Name
Section
MATH 152 Honors
FINAL EXAM
Sections 201-202
Solutions
Spring 2014
1-14
/56
15
/22
16
/20
17
/4
18
/4
P. Yasskin
Multiple Choice: (14 problems, 4 points each)
Total
1.
a.
b.
c.
d.
e.
x2
above the interval
1
0, 1 .
1
4
CORRECT
4
1
2
1
2
1
Solution: A
1
x
0
2.
1
Find the area under the curve y
/106
2
1
dx
arctan x
1
arctan 1
0
arctan 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
4
1
2 rh dx
r
x
0
1
0
is revolved about the
0, 1
CORRECT
Solution: A
2 x
4
1
x
2
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
1
3.
A plate with constant density
interval
b.
c.
above the
1
CORRECT
2 ln 2
2
ln 2
d.
e.
2
ln 2
1
Solution: M
A
M1
4
ln 2 4
2
M1
M
x
4.
x2
0, 1 . Find the x-coordinate of its center of mass.
ln 2
2
2 ln 2
a.
1
has the shape of the region below y
/4
Compute
0
0
x
x
2
1
dx
2
ln x 2
1
1
2
0
ln 2
2 ln 2
sec 2 d .
tan 2
a.
b.
c.
d.
1
1
2
1
CORRECT
e.
/4
Solution:
0
OR:
u tan
du
5.
sec
/4
2
x2
1
a.
b.
c.
d.
e.
1
2
1
2
1
2
1
2
1
2
0
arcsin x
x 1 x2
3
x 1 x2
arcsin x
x 1
arcsin x
x 1
2
x 1
2
arcsin x
arcsin x
cot
du
u2
3/2
/4
0
1
u
HINT: sin 2
dx.
x2
csc 2 d
sec 2 d
tan 2
0
d
Compute
/4
sec 2 d
tan 2
1
tan
cot
lim cot
4
/4
0
0
1
tan /4
lim
0
1
tan
2 sin cos
C
C
x2
C
x2
C
x2
C
CORRECT
Solution:
x
dx
x2
sin
cos d
1
x
1
2
2
sin 2
1 sin 2
dx
sin 2
2
C
cos d
1
2
1
sin 2 d
sin cos
C
cos 2
2
1 arcsin x
2
d
x 1
x2
C
2
6.
ln x 2 dx.
Compute
2x ln x
1 x2 C
2
1 x2 C
2
2x C
2
2x ln x
2x
C
2
2x ln x
4x
C
a.
x ln 2 x
x 2 ln x
b.
x ln 2 x
x 2 ln x
c.
x ln 2 x
d.
e.
x ln x
x ln x
Solution:
u
ln x 2
du
7.
dv
2 ln x 1x dx
ln x 2 .
HINT: u
v
CORRECT
dx
ln x 2 dx
x
x ln x
2
2 ln x dx
x ln x
Which of the following is the general partial fraction expansion of
a.
b.
c.
d.
e.
A
x 2 2
Ax B
x 2 2
A
x 2
A
x
2
A
x
2
Bx C
2
x2 9
Cx D
x2 9 2
Cx D
B
2
x2 9
x 2
Cx
B
x2 9
x 2 2
Bx C
D
x2 9
x 2 2
Ex F
x2 9 2
Dx
2
x 9 2
Ex F
x2 9 2
x
2
2 x ln x
4x 2 5
2 2 x2 9
x
2
C
?
CORRECT
Solution: For each linear factor to the p, you need a linear denominator for each power up to p.
For each quadratic factor to the p, you need a quadratic denominator for each power up to p.
The linear denominators get a constant on top. The quadratic denominators get a linear on top.
Cx D
4x 2 5
A
B
Ex F
So the correct expansion is:
x 2
x2 9
x 2 2
x 2 2 x2 9 2
x2 9 2
8.
The curve x y 3 between y
the resulting surface.
d.
10 3/2
27
48 10 3/2
7
27
336
e.
48 10 3/2
a.
b.
c.
Solution: A
1 is rotated about the y-axis. Find the area of
CORRECT
1
1
1
1
2 r ds
y 0
1
0 and y
2 y 3 9y 4
dx
dy
2 x
0
1 dy
9y 4
u
2
1
1 dy
18
u du
1
2u 3/2
18 3
1
10
1
27
3y 2
2
1 dy
0
36y 3 dy
du
0
10
2 y3
10 3/2
1 du
36
y 3 dy
1
3
9.
Solve the differential equation
b.
1
3
1
c.
3
d.
6
e.
9
a.
dy
dx
y
x
with y 1
CORRECT
dy
y
dy
dx
dx
x
y
x
Use the initial condition: ln 3 ln 1 C
C ln 3
Solve: y e ln x
Substitute back: ln y ln x ln 3
Solution: Separate:
10. Solve the differential equation x
d.
1
2
2
13
6
3
e.
6
a.
b.
c.
3. Then y 3
dy
dx
y
ln y
ln 3
x 2 with y 1
ln x
C
So: y 3
3x
9
2. Then y 2
CORRECT
1 dx
dy
1y x
x
Solution: Standard form:
Integrating
factor:
I
e
x
dx
1 y 1
1 dy
Multiply by I:
Rewrite: d 1x y
1
x dx
dx
x2
y
Integrate:
1 dx x C
Use the initial condition: 2
1 C
C
x
1
y
Substitute back: x
x 1
Solve: y x 2 x
So: y 2
6
ln x
e
1
x
1
2 of its previous height on each bounce.
3
Find the total length travelled during an infinite number of bounces.
11. A ball is dropped from 27 feet and bounces to
a.
54
b.
81
c.
108
d.
135
e.
162
CORRECT
Solution: L
27
2
n 1
27 2
3
n
27
2
3
54
1
2
3
27
54
2
3
2
27
108
135
4
12. Compute lim sin x
x cos x
x3
x 0
a.
b.
c.
1
3
1
6
0
CORRECT
d.
e.
Solution:
lim sin x
x 0
x cos x
x3
n 0
b.
c.
d.
e.
3
2
3 3
2
3
6
3
2
lim
x3
x 0
x3
6
x3
2
1
2
x3
x 0
1
6
1
3
CORRECT
0
1
Solution:
n 0
2n 1
n 0
lim
x2
2
x 1
1 n 2n 1
9 n 2n 1 !
13. Compute
a.
x3
6
x
2n
1 n
9 n 2n 1 !
n
1 !
x 2n
1
1
3
n 0
2n
So
sin x
n
1 !
2n 1
3
3 sin
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
3
c.
center:
2, 0, 3
radius: R
3
d.
center: 2, 0, 3
radius: R
9
e.
center:
radius: R
9
2, 0, 3
Solution: x 2 4x 4 y 2 z 2 6z
center: 2, 0, 3
radius: R 3
4x
3
y2
3 3
2
z2
6z
4
0
2
y2
z
CORRECT
9
9
0
x
2
3
2
9
5
Work Out (3 questions, Points indicated)
Show all you work.
15.
22 points
1 n 1n
.
2n 1
Consider the series S
n 1
a.
4 Determine whether the series is absolutely convergent, convergent but not
absolutely convergent or divergent.
Solution:
Method 1: Apply the Ratio Test
n 1
lim aann1
lim n n 1 2 n
n
n
2
1
2
So the series is absolutely convergent.
1
Method 2: The related absolute series is
n 1
For n
12,
2
n 1
n
3
1
So
2n
1
n3
1
1 converges (p-series with p
2
n
n 1
the comparison test.
Since
b.
1
7
n 1
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 |
d.
1
n2
1) the absolute series converges by
Compute S 7 , the 7 th partial sum for S. Do not simplify.
Solution: S 7
c.
2
n .
2n 1
n
2n 1
1 Real easy
|a 8 |
8
27
8
128
1
16
Find a power series S x
is the given series S, i.e S
nx n
Solution: S x
n 1
1
1
2
So S
n 1
1
2
4
8
|S
5
16
6
32
7
64
S 7 | when S 7 is used to approximate
by the alternating series bound theorem
centered at 0 whose value at x
1 n 1n
.
2n 1
n
n 1
1
2
n 1
n 1
1
2
1 n 1n
.
2n 1
6
#15 continued.
1 n 1n
.
2n 1
Recall S
n 1
e.
10 Find the interval of convergence of the power series S x
the radius and check the endpoints.
nx n
Solution: S x
1
from part (d). Give
Apply the ratio test:
n 1
lim aann1
n
Endpoint x
n 1 xn
nx n 1
lim
n
n
|x| nlim
1
1
|x|
n
radius:
R
1
n which diverges by the n th term
1: The series becomes
n 1
divergence test since nlim n
Endpoint x
0
1: The series becomes
divergence test since nlim
1
n 1
2 Find a function f x
part (d).
n 1
n which diverges by the n th term
n 1
n diverges.
So the interval of convergence is
f.
1
1, 1 .
whose Maclaurin series is the power series S x
from
Solution:
nx n
Sx
d
dx
1
n 1
g.
1
Use f x
n 1
S
1
2
1
x
x
1
x
1
x 1
x
2
1
1
x
2
fx
to find the sum of the series S.
nx n
Solution: Since S x
S
d
dx
xn
n 1
1
1
1
2
2
1
1
1
x
2
, we have
4
9
7
16.
20 points
Let f x
ln x .
a.
6 Find the Taylor series for f x
centered at x
4.
Solution:
fx
f4
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 4
1
4
4
1
4
42
2
4
43
3!
4
4
44
1
n
4
Tx
1
ln 4
T0
ln 3
3 is T x
n 1
3|
|x
3
1 n
3nn
1
1 n
3nn
1
1
x
4
4
n
n
1 !
1 n
3nn
ln 3
n 1
Solution: Apply the Ratio Test:
|x 3| n 1
an 1
3nn
lim
lim
n 1
an
n
n
3
n 1 |x 3| n
R 3.
n
n
Find the interval of convergence for the Taylor series centered at x
Give the radius and check the endpoints.
0:
n 1 !
x
4 n!
1 n
4nn
n 1
4
4
n
ln 4
n
x
n 1
n 1
n 1
4
n!
n 1
11 The Taylor series for f x centered at x
endpoint: x
n
f
ln 4
3
n
1
|x
3|
1
x
3 n.
3
3
radius:
1
n
ln 3
n 1
diverges (harmonic series)
endpoint: x
6:
T6
ln 3
n 1
3
n
1 n
n
ln 3
1
n 1
converges (alternating series test)
Interval of Convergence: 0, 6
c.
3 If the cubic Taylor polynomial centered at x 3 is used to approximate ln x
on the interval 2, 5 , 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 3. Since c is between 3 and x, while x
can be anything in 2, 5 , then c can be anything in 2, 5 .
6
6
3
is largest on 2, 5 when x 2. So we take M
|f 4 c |
8
c4
24
For x in 2, 5 , the largest value of |x 3| is |5 3| 2.
M |x 4| 4
3 |x 4| 4
3 24
1
So |R 3 x |
4!
8 4!
8 4!
4
8
17.
4 points
When a ball with mass, m, is dropped from a height, h, and falls under the
forces of gravity with acceleration, g, and air resistance with drag coefficient, k, the altitude,
y t , satisfies the differential equation,
m
d2y
dt 2
mg
k
dy
dt
The solution is
yt
m 2 g e kt
m
k2
h
m2 g
k2
m gt
k
Verify that this solution reduces to the standard freefall formula (no air resistance) by taking the
limit of the solution as k approaches 0. (DO NOT SOLVE ANY DIFFERENTIAL
EQUATIONS.)
Solution: We use the Maclaurin series e x 1 x 1 x 2
2
kt
2
2
m g m gt
lim y t
lim h m2 g e m
k 0
k 0
k
k
k2
2
1 kt 2 1 kt 3
lim h m2 g 1 kt
m
k 0
2 m
6 m
k
2
3
2
1 kt
lim h m2 g 1 kt
k 0
2 m
6 m
k
3
lim h 1 gt 2 1 g kt
h 1 gt 2
k 0
2
6 m
2
18.
y 2 for 0
4 points
The curve x
of the resulting surface.
dx
dy
Solution:
2
2 r ds
2 y
0
Let u
A
1
8
0
4y 2
1, du
9
2
1
2
with x
m2 g
k2
kt :
m
m gt
k
is revolved around the x-axis. Find the area
2y
2
A
y
1 x3
6
u du
dx
dy
8y dy
1 2u 3/2
4 3
2
2
1 dy
2 y 4y 2
1 dy
0
1 du
8
9
1
y dy
1 9 3/2
6
1
1 26
6
13
3
9