Name
Section
MATH 152
FINAL EXAM
Spring 2014
Solutions
P. Yasskin
Sections 549-551
Multiple Choice: (14 problems, 4 points each)
1-14
/56
15
/15
16
/20
17
/10
18
/5
Total
1.
Find the area between the curves y
a.
b.
c.
d.
e.
4
3
8
3
16
3
32
3
64
3
A
4x
x2
4x
2x 2 dx
2x 2
0
2.
2x
3
2x
3
x
0, 2
2
8
0
The line is above the parabola.
16
8
3
3
e x on the interval
Find the average value of the function f x
a.
b.
c.
d.
e.
e 1e
1 e
2
e 1e
1 e
2
1
4x.
CORRECT
Solution: 2x 2
2
2x 2 and y
/106
1, 1 .
1
e
1
e
Solution: f ave
CORRECT
1
1
1
1
x
e dx
1
1 ex
2
1
e
e
1
1
2
1
3.
/2
Compute
a.
b.
c.
d.
e.
6
2
5
2
5
2
3
6
/2
CORRECT
du
b.
c.
d.
e.
/2
cos d
1 x2
2
1 x2
2
1 x2
2
1 x2
2
1 x2
2
1x
2
1x
2
1x
2
1x
2
1x
2
1
4
1
4
1
4
1
2
1
2
e 2x
C
e 2x
C
e 2x
C
e 2x
C
e 2x
C
Solution: Use parts twice:
1 x 2 e 2x
2
x 2 e 2x dx
b.
c.
d.
e.
x2
arcsec x
x2
xe 2x dx
x2 1
dx.
x
Compute
a.
1
1
C
arcsec x
C
x 2 1 arcsec x C
1 x 2 1 3/2 x 2 1
3
1 x 2 1 3/2 x 2 1
3
Solution:
u 4 du
/2
sin 5
5
u5
5
/2
2
5
CORRECT
x2
u
du
5.
sin 4 cos d
x 2 e 2x dx.
Compute
a.
/2
sin
u
Solution:
4.
sin 4 cos d .
x
dx
e 2x dx
1 e 2x
v
2
1 xe 2x 1 e 2x dx
2
2
dv
2x dx
1 x 2 e 2x
2
HINT: tan 2
sec 2
u
x
du
e 2x dx
1 e 2x
v
2
1 xe 2x 1 e 2x
2
4
dv
dx
1 x 2 e 2x
2
C
1
CORRECT
1/2
C
1/2
C
sec
sec tan d
sec 2
sec
x2 1
dx
x
sec 2
1d
tan
1
tan 2 d
sec tan d
C
x2
1
arcsec x
C
2
6.
c.
1
4
1
2
2
d.
8
e.
16
a.
b.
Set x
0:
4x 5 2x 2
x2 4
2
c.
d.
e.
Bx
C
C
x3
Fx G , find C.
2
x2 4
Dx E
x2 4
8
Ax 2
2
Fx G x 3
2
x2 4
Dx E x 3
x2 4
1
2
1
4
1
8
0
1
4
undefined
Solution:
8.
B
x2
1 dx.
x3
2
b.
8
16
C
Compute
a.
A
x
CORRECT
Solution: Multiply by x 3 :
7.
4x 5 2x 2 8
2
x3 x2 4
In the partial fraction expansion,
CORRECT
2
1 dx
3
2 x
1
lim
b 0
2b 2
0
2
2
1 dx
x3
1
8
0
1 dx
x3
1
8
x
2
0
2
2
0
1
above the interval
x2 1
y-axis. Find the volume of the resulting solid.
The region under the curve y
a.
ln 2
b.
ln 2
c.
2 ln 2
d.
4 ln 2
e.
4 ln 2
0
0
1
2x 2
2
undefined
2
0
is revolved about the
0, 1
4
1
2 rh dx
r
x
0
1
2 x
2
1
2x 2
CORRECT
Solution: A
0
2
1
2a 2
lim
a
2
x
1
x2
1
dx
1 du
u
ln u
h
1
x2
ln x 2
u
1
x2
1
du
2x dx
1
1
ln 2
0
3
9.
2t 4 ,
Find the arclength of the curve x
a.
b.
c.
d.
e.
1
27
1
3
61
12
61
27
61
54
t 6 between t
y
0 and t
1.
CORRECT
1
Solution: L
0
1
64t 6
2
dx
dt
2
dy
dt
1
36t 10 dt
0
1
8t 3
dt
2
6t 5
2
dt
0
2t 3 16
9t 4 dt
u
9t 4
16
du
36
36t 3 dt
du
0
1
18
25
25
1 2u 3/2
18 3
u du
16
1 25 3/2
27
16
16 3/2
1 125
27
64
t 3 dt
61
27
2t 4 , y t 6 between t 0 and t 1 is rotated about the x-axis.
Which integral gives the area of the resulting surface?
10. The curve x
1
a.
0
1
b.
0
1
c.
0
1
d.
0
1
e.
4 t 9 16
9t 4 dt
8 t 7 16
9t 4 dt
4 t 7 16
9t 4 dt
2 t 7 16
9t 4 dt
4 t 3 16
9t 4 dt
CORRECT
0
1
Solution: A
1
2 r ds
2 y
t 0
1
2 t 6 64t 6
0
36t 10 dt
0
n 0
b.
c.
d.
e.
n 0
S
dy
dt
2
1
dt
2 t6
8t 3
2
6t 5
2
dt
0
9t 4 dt
1 n3n
2 2n
converges to 4
converges to 4
3
converges to 4
7
converges to 2
5
diverges
Solution: S
So
4 t 9 16
2
0
11. The series S
a.
1
dx
dt
CORRECT
3
4
n
1
1
is geometric.
4
3/4
4
3
a
1
r
3
4
and |r|
1
4
7
4
x 3 6 sin x
x5
12. Compute lim 6x
x 0
a.
b.
c.
d.
e.
6
5!
6
6
2
3!
1
20
CORRECT
Solution:
lim 6x
x
3
6 sin x
1
n 0
b.
c.
0
d.
1
e.
6 x
x 0
13. Compute
1
2
1
x3
lim
x5
x 0
a.
6x
n
x3
6
x5
x5
120
6
lim
x 0
x5
120
x5
1
20
2n
9 n 2n !
CORRECT
1
2
Solution:
n 0
1 n 2n
x
2n !
cos x
1
So
n 0
14. Find the center and radius of the sphere x 2
a.
center:
2, 0, 3
radius: R
2
b.
center: 2, 0, 3
radius: R
9
c.
center:
2, 0, 3
radius: R
9
d.
center: 2, 0, 3
radius: R
3
e.
center:
radius: R
3
2, 0, 3
Solution: x 2 4x
center:
2, 0, 3
4
y 2 z 2 6z 9
radius: R 3
n
2n
n
9 2n !
4x
y2
n 0
z2
1 n
2n !
2n
3
6z
4
0
2
y2
z
cos
3
1
2
CORRECT
9
0
x
2
3
2
9
5
Work Out (4 questions, Points indicated)
Show all you work.
15.
15 points
1 n 1n
.
2n 1
Consider the series S
n 1
a.
3
Show the series is convergent.
n
decreases for n
2n 1
So it converges by the alternating series test.
Solution: The series alternates,
b.
6 Extra Credit
Show the series is absolutely convergent.
HINT: You may use these formulas without proof, if needed:
bx
2 n 1 n 3 for n 12
b x dx
C
x b x dx
ln b
Solution: The related absolute series is
n 1
Method 1: Apply the Ratio Test
n 1
lim aann1
lim n n 1 2 n
n
n
2
1
2
C
1 .
n2
Method 3: Apply the integral test with the integral
n
1
1
2
n 1
dx.
Compute S 7 , the 7 th partial sum for S. Do not simplify.
7
Solution: S 7
n 1
d.
bx
ln 2 b
0.
So the series is absolutely convergent.
1
n 1
3
xb x
ln b
n
2n 1
n .
2n 1
Method 2: Apply the comparison test with
c.
2 and nlim
1 n 1n
2n 1
1
2
2
3
4
3 Find a bound on the remainder |R 7 |
S. Name the theorem you used.
Solution: |R 7 |
|a 8 |
8
27
8
128
4
8
|S
5
16
6
32
7
64
S 7 | when S 7 is used to approximate
1 by the alternating series bound theorem.
16
6
16.
20 points
Let f x
ln x .
a.
6 Find the Taylor series for f x
centered at x
3.
Solution:
fx
f3
f
f
f
f
f
f
b.
ln x
1
x
x
1
x
x2
2
x
x3
3!
4
x
x4
1
n
x
f
f
f
n 1
x
n
1 !
n
f
ln 3
1
3
3
1
3
32
2
3
33
3!
4
3
34
1
n
3
Tx
1
ln 3
n 1 !
x
3 n!
1 n
3nn
n 1
3
T0
2 ln 2
n 1
3
1
x
3
n
n
1 !
4 is T x
1 n
4nn
2 ln 2
n 1
0:
n
n
Find the interval of convergence for the Taylor series centered at x
radius and check the endpoints.
endpoint: x
3
n
ln 3
n
x
n 1
n 1
n 1
3
n!
n 1
11 The Taylor series for f x centered at x
Solution: Apply the Ratio Test:
|x 4| n 1
4nn
lim aann1
lim n 1
n
n
4
n 1 |x 4| n
n
f
ln 3
4|
|x
1
4
1 n
4nn
1
1 n
4nn
1
4
n
|x
4|
1
x
4 n.
4. Give the
radius R
4
4.
1
n
2 ln 2
n 1
diverges (harmonic series)
endpoint: x
8:
T8
2 ln 2
n 1
converges (alternating series test)
c.
4
n
1 n
n
2 ln 2
1
n 1
Interval of Convergence:
0, 8
3 If the cubic Taylor polynomial centered at x 4 is used to approximate ln x
on the interval 3, 6 , use the Taylor’s Inequality to bound the error.
Taylor’s Inequality:
Let T n x be the n th -degree Taylor polynomial for f x centered at x a
and let R n x
f x T n x be the remainder. Then
M |x a| n 1
|R n x |
n 1 !
provided M |f n 1 c | for all c between a and x.
Solution: Here n 3 and a 4. Since c is between 4 and x, while x
can be anything in 3, 6 , then c can be anything in 3, 6 .
6
6
2
is largest on 3, 6 when x 3. So we take M
|f 4 c |
27
c4
34
For x in 3, 6 , the largest value of |x 4| is |6 4| 2.
M |x 4| 4
2 |x 4| 4
2 24
4
So |R 3 x |
4!
27 4!
27 4!
81
7
10 points
17.
y
A conical tank with vertex down, radius 4,
8
6
and height 8, is filled with water.
Find the work done to pump the water out the top of the tank.
4
Take the density of water to be
2
and the acceleration of gravity to be g.
-4
-2
Solution: Put the origin of the y-axis at the vertex of the
y
cone at the bottom of the tank.
The slice at height y is a circle of radius r where yr
y
, the area is A y
r2
y2,
So r
2
4
the volume is dV A y dy
y 2 dy
4
and the force is dF
g dV
g y 2 dy.
4
This slice needs to be lifted a distance D 8 y.
0
2
4
0
2
4
8
6
4.
8
4
2
-4
-2
8y 3
4
3
256
g
3
y4
4
So the work is
8
W
8
D dF
8
0
g
0
g
18.
y
4
84
3
84
4
4
8
y 2 dy
g
4
84
g
4
1
3
8y 2
y 3 dy
0
1
4
g
5 points
The region between the curves y
Find the volume of the resulting solid.
84
4 12
2x 2 and y
Solution: x integral
2x 2 4x
x 2 2x
x
Rotating about the x-axis gives thin washers.
2
V
R2
r 2 dx
0
2
4x
2
2x 2
2
2
dx
0
3
16 2
3
5
42
5
0
2
7
1
3
1
5
g
8
0
4x is rotated about the x-axis.
0, 2
The line is above the parabola.
16x 2
4x 4 dx
3
16 x
3
5
4x
5
2
0
28
15
8