MATH 172.507
Exam 2 Solutions
April 7, 2016
1.
After trigonometric substitution, the integral
Z √ 2
x −4
dx
x
becomes
Z
tan3 θ
(A) 4
dθ
sec θ
Z
(B) 2 tan2 θ dθ
Z
sec3 θ
(C) 8
dθ
tan θ
Z
cos θ
dθ
(D) 4
sin θ
Z
(E) 2 cos2 θ sin θ dθ
Solution: B Let x = 2 sec θ so that x2 − 4 = 4 tan2 θ and dx = 2 sec θ tan θ dθ. We
substitute:
Z √ 2
Z √
Z
x −4
4 tan2 θ
dx =
2 sec θ tan θ dθ = 2 tan2 θ dθ .
x
2 sec θ
2.
Consider the following improper integrals:
Z ∞
Z 1
1
1
I.
dx
II.
dx
2
x2
0
0 x
Z
III.
1
∞
1
dx
x2
Which of these integrals converges?
(A) I, II, and III
(B) I and III only
(C) I only
(D) II only
(E) III only
Solution: E We compute
Z
0
1
1
1
1
1
dx = −
= lim −1 +
= ∞,
x2
x 0 a→0+
a
1
so I diverges . Since the interval of integration for II contains [0, 1], II also diverges .
By process of elimination we rule out A–D, so the answer must be E. However, we
also compute
∞
Z ∞
1
1
1
dx = −
= lim − + 1 = 1,
b→∞
x2
x 1
b
1
and III converges .
3.
The partial fraction decomposition of
1
(x2 − 1)2
has the form
A
B
(A)
+
x−1 x+1
Ax + B
Cx + D
(B) 2
+ 2
x −1
(x − 1)2
A
B
C
D
(C)
+
+
+
2
x − 1 (x − 1)
x + 1 (x + 1)2
A
Bx + C
D
Ex + F
(D)
+
+
+
x − 1 (x − 1)2 x + 1 (x + 1)2
Ax + B
Cx + D
(E)
+
2
(x − 1)
(x + 1)2
Solution: C We factor the denominator as
(x2
1
1
1
=
=
.
2
2
2
− 1)
((x − 1)(x + 1))
(x − 1) (x + 1)2
Thus the partial fraction decomposition is C.
4.
If we use Simpson’s rule to approximate
Z 2
sin(x) dx
0
with 6 equal subintervals, we obtain which expression below?
1
1
2
4
5
(A)
sin(0) + 4 sin( 3 ) + 2 sin( 3 ) + 4 sin(1) + 2 sin( 3 ) + 4 sin( 3 ) + sin(2)
9
1
1
2
4
5
(B)
sin(0) + 2 sin( 3 ) + 4 sin( 3 ) + 2 sin(1) + 4 sin( 3 ) + 2 sin( 3 ) + sin(2)
9
1
1
2
4
5
(C)
sin(0) + 2 sin( 3 ) + 2 sin( 3 ) + 2 sin(1) + 2 sin( 3 ) + 2 sin( 3 ) + sin(2)
6
1
1
2
4
5
(D)
sin(0) + 2 sin( 3 ) + 4 sin( 3 ) + 4 sin(1) + 4 sin( 3 ) + 2 sin( 3 ) + sin(2)
6
2
1
2
4
5
1
(E)
sin(0) + sin( 3 ) + sin( 3 ) + sin(1) + sin( 3 ) + sin( 3 ) + sin(2)
3
Solution: A It is a matter of checking the formula for Simpson’s rule against the
choices given.
5.
n cos((2n + 1)π)
.
n→∞
2n − 1
Evaluate lim
1
2
(B) 1
(A)
(C) −1
1
(D) −
2
(E) does not exist
Solution: D We note that for n an integer, the expression 2n + 1 is always an odd
integer. Also,
cos(π) = cos(3π) = cos(5π) = · · · = cos((2n + 1)π) = −1,
and thus
n cos((2n + 1)π)
−n
1
= lim
= − .
n→∞ 2n − 1
n→∞
2n − 1
2
lim
6.
The series
∞
X
3n − 2
n=0
4n
(A) converges to 0
(B) converges to 2
4
(C) converges to
3
8
(D) converges to
3
(E) diverges
Solution: We rewrite the series
∞
X
3n − 2
n=0
4n
=
∞
X
3n
n=0
∞
X
2
−
.
4n n=0 4n
Both of the series on the right are geometric series, and continuing we find
∞
X
3n
n=0
n
∞
∞ n
∞
X
X
X
2
1
3
−
=
−
2
.
n
n
4
4
4
4
n=0
n=0
n=0
3
Each series converges since |r| = 34 < 1 for the first and |r| =
and by the formula for a convergent geometric series,
∞
X
3n − 2
n=0
7.
4n
=
1
1−
3
4
−
2
1−
1
4
=
1
1
4
−
2
3
4
=4−
1
2
< 1 for the second,
4
8
=
.
3
3
Evaluate the following integrals.
Z
3x + 9
dx
(a)
x2 + x − 2
Z
1
(b)
dx
(4 − 9x2 )3/2
Z ∞
2
(c)
xe1−x dx
2
Solution: (a) We use the method of partial fractions. The denominator factors as
x2 + x − 2 = (x + 2)(x − 1),
so the partial fraction decomposition will be of the form
A
B
3x + 9
=
+
.
(x + 2)(x − 1)
x+2 x−1
To find A and B we recombine to find
B
A(x − 1) + B(x + 2)
A
+
=
,
x+2 x−1
(x + 2)(x − 1)
and so we must have
A(x − 1) + B(x + 2) = 3x + 9.
Substituting in x = 1, we see that 3B = 12, so B = 4. Substituting in x = −2, we
see that −3A = 3, so A = −1. Therefore,
Z
Z 3x + 9
−1
4
dx =
+
dx = − ln |x + 2| + 4 ln |x − 1| + C .
x2 + x − 2
x+2 x−1
(b) We use trigonometric substitution. Take x = 23 sin θ so that 4−9x2 = 4−4 sin2 θ =
4 cos2 θ and dx = 32 cos θ dθ. Substituting we find
Z
Z
Z
Z
1
1
2
2
cos θ
1
1
dx =
· cos θ dθ =
dθ =
dθ.
2
3/2
2
3/2
3
(4 − 9x )
(4 cos θ)
3
3
8 cos θ
12
cos2 θ
To integrate this we recall that cos12 θ = sec θ, so we now have
Z
Z
1
1
1
2
dx
=
sec
θ
dθ
=
tan θ + C.
(4 − 9x2 )3/2
12
12
4
Since sin θ = 3x
, we consider a right triangle with angle θ, opposite side 3x, and
2
√
hypotenuse 2. The third side then has length 4 − 9x2 , and so
3x
tan θ = √
.
4 − 9x2
From this we see that
Z
1
3x
1
x
·√
dx =
+C = √
+C .
2
3/2
(4 − 9x )
12
4 − 9x2
4 4 − 9x2
(c) We use a u-substitution. Taking u = 1 − x2 , du = −2x dx, we have
Z
Z
1
1
1
2
1−x2
xe
dx = −
eu du = − eu + C = − e1−x + C.
2
2
2
Thus
Z
∞
xe
2
8.
1−x2
1
2
dx = − e1−x
2
∞
1 1−b2 1 −3
1
1
.
= lim − e
+ e
= 0 + e−3 =
b→∞
2
2
2
2e3
2
Write down an integral for the length of the curve defined by y =
points (0, 0) and (4, 2). (You need not evaluate the integeral.)
√
x between the
Solution: Using the arclength formula, we have
s
s
2
2
Z 4
Z 4r
Z b
dy
1
1
√
1+
dx =
1+
1+
dx =
dx .
L=
dx
4x
2 x
0
0
a
9.
Solve the following initial value problems.
dy
(a)
= x + xy 2 , y(0) = 1
dx
(b) y 0 = e2x − 3y, y(0) = 1
Solution: (a) We use separation of variables. We see that
dy
= x(1 + y 2 ),
dx
so
dy
= x dx.
1 + y2
Integrating both sides,
Z
dy
=
1 + y2
Z
⇒
x dx
tan−1 (y) =
Thus taking tangent of both sides,
x2
y(x) = tan
+C .
2
5
x2
+ C.
2
Now y(0) = 1 implies taht
⇔
1 = tan(C)
C=
π
,
4
so
x2 π
y(x) = tan
+
2
4
.
(b) The equation is first order linear. We rewrite it as
y 0 + 3y = e2x ,
and so P = 3 and Q = e2x . The integrating factor I(x) is
I(x) = e
R
3 dx
= e3x .
Therefore, from the formula for first order linear equations,
Z
Z
Z
1
1
3x
2x
−3x
5x
−3x 1 5x
I(x)Q(x) dx = 3x e · e dx = e
e +C .
y(x) =
e dx = e
I(x)
e
5
Since y(0) = 1, we have
1=
1
+C
5
⇔
4
C= ,
5
and so
1
4
y(x) = e2x + e−3x .
5
5
10.
For each of the following sequences, find its limit or show that it does not exist.
Justify your work.
e2n
(a) an = n
4
cos(nπ)
(b) bn =
n2
n/2
2
(c) cn = 1 −
n
Solution: (a) We observe that
2 n
e2n
e
lim n = lim
.
n→∞
n→∞ 4
4
Since e2 > 4, the limit on the right will be ∞ and so the limit does not exist .
(b) For different integers n, we know that cos(nπ) = ±1, so the denominator should
overwhelm the numerator giving a value of 0. To see this, we observe that
−1 ≤ cos(nπ) ≤ 1
⇔
6
−
1
cos(nπ)
1
≤
≤ 2.
2
2
n
n
n
Both the left- and right-hand sides go to 0, so by the squeeze theorem
cos(nπ)
= 0.
n→∞
n2
lim
(c) We recall that for any real number c,
n
c
lim 1 +
= ec .
n→∞
n
Thus
n/2
1/2
2
1
=
.
lim 1 −
= e−2
n→∞
n
e
Or we calculate:
n/2
2
n
2
lim ln 1 −
= lim · ln 1 −
n→∞
n→∞ 2
n
n
2
ln 1 − n
= lim
2
n→∞
n
2
1
2 ·
2
n
1− n
= lim
2
n→∞
− n2
= −1.
(In the third equality we have applied l’Hôpital’s rule.) Thus,
n/2
2 n/2
2
lim 1 −
= lim eln(1− n ) = e−1 .
n→∞
n→∞
n
7