Math 2260 Exam #2 Solutions - Colorado State University

Math 2260 Exam #2 Solutions

1. Evaluate the indefinite integral

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

du = 2 dx v = sin(3x). 3

Then the above integral is equal to

2

2

2

2

x sin(3x) - sin(3x) dx = x sin(3x) + cos(3x) + C.

3

3

3

9

2.

Shown below is the graph of the function f (x) =

13-2x 2x2 -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

2 13 - 2x

2 2x - 13

- 0 2x2 - 5x - 3 dx = 0 2x2 - 5x - 3 dx

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

2x2

-

5x

-3

=

(2x

+ 1)(x -

3)

=

2x +

1

+

. x-3

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,

2 2x - 13

2

4

1

0 2x2 - 5x - 3 dx = 0

-

dx

2x + 1 x - 3

2

= 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 converge or diverge?

1 1 (x + 1)3 dx

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

(x + 1)3 < x3 .

Moreover,

1

1

x3

dx

=

lim

b

= lim

b

= lim

b

b1 1 x3 dx

b

x-3 dx

1

x-2 b

-2 1

= lim

b

1 =.

2

-1 1 2b2 + 2

Therefore, the improper integral

1

1 x3

dx

converges.

Since

1 (x+1)3

<

1 x3

for

all

x > 1,

the

Direct

Comparison Test implies that the improper integral

1

1 (x+1)3

dx

converges

as

well.

2

Limit

Comparison:

Again,

we

want

to

compare

to

1 x3

,

so

compute

the

limit

1

lim

x

(x+1)3

1 x3

x3

=

lim

x

(x

+

1)3

= lim

x

x3 =1

x+1

since

limx

x x+1

=

limx

x x

?

1 1+1/x

=

1

and

13

=

1.

Since 0 < 1 < and we already saw above that

implies that

1

1 (x+1)3

dx

also

converges.

1

1 x3

dx

converges,

the

Limit

Comparison

Test

Computing the Integral: We could also just compute the value of this definite integral:

1

b1

1

(x

+

1)3

dx

=

lim

b

1

(x + 1)3 dx

b

= lim (x + 1)-3 dx

b 1

= lim

b

(x + 1)-2 b -2 1

-1

1

= lim

b

2(b + 1)2 + 8

1 =.

8

4. Evaluate the definite integral

1x

dx.

0 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,

1x

bx

dx = lim

dx.

0 1-x

b1- 0 x - 1

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

better choice than u = 1 . Therefore, the good choices for u and dv are

x-1

u=x du = dx

dx dv =

= (1 - x)-1/2 dx

1-x

v = -2(1 - x)1/2 = -2 1 - x.

Hence, the above integral is equal to

lim

b1-

b

b

-2x 1 - x 0 + 2

0

1 - x dx = lim

b1-

= lim

b1-

4 =.

3

b

-2x 1 - x 0 -

4 (1 - x)3/2 b

3

0

-2b 1 - b + 0

-

4 (1

-

b)3/2

-

4

3

3

5. A research group publishes a paper claiming that the world wombat population is governed by the

differential equation

(4 - t2)3/2 dW = 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

W = (4 - t2)3/2 dt.

Now, integrate both sides:

ln W =

dW =

W 2 (4 - t2)3/2 dt

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 , so that dt = 2 cos d and we can re-write as

2

4 cos d

(4 - 4 sin2 )3/2 2 cos d = 43/2(1 - sin2 )3/2

4 cos d =

8(cos2 )3/2

1 cos d

= 2

cos3

1 d

= 2

cos2

1 =

sec2 d

2

1 = tan + C.

2

Now, we have to translate from back into the variable t. Since we made the substitution

t

=

2 sin ,

we

know

that

sin

=

t 2

,

and

so

we

can

realize

as

the

indicated

angle

in

the

following

triangle:

Therefore, and so we conclude that

2 t

4 t2

t

tan =

,

4 - t2

2 ln W = (4 - t2)3/2 dt

1t

ln W =

+ C.

2 4 - t2

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) = 1 0 + C = C, 2 4-0

so we know that C = ln 100 and we know that

1t

ln W =

+ ln 100.

2 4 - t2

Now, exponentiating both sides gives us the wombat population as a function of t:

W

= e1 2

t

4-t2

+ln

100

W

=

1

eln 100e 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.

5

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