Math 2260 Exam #2 Solutions
Math 2260 Exam #2 Solutions
1. Integrate
x x2 - 5x + 4 dx.
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:
x
A
B
=
+
.
(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,
x x2 - 5x - 4 dx =
4/3 1/3
-
dx
x-4 x-1
4
1
= ln |x - 4| - ln |x - 1| + C.
3
3
2. Does the improper integral
xe-2x dx
0
converge or diverge? If it converges, find the value of the integral.
Answer: By definition
b
xe-2x dx = lim
xe-2x dx.
0
b+ 0
To evaluate this integral, I want to use integration by parts, letting
u=x du = dx
dv = e-2x dx
v
=
- 1 e-2x 2
=
-1 2e2x .
1
Then the above limit is equal to
lim
b+
xb 1
- 2e2x
+ 02
b
e-2x dx
0
x
1b
= lim
b+
- 2e2x - 4e2x
0
= lim
b+
= lim
b+
-2x - 1 b
4e2x 0
-2b - 1 -1
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'H^opital's Rule to see that
2b + 1
2
lim
b+
4e2b
=
lim
b+
8e2b
=
0,
so we conclude, finally that
xe-2x
dx
=
1 .
0
4
3. Evaluate the definite integral
1 x2
dx.
0 1 - x2
Answer: First, notice that the denominator is undefined when x = 1, so this is an improper integral.
Therefore, by definition,
1 x2
b x2
dx = lim
dx.
0 1 - x2
b1- 0 1 - x2
Make the substitution x = sin . Then dx = cos d. Also, when 0 = x = sin we have that = 0,
and when b = x = sin we have that = arcsin(b). Hence, the above limit is equal to
arcsin(b)
lim
b1- 0
sin2
cos d = lim
1 - sin2
b1-
= lim
b1-
= lim
b1-
arcsin(b) sin2
cos d
0
cos
arcsin(b)
sin2 d
0
arcsin(b) 1 - cos(2)
d
0
2
2
Then this limit is equal to
lim
b1-
sin(2) arcsin(b)
-
= lim
2
40
b1-
= lim
b1-
arcsin(b) sin(2 arcsin(b))
-
2
4
arcsin(b) sin(2 arcsin(b))
-
2
4
/2 sin(2/2)
=-
2
4
= -0
4
=.
4
- (0 - 0)
4. Find an antiderivative for the function
f (x) = e3x cos(x).
Answer: Integrate by parts with u = e3x dv = cos(x) dx
du = 3e3x dx v = sin(x)
Then the given integral is equal to
e3x sin(x) - 3 e3x sin(x) dx.
For the remaining integral, integrate by parts again:
u = e3x dv = sin(x) dx du = 3e3x v = - cos(x)
Substituting this into the above yields
e3x cos(x) dx = e3x sin(x) - 3 -e3x cos(x) + 3 e3x cos(x) dx
e3x cos(x) dx = e3x sin(x) + 3e3x cos(x) - 9 e3x cos(x) dx.
Adding 9 e3x cos(x) dx to both sides yields
10 e3x cos(x) dx = e3x sin(x) + 3e3x cos(x),
and so is an antiderivative for f (x).
e3x sin(x) + 3e3x cos(x) F (x) =
10
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:
dP P 2 = k dt -1
= kt + C. P
Therefore,
-1
P (t) =
.
kt + C
We can determine C by using the initial condition:
-1 P (0) =
k(0) + C -1 100 = , C
so
C
=
-1 100
and
we
can
re-write
the
expression
for
P (t):
-1
P (t)
=
kt
-
1 100
.
Now, we use the fact that P (3) = 200 to solve for k:
and so
-1
P (3)
=
k(3)
-
1 100
-1
200
=
3k
-
1 100
1
200 3k -
= -1
100
600k - 2 = -1,
meaning
that
600k
=
1
and
so
k
=
1 600
.
Therefore,
-1
-1
-600 600
P (t) =
t 600
-
1 100
=
1 600
(t
-
6)
=
t-6
=
. 6-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).
4
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