Math 172.503
Exam 2 Solutions
March 23, 2010
1. E : Use the trigonometric substitution x =
Z
Z
2. A :
1
∞
1
dx = lim
b→∞
e2x
e
1
−2x
tan θ, dx =
2
3
sec2 θ dθ. Then
( 2 tan θ)2
2
√ 3
· sec2 θ dθ
2
3
4 tan θ + 4
Z 4
2
tan
θ 2
9
· sec2 θ dθ
=
2 sec θ 3
Z
4
=
tan2 θ sec θ dθ.
27
x2
√
dx =
9x2 + 4
b
Z
2
3
Z
b
1 −2x
1
1
1
1
dx = lim − e
= lim − 2b + 2 = 0 + 2 = 2 .
b→∞
b→∞
2
2e
2e
2e
2e
1
3. B : Because x4 − 1 = (x2 − 1)(x2 + 1) = (x − 1)(x + 1)(x2 + 1), we see that the denominator
has two distinct linear factors and one irreducible quadratic factor. Therefore, the partial fraction
decomposition is
1
A
B
Cx + D
=
+
+ 2
.
4
x −1
x−1 x+1
x +1
4. A : Splitting [1, 3] into n = 6 equal subintervals, we find width of each subinterval is
the end-points of the subintervals are
3−1
6
5
7
8
4
, x2 = , x3 = 2, x4 = , x5 = , x6 = 3.
3
3
3
3
x0 = 1, x1 =
Checking with the formula for the trapezoidal rule, we see that the answer must be A.
5. C : We quickly evaluate the three integrals:
∞
Z
Z
b
h
ib
1
dx = lim ln x = lim [ln b − 0] = ∞,
b→∞
b→∞
1
1
1 x
Z ∞
Z 0
Z b
sin x dx = lim
sin x dx + lim
sin x dx
−∞
1
dx = lim
b→∞
x
a→−∞
= lim
a→−∞
Z
0
1
b→∞
a
h
− cos x
i0
h
0
+ lim − cos x
a
b→∞
ib
0
= lim [−1 − cos a] + lim [cos b + 1]
(neither limit exists),
a→−∞
b→∞
Z 1
h √ i1
√
1
1
√ dx = lim
√ dx = lim 2 x = lim [2 − 2 a] = 2.
+
+
+
x
x
a→0
a→0
a
a→0
a
Thus, I and II both diverge, and III converges.
= 23 . Thus
Z
6. To find the arc length of the curve, we use the arc length formula L =
Z
2
q
3x1/2
2
0
q
y0
2
+ 1 dx. We use
a
y = 2x3/2 and [a, b] = [0, 2]. Since y 0 = 3x1/2 , we have
L=
b
Z
2
+ 1 dx =
√
9x + 1 dx
0
19
1√
u du
(using u = 9x + 1, du = 9 dx)
9
1
19
2 3/2
2 3/2
2
.
=
u
19 −
=
27
27
27
1
Z
=
7. (a) For this integral, we first perform long division and find
x4
16
dx = x2 − 4 + 2
,
+4
x +4
x2
so
Z
x4
dx =
2
x +4
Z 16
dx
x2 + 4
Z
x3
1
=
− 4x + 16
· 2 sec2 θ dθ
3
4 sec2 θ
x3
− 4x + 8θ + C
=
3
x3
x
=
− 4x + 8 tan−1
+C .
3
2
x2 − 4 +
(using x = 2 tan θ, dx = 2 sec2 θ dθ)
R
It is also possible to use the formula x21+1 = tan−1 (x) + C by pulling a 4 out of the denominator of
1
x2 +4 and using an appropriate u-substitution.
(b) We factor x3 − x = x(x2 − 1) = x(x − 1)(x + 1). So we make the partial fraction decomposition
6x − 4
A
B
C
= +
.
3
x −x
x
x+1x−1
Adding the right-hand side together we find,
6x − 4
(A + B + C)x2 + (−B + C)x − A
=
,
3
x −x
x3 − x
and so we obtain the system of equations
A+B+C =0
−B + C = 6
−A = 4.
Solving this system, we find A = 4, B = −5, and C = 1. Therefore,
Z
Z 6x − 4
4
5
1
dx =
−
+
dx = 4 ln |x| − 5 ln |x + 1| + ln |x − 1| + C .
x3 − x
x x+1 x−1
dy
−2 dx
. Integrating both sides we find
=√
y
1 − x2
Z
Z
dy
dx
= −2 √
y
1 − x2
ln |y| = −2 sin−1 (x) + C.
8. We use separation of variables and find
−1
Then exponentiating both sides, we find y = Ae−2 sin
this equation as a linear first order equation.
(x)
for a constant A. It is also possible to solve
9. This is a linear first order equation, so we first divide by x2 to put it in the normal form,
dy
2
sin(3x)
.
+ y=
dx x
x2
R
2
x
1
x2
Z
The integrating factor v(x) = e
y=
1
v(x)
Z
v(x)Q(x) dx =
dx
= e2 ln x = x2 . Therefore, with Q(x) =
x2 ·
1
sin(3x)
dx = 2
2
x
x
sin(3x)
x2 ,
we have
Z
sin(3x) dx
1
1
C
cos(3x)
= 2 − cos(3x) + C = −
+ 2.
2
x
3
3x
x
Now using the initial condition that y(π/6) = 2, we substitute x = π/6 into this expression to solve
2
for C. We find that C = π18 (recalling that cos(π/2) = 0). Therefore,
y=−
cos(3x)
π2
.
+
2
3x
18x2