MATH 172.200
Exam 2 Solutions
March 20, 2012
π
Z
1.
x2 sin x dx.
Evaluate
0
(A) π
(B) π 2
(C) π 2 + 4
(D) π 2 − 4
(E) 0
Solution: D We use integration by parts, taking u = x2 and dv = sin x dx. Then
du = dx and v = − cos x, from which we have
Z
Z
2
2
x sin x dx = −x cos x + 2 x cos x dx.
We apply integration by parts again, with u = x, dv = cos x dx, and du = dx,
v = sin x:
Z
Z
2
2
x sin x dx = −x cos x + 2 x sin x − sin x dx .
Therefore
Z
π
h
iπ
x2 sin x dx = −x2 cos x + 2x sin x + 2 cos x
0
0
= (−π 2 (−1) + 0 + 2(−1)) − (0 + 0 + 2) = π 2 − 4 .
2.
After trigonometric substitution, the integral
Z
√
x4 x2 − 4 dx
becomes
Z
(A) 64
Z
(B) 16
Z
(C) 32
Z
(D) 64
Z
(E) 32
sin4 θ cos2 θ dθ
tan4 θ sec3 θ dθ
tan θ sec4 θ dθ
tan2 θ sec5 θ dθ
sin4 θ cos θ dθ
1
Solution: D 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 .
Z
3.
∞
Evaluate
2
x2
1
dx.
+4
π
(A)
2
π
(B)
4
π
(C)
8
π
(D)
16
(E) The integral diverges.
Solution: C We calculate
Z
∞
2
4.
x t
1
1
dx = lim
arctan
2
t→∞ 2
2 2
2 x +4
1
t
1
= lim
arctan
− arctan(1)
t→∞ 2
2
2
1 π 1 π
π
= · − · =
.
2 2 2 4
8
1
dx = lim
2
t→∞
x +4
Z
t
Consider the following improper integrals:
Z 0
Z 1
1
x
I.
e dx
II.
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) II and III only
Solution: A We consider each integral individually. First,
Z 0
Z 0
h i0
x
e dx = lim
ex dx = lim ex = lim [1 − et ] = 1 − 0 = 1.
−∞
t→−∞
t
t→−∞
2
t
t→−∞
Therefore the
R0
−∞
ex dx converges . Second,
1
Z
−1
1
dx = lim−
t→0
x2
Z
t
−1
1
dx + lim+
s→0
x2
Z
s
1
1
dx.
x2
We calculate that
Z
lim
s→0+
s
1
1
1
1
1
= lim −1 +
= ∞.
dx = lim+ −
s→0
x2
x s s→0+
s
R1
Since one of the integrals making up −1 x12 dx diverges, the entire integral diverges .
Finally,
Z ∞
Z t
h √ it
√
1
√ dx = lim
x−1/2 dx = lim 2 x = lim [2 t − 2] = ∞,
t→∞
t→∞ 1
t→∞
x
1
1
R∞
and so 1 √1x dx diverges . Thus the answer is A.
5.
If we use Simpson’s rule to approximate
Z 2
f (x) dx
0
with 4 equal subintervals, we obtain which expression below?
1 (A)
f (0) + 4f ( 12 ) + 2f (1) + 4f ( 23 ) + f (2)
12
1
(B)
f (0) + 4f ( 21 ) + 2f (1) + 4f ( 23 ) + f (2)
6
1 (C)
f (0) + 2f ( 12 ) + 4f (1) + 2f ( 23 ) + f (2)
12
1
(D)
f (0) + 2f ( 12 ) + 2f (1) + 2f ( 32 ) + f (2)
8
1
(E)
f (0) + 2f ( 12 ) + 2f (1) + 2f ( 23 ) + f (2)
2
Solution: B 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 B is the correct answer.
6.
Evaluate the following integrals.
Z
2x − 1
(a)
dx
3
x + 2x2 + x
3
Z
(b)
sin(x)e2x dx
x2 − 9
√
dx
6x − x2
Z
√
(d) sin2 ( x) dx
Z
(c)
Solution: (a) We use partial fractions: since x3 + 2x2 + x = x(x + 1)2 ,
x3
2x − 1
A
B
C
= +
+
2
+ 2x + x
x x + 1 (x + 1)2
A(x + 1)2 + Bx(x + 1) + Cx
(A + B)x2 + (2A + B + C)x + A
=
.
=
x(x + 1)2
x3 + 2x2 + x
Therefore,
A + B = 0,
2A + B + C = 2,
A = −1.
From this we see that
A = −1,
B = 1,
C = 3,
and so
Z
2x − 1
dx =
2
x + 2x2 + x
Z −1
1
3
+
+
x
x + 1 (x + 1)2
= − ln |x| + ln |x + 1| −
dx
3
+C .
x+1
(b) We use integration by parts starting with u = sin x, dv = e2x dx, which yields
du = cos x dx, v = 12 e2x and thus
Z
Z
1
1
2x
2x
sin(x)e dx = sin(x)e −
cos(x)e2x dx.
2
2
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
2x
2x
= sin(x)e − cos(x)e −
sin(x)e2x dx.
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 =
2
1
sin(x)e2x − cos(x)e2x .
5
5
4
(c) We want to use trigonometric substitution, but we need to put the integral in the
proper form. By completing the square, we see that
6x − x2 = 9 − (x2 − 6x + 9) = 9 − (x − 3)2 ,
so really we want to consider
Z
Z
x2 − 9
x2 − 9
√
p
dx =
dx.
6x − x2
9 − (x − 3)2
We make the substitution x − 3 = 3 sin θ, dx = 3 cos θ dθ:
Z
Z
x2 − 9
(3 sin θ + 3)2 − 9
p
p
dx =
· 3 cos θ dθ
9 − (x − 3)2
9 − 9 sin2 θ
Z
(9 sin2 θ + 18 sin θ + 9 − 9) · 3 cos θ
dθ
=
3 cos θ
Z
Z
1
=9
(1 − cos(2θ)) dθ + 18 sin θ dθ
2
9θ 9
=
− sin(2θ) − 18 cos θ + C.
2
4
We now substitute back to x. The relationship between x and θ is determined by a
right triangle with angle θ, opposite side x − 3 and hypotenuse 3. Thus sin θ = x−3
3
√
√
9−(x−3)2
6x−x2
and cos θ =
=
, which yields (using sin(2θ) = 2 sin(θ) cos(θ))
3
3
Z
9θ 9
x2 − 9
√
dx =
− (2 sin θ cos θ) − 18 cos θ + C
2
4
6x − x2
√
√
9
x−3
1
= arcsin
− (x − 3) 6x − x2 − 6 6x − x2 + C .
2
3
2
√
√
(d) We make the substitution u = x, du = 2√1 x dx. We observe that dx = 2 x du =
2u du, so
Z
Z
√
2
sin ( x) dx = sin2 (u) · 2u du.
We then apply our usual identity for sin2 (u) here:
Z
Z
Z
u
2
2 u sin (u) du = 2
(1 − cos(2u)) du = (u − u cos(2u)) du.
2
The second part of the integral requires integration by parts, using w = u, dv =
cos(2u) du; dw = du, v = 12 sin(2u) du:
Z
Z
u2
u
1
(u − u cos(2u)) du =
−
sin(2u) −
sin(2u) du
2
2
2
u2 u sin(2u) 1
=
−
− cos(2u) + C.
2
2
4
5
Therefore substituting back to x,
Z
7.
√
√
√
x 1√
1
sin2 ( x) dx =
−
x sin(2 x) − cos(2 x) + C .
2 2
4
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
=
sin3 x cos4 x dx
y dy =
1 2
y =
2
dy
dx
Z
sin x(1 − cos2 x) cos4 x dx.
Now taking u = cos x, du = − sin x dx, we se that
Z
Z
1
1
1 2
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,
x2 ln x + y − xy 0 = 0,
y(e) = 2e.
Solution: We use the method of first order linear equations. First we transform the
equation into the proper form:
y0 −
y
= x ln x.
x
This makes P (x) = − x1 and Q(x) = 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
1
.
x
Z
v(x)Q(x) dx = x
6
ln x dx.
To evaluate this integral we use integration by parts with u = ln x, dv = dx; du =
1
dx, v = x:
x
Z
y(x) = x x ln x − dx = x2 ln x − x2 + Cx.
Now we want y(e) = 2e, so plugging in we have
y(e) = e2 ln(e) − e2 + Ce = 2e.
Since ln e = 1, we see that C = 2. Therefore,
y(x) = x2 ln x − x2 + 2x .
7