Math 2260 Exam #2 Solutions
1. Integrate
Z
x
dx.
x2 − 5x + 4
Answer: I will use the method of partial fractions. First, I factor the denominator:
x
x
=
.
x2 − 5x + 4
(x − 4)(x − 1)
Now, split this into two unknown fractions:
A
B
x
=
+
.
(x − 4)(x − 1)
x−4 x−1
To solve for A and B, clear the denominators to get
x = A(x − 1) + B(x − 4)
or, equivalently,
x = (A + B)x − A − 4B.
Equating coefficients on both sides yields the system of equations
1=A+B
0 = −A − 4B.
The first tells us that A = 1 − B, while the second implies that A = −4B. Therefore, since A = A, we
know that
1 − B = −4B or 1 = −3B,
so B = −1/3 and hence A = 1 − B = 1 − (−1/3) = 4/3. Therefore,
Z
Z x
4/3
1/3
dx =
−
dx
x2 − 5x − 4
x−4 x−1
4
1
= ln |x − 4| − ln |x − 1| + C.
3
3
2. Does the improper integral
Z
∞
xe−2x dx
0
converge or diverge? If it converges, find the value of the integral.
Answer: By definition
Z
∞
xe−2x dx = lim
b→+∞
0
Z
b
xe−2x dx.
0
To evaluate this integral, I want to use integration by parts, letting
dv = e−2x dx
−1
1
du = dx v = − e−2x = 2x .
2
2e
u=x
1
Then the above limit is equal to
lim
b→+∞
!
b
Z
h
x
1
x ib 1 b −2x
e
dx = lim − 2x − 2x
− 2x +
b→+∞
2e 0 2 0
2e
4e
0
b
−2x − 1
= lim
b→+∞
4e2x
0
−2b − 1 −1
−
= lim
b→+∞
4e2b
4
1 2b + 1
= lim
−
b→+∞ 4
4e2b
1
2b + 1
= − lim
.
4 b→+∞ 4e2b
Now, both the numerator 2b + 1 and the denominator 4e2b are going to zero as b → +∞, so we can
apply L’Hôpital’s Rule to see that
lim
b→+∞
2b + 1
2
= lim
= 0,
b→+∞ 8e2b
4e2b
so we conclude, finally that
∞
Z
xe−2x dx =
0
1
.
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3. Evaluate the definite integral
Z
1
√
0
x2
dx.
1 − x2
Answer: First, notice that the denominator is undefined when x = 1, so this is an improper integral.
Therefore, by definition,
Z 1
Z b
x2
x2
√
√
dx = lim−
dx.
b→1
1 − x2
1 − x2
0
0
Make the substitution x = sin θ. Then dx = cos θ dθ. Also, when 0 = x = sin θ we have that θ = 0,
and when b = x = sin θ we have that θ = arcsin(b). Hence, the above limit is equal to
Z
lim−
b→1
0
arcsin(b)
sin2 θ
p
cos θ dθ = lim−
b→1
1 − sin2 θ
Z
0
Z
= lim−
b→1
b→1
2
sin2 θ
cos θ dθ
cos θ
arcsin(b)
sin2 θ dθ
0
Z
= lim−
arcsin(b)
0
arcsin(b)
1 − cos(2θ)
dθ
2
Then this limit is equal to
lim−
b→1
θ sin(2θ)
−
2
4
arcsin(b)
arcsin(b) sin(2 arcsin(b))
−
− (0 − 0)
2
4
b→1
arcsin(b) sin(2 arcsin(b))
= lim−
−
2
4
b→1
π/2 sin(2π/2)
=
−
2
4
π
= −0
4
π
= .
4
= lim−
0
4. Find an antiderivative for the function
f (x) = e3x cos(x).
Answer: Integrate by parts with
u = e3x
du = 3e
3x
dx
dv = cos(x) dx
v = sin(x)
Then the given integral is equal to
e3x sin(x) − 3
Z
e3x sin(x) dx.
For the remaining integral, integrate by parts again:
u = e3x
du = 3e
dv = sin(x) dx
3x
v = − cos(x)
Substituting this into the above yields
Z
Z
e3x cos(x) dx = e3x sin(x) − 3 −e3x cos(x) + 3 e3x cos(x) dx
Z
Z
e3x cos(x) dx = e3x sin(x) + 3e3x cos(x) − 9 e3x cos(x) dx.
Adding 9
R
e3x cos(x) dx to both sides yields
Z
10 e3x cos(x) dx = e3x sin(x) + 3e3x cos(x),
and so
F (x) =
e3x sin(x) + 3e3x cos(x)
10
is an antiderivative for f (x).
5. A super-fast-growing bacteria reproduces so quickly that the rate of production of new bacteria is
proportional to the square of the number already present. If a sample starts with 100 bacteria, and
after 3 hours there are 200 bacteria, how long after the starting time will it take until there are
(theoretically) an infinite number of bacteria?
3
Answer: If P (t) is the number of bacteria present after t hours, then we have that
dP
= kP 2
dt
for some constant k and that P (0) = 100 and P (3) = 200. First, notice that we can separate variables
and integrate to solve the differential equation:
Z
Z
dP
= k dt
P2
−1
= kt + C.
P
Therefore,
−1
.
kt + C
P (t) =
We can determine C by using the initial condition:
−1
k(0) + C
−1
100 =
,
C
P (0) =
so C =
−1
100
and we can re-write the expression for P (t):
P (t) =
−1
1 .
kt − 100
Now, we use the fact that P (3) = 200 to solve for k:
−1
1
k(3) − 100
−1
200 =
1
3k − 100
P (3) =
and so
1
200 3k −
100
= −1
600k − 2 = −1,
meaning that 600k = 1 and so k =
P (t) =
1
600 .
t
600
Therefore,
−1
1 =
− 100
−1
−600
600
=
=
.
t−6
6−t
− 6)
1
600 (t
But now notice that this expression goes to +∞ as t → 6− , so the model says that the bacteria
population will become infinite after 6 hours (which of course means that this isn’t a realistic model).
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