MATH 152, SPRING 2012
HONORS EXAM II - SOLUTIONS
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Section No:
PART I: Multiple Choice (4 pts each)
1. Compute
1
a. 3
4
b.
2
c.
4
d.
e. 0
1
dx .
x2
Correct Choice
Solution:
1
1
dx
x2
2. The improper integral
arctan x
1
e
1
2
3
4
4
dx
x ln x
a.
b.
c.
d.
e.
diverges to .
Correct Choice
.
diverges to
converges to 1.
converges to 1.
converges to 1e 1.
e
dx
dx
du
Solution: u ln x
x
x
ln x
1
3.
1
0
du
u
ln 1
ln 0
0
1
dx
x2 x 1
a. ln|x 1| 1x C
b. ln|x 2 x 1 | C
c. ln|x| 1x ln|x 1| C
d. ln|x| 1x ln|x 1| C
Correct Choice
1
e. ln|x 1| x C
C
1
A
B
Solution:
1 Ax x 1 B x 1 Cx 2
x
x 1
x2 x 1
x2
x 0:
B
1
x 1:
C 1
Coeff of x 2 : 0 A C
1
1
1
1
dx
dx
ln|x| 1x ln|x 1| C
x
x 1
x2 x 1
x2
A
1
1
3
4. By substituting x
x2 x2
3 tan , the integral
9 dx becomes
0
/4
a.
0
3
b.
27 tan 2 sec d
27 tan 2 sec 3 d
0
/4
c.
81 tan 3 sec 2 d
0
/4
d.
81 tan 2 sec 2 d
0
/4
e.
81 tan 2 sec 3 d
Correct Choice
0
Solution: x 2
/4
9 tan 2
9 tan 2
x2
/4
3 sec 2 d
3 sec
0
b.
c.
d.
e.
1
27
2
27
1
27
1
27
2
27
13 13
8
13 13
8
13 13
1
13 13
1
1
3 sec
3 sec 2 d
dx
81 tan 2 sec 3 d
0
9t 2
4
L
t2, y
t 3 , for 0
t
1.
Correct Choice
dx
dt
1
18
Solution: L
u
9
0
5. Find the length of the curve x
a.
9 tan 2
9
2
dy
dt
13
u du
4
2
1
dt
2t
2
3t 2
2
1
dt
1 2 u 3/2
18 3
9t 2 dt
t 4
0
0
13
4
1 13 13
27
8
6. Which of the following integrals gives the surface area obtained by rotating the curve y
for 0 x 1, about the y-axis?
a.
b.
c.
1
2 e
4x
1
2 x 1
0
e
4
2 y 1
1
d.
e.
16e
0
1
1
0
1
0
2
16y 2
16e
8x
8x
dx
dx
Correct Choice
1 dy
16y 2
1 dy
ln y
16y 2
8 y
1 dy
Solution: x-integral because y f x and 0 x 1.
1
1
dy 2
A
2 r 1
dx
2 x 1
4e 4x 2 dx
dx
0
0
2
r
x because y-axis.
1
2 x 1
0
16e
8x
dx
e
4x
,
dx
7. Which of the following statements is true regarding the improper integral
1
dx
ex
x
dx
ex
x
dx
ex
x
dx
x
x
1 e
a. The integral converges because
1
b. The integral diverges because
1
c. The integral diverges because
1
d. The integral converges because
1
1
1
dx
x
and
1
dx and
x
dx and
ex
dx and
x
1 e
dx
x
ex
?
x
converges.
dx diverges.
x
dx diverges.
x
1 e
dx converges.
x
1 e
1
Correct Choice
e. The integral converges to 0.
1
dx
x
1
ex
1
x
1
Solution:
ex
x
diverges and
1
0
x
1
and
dx
ex
ex
1
ex
x
converges
So (e) is wrong.
So (b) and (c) are wrong.
So (a) and (c) are wrong.
(d) is correct by the Comparison Test.
8. Find the integrating factor for the differential equation
1
x2 y
2xy
1
x2.
x2
1
Correct Choice
1 x2
c. I x
2 1 x2
2
d. I x
1 x2
e. The equation is not linear, so there is no integrating factor.
a. I x
b. I x
1
Solution: Standard form:
Ix
e
9. If y
P x dx
fx
e
ln 1 x 2
1
a.
b.
c.
d.
e.
Correct Choice
f x
dy
dx
2x y
1 x2
1 .
1 x2
1
Px
is a solution of the Initial Value Problem:
then f
1
2
3
4
5
dy
dx
x
y2
f 2
1
22
1
dy
dx
2x
x2
x
y 2 with f 1
2,
5
3
10. Solve the Initial Value Problem:
a.
b.
c.
d.
e.
1
2
3
4
5
11.
y
x
1
1
ln|y
1|
ln|x
1|
2
1
A0
1
with y 0
2. Then y 1
Correct Choice
Solution:
1
y
y
dy
dx
x
dy
dx
x 1
Ax 1
3
y 1
C
e x 1
2
y1
C
A
e C |x
1|
|y
1
y
1|
1
x
1
The plot at the right is the direction field
for which differential equation?
dy
a.
y x2
Correct Choice
dx
dy
b.
y x2
dx
dy
c.
x y2
dx
dy
d.
x y2
dx
dy
e.
y2 x
dx
Solution: The slope is zero when the right side is zero. Each right side is a parabola and the
plot has zero slope on the parabola y x 2 0.
PART II: WORK OUT (10 pts each)
Show all your work neatly and concisely and box your final answer. You will be graded not
merely on the final answer, but also on the quality and correctness of the work leading up to it.
x2
1 ln x, for 1 x 2,
12. Find the surface area obtained by rotating the curve y
4
2
about the y-axis.
Solution: x-integral because y f x and 1 x 2.
r x because y-axis.
2
A
2 r 1
1
2
2
2 x x
4
1
8
3
4
2
dy
dx
2
2
dx
x
2
1
2x
2 x x
2
1
1
2x
2 x 1
1
1
2
2
1 dx
4x 2
1
3
1
10
3
2
2
dx
2 x 1
1
2
dx
1
x2
1 dx
x2
4
1
2
x3
3
1
4x 2
2
x
1
13. Integrate
9x 2 dx.
16
Solution: 3x
4 sin
Draw a triangle or:
cos 2 d
2
8
3
sin 2
1
1
16 sin 2
4 cos
sin 2
2
C
9x 2
16
9x 2
16
3x 1
4
9x 2
16
3x 1
4
16
C
4x 2 1
dx.
x 1 x 2
14. Integrate
2
Solution:
4x 2
8 arcsin 3x
3
4
9x 2 dx
1
cos
sin cos
9x 2
16
16
3
3x
4
sin
sin 2
2
arcsin 3x
4
16
4 cos d
4 cos 4 cos d
3
9x 2 dx
16
3 dx
1
4x 2 1
x 1 x 2
2
Ax
B x
A
C
4
B
2A
C
4
A
B
2A
4x 2 1
dx
x 1 x 2
2
1 ln x 2
2
1
2
0
C
Ax
x2
B
1
x
C x2
1
A
C
2B
1
A
4A
1
x 2
x2 1
x
4
2 arctan x
3 ln|x
3
2
2|
dx
4x 2
2
C x2
5A
B
1
Ax
2A x
5
A
1
x2
x
x2
B x
C
C x2
1
2B
1
B
1
x
2
2
2
3
2
C
3
dx
C
5
15. When a dead body was found by the police its temperature was 85°F. After 2 hours its
temperature dropped to 75°F. How many hours before the body was found was its temperature
95° F, assuming the room temperature was a constant 65° F?
Solution:
d T T0
dt
kT
T0
T
T0
t
0:
85
65
Ae 0
t
2:
75
65
20e
2k
95
65
20e
ln 2 t/2
t
2
A
Ae
kt
T0
65,
T0
85,
T2
75
20
e
1
2
2k
e
2k
3
2
ln 2 t/2
ln 1
2
ln 2 t
2
1 ln 2
2
k
ln 3
2
ln 3/2
hours
ln 2
16. A bathtub starts out with 10 gallons of water containing 5 pounds of bath salts. Pure water is
entering the bathtub at 0. 2 gallons per minute and bath salts are poured in at 3 pound per
minute. The water is kept well mixed and drains out at 0. 2 gallons per minute. How long will it
take until there are 10 pounds of bath salts in the bathtub?
dS lb
dt min
Solution:
3
Standard Linear form:
e .02t dS
dt
e .02t S
S0
e05
10
6
. 02e .02t S
150e .02t
5:
150e 0
150
C
0. 2 gal S lb
min 10 gal
lb
min
dS
dt
. 02S
3e .02t
3
Pt
d e .02t S
dt
S0
. 02
I
3e .02t
5
e
P t dt
e .02t S
e .02t
3e .02t dt
C
C
145e .02t
145
e .02t S
145e .02t
140
150e .02t
. 02t
145
ln 140
145
S
150
t
145e .02t
50 ln 29
28
.02t
3e
. 02
C
17. A recursive sequence is defined by a 1
2, a n
4
an .
5
1
a. Use Mathematical Induction to prove:
The sequence is increasing and bounded between 0 and 5, i.e.
Pn :
an 1 an
and
0 a n 5.
i. Initialization Step: Find some terms and verify P 1
Solution: a 1
2, a 2
P1 :
a2
3
P2 :
a3
11
3
ii. Write out P k
4
2
5
a1
2
3, a 3
and
a2
3
and P k
and
4
3
5
0
and P 2
a1
0
are true:
11 ,
3
2
5.
a2
3
True
5.
True
1 :
Solution:
Pk :
Pk
ak
1 :
ak
1
ak
and
ak
2
1
0
ak
5.
and
0
ak
iii. Induction Step: Assuming P k
5.
1
prove P k
1 :
Solution:
ak
ak
1
0
implies
1
ak 1
5 a4k
implies
5
4
ak 1
1
ak
4
ak 1
implies
implies
ak
ak
2
4
ak
1
So the sequence is increasing.
0
ak
5
implies
1
ak
implies
5
4
ak
1
5
4
5
5
4
ak
implies
implies
ak
Also since the sequence is increasing from a 1
1
4
5
5
2, we have 0
ak 1.
So the sequence is bounded between 0 and 5.
b. What Theorem guarantees the sequence converges? Find the limit.
Solution: The Bounded Monotonic Sequence Theorem says the sequence converges.
Let L
So L
lim a n . Then nlim a n
n
5
4
L
L2
5L
4
an
lim 5
1
n
4
0
L
4 L
5
1
4
lim
an
n
0
Limit must be 1 or 4.
Since a n is increasing from 2, the limit must be 4.
7