Math 2260 Exam #2 Solutions - Colorado State University

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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