Name
MATH 152H
Exam 2
Sections 201/202 (circle one)
1-14
/56
Spring 2016
15
/15
P. Yasskin
16
/15
17
/18
Total
/104
Solutions
Multiple Choice: (4 points each. No part credit.)
1. Find the general partial fraction expansion of f x
a. A
x
Dx E
x2 4 2
Bx C
x2 4
b. A
x
d. A
x
Bx
x2
Bx
x2
e. A
x
Bx
x2 4
c. A
x
C
4
C
4
Solution: f x
fx
Bx
x2
A
x
Dx E
x2 4 2
Dx E
x2 4
Dx E
x2 4 2
Dx
2
x2 4
xx
C
4
16
.
correct choice
x2 4
4 x2 4 x2
2
x
x2 4
4x x 4
3
1
4
x x2
2
4
Dx E
x2 4 2
x3
2. In the partial fraction expansion
4x
x4
32
Ax
x2
16
C
B
4
x
2
x
D ,
2
which coefficient is INCORRECT?
a. A
1
b. B
4
c. C
1
d. D
1
correct choice
e. All of the above are correct.
Solution: Clear the denominator:
x 3 4x 32
Ax B x 2
Plug in x
2:
32 C 32
C
Plug in x 2 :
32 D 32
D
8
8
32
1
32
A2
B 0
4 x
2
D x2
4 x
2
C4
4
2
D4
4 0
2
1
8
1
Plug in x 0 :
4B 16
B 4
Coeff of x 3 :
8
C x2
4
32
1
A
A2
B 0
B 4
4
2
C
A
1
D
C4 4 0
incorrect
14 2
1
D4
4 2
32
4B
2
16
A
1
4
3. Compute
1 dx
x2
1
a. 5
4
b.
5
4
c. diverges to
correct choice
d. diverges to
e. diverges but not to
4
Solution:
1
1 dx
x2
0
0
1
x
1
x
“
1
1
1
0
4. The integral
”
3
x 2 dx
1
4
x
x 2 dx
1
0
4
0
1
4
1
x
lim
x 0
“
1
0
x
1
1
4
1
0
1
4
lim
x 0
1
x
5
4
”
sin x
dx is
x
2 dx.
x
b. divergent by comparison with
3 dx.
x
c. divergent by comparison with
4 dx.
x
correct choice
d. convergent by comparison with
2 dx.
x
e. convergent by comparison with
4 dx.
x
Solution: The integrals
4 dx and
x
2 dx both diverge to
x
k dx
x
Since
0
1
1
a. divergent by comparison with
Since
1
because
k ln x
3
sin x
x
2 , we have
x
3
sin x
dx
x
and so diverges.
3
sin x
x
4 , we have
x
3
sin x
dx
x
and cannot conclude anything.
2
4/3
5. Compute
1
9x 2
0
a. 16
16
dx
9
1
b.
9
c.
d.
e.
12
24
correct choice
48
Solution: 3x
4/3
0
4 tan
3
x
4/3
1
9x
4 tan
2
1
dx
16
x 0
1 arctan 3x
12
4
dx
4 sec 2
16 3
16 tan
2
4/3
1 arctan 1
12
0
4 sec 2 d
tan 2
1
3
1 4/3 1 d
d
12 x 0
12
arctan 0
sec 2
4/3
x 0
48
6. Which of the following integrals results after performing an appropriate trig substitution for
4/3
1
a.
b.
c.
d.
e.
4/3
1
dx
16
csc d
arcsec 5/3
arcsec 5/3
1
20
correct choice
csc d
arcsec 5/4
arcsec 5/4
5
64
sec d
arcsec 5/3
arcsec 5/3
5
64
sec d
arcsec 5/4
arcsec 5/3
1
16
cot 2 d
arcsec 5/4
4 sec
5
x
1
20
dx
16
4 sec tan d
5
arcsec 5
4
arcsec 5/3
1
25x 2
because from the limits x
4 sec
1 when
x
25x
arcsec 5/4
1
20
Solution: 5x
So
1
2
dx
arcsec 5/4
x
1
16 sec 2
sec 2
4
3
4/5.
1
when
4 sec tan d
16 5
tan 2
arcsec 5
3
1
20
arcsec 5/3
arcsec 5/4
sec
tan
d
arcsec 5/3
csc d
arcsec 5/4
3
3/5
7. Compute
0
a.
b.
c.
d.
e.
1
dx
25x 2 16
1
20
1
20
1 ln 7
correct choice
40
1
1 ln 7
20
40
1 ln 7
1
40
20
Solution: 5x 4 sin because from the limits x 4/5.
4 sin
4 cos d
So
x
dx
sin 2
1
cos 2
5
5
3/5
3/5
1
1
4 cos d
1 3/5 cos d
dx
2
2
20 x 0 cos 2
0
25x 16
x 0 16 sin
16 5
3/5
1 3/5 sec d
1 ln|sec
tan |
20 x 0
20
x 0
5x : Opp 5x, Hyp 4, Adj
Draw a triangle with sin
16 25x 2 .
4
4
5x
Then sec
tan
16 25x 2
16 25x 2
3/5
3/5
0
1
25x 2
dx
16
1
20
1 ln
20
ln 7
7
ln 1
4
16
5x
25x 2
1
20
x 0
ln
4 3
16 9
ln
4
16
1 ln 7
40
The integral can also be done by partial fractions.
8. Find the length of the curve x
2t 2 , and y
t 3 for 0
t
1.
a. 732
b. 108
c. 7
d. 61
correct choice
27
e. 1
3
Solution: As a t-integral, the differential of arc length is
2
dx
dt
ds
dy
dt
2
dt
2
4t
3t 2
2
16t 2
dt
9t 4 dt
t 16
9t 2 dt
So the arc length is
1
L
ds
t 16
9t 2 dt
u
16
9t 2
du
18t dt
0
L
1
18
25
u du
16
1 2u 3/2
18 3
25
16
1 125
27
64
61
27
4
9. The curve y
sin x for 0
is rotated about the x-axis.
x
Which integral gives the area of the resulting surface?
2 x 1
a.
cos 2 x dx
0
b.
2 sin x 1
cos 2 x dx
2 cos x 1
sin 2 x dx
correct choice
0
c.
0
sin 2 x dx
cos x 1
d.
0
sin x dx
e.
0
Solution: As an x-integral, the differential of arc length is
ds
dy
dx
1
2
dx
cos 2 x dx
1
The radius of revolution is r
A
2 r ds
y
2 sin x 1
0
sin x. So the surface area is
cos 2 x dx
0
10. The curve y
sin x for 0
x
is rotated about the y-axis.
Which integral gives the area of the resulting surface?
2 x 1
a.
cos 2 x dx
correct choice
0
b.
2 sin x 1
cos 2 x dx
2 cos x 1
sin 2 x dx
0
c.
0
sin 2 x dx
cos x 1
d.
0
sin x dx
e.
0
Solution: As an x-integral, the differential of arc length is
ds
1
dy
dx
2
dx
The radius of revolution is r
A
2 r ds
0
2 x 1
1
cos 2 x dx
x. So the surface area is
cos 2 x dx
0
5
11. Consider the sequence a n
3n 6
3n 6
arctan
. Compute lim
an.
n
a.
b.
c.
d.
e.
2
correct choice
3
4
6
Solution: lim
an
n
3n 6
3n 6
lim
arctan
n
3
3
arctan
3n 6
arctan lim
n
3n 6
arctan 3
3
2n
12.
n 1
32n
3
12
a.
b. 12
9
c.
d. 9
e. diverges
correct choice
2
32
3
Solution: a
4
22
3
r
4
3
1
diverges
n
13.
n 1
3 2n
3
a. 3
correct choice
b. 6
c. 9
d. 12
e. diverges
32
3
Solution: a
n
n 1
3 2n
3
2
a
1
2
r
1
2
3
r
2
3
1
6
6
n2
ln
14.
n2
n 3
1
a. ln 3
b.
c.
d.
e.
4
ln 3
2
ln 3
ln 1
4
ln 1
2
correct choice
n2
Solution: ln
n
2
n
ln
1
n
n
1
n
n
ln
1
n
ln n n 1
1
k
Sk
ln
n 3
ln 3
2
S
lim S k
k
n
n
1
ln k
ln n n 1
ln 3
2
ln k
ln 3
2
ln 4
3
ln 4
3
ln 5
3
ln
k
k
1
ln k
1
k
1
k
lim ln 3
k
2
1
k
Work Out: (Points indicated. Part credit possible. Show all work.)
15.
15 points
Solution:
3x
x4
3 2
81
3x 3
x 9 x 3
3x
3:
3
18
Plug in x
0:
Plug in x
3:
0
3x
x4
3 2
81
x2
So
Ax
x2
2
Clear the denominator:
Plug in x
3 2
dx. Simplify to a single ln.
81
3x
x4
Compute
9
Ax
Ax
B3
A3
C9
B 6
x
B x
B 0
C
B
9
x
3B
C9
C
9
9
C x2
3
C 18
3
9
1
B
0
18A
18
A
1
C
ln
1
9
x
3
So the integral is
3x
x4
3 2
dx
81
x
x2
1
9
x
3
dx
1 ln|x 2
2
9|
ln|x
3|
x2 9
x 3
C
7
16.
15 points
4/3
Compute
1
1
dx. Simplify to a single ln.
25x 2 16
Solution: 5x 4 sec because from the square root x 4/5.
4 sec
4 sec tan d
So
x
dx
sec 2
1 tan 2
5
5
4/3
4/3
1
1
4 sec tan d
1 4/3 sec d
dx
5 x1
1
x 1
25x 2 16
16 sec 2
16 5
1 ln|sec
5
tan |
Draw a triangle with sec
1
1
25x 2
x 1
5x : Hyp
4
5x, Adj
4, Opp
25x 2
16 .
25x 2 16
4
Then tan
4/3
4/3
dx
16
1
5
ln 5
4
1 ln 5
5
3
25x 2 16
4
1 ln 5x
5
4
4
3
4
3
2
25 4
3
16
4
ln 2
4/3
x 1
ln 5
4
3
4
1 ln 3
5
2
8
17.
18 points
Consider the sequence defined recurrsively by
a1
a.
6 pts
i.e. a n
1
a1
Solution: We compute
ii. Assume a k
Therefore, a n
4
a1
5
4
ak
5
ak
1
4
2
5
ak
2
1
ak 1
4
an
1
ak
1
3
2
a1
0
0:
since a k
1
and a k are positive.
since multiplying by a negative reverses an inequality.
4
a k since adding 5 preserves the direction of the inequality.
and we already know a k 1 0
Use mathematical induction to show a n is bounded above by 8.
i. To get started, show a 1
Solution: a 1
ii. Assume a k
3
8 and prove a k
but this says
1
1
ak
8
4
ak
5 a4k
Therefore, a n
8 and a 2
8 and a 2
2
Solution: a k
6 pts
5
0 for all n.
an
1
a2
ak
1
4
ak 1
5 a 4k 1
but this says
ak 2
c.
1
0:
0 and prove a k
ak
1
Solution: a k
6 pts
an
Use mathematical induction to show a n is positive and increasing for all n,
a n 0.
i. To get started, show a 2
b.
and
2
4
8
5
ak
1
1
2
8:
8
8:
1 since a k is positive.
8
since multiplying by a negative reverses an inequality.
1
4. 5 since adding 5 preserves the direction of the inequality.
2
4. 5 8.
8 for all n.
What do you conclude about lim
an?
n
Find lim
an.
n
Solution: Since a n is increasing and bounded above, we conclude lim
a n exists.
n
Let lim
an
n
L. Then we also have lim
an
n
We compute the limit of a n
lim
an
n
1
1
4
lim
an
n
5
L
1 L
4
Since a n starts at a 1
0
5
4
an :
L
5
1
L.
4
L
L
L2
5L
4
0
1, 4
2 and increases, and the limit must exist, we must have L
4.
9