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MATH 152
Final Exam
Sections 513,514
Spring 2002
Solutions
1-14
/70
15
/10
16
/10
17
/10
P. Yasskin
Multiple Choice: (5 points each)
∑
∞
1.
n=2
3n =
4 n−1
a. 4
b. 9
CORRECT
9
c.
7
d. 3
e. Diverges
2
a = 32−1 = 9
4
4
r= 3
4
∑
∞
|r| < 1
n=2
2. Find the angle between the vectors ⃗
u
a. 0°
b. 30°
c. 45°
d. 60°
e. 90°
CORRECT
⃗u ⋅ ⃗v = 1 − 2 + 1 = 0
9
4
3n =
= 9 =9
4−3
3
4 n−1
1−
4
= ⟨1, 1, −1 ⟩ and ⃗v = ⟨1, −2, −1 ⟩.
u ⋅ ⃗v = 0
cos θ = ⃗
⃗ ||v⃗|
|u
θ = 90°
y = 4 sin 2 t for 0 ≤ t ≤ π .
4
HINT: When you differentiate, remember the chain rule.
a. 5
CORRECT
2
b. 5
c. 6
d. 12
e. 10π
3. Find the arc length of the curve x
= 3 cos 2 t
dy
= 8 sin t cos t
dt
π/4
dy 2
dt =
36 cos 2 t sin 2 t + 64 sin 2 t cos 2 t dt =
dt
0
dx = −6 cos t sin t
dt
π/4
dx 2 +
L=
dt
0
∫
= 5sin 2 t|π0 /4 = 5
∫
∫
π/4
10 sin t cos t dt
0
2
1
4. Find an integrating factor for the differential equation
a. e −cosx
b. e −sinx
dy
= 2xy + sin x.
dx
c. e cosx
2
d. e x
2
e. e −x
CORRECT
dy
− 2xy = sin x  P = −2x
dx
∫
∫
∫
∫
∫

5. The area between the curves y
a.
b.
c.
d.
e.
6
0
6
0
2
0
2
0
3
0
2
+ 2  − 2x + 5  dx
2x
+ 5  − x 2 + 2  dx
x
x
2
∫
+∫
+∫
+ 2  − 2x + 5  dx +
2x
+ 5  − x 2 + 2  dx
2x
+ 5  − x 2 + 2  dx
x 2 + 2 = 2x + 5

2
6
3
∫ P dx
2
= e −x
= x 2 + 2 and y = 2x + 5 for 0 ≤ x ≤ 6 is given by the integral:
6
2
6
I=e
2x
+ 5  − x 2 + 2  dx
x
2
+ 2  − 2x + 5  dx
x
2
+ 2  − 2x + 5  dx
CORRECT
y
x 2 − 2x − 3 = 0
+ 1 x − 3  = 0  x = −1, 3
For 0 ≤ x ≤ 3,
2x + 5 ≥ x 2 + 2
For 3 ≤ x ≤ 6,
x 2 + 2 ≥ 2x + 5
x

A=
∫
40
3
0
2x
+ 5  − x 2 + 2  dx +
∫
6
3
x
2
20
+ 2  − 2x + 5  dx
0
0
1
2
3
4
5
6
x
6. If ⃗
u points North and ⃗v points South-East, then ⃗
u × ⃗v points
a. Up
b. Down
CORRECT
c. East-North-East
d. West-South-West
e. North-West
Hold your fingers North with your palm facing East. Rotate your fingers East and then South, until
they point South-East. While your fingers rotate, your thumb points Down.
7.
∫
e
1
9x 2 ln x dx =
a.
b.
c.
d.
e.
2e 3
2e 3
2e 3
3e 3
3e 3
+1
−2
CORRECT
− 3e 2
− 3e 2 + 3
u = ln x dv = 9x 2 dx
du = 1x dx
v = 3x 3
∫
e
1
∫
3x 2 ln x dx = 3x 3 ln x − 3x 2 dx = 3x 3 ln x − x 3
e
1
= 2e 3 + 1
2
= 0, x = cos y, y = 0, y = π
8. The region bounded by x
4
Which integral gives the volume of the solid of revolution?
a.
b.
c.
d.
e.
∫
∫
∫
∫
∫
π/4
0
2π cos 2 y dy
2 /2
0
π/4
0
0
0
2πx arccos x dx
2πy cos y dy
2 /2
π/4
is rotated about the x-axis.
CORRECT
πcos 2 x − x 2  dx
2πy 2 dy
y
1
cylinders
y-integral
r=y
V=
∫
π/4
0
h = x = cos y
2πrh dy =
∫
π/4
0
2πy cos y dy
0
0
1
∫
x
9. With error |E|
< 0. 0001, evaluate
0. 1 
a. 0. 1 −
3
0.1
0
sinx 2  dx. HINT: Use a Maclaurin series.
6
b. 0. 1
c.
0. 1 
3
CORRECT
3
d.
0. 1 
2
e.
0. 1 
2
−
0. 1 
6
6
∫
x
6
3
7
sin x 2 = x 2 − x + ⋯
sinx 2  dx = x − x + ⋯
6
3
42
0
3
0.1
0.
1


sinx 2  dx ≈
.
The error is less than the next term |E| <
3
0
3
sin x = x − x + ⋯
6
∫
Using 1 term:
10. Using a trig substitution,
a. 1
b.
c.
d.
e.
3
1
3
3
4
1
4
1
4
∫
∫
∫
∫
∫
∫
dx
9 + 16x 2
0. 1 
42
7
< 10 −7
becomes
cos θ dθ
tan θ dθ
sec 2 θ dθ
sec θ dθ
CORRECT
sin θ dθ
4x = 3 tan θ
4dx = 3 sec θ dθ
2
∫
dx
=
9 + 16x 2
∫
3 sec 2 θ dθ
4
= 3
4
2
9 + 9 tan θ
∫
sec 2 θ dθ = 1
4
3 sec θ
∫
sec θ dθ
3
11. The partial fraction decomposition of
1 + 1
x
x−1
1 − 1
x
x−1
1 − 1
x
x−1
1 + 1
x
x+1
1 − 1
x
x+1
a.
b.
c.
d.
e.
1
x2 − x
is
CORRECT
1
1
=
= Ax + B
x−1
xx − 1 
x2 − x

A + B = 0,
−A = 1
12. A 4 cm bar has density ρ


1 = Ax − 1  + Bx = A + B x − A
A = −1
B=1

1
= −x1 + 1
x−1
x2 − x
gm
= 1 + 5x 3 cm where x is measured from one end.
Find its center of mass.
a. x̄ = 324
b. x̄ = 1032
c. x̄ = 27
86
d. x̄ = 86
CORRECT
27
e. x̄ = 321
4
∫
=∫
M=
M1
4
0
1
4
0
4
+ 5x 3  dx = x + 5x
4
x1 + 5x 3  dx =
13. lim sin x − 3x cos x
x→0
x
1
a.
6
b. 1
3
c. 1
2
d. 2
3
4
0
x2 + x5
2
= 4 + 320 = 324
4
0
= 8 + 1024 = 1032
x̄ = M 1 = 1032 = 86 ≈ 3. 2
M
324
27
=
CORRECT
e. ∞
3
3
2
sin x = x − x + ⋯ = x − x + ⋯
cos x = 1 − x + ⋯
x cos x = x −
3!
6
2
3
3
3
3
x− x +⋯ − x− x +⋯
−x + x +⋯
6
2
2
lim sin x − 3x cos x = lim
= lim 6
=
x→0
x→0
x→0
x
x3
x3
x3 + ⋯
2
1 − 1 = 1
2
6
3
4
dy
= xy
dx
14. If yx  satisfies the differential equation
a.
b.
c.
d.
e.
∫
1
2
3
4
5
and the initial condition y0  = 3, find y4 .
CORRECT
y dy =
∫
x dx

x=0 & y=3
2
y2
= x +C
2
2
9 = 0 +C

2
2


y = ± x 2 + 2C
C= 9

y=
2
x2 + 9
y4  = 5

Work Out Problems: (10 points each)
15.
Find the work done to pump the water
out the top of a hemispherical bowl of
radius 5 cm if it is filled to the top.
gm
The density of water is ρ = 1
.
cm 3
The acceleration of gravity is g = 980 cm2 .
sec
Measure y down from the top. The slice at depth y must be lifted a distance D = y, and has
radius r which satisfies r 2 + y 2 = 25. So this slice has volume dV = πr 2 dy = π25 − y 2  dy. The
force to lift this slice is its weight dF = ρg dV = ρgπ25 − y 2  dy. So the work is
∫
∫
∫
y4
y2
D dF =
yρgπ25 − y  dy = ρgπ 25y − y  dy = ρgπ 25
−
W=
2
4
0
0
0
4
ρgπ5
= ρgπ5 4 1 − 1 =
= 980π625
2
4
4
4
5
5
2
5
5
3
0
5
∑
∞
16. Find the radius and interval of convergence of
x
n=2
Ratio Test:
a n+1
lim
an
n→∞
an =
x
− 4n
a n+1 =
3 ln n
− 4  n+1 3 ln n
= lim
n→∞ 3 lnn + 1  x − 4  n
x
− 4n
3 ln n
. Be sure to check the endpoints.
− 4  n+1
3 lnn + 1 
x
ln n
= |x − 4|lim
n→∞ lnn + 1 
= |x − 4|lim
n→∞
Radius of convergence is R = 1.
Converges if 3 < x < 5.
∑
∑
∞
At x = 3 the series is

At x = 5 the series is
n=2
= |x − 4| < 1
n+1
which converges by the Alternating Series Test.
3 ln n
n=2
∞
−1  n
1
n
1
1 . We apply the Comparison Test with
3 ln n
divergent harmonic series. Since n > ln n we have
1 < 1
3n
3 ln n
∑
∞
n=2
1
3n
and hence
which is a
∑
n=2
diverges.
So the interval of convergence is
∑
∞
17. Determine if the series

n=0
also
converges absolutely, converges conditionally or diverges.
If it diverges, does it diverge to +∞,
If it converges, find the sum.
1
3 ln n
3, 5 .
−1  n 2 n
n!
∞
− ∞ or neither?
Show all work and name any tests you use.
Circle One:
Converges Absolutely
Fill in the Blank:
Converges Conditionally
∑
∑
∞
e −2
Converges to
Or Circle One:
Diverges to
Diverges
+∞
−∞
Neither
∑
∞ n
∞ n
2 n . Since e x =
x , we conclude
2 = e2.
n!
n!
n!
n=0
n=0
n=0
n
n+1
a n+1 = 2
Alternatively, apply the ratio test:
an = 2
n!
n + 1 !
n+1
a
n+1
2
n!
2
lim
= lim
= lim
=0<1
n→∞ a n
n→∞ n + 1 ! 2 n
n→∞ n + 1 

Absolute series converges and original series converges absolutely.
∞ n
∞
n n
 −1  2
x
x
, we conclude
= e −2 .
Similarly, since e =
n!
n!
n=0
n=0

Original series converges to e −2 .
The related absolute series is
∑
∑
6