Math 2260 Exam #2 Solutions
1. Evaluate the indefinite integral
Z
2x cos(3x) dx.
Answer: The plan is to use integration by parts with u = 2x and dv = cos(3x) dx:
u = 2x
dv = cos(3x) dx
1
v = sin(3x).
3
du = 2 dx
Then the above integral is equal to
2
2
x sin(3x) −
3
3
Z
sin(3x) dx =
2
2
x sin(3x) + cos(3x) + C.
3
9
2. Shown below is the graph of the function f (x) = 2x13−2x
2 −5x−3 between x = 0 and x = 2. What is the area
of the shaded region? Simplify as much as possible.
1
2
-1
-2
-3
-4
-5
Answer: Notice, first of all, that since the region lies entirely below the axis, the area of the region is
−1 times the integral of the function from 0 to 2. In other words, the goal is to compute
Z 2
Z 2
13 − 2x
2x − 13
−
dx =
dx
2 − 5x − 3
2 − 5x − 3
2x
2x
0
0
after bringing the minus sign inside the integral. Now, this is a classic setup for using partial fractions.
Notice that the denominator factors as
2x2 − 5x − 3 = (2x + 1)(x − 3),
so we write the fraction as the sum of fractions
2x − 13
2x − 13
A
B
=
=
+
.
− 5x − 3
(2x + 1)(x − 3)
2x + 1 x − 3
2x2
Clearing denominators yields the equation
2x − 13 = A(x − 3) + B(2x + 1)
2x − 13 = (A + 2B)x + (B − 3A).
Equating like terms then yields the system of equations
2 = A + 2B
−13 = B − 3A,
1
which we now solve for A and B. From the first equation, we know that A = 2 − 2B; substituting into
the second equation yields
−13 = B − 3(2 − 2B) = B − 6 + 6B = 7B − 6.
Therefore, after adding 6 to both sides, we see that
−7 = 7B,
and so B = −1. In turn, this means that A = 2 − 2B = 2 − 2(−1) = 4.
Therefore,
Z
0
2
2x − 13
dx =
2
2x − 5x − 3
2
Z
0
4
1
−
2x + 1 x − 3
dx
h
i2
= 2 ln |2x + 1| − ln |x − 3|
0
= (2 ln 5 − ln 1) − (2 ln 1 − ln 3)
= 2 ln 5 + ln 3.
Now, we can simplify this using the rules of logarithms to
2 ln 5 + ln 3 = ln(52 ) + ln 3 = ln(3 · 52 ) = ln 75.
Therefore, the area of the region is ln 75.
3. Does the improper integral
Z
1
∞
1
dx
(x + 1)3
converge or diverge?
Answer: There are a variety of ways to approach this problem. Here are three:
Direct Comparison: Notice that x + 1 > x for any x. Then so long as x > 1 (which is the only
region we care about), (x + 1)3 > x3 , and so
1
1
< 3.
3
(x + 1)
x
Moreover,
Z
1
∞
1
dx = lim
b→∞
x3
Z
b
1
Z
1
dx
x3
b
x−3 dx
−2 b !
x
= lim
b→∞
−2 1
−1
1
= lim
+
b→∞ 2b2
2
1
= .
2
= lim
b→∞
1
R∞
1
1
Therefore, the improper integral 1 x13 dx converges. Since (x+1)
3 < x3 for all x > 1, the Direct
R∞ 1
Comparison Test implies that the improper integral 1 (x+1)3 dx converges as well.
2
Limit Comparison: Again, we want to compare to
1
(x+1)3
1
x→∞
x3
lim
x
x+1
x
x
1
x3 ,
so compute the limit
3
x
x3
=
lim
=1
= lim
x→∞ x + 1
x→∞ (x + 1)3
1
1+1/x
= 1 and 13 = 1.
R∞
Since 0 < 1 < ∞ and we already saw above that 1 x13 dx converges, the Limit Comparison Test
R∞ 1
implies that 1 (x+1)3 dx also converges.
since limx→∞
= limx→∞
·
Computing the Integral: We could also just compute the value of this definite integral:
Z b
Z ∞
1
1
dx
=
lim
dx
3
b→∞
(x
+
1)
(x
+
1)3
1
1
Z b
= lim
(x + 1)−3 dx
b→∞ 1
b !
(x + 1)−2
= lim
b→∞
−2
1
−1
1
= lim
+
b→∞ 2(b + 1)2
8
1
= .
8
4. Evaluate the definite integral
Z
1
√
0
x
dx.
1−x
Answer: Notice, first of all, that this is an improper integral since the denominator goes to zero as x
goes to 1. Therefore,
Z 1
Z b
x
x
√
√
dx = lim−
dx.
b→1
1−x
x−1
0
0
Now, trig substitution and u-substitution both look unpromising, so integration by parts seems like
the best bet. Since I want to choose a u that gets nicer when it gets differentiated, u = x is a much
1
better choice than u = √x−1
. Therefore, the good choices for u and dv are
u=x
du = dx
dx
= (1 − x)−1/2 dx
1−x
√
v = −2(1 − x)1/2 = −2 1 − x.
dv = √
Hence, the above integral is equal to
!
Z b
√
√
b
lim− −2x 1 − x 0 + 2
1 − x dx = lim−
b→1
0
b !
4
−2x 1 − x 0 − (1 − x)3/2
3
b→1
0
h
i 4
√
4
= lim
−2b 1 − b + 0 − (1 − b)3/2 −
3
3
b→1−
4
= .
3
√
b
5. A research group publishes a paper claiming that the world wombat population is governed by the
differential equation
dW
(4 − t2 )3/2
= 2W,
dt
where W = W (t) is the wombat population after t years.
3
(a) If there are currently 100 wombats in the world, what is the wombat population as a function of
time according to this group’s model?
Answer: The goal is to solve the differential equation for W , so first we separate variables:
dW
2
dt.
=
W
(4 − t2 )3/2
Now, integrate both sides:
Z
Z
ln W =
dW
=
W
Z
2
dt
(4 − t2 )3/2
2
dt
(4 − t2 )3/2
(I don’t have to worry about absolute value signs in the natural log because the wombat population
can never be negative). Now, to evaluate the integral on the right hand side, the most promising
approach is to use the trig substitution t = 2 sin θ, so that dt = 2 cos θ dθ and we can re-write as
Z
Z
2
4 cos θ dθ
2
cos
θ
dθ
=
2
3/2
3/2
(4 − 4 sin θ)
4 (1 − sin2 θ)3/2
Z
4 cos θ dθ
=
8(cos2 θ)3/2
Z
1
cos θ dθ
=
2
cos3 θ
Z
1
dθ
=
2
cos2 θ
Z
1
sec2 θ dθ
=
2
1
= tan θ + C.
2
Now, we have to translate from θ back into the variable t. Since we made the substitution
t = 2 sin θ, we know that sin θ = 2t , and so we can realize θ as the indicated angle in the following
triangle:
2
t
Θ
4 - t2
Therefore,
tan θ = √
t
,
4 − t2
and so we conclude that
Z
2
dt
(4 − t2 )3/2
1
t
ln W = √
+ C.
2 4 − t2
ln W =
4
Now, we know that the wombat population at time t = 0 is equal to 100, so substituting t = 0
and W = 100 into the above equation yields
ln(100) =
0
1
√
+ C = C,
2 4−0
so we know that C = ln 100 and we know that
ln W =
1
t
√
+ ln 100.
2 4 − t2
Now, exponentiating both sides gives us the wombat population as a function of t:
1
W = e2
√
t
4−t2
+ln 100
1
W = eln 100 e 2
√
t
4−t2
√t
W = 100e 2
4−t2
(b) What does this model say is going to happen to the wombat population in 4 years?
Answer: Nothing very sensible. As t goes to 2 the model predicts that the wombat population
is
√ going to infinity. Past that point, the model predicts an imaginary number of wombats, since
4 − t2 is an imaginary number when t > 2.
While it is possible that this model is making a rather profound philosophical or psychological
point (“any wombat you happen meet four years from now must be a product of a deluded
imagination”), it’s more likely that it’s just not a very good model.
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