Z ∞
x
5. (a) The integral
d x converges by comparison to
1 + x4
1
Z ∞
x
1
d x. First note that the integrand
is positive
3
x
1 + x4
1
on [1, ∞). We then have
Z ∞
Z ∞
Z ∞
x
x
1
dx ≤
dx =
dx
4
4
3
1
+
x
x
x
1
1
1
Z t
= lim
x −3 d x
t→∞ 1
t = lim − 12 x −2 t→∞
1
1
1
−1
= = 0.50.
−
−
= lim
t→∞ 2t 2
2
2
Z ∞
x
Hence
d x converges by the Comparison
1
+
x4
1
Theorem to a value L ≤ 12 . [NOTE: This integral is easy to
compute directly as follows. Although you were not required
to do this, it provides a nice independent check!]
Z ∞
Z t
x
x
d x = lim
2 d x
4
t→∞
1+x
1
1 1 + x2
t
= lim 12 tan−1 x 2 t→∞
1
1 π 1
−1
2
= lim 2 tan
t −2 4
t→∞
= 12 π2 − 12 π4 = π4 − π8 = π8
≈ 0.39 ≤ 0.50
Spring 2006 Math 152
Exam 2A: Solutions
c
Mon, 27/Mar
2006,
Art Belmonte
1. (c) The arc length of the curve
i
h
r(t) = [x(t), y(t)] = t 2 , t 2 + t , 0 ≤ t ≤ 1,
is represented by the following integral.
Z b
Z 1
0 k[2t, 2t + 1]k dt
L=
r (t) dt =
a
Z
0
Z
0
=
=
1p
1p
4t 2 + 4t 2 + 4t + 1 dt
1 + 4t + 8t 2 dt
0
dy
= x y 2 + x 2 y = x y (y + x)
dx
is not separable. Other choices are separable, as follows.
2. (c) The differential equation
dy
= sin x cos y as sec y d y = sin x d x.
dx
dy
• (b) Rewrite
= x y + x 2 y = y x + x 2 as
dx 1
d y = x + x 2 d x.
y
dy
= e x+y = e x e y as e−y d y = e x d x.
• (d) Rewrite
dx
• (e) Repeated factoring gives
• (a) Rewrite
dy
dx
dy
dx
dy
dx
whence
=
1 + x + y + xy
=
(1 + x) + y (1 + x)
=
(1 + x) (1 + y) ,
6. (b) The plate has constant density. Its semicircular area is
A = 12 πr 2 = 12 π (4)2 = 8π. Accordingly, the x-coordinate
of the center of mass is given by
Z
1 b
x ( f (x) − g (x)) d x
x̄ =
A a
Z 4 p
p
1
x
16 − x 2 − − 16 − x 2 d x
=
8π 0
Z 4
1/2
1
16 − x 2
· 2x d x (Sub: u = 16 − x 2 )
=
8π 0
Z
Z 16
1
−1 0 1/2
u du =
u 1/2 du
=
8π 16
8π 0
16
16
1 2
16
u 3/2 =
−0=
≈ 1.70.
=
0
8π 3
3π
3π
1
d y = (1 + x) d x.
1+y
3. (d) The linear differential equation
y 0 + (2 sin 2x) y = cos 4x
is already in standard linear form
an
R (SLF). Accordingly,
integrating factor is µ = exp 2 sin 2x d x = e− cos 2x .
7. (e) The step size is h =
b−a = 2−0 = 1 . Hence
n
4
2
Tn = step size × (average of endpoint func vals + sum of interior func vals)
1 1.00 + 0.70
+ (0.25 + 0.40 + 0.20)
T4 =
2
2
1
(0.85 + 0.85) = 0.85.
=
2
4. (b) Let y = y(t) be the amount of salt in the tank at time t.
The classical balance law gives
dy
dt
dy
dt
dy
dt
=
=
=
rate in − rate out
kg
L
L
y kg
0.1
−
× 10
× 10
L
min
100 L
min
y
kg
1−
.
10
min
b−a
2−0
1
n = 4 = 2 . Hence
Mn = step size × (sum of midpoint function values)
8. (c) The step size is h =
Since the tank initially contains pure water, we have
y(0) = 0 kg of salt in the tank at the start. Therefore,
y
dy
= 1 − , y (0) = 0.
dt
10
M4
=
=
1
1
(0.50 + 0.75 + 0.30 + 0.10)
2
1
(1.65) = 0.825.
2
9. (e) The differential equation
separable.
1
dy
1 + y2
tan
−1
=
y
=
y
=
dy
= 2 + 3x 2 y 2 + 1 is
dx
• Multiply the SLF by µ.
x 2 y 0 + 2x y = x 4
2 + 3x 2 d x
or
x2y
0
= x4
• Integrate to obtain x 2 y = 15 x 5 + C. Therefore,
3
2x + x + C
tan 2x + x 3 + C
y = 15 x 3 + C x −2
is a general solution.
dy
= 2y is separable. Find a
10. (d) The differential equation
dx
general solution, then resolve the constant of integration
using the initial condition y (0) = 4.
1
dy =
y
ln |y| =
±y = |y| =
y =
Substitute data: 4 =
y =
Thus y (1) =
• Use the initial condition y (−1) = 2 to determine C and
thus the unique solution to the initial value problem.
2 dx
2x + A
13.
Ce0 = C
4e2x
C
=
y
=
10
1
11
5 + 5 = 5
1 3
11 −2
5x + 5 x
Z t
x
x
d
x
=
lim
dx
2
t→∞ 0 1 + x 2
1+x
0
diverges to ∞ via direct computation.
Z
e2x+ A = e2x e A = Be2x
Ce2x
∞
(a) The integral
t
lim 12 ln 1 + x 2 = lim 12 ln 1 + t 2 − 0 = ∞
t →∞
4e2 ≈ 29.56.
0
Z
2
(b) The integral
x2
ln x
−
, 1 ≤ x ≤ 2, is
11. The arc length of the curve y =
4
2
s
2
Z
Z b
dy
L =
ds =
1+
dx
dx
a
s
Z 2
1 2
x
−
1+
dx
=
2 2x
1
s
Z 2
x 2 1 1 2
1+
− +
dx
=
2
2
2x
1
s
2
Z 2 x 2 1
1
+ +
dx
=
2
2
2x
1
s
Z 2 1 2
x
+
dx
=
2
2x
1
Z 2
1 1
1
x+
dx
=
2
2 x
1
2
1 2
1
=
x
+
ln
x
4
2
1 =
1 + 12 ln 2 − 14 = 34 + 12 ln 2 ≈ 1.10.
0
1π
2
converges to
Z t
lim
t →2−
0
p
p
t →∞
Z
1
1
4 − x2
dx
=
=
t→2−
0
• Put the equation into standard linear form (SLF).
=
2
y = x2
x
=
=
2
, the coefficient of y in the SLF.
x
• Construct an integrating factor.
R
R
µ = exp P(x) d x = exp( 2x d x) = e2 ln x = x 2
Here P(x) =
2
1
1 + 9x 4
2π
1/2
1
4 − x2
dx
1
2 d x
1 − 12 x
x t
lim sin−1
0
2
t →2−
π
t
−1
−0 =
lim sin
2
2
t →2−
surface area obtained by rotating the
0 ≤ x ≤ 1, about the x-axis is
Z
S =
2πr ds
Z
=
2π y ds
Z 1 q
2
x 3 1 + 3x 2 d x
= 2π
=
0
Z
1 t
r
t →2− 2 0
14. The arc length differential is ds =
0
p
lim
r
Z
t
d x = lim
4−
as follows.
x2
=
12. The differential equation x y 0 + 2y = x 3 is linear.
y0 +
− 15 + C
2 = y (−1) =
dy 2
dx
dx
curve y = x 3 ,
1+
x3 dx
. The
(Sub: u = 1 + 9x 4 )
Z
Z 10
π
2π 10 1/2
u du =
u 1/2 du
36 1
18 1
10
π 2
π √ π
u 3/2 =
10 10 −
1
18 3
27
27
√
π
10 10 − 1 ≈ 3.56.
27