
p
f 0 .x/ D u0 1 C u3 D sin x 1 C cos3 x
==
3. Let C be the arc y D 12 .e x C e x / 0 x ln 2. Let A denote the area enclosed between C ,
the x and y axes, and the line x D ln 2.
a) Compute A.
2
Solution. Through the standard method (note that 12 .e x C e
e x > 0 for all x),
ln 2
Z
AD
0
1 x
.e C e
2
x
1
/ dx D
2
x
/ > 0 for all x because
ln 2
Z
Z
1 ln 2 x
e dx C
e dx
2 0
1
1 1
3
1
2
2
Πe x jln
C D :
D Œe x jln
0 C
0 D
2
2
2 4
4
x
0
==
b) Find the volume generated when A is revolved around the x-axis.
Solution. By the standard formula,
Z
V D
0
ln 2
2
Z
1 x
ln 2 2x
x
.e C e / dx D
e C e 2x C 2 dx
2
4 0
ˇln 2
ˇ
1 2x 1 2x
15
15
ln 2
ˇ
D
e
e
C 2x ˇ D
C 2 ln 2 D
C
:
4 2
2
4 8
32
2
0
==
c) Find the length of the arc C .
Solution. Note dy=dx D
LD
ˇ
0
ln 2
s
dy
1C
dx
ˇ
D
1
2
.e x
x
e
2
dx D
ˇ
/, so
ln 2
r
1C
1 x
e
4
e
x
2
dx
0
ln 2
r
1 x
.e C e
4
x /2 dx
1
D
2
ln 2
Z
.e x C e
0
x
3
/ dx D A D :
4
0
It is purely coincidence that L D A, but it surely saves a bunch of work.
==
d) Use Pappus Theorem (or otherwise) to find the y coordinate yN of the centroid of the area
A.
Solution. Pappus Theorem is as follows:
“Let R be a plane region that lies entirely on one side of a line ` in the plane. If
R is rotated about `, then the volume of the resulting solid is the product of the
area A of R and the distance d traveled by the centroid of R.”
C 2 ln 2 . Pappus Theorem then
From (a) and (b), we know that A D 3=4 and V D 4 15
8
asserts that
15
3d
15
C 2 ln 2 D
and so
dD
C 2 ln 2 :
4 8
4
3 8
We also know that d D 2 r, where r is the height of the centroid (i.e., the y coordinate).
Thus
15
1 15
2 r D
C 2 ln 2 ;
and we conclude
yN D r D
C 2 ln 2 :
3 8
6 8
==
P
n
.x
p 3/
4. Find the radius of convergence and the interval of convergence of the series 1
nD1 2n n.nC1/ .
3
Solution. Note,
ˇ
ˇ
p
r
ˇ
ˇ
n
nC1
2
n.n
C
1/
.x
3/
jx 3j
jx 3j
n
ˇ
ˇ
D
lim
D
;
lim ˇ
p
ˇ
n!1 ˇ 2nC1 .n C 1/.n C 2/
.x 3/n ˇ n!1 2
nC2
2
so the ratio test says that we must have jx 3j=2 < 1 for convergence of the series. Hence
jx 3j < 2 and so 1 < x < 5. We must check x D 1 and x D 5 individually. If x D 1, the
series becomes
1
X
. 1/n
;
p
n.n
C
1/
nD1
which converges by The Alternating Series Test. If x D 5, the series becomes
1
X
1
:
p
n.n C 1/
nD1
(1)
Through the limit comparison test with 1=n,
p
p
n.n C 1/
1 C 1=n
1=n
lim
D lim
D 1;
D lim
p
n!1 1= n.n C 1/
n!1
n!1
n
1
P
1
so (1) converges if and only if 1
nD1 n converges. But this is the harmonic series, which is
known to diverge. Therefore (1) diverges if x D 5. We conclude that the interval of convergence
is I D Œ1; 5/.
==
5.
a) Find the Partial Fraction Decomposition of an D
1
.
4n2 C8nC3
Solution. Consider 4n2 C 8n C 3. We use the “AC method” (the quadratic formula would
work as well): 4 3 D 12: What two numbers add to give me 8 and multiply to give me
12? 6 and 2. So
4n2 C 8n C 3 D 4n2 C 2n C 6n C 3 D 2n.2n C 1/ C 3.2n C 1/ D .2n C 1/.2n C 3/
and an D
1
.
.2nC1/.2nC3/
Set
1
A
B
D
C
.2n C 3/.2n C 3/
2n C 1 2n C 3
Multiplying both sides by the denominator gives
1 D A.2n C 3/ C B.2n C 1/
D .2A C 2B/n C .3A C B/:
Equating coefficients yields the system
2A C 2B D 0;
3A C B D 1:
From (2) we see that A D
A D 1=2. Therefore
B. Using this information in (3) shows B D
an D
1
4n C 2
4
1
:
4n C 6
(2)
(3)
1=2 and so
==
b) Show that the partial sum sn of the series
telescope.)
P1
nD1
an is sn D
n
.
6nC9
(Hint: Use part a) and
Solution. Note
an D
where bn D
1
4n C 2
1
.
4nC2
sn D
n
X
1
1
D
4n C 6
4n C 2
1
D bn
4.n C 1/ C 2
So
n X
ak D
bk
kD1
bkC1
kD1
D .b1 b2 / C .b2 b3 / C .b3
D b1 bnC1
1
1
n
D
D
:
6 4n C 6
6n C 9
Does the series
bnC1 ;
P1
nD1
bnC1 /
==
an converge? If yes, compute its value. If not, explain.
Solution. From the previous part, sn D
1
X
b4 / C C .bn
n
.
6nC9
So
n
1
1
D lim
D :
n!1 6n C 9
n!1 6 C 9=n
6
an D lim sn D lim
n!1
nD1
==
6. Let f .x/ D 1=x; 1 x 3. Let Pn be the partition of the interval Œ1; 3 into n subintervals
of equal length. Let xi denote the left end of the i th subinterval for each i. Finally, let Sn be
Riemann sum for f .x/ using partition Pn and the points xi .
a) Compute S4 . Also make a graph showing f .x/ as well as the rectangle whose total area
is S4 .
Solution. The interval width is 2, so the subinterval length is 2=n. Also xi D 1 C
for the i th rectangle. If n D 4, then the subinterval length is 1=2 and
S4 D
f .1/ f .1:5/ f .2/ f .2:5/
1 1 1 1
77
C
C
C
D C C C D :
2
2
2
2
2 3 4 5
66
We also have this pretty graph:
1
1 1:5 2 2:5 3
5
5
2.i 1/
n
==
b) Write Sn using the † notation.
Solution. From the previous, xi D 1C 2.in 1/ and the subinterval length is 2=n. Therefore
Sn D
n
X
n
f .xi /x
i D1
2X
n
D
n i D1 n C 2i
2
D
n
X
i D1
2
n C 2i
2
:
==
c) Compute limn!1 Sn .
7.
R3
Solution. We know that we should have limn!1 Sn D ln 3 because 1 x1 dx D ln 3, but
that is going to be hard to get out of an algebraic series like this. I’m lead to believe that
this is some sort of typo, as it’s unreasonable for to expect calculus II students toRdeduce
3
limn!1 Sn D ln 2, unless it is expected that you simply notice limn!1 Sn D 1 x1 dx
and compute
the integral. (I actually think that this is the main way one would prove that
R3
Sn ! 1 x1 dx D ln 3.)
==
R
a) Find a reduction formula for In D x n e x dx: (Hint: Use Integration by Parts.)
Solution. We will use Integration by Parts:
Z
u.x/v 0 .x/dx D u.x/v.x/
Z
u0 .x/v.x/dx;
where u.x/; v.x/ denotes that u and v are functions of x, not multiplication. Let u D x n
and v 0 D e x , so that u0 D nx n 1 and v D e x . Therefore
Z
Z
n
x
n
x
In D x e dx D x e C nx n 1 e x D x n e x C nIn 1
==
b) Using part a) or otherwise compute I2 D
R
x2e
x
dx.
Solution. From part a),
x2e
x
C 2I1
D
2
x e
x
C 2 . xe
D
x2e
x
I2 D
8.
x
2xe
x
C I0 /
2e
x
C C:
==
2
a) Find the Taylor polynomial T2 .x/ of degree 2 for the function f .x/ D x 3 at x D 8.
Solution. Recall the Taylor polynomial centered at x D a:
n
X
f .n/ .a/
Tn .x/ D
nŠ
kD0
a/n :
.x
Let n D 2 and a D 8 so that
T2 .x/ D f .8/ C f 0 .8/.x
D4C
.x
8/
3
6
.x
8/ C
f 00 .8/
.x
2
8/2
:
144
8/2
==
b) If T2 is used to approximate .7:5/2=3 estimate the error.
Solution. Recall Taylor’s Inequality:
“If jf .nC1 .x/j M for jx aj < R, then the remainder Rn .x/ of the Taylor
series satisfies the inequality
jRn .x/j M
jx
.n C 1/Š
ajnC1
for jx
aj < R:”
8
,
27x 7=3
Let n D 2; a D 8; x D 7:5; and choose R D 1. Also, f 000 .x/ D
and jf 000 .x/j :0032 for jx 8j < 1. Therefore
:0032
j7:5
6
jR2 .7:5/j 9.
so f 000 .7/ :0032
8j3 .:0005/.:125/ :0000625:
Since T2 .7:5/ D 3:8315972 and .7:5/2=3 3:83154716, so the actual remainder is
jT2 .7:5/ .7:5/2=3 j :00005004. Not a bad estimate.
==
R1
2
a) Show that the improper integral 0 xe x dx converges and compute its value.
Solution.
Z 1
xe
x2
1h
e
2
dx D
0
b) Does the series
P1
nD1
ne
n2
ˇ1
ˇ D
x2 ˇ
0
1
lim e
2 x!1
x2
1
.0
2
1 D
1
1/ D :
2
==
converge? Justify your answer.
Solution. Consider f .x/ D xe
x2
. Recall The Integral Test:
“Suppose f is continuous,
P1 positive, decreasing function on Œ1; 1/ and let an D
f
.n/.
Then
the
series
nD1 an is convergent if and only if the improper integral
R1
1 f .x/dx is convergent.”
2
On Œ1; 1/, xe x is clearly positive and continuous. We want to show that f .x/ is decreasing, so set
2
2
2
f 0 .x/ D e x
2x 2 e x D e x .1 2x 2 / < 0:
2
e x > 0 for all x, we only need to find when 1
solving the inequality 1 2x 2 < 0:
2x 2 is negative. This is equivalent to
1
1
< x2
)
< x:
2
4
So
.1=4; 1/, and hence decreasing on Œ1; 1/. Part a) P
shows that
R 1f .x/x 2is decreasing onR 1
x2
n2
converges, so 1 xe
converges and The Integral Test asserts that 1
nD1 ne
0 xe
converges.
==
P
n2
c) If the partial s5 is used to approximate s D 1
, give an estimate for js sn j.
nD1 ne
2x 2 < 0
1
1 < 2x 2
)
)
Solution. Recall the Remainder Estimate for The Integral Test:
“If †an converges by The Integral Test and Rn D s sn , then
Z 1
Z 1
f .x/dx Rn f .x/dx:”
nC1
Applying this gives
Z 1
2
xe x dx D
R5 5
1h
e
2
n
ˇ1
x2 ˇ
ˇ
5
D
7
1
lim e
2 x!1
x2
e
52
1
D e
2
25
6:94:
==