Name
Sec
1-11
/55
Fall 2008
12
/12
P. Yasskin
13
/22
14
/16
Total
/105
ID
MATH 152 Honors
Final Exam
Sections 201,202
Solutions
Multiple Choice: (5 points each)
1. Find the average value of f(x) = sin x on the interval 0 ≤ x ≤ π.
1
a. π
2
b. π
correct choice
3
c. π
d. 1
2
e. 1
3
1
f ave = π
2. Compute
a.
b.
c.
d.
e.
Let
4
15
4
75
3
20
3
100
3
500
π
∫ 0 sin x dx =
3
∫0
1 (− −1 − −1) = 2
= π
π
correct choice
x = 5 sin θ.
sin θ = x
5
0
1
dx.
(25 − x 2 ) 3/2
1
dx =
(25 − x 2 ) 3/2
∫
π
1
π (− cos x)
dx = 5 cos θ dθ.
Then
∫
5 cos θ
dθ = 1
2 3/2
25
(25 − 25 sin θ)
cos θ =
1−
x
5
2
25 − x 2
5
=
∫
cos θ dθ = 1
25
cos 3 θ
tan θ =
∫ sec 2 θ dθ =
1 tan θ + C
25
x
25 − x 2
3
3
∫0
1
dx = 1
3/2
2
25
(25 − x )
x
25 − x 2
=
0
3
= 3
25 ⋅ 4
100
1
3. The region below y = x 2 for 0 ≤ x ≤ 2 is rotated about the y-axis.
Find the volume of the solid swept out.
correct choice
a. 8π
b. 4π
c. 32π
5
32π
d.
3
16π
e.
3
V=
cylinders
x-integral
∫
4. Compute
x + 2 dx.
x3 + x
2
2
4
∫ 0 2πrh dx = ∫ 0 2πxx 2 dx = 2π x4
2
0
= 8π
HINT: Equate coefficients.
a. ln
x2
+ arctan(x) + C
(x 2 + 1) 2
b. ln
x2
− arctan(x) + C
(x 2 + 1) 2
x
+ arctan(x) + C
x +1
d. ln 2 x
− arctan(x) + C
x +1
2
e. ln 2x
+ arctan(x) + C
correct choice
x +1
c. ln
2
x + 2 = A + Bx + C
x
x(x 2 + 1)
x2 + 1
x + 2 = A(x 2 + 1) + (Bx + C)x = (A + B)x 2 + Cx + A
A=2
B = −2
∫
C=1
x + 2 dx =
x3 + x
A+B = 0
2 + −2x + 1 dx =
x
x2 + 1
2 − 2x + 1 dx
x
x2 + 1
x2 + 1
2
= 2 ln|x| − ln(x 2 + 1) + arctan(x) + C = ln 2x
+ arctan(x) + C
x +1
∫
∫
5. Which of the following differential equations is NOT separable?
a.
b.
c.
d.
e.
dy
dx
dy
dx
dy
dx
dy
dx
dy
dx
= x + xy
= xy + xy 2 + x 2 y + x 2 y 2
1 + 1 + 1
= xy
x
y
correct choice
= xy + x + 1y + 1
y
= x −y
x + xy − x(1 + y)
xy + xy 2 + x 2 y + x 2 y 2 = (x + x 2 )(y + y 2 )
y
x + x + 1 + 1 = (x + 1) 1 + 1
1
y
y
y
x −y = y x −1
2
6. Suppose y = f(x) is the solution of the differential equation
initial condition f(1) = 2. Find f ′ (1).
dy
= xy 2 + x 2 satisfying the
dx
a. 1
b. 2
c. 3
d. 4
correct choice
e. 5
f ′ (x) =
dy
= xy 2 + x 2
dx
So f ′ (1) = (1)(2) 2 + (1) 2 = 5
When x = 1, we have y = f(1) = 2.
∞
7. Compute
∑ (e 1/n − e 1/(n+1) ).
n=1
a. e
b. 1 − e
c. e − e 2
d. e − 1
correct choice
e. e 2 − e
k
S k = ∑ (e 1/n − e 1/(n+1) ) = (e − e 1/2 ) + (e 1/2 − e 1/3 ) + ⋯ + (e 1/k − e 1/(k+1) ) = e − e 1/(k+1)
n=1
S = n∞
lim S k = n∞
lim (e − e 1/(k+1) ) = e − e 0 = e − 1
∞
8. The series
∑
n=1
1
n +n
a. Converges by the n th -Term Divergence Test.
b. Diverges by the Comparison Test with
∞
∑ 1n .
n=1
c. Diverges by the Comparison Test with
∞
∑ 1n .
n=1
d. Diverges by the Limit Comparison Test with
∞
∑ 1n .
correct choice
n=1
e. Diverges by the Limit Comparison Test with
∞
∑ 1n .
n=1
n th -Term Divergence Test fails.
lim
n∞
1
n =1
n +n 1
and
1
< 1n ≤
n +n
1
n
lim
n∞
n
1
=0
n +n 1
∞
∑ 1n is the harmonic series which is divergent.
n=1
3
∞
9. The series
∑
n=1
n
n +1
2
a. Converges by the Ratio Test.
b. Diverges by the Integral Test.
correct choice
c. Converges because it is a p-series with p = 2 > 1.
d. Converges because it is a p-series with p = 1.
e. Diverges by the Limit Comparison Test with
∞
∑ 12 .
n=1
n
2
n +1
n dn = 1 ln(n 2 + 1) ∞ = ∞
2
1
n2 + 1
∞
∞
∑ 2 n is not a p-series.
∑ 12 is convergent.
n=1 n + 1
n=1 n
is positive and decreasing, and
Ratio Test fails.
10. Compute lim
x0
n
∞
∫1
sin(x 2 ) − x 2
x5
a. − 1
b.
c.
d.
e.
3
−1
6
0
correct choice
1
6
1
3
3
sin(t) = t − t + ⋯
3!
6
sin(x 2 ) = x 2 − x + ⋯
3!
6
sin(x 2 ) − x 2 = − x + ⋯
3!
6
−x +⋯
sin(x 2 ) − x 2
3!
lim
= lim
= lim − x + ⋯ = 0
5
5
x0
x0
x0
3!
x
x
11. Find the volume of the parallelepiped with edge vectors
⃗
a = (2, 1, 4), ⃗
b = (0, 3, 2) and ⃗c = (−2, 1, 3).
a. 42
b. 34
correct choice
c. 28
d. 21
e. 17
The is the absolute value of the triple product ⃗
a×⃗
b ⋅ ⃗c which is a determinant:
V= ⃗
a×⃗
b ⋅ ⃗c =
2
1 4
0
3 2
= |2(9 − 2) − 1(0 + 4) + 4(0 + 6)| = 34
−2 1 3
4
Work Out: (Points indicated. Part credit possible.)
12. (12 points) The curve
x = 2t 2 ,
y = t3
for 0 ≤ t ≤ 1 is rotated about the y-axis.
Find the surface area of the surface swept out. Don’t simplify your numbers.
dy
= 3t 2
dt
dx = 4t
dt
A = ∫ 2πr ds =
u = 16 + 9t 2
A = 4π ∫
25
16
= 2π
81
1
∫0
2πx
The radius of revolution is r = x = t 2 .
2
dx
dt
+
u − 16 u du = 2π
9
18
81
2(5)
32(5)
−
5
3
2
dt =
3
1
∫ 0 2π2t 2
16t 2 + 9t 4 dt =
25
2π
81
∫ 16 (u 3/2 − 16u 1/2 ) du =
− 2π
81
1
∫ 0 4πt 3
16 + 9t 2 dt
t 2 = u − 16
9
du = t dt
18
du = 18t dt
5
dy
dt
5
2(4)
32(4)
−
5
3
3
2u 5/2 − 32u 3/2
5
3
25
16
= 5692 π
1215
5
13. (22 points) A bucket starts out containing 4 liters of salt water with concentration of 60 gm of
salt per liter. Salt water is added to the bucket at 3 liters per minute with concentration of 50
gm of salt per liter. The water in the bucket is kept mixed but is leaking out at the rate of 1
liter per minute. (Note the amount of water in the bucket is not constant.)
Let S(t) denote the amount of salt in the bucket at time t.
a. (7 pt) Write a differential equation for S(t) and an initial condition for S(t).
HINT: How much water is in the bucket at time t ?
The volume of water at time t is V = (4 + 2t) L.
dS = 50 gm 3 L − S(t) gm 1 L = 150 − S(t)
(4 + 2t)
dt
(4 + 2t) L min
L
min
IN
S(0) =
60 gm
⋅ 4 L = 240
L
OUT
b. (2 pt) Select one: The differential equation is
□ separable
■ linear
□ both
□ neither
c. (9 pt) Solve the initial value problem for S(t).
dS + S(t) = 150
(4 + 2t)
dt
P(t) =
1
(4 + 2t)
P(t) dt
I = e∫
= e ln
4 + 2t dS +
dt
∫ P(t) dt = ∫
4+2t
=
1
dt = 1 ln(4 + 2t) = ln 4 + 2t
2
(4 + 2t)
4 + 2t
S(t)
= 150 4 + 2t
4 + 2t
d
dt
4 + 2t S
= 150 4 + 2t
4 + 2t S = ∫ 150 4 + 2t dt = 50(4 + 2t) 3/2 + C
At t = 0, we have S = 240. So
4 + 2t S = 50(4 + 2t) 3/2 + 80
4 240 = 50(4) 3/2 + C
S = 50(4 + 2t) +
C = 480 − 400 = 80
80
= 200 + 100t +
4 + 2t
80
4 + 2t
d. (4 pt) How much salt is in the bucket after 6 minutes?
What is the concentration after 6 minutes?
S(6) = 200 + 100 ⋅ 6 +
Concentration =
80
= 820 gm
4+2⋅6
gm
S(6)
= 820 = 205 = 51. 25
16
4
V(6)
L
6
14.
(16 points) The "Gaussian bell curve" used in statistics is
2 e −t 2
π
y 1.0
as shown in the graph. The "error function" is its integral:
x
2
erf(x) = 2 ∫ e −t dt
0
π
0.8
0.6
0.4
which gives the probability that the value of t is between 0 and x.
0.2
For this problem we will ignore the coefficient and study the
function
-3
f(x) =
x
∫0 e
−t 2
-2
-1
0
1
2
3
x
dt
a. (6 pt) Find a Maclaurin series for f(x) in summation notation.
ex =
∞
xn
n!
∑
n=0
f(x) =
x
∫0
e −t =
2
∞
(−t 2 ) n
=
n!
∑
n=0
e −t dt =
2
∞
∑
n=0
(−1) n
n!
x
∫0
t 2n dt =
∞
∑
n=0
∞
∑
n=0
(−1) n t 2n
n!
(−1) n x 2n+1
n!(2n + 1)
b. (6 pt) Use the first 3 terms of the Maclaurin series for f(x) to approximate f(1).
(n = 0, 1, 2) Do not simplify. Find a bound on the error in this approximation. Explain why.
f(x) ≈
2
∑
n=0
3
5
(−1) n x 2n+1
= x− x + x
3
10
n!(2n + 1)
f(1) ≈ 1 − 1 + 1 ≈ 0. 766 67
3
10
Since this series is alternating, the error is bounded by the absolute value of the next term.
|E| < |a 3 | =
(−1) 3 x 7
3!(7)
x=0.1
= 1 =. 23810 × 10 −1
42
c. (4 pt) How many terms of the Maclaurin series for f(x) would you need to approximate
f(1) to within 10 −3 ? Why?
(−1) n x 2n+1
1
=
< 10 −3
n!(2n + 1) x=1
n!(2n + 1)
1
|E| < 1 =
< 10 −2
24 ⋅ 9
4!(9)
1
1
|E| <
=
< 10 −3
So 5 terms are needed.
120 ⋅ 11
5!(11)
|E| < |a n | =
n = 4:
n = 5:
7