Math 2260 Exam #2 Practice Problem Solutions

1. Evaluate

Math 2260 Exam #2 Practice Problem Solutions tan3(x) dx.

Answer:

Recall

that

tan(x) =

sin(x) cos(x)

,

so

the

above

integral

is

equal

to

sin3(x) cos3(x) dx.

Now, use the trig identity sin2(x) = 1 - cos2(x) to re-write this as

sin2(x) ? sin(x) cos3(x) dx = = =

(1 - cos2(x)) sin(x)

cos3(x)

dx

sin(x) sin(x) cos3(x) - cos(x) dx

tan(x) sec2(x) dx -

sin(x) dx.

cos(x)

For the first, let u = tan(x). Then du = sec2(x) dx and so

tan(x) sec2(x) dx =

u2

tan2(x)

u du = 2 + C1 = 2 + C1.

For the second integral above, let u = cos(x). Then du = - sin(x) dx, and so

sin(x)

du

=- cos(x)

u = - ln |u| + C2 = - ln | cos(x)| + C2.

Combining these, then, we see that the given integral is equal to

where I let C = C1 + C2. 2. Integrate

tan2(x) + ln | cos(x)| + C,

2 sin4(2x) dx.

Answer:

Using

the

trig

identity

sin2

=

1-cos(2) 2

,

we

know

that

sin4(2x) dx = sin2(2x) 2 dx =

1 - cos(4x) 2 dx =

2

1 - 2 cos(4x) + cos2(4x) dx.

4

The first two terms will be easy to integrate, but for the third I need to use the power-reduction formula again, so the above integral is equal to

1

1 + cos(8x)

1

sin(4x) x sin(8x)

1 - 2 cos(4x) +

dx = x -

++

+C

4

2

4

2

2 16

3x sin(4x) sin(8x)

=-

+

+ C.

8

8

64

1

3. Integrate

x2 - 25 dx. x

Answer: Use the trig substitution x = 5 sec . Then dx = 5 sec tan d, and so the above integral is

equal to

25 sec2 - 25 5 sec tan d =

5 sec2 - 1 tan d

5 sec

= 5 tan2 tan d

= 5 tan2 d.

sin2 = 5 cos2 d.

This is a little annoying, but we can help ourselves out by using the trig identity sin2 = 1 - cos2 to re-write it as

1 - cos2

5

cos2 d = 5

1 cos2 - 1 d

= 5 sec2 - 1 d

= 5 [tan - ] + C.

Now, since x = 5 sec , we know that sec = x/5, meaning that = sec-1(x/5) and, after using the

Pythagorean Theorem, tan =

x2 -25 5

.

Therefore,

our

integral

is

equal

to

5 x2 - 25 - sec-1(x/5) + C = x2 - 25 - 5 sec-1(x/5) + C.

5

4. Evaluate

1 e x(ln x)3 dx.

Answer: By definition

1

b1

e

x(ln

x)3

dx

=

lim

b+

e

x(ln x)3 dx,

which we can evaluate using the substitution u = ln x.

Then

du

=

1 x

dx,

u(e)

=

ln(e)

=

1,

and

u(b) = ln(b), so the above limit is equal to

ln(b) du

u-2 ln(b)

lim

b+ 1

u3

= lim

b+

-2 1

1 ln(b)

= lim

b+

- 2u2

1

1

1

= lim

b+

- 2(ln b)2 + 2(1)2

1 =

2

since the first term goes to zero as b +.

2

5. Suppose a 20 foot chain which weighs 5 pounds per foot is coiled on the ground. One end of the chain is attached to a small crane. How much work does it take to lift this end 20 feet off the ground, so that the chain is fully extended in the air?

Answer: Since the chain weighs 5 pounds per foot, when the end of the crane is x feet in the air, it is supporting a weight of 5x, which is to say, exerting a force of F (x) = 5x. Hence,

20

20

5x2 20 5(400)

W = F (x) dx = 5x dx =

=

- 0 = 1000,

0

0

20

2

so the crane is doing 1000 foot-pounds of work in lifting the chain until it is fully extended.

(Not accidentally, the chain weighs 5 ? 20 = 100 pounds, and points on the chain are being lifted an average of 10 feet off the ground; notice that 100 ? 10 = 1000, which is the amount of work that we calculated was done. Can you figure out why these two numbers coincide?)

6. Evaluate

dx

.

0 x(x + 4)

Answer: By definition,

dx

b dx

= lim

.

0 x(x + 4) b+ 0 x(x + 4)

Let

u

=

x.

Then

du

=

1 2x

dx,

u(0)

=

0

=

0,

and

u(b)

=

b,

so

the

above

limit

is

equal

to

b1

1

b1

lim 2

? dx = lim 2

b+ 0 x + 4 2 x

b+ 0

u2 + 4 du.

Now, let u = 2 tan , meaning that = arctan(u/2). Therefore, (0) = arctan(0/2) = 0, (b) = arctan(b/2), and du = 2 sec2 d, so the above is equal to

lim 2

b+

arctan(b/2) 0

1 4 tan2

+

4

?

2 sec2

d

=

lim

b+

arctan(b/2) sec2

0

sec2 d

=

lim

b+

[]a0rctan(b/2)

= lim (arctan(b/2) - 0)

b+

=,

2

since tan + as b /2-.

7. Solve the differential equation

(x2 + 1) dy = y. dx

Answer: We want to separate variables and integrate, so we have

dy

dx

= y

x2 + 1 .

The left hand side is easy; for the right hand side, let x = tan . Then dx = sec2 d, so

dx

sec2 d

sec2 d

x2 + 1 = tan2 + 1 =

sec2 = + C.

3

Therefore, since = arctan(x), we see that the above equation between integrals is equivalent to

ln y = arctan(x) + C.

Exponentiating both sides, we have that y = earctan(x)+C = eC earctan(x) = Aearctan(x),

where I let A = eC .

8. Solve the differential equation

dy = xy + x,

dx

y(0) = 10.

Answer: Notice that xy + x = x(y + 1), so separating variables and integrating yields

dy = x dx

y+1 x2

ln(y + 1) = + C. 2

We can solve for y by exponentiating both sides

y + 1 = ex2/2+C = eC ex2/2 = Aex2/2

(where A = eC), and then subtracting 1 from both sides:

y = Aex2/2 - 1.

To determine A, we use the initial condition y(0) = 10: y(0) = Ae02/2 - 1 10 = A - 1 11 = A.

Therefore,

y = 11ex2/2 - 1.

9. In a second-order chemical reaction, the reactant A is used up in such a way that the amount of it present decreases at a rate proportional to the square of the amount present. Suppose this reaction begins with 50 grams of A present, and after 10 seconds there are only 25 grams left. How long after the beginning of the reaction will there be only 10 grams left? Will all of the A disappear in a finite time, or will there always be a little bit present? Answer: Since the rate of change of A is proportional to A2, we have

dA = kA2. dt

Since the reaction begins with 50 grams of A, we have A(0) = 50. Likewise, since there are 25 grams left after 10 seconds, A(10) = 25. Then the first question is to determine the time t0 when A(t0) = 10, and the second is to determine whether A(t) ever gets to zero.

To solve the differential equation, we separate variables and integrate:

dA A2 = k dt

1 - = kt + C.

A

4

Therefore,

-1

A(t) =

.

kt + C

Now, we use the fact that A(0) = 50 to determine C:

-1 A(0) =

k(0) + C -1 50 = , C

so

C

=

-

1 50

and

thus

A(t)

=

. -1

kt-

1 50

Now we can determine the constant k using the fact that

A(10) = 25:

Therefore,

-1

A(10)

=

k(10)

-

1 50

-1

25

=

10k

-

1 50

.

1 25 10k - = -1

50 1

250k - = -1. 2

Hence

1 250k = -

2

and

so

k

=

-

1 500

.

Therefore,

the

amount

of

A

is

determined

by

the

equation

-1

-1

500

A(t) =

-

t 500

-

1 50

=

-

1 500

(t

+

10)

=

. t + 10

Now, we want to determine t0 so that A(t0) = 10:

500 A(t0) = t0 + 10

500 10 =

t0 + 10

Therefore,

500 t0 + 10 = 10 = 50,

so t0 = 50 - 10 = 40, meaning that there will be 10 grams left after 40 seconds.

Finally,

500 A(t) =

t + 10

is always positive, so there will always be a little bit of A present.

10. On a hot day, a thermometer was brought outdoors from an air-conditioned building. The temperature inside the building was 21C, and so this is what the thermometer read at the moment it was brought outside. One minute later the thermometer read 27C, and a minute after that it read 31C. What

was the temperature outside? (Impress us and express the answer without using logarithms or the

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