MATH 172.504
Exam 2 Solutions
March 20, 2012
Z
1.
π
x cos(x) dx.
Evaluate
0
(A) 2
(B) 1
(C) 0
(D) −1
(E) −2
Solution: E We use integration by parts, taking u = x and dv = cos(x) dx. Then
du = dx and v = sin(x), from which we have
Z π
h
iπ Z π
x cos(x) dx = −x sin(x) −
sin(x) dx
0
0
0
h
iπ
= −x sin(x) + cos(x) = (0 − 1) − (0 + 1) = −2 .
0
2.
After trigonometric substitution, the integral
Z
√
x4 x2 − 4 dx
becomes
Z
(A) 16
Z
(B) 32
Z
(C) 64
Z
(D) 32
Z
(E) 64
tan4 θ sec3 θ dθ
tan θ sec4 θ dθ
tan2 θ sec5 θ dθ
sin4 θ cos θ dθ
sin4 θ cos2 θ dθ
Solution: C We take x = 2 sec θ, and so dx = 2 sec θ tan θ dθ. Substituting we find
Z
Z
√
√
4
2
x x − 4 dx = 16 sec4 θ( 4 sec2 θ − 4)2 sec θ tan θ dθ
Z
Z
√
5
2
= 32 sec θ tan θ 4 tan θ dθ = 64 sec5 θ tan2 θ dθ .
1
∞
Z
3.
Evaluate
(x2
1
4x
dx.
+ 1)2
(A) 0
(B) 1
(C) 2
(D) 4
(E) The integral diverges.
Solution: B Letting u = x2 + 1, du = 2x dx, we find that
Z
Z
4x
2
2
2
+ C.
dx =
du = − + C = − 2
2
2
2
(x + 1)
u
u
x +1
Therefore,
Z
1
4.
∞
4x
dx = lim
2
t→∞
(x + 1)2
Z
t
4x
dx
+ 1)2
1
t
2
2
2
= lim − 2
+
= 1.
= lim − 2
t→∞
x + 1 1 t→∞
t +1 2
(x2
Consider the following improper integrals:
Z 1
Z 0
1
I.
dx
II.
ex dx
2
−1 x
−∞
Z
III.
1
∞
1
√ dx
x
Which of these integrals converge?
(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I and III only
Solution: B We consider each integral individually. First,
Z 1
Z t
Z 1
1
1
1
dx = lim−
dx + lim+
dx.
2
2
2
t→0
s→0
−1 x
−1 x
s x
We calculate that
1
1
1
1
lim
dx = lim+ −
= lim −1 +
= ∞.
s→0+ s x2
s→0
x s s→0+
s
R1
Since one of the integrals making up −1 x12 dx diverges, the entire integral diverges .
Second,
Z 0
Z 0
h i0
x
e dx = lim
ex dx = lim ex = lim [1 − et ] = 1 − 0 = 1.
Z
−∞
1
t→−∞
t
t→−∞
2
t
t→−∞
Therefore the
Z
∞
1
and so
5.
R∞
1
√1
x
R0
−∞
ex dx converges . Finally,
1
√ dx = lim
t→∞
x
Z
t
x
−1/2
h √ it
√
dx = lim 2 x = lim [2 t − 2] = ∞,
t→∞
1
1
t→∞
dx diverges . Thus the answer is B.
If we use Simpson’s rule to approximate
Z 2
f (x) dx
0
with 4 equal subintervals, we obtain which expression below?
1
f (0) + 4f ( 12 ) + 2f (1) + 4f ( 23 ) + f (2)
(A)
6
1 f (0) + 4f ( 12 ) + 2f (1) + 4f ( 32 ) + f (2)
(B)
12
1 (C)
f (0) + 2f ( 12 ) + 4f (1) + 2f ( 23 ) + f (2)
12
1
f (0) + 2f ( 12 ) + 2f (1) + 2f ( 32 ) + f (2)
(D)
8
1
(E)
f (0) + 2f ( 12 ) + 2f (1) + 2f ( 23 ) + f (2)
2
Solution: A The formula for Simpson’s rule, using n equal subintervals is
Z
b
f (x) dx =
a
b − a
f (x0 ) + 4f (x1 ) + 2f (x2 ) + 4f (x3 ) + 2f (x4 )
3n
+ · · · + 2f (xn−2 ) + 4f (xn−1 ) + f (xn ) .
Taking [a, b] = [0, 2] and n = 4, we see that A is the correct answer.
6.
Evaluate the following integrals.
Z
(a) sin(x)e2x dx
Z
(b) sin(7x) sin(3x) dx
Z
3x2
√
(c)
dx
1 − 9x2
Z 2
x −x+3
(d)
dx
x3 + x
Solution: (a) We use integration by parts starting with u = sin x, dv = e2x dx, which
yields du = cos x dx, v = 21 e2x and thus
Z
Z
1
1
2x
2x
cos(x)e2x dx.
sin(x)e dx = sin(x)e −
2
2
3
We apply integration by parts again starting with u = cos x, du = e2x dx, which yields
du = − sin x dx, v = 12 e2x and thus
Z
Z
1
1 1
1
2x
2x
2x
2x
sin(x)e dx = sin(x)e −
cos(x)e +
sin(x)e dx
2
2 2
2
Z
1
1
1
sin(x)e2x dx.
= sin(x)e2x − cos(x)e2x −
2
4
4
Therefore
5
4
Z
Multiplying through by
Z
sin(x)e2x dx =
4
5
1
1
sin(x)e2x − cos(x)e2x .
2
4
we see that
sin(x)e2x dx =
1
2
sin(x)e2x − cos(x)e2x .
5
5
(b) We use the identity sin A sin B = 12 [cos(A − B) − cos(A + B)]:
Z
Z
1
sin(7x) sin(3x) dx =
[cos(4x) − cos(10x)] dx
2
1 sin(4x) sin(10x)
sin(4x) sin(10x)
=
.
−
=
−
2
4
10
8
20
(c) We use trigonometric substitution with 3x = sin θ (so x =
cos θ dθ:
Z
Z 3 sin2 θ · 1 cos θ dθ
2
9
3
3x
√
r
dx =
1 − 9x2
sin2 θ
1−9
9
Z
1
sin2 θ cos θ dθ
√
=
9
cos2 θ
Z
1
=
sin2 θ dθ.
9
1
3
sin θ) and 3 dx =
Now to evaluate this integral we use the identity sin2 θ = 21 (1 − cos(2θ)), and so
Z
Z
1
1
1
sin(2θ)
θ
sin(2θ)
2
sin θ dθ =
(1 − cos(2θ)) dθ =
θ−
=
−
.
9
18
18
2
18
36
Now to substitute back to x, we use the identity sin(2θ) = 2 sin θ cos θ. Since sin θ =
3x, it follows using
√ a right triangle with opposite side 3x (opposite θ) and hypotenuse
1 that cos θ = 1 − 9x2 . Therefore,
Z
√
3x2
θ
2 sin(θ) cos θ
arcsin(3x) x 1 − 9x2
√
dx =
−
=
−
.
18
36
18
6
1 − 9x2
4
(d) We use partial fractions: since x3 + x = x(x2 + 1),
x2 − x + 3
A Bx + C
Ax2 + A + Bx2 + Cx
=
+
=
.
x3 + x
x
x2 + 1
x3 + x
This yields the equations
C = −1,
A + B = 1,
A = 3,
⇒ B = −2.
Therefore,
Z
x2 − x + 3
dx =
x3 + x
Z 3 2x + 1
−
dx
x x2 + 1
Z
Z
Z
3
2x
1
=
dx −
dx
−
dx
x
x2 + 1
x2 + 1
= 3 ln |x| − ln |x2 + 1| − arctan(x) + C ,
where the middle integral is obtained by u-substitution with u = x2 + 1, du = 2x dx.
7.
Find all solutions of the differential equation
y
dy
·
− sin3 (x) = 0.
4
cos (x) dx
Solution: We solve by separation of variables: the equation transforms to y
sin3 x cos4 x, which yields
y dy = sin3 x cos4 x dx.
We integrate both sides:
Z
Z
y dy =
1 2
y =
2
Z
sin3 x cos4 x dx
sin x(1 − cos2 x) cos4 x dx.
Now taking u = cos x, du = − sin x dx, we se that
Z
Z
1 2
1
1
2 4
y = − (1 − u )u du = − (u4 − u6 ) du = u7 − u5 + C.
2
7
5
Substituting back to x and solving for y, we see that
q
y(x) = ± 27 cos7 x − 52 cos5 x + C .
8.
Solve the initial value problem,
x
dy
− y = x ln x,
dx
5
y(e) = 2e.
dy
dx
=
Solution: We use the method of first order linear equations. First we transform the
equation into the proper form:
dy
y
− = ln x.
dx x
This makes P (x) = − x1 and Q(x) = ln x, and thus
v(x) = e
We then have
R
−1
x
1
y(x) =
v(x)
dx
= e− ln x = (eln x )−1 =
Z
Z
v(x)Q(x) dx = x
1
.
x
ln x
dx.
x
By u-substitution with u = ln x, du = x1 dx, we then have
Z
1 2
(ln x)2
x(ln x)2
y(x) = x u du = x u + C = x
+C =
+ Cx.
2
2
2
Now we want y(e) = 2e, so plugging in we have
y(e) =
e(ln e)2
+ Ce = 2e.
2
Since ln e = 1, we see that C = 23 . Therefore,
y(x) =
x(ln x)2 3x
+
.
2
2
6