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[answered] Calculus II Materials Math 252 (13768) This document will e

This is Calculus problem that i need Homework due November 18th, problem 1 to 7

Calculus II Materials Math 252 (13768) This document will evolve as the term progresses. It currently contains the following:

Homework: (Starting on page 2.) Homework will be assigned every day shortly after class. It is due in my mailbox by 1:50pm

every Friday. Solutions to homework problems will be included eventually. Sometimes I?ll give a lot of information and

sometimes I?ll give less, but I?ll try to include some explanation of how the problem is done. The solutions will be in boxes

and within them you?ll find something written in red; that?s the actual answer to the problem. Here are some notes:

I. Sometimes I won?t be able to put the whole answer in red (for example, if the answer is a figure). In these cases I?ll

just explain what the answer is supposed to look like.

II. In my .pdf I?m able to draw graphs and things very accurately. I don?t expect perfectly drawn figures on your

homework, though; just do your best.

III. Please check the syllabus to make sure that you?re formatting your homework correctly. It?s important.

IV. Please do not print out any pages from this document to include with your assignment. If you need to use a figure

then I expect you to re-draw the figure yourself. (This is good practice for exams.)

Quiz and Exam Solutions: (Starting on page 25.) At some point after we take a quiz or exam I will try to add the solutions.

Exam Review Problems: (Starting on page 28.) I will put some problems from each section that are meant for exam review.

Note that these are NOT always problems which will make good exam questions. Instead, they are problems which I think

you should capable of doing based on the skills that I?ve tried to teach you. The homework problems, quiz problems, and

examples done in class are a closer approximation of the exam problems. The book problems that I list are meant to

practice skills; some of them will be much too difficult for an exam. Last Updated: November 14, 2016 Page 1 of 28 Calculus II Materials Math 252 (13768) Homework

Homework Due November 18th

1. Let x be a continuous random variable with ?1 ? x ? 1. One (and only one) of the following functions is a probability

density function for x. Which is it? (This is a multiple choices question; your answer should be ?A,? ?B,? or ?C.?)

(a) f (x) = 2 ? 4.5x2

(b) f (x) = (1 ? x)ex

(c) f (x) = 1

(x + 3) ln(2) 2. Joan has been tracking the batting averages of all the players in the little league in which her son plays. (In baseball, a

player?s ?batting average? is their percentage of hits represented as a decimal. For example, if Dave gets 3 hits in 10 at

bats then his batting average is 0.300.) If x represents a randomly selected player?s batting average then the probability

density function she discovered for x is f (x) = 18.75x(2 ? 5x). Note that she assumed 0 ? x ? 0.4.

(a) What is the probability that a randomly selected player has a batting average above 0.3?

(b) What is the expected value of the batting average of a player in this league?

3. A meteorologist has been studying the snowfall in a particular city in Japan during the month of January. Let x represent

the inches of snow that this city gets on a randomly selected day in January and assume 0 ? x ? 5. The meteorologist

finds that the probability density function for x is f (x) = 0.0006x4 ? 0.03x2 + 0.375.

(a) On a randomly selected day in January, what is the probability that this city receives between 1 and 5 inches of snow?

(b) What is the expected value of the amount of snowfall that a city gets on a randomly selected day in January?

4. A mathematician has been watching a game of poker and studying the size of each bet that a player makes. Let x be the

size of a randomly chosen bet in dollars. There is no minimum or maximum bet in this game so x ? 0. The mathematician

has found that f (x) = 0.01xe?0.1x is a reasonable approximation for a probability density function.

(a) What is the probability that a randomly chosen bet is exactly $30?

(b) What is the probability that a randomly chosen bet is between $20 and $40?

(c) What is the probability that a randomly chosen bet is less than $15 (i.e. that x < 15)?

(d) What is the probability that a randomly chosen bet is greater than $15 (i.e. that x > 15)?

(e) What is the expected value of a randomly chosen bet?

Here are some notes about this problem:

R

? You might want to compute f (x) dx before you start the problem; you?ll need it several times.

? While I did not make it a part of this homework problem, you should be able to verify that f (x) = 0.01xe?0.1x is a

probability density function for x ? 0.

? If you did part (c) correctly then it is possible to do part (d) easily without evaluating an integral. Can you see why?

? In general ?at least 15? or ?greater than 15? should mean ?x > 15.? Similarly, ?at most 15? or ?less than 15? should

mean ?x < 15? or, in this example ?0 ? x < 15.?

5. Phoebe predicts that the price of a stock will oscillate over the course of the next year. She predicts that t days from now

?

t) + 10 dollars per share.

the price of the stock will be P (t) = 5 sin( 10

(a) What is the average value of the stock between t = 22 and t = 25?

(b) There is an occasion 25 days from now when she will need some extra money and she is thinking about selling her

shares in this stock. One option that she has is to wait until moments before (i.e. between t = 22 and t = 25) to sell

her stock. There is another window of time a week before that ? specifically between t = 15 and t = 18 ? during

which she has the option to sell, as well. Use averaging to determine when she should sell her stock (between those

two windows of time) if she wants to get the most money that she can out of the shares. Explain your answer.

Last Updated: November 14, 2016 Page 2 of 28 Calculus II Materials Math 252 (13768) 6. Consider the region shown below. Let (?

x, y?) be the center of mass of a thin laminar solid with this area.

y (0, 1) y= (?1, 0) ? 1 ? x2 (1, 0)

x (a) Set up but do not solve an equation for x

?.

(b) Set up but do not solve an equation for y?.

(c) Find the center of mass. (That is, solve the equations that you set up and write them as a point.)

7. Consider the region shown below. The two prabolas are y = ?x2 + 6x ? 4 and y = ?2x2 + 12x ? 9. Let (?

x, y?) be the center

of mass of a thin laminar solid with this area.

y (3, 9) (3, 5)

(1, 1) (5, 1)

x (a) Set up but do not solve an equation for x

?.

(b) Set up but do not solve an equation for y?.

(c) Find the center of mass. (That is, solve the integrals that you set up and write them as a point.)

X

X

X 16th

11th

Homework Due November

Note that this assignment is due on Wednesday, November 16th. Put it in my mailbox by 3pm; I will not accept it

any later than that, meaning you cannot turn it in during class.

1. Consider the region in the first quadrant enclosed by y = ex , y = 15, and the y-axis.

(a) A solid has this region as a base and each cross-section perpendicular to the x-axis is a square. Set up (but do not

solve) a definite integral which computes the volume of this solid.

(b) A different solid has the same region as a base and each cross-section perpendicular to the y-axis is an equilateral

triangle. Set up (but do not solve) a definite integral which computes the volume of this solid.

2. Consider the region in the first quadrant bound by y = 8, y = x3 , and the y-axis.

(a) Set up (but do not solve) a definite integral which computes the volume of the solid obtained by revolving the region

about the y-axis.

(b) Set up (but do not solve) a definite integral which computes the volume of the solid obtained by revolving the region

about the x-axis.

?

3. Consider the region in the first quadrant bound by y = x and y = 21 x. Compute the volume of the solid obtained by

revolving this region about the (horizontal) line y = 3. (In this problem you need to evaluate the definite integral after

you set it up.) This one can be tricky so think about it carefully.

4. Consider the region (in the first quadrant) enclosed by y = ?x2 + 6x ? 4 and y = ?2x2 + 12x ? 9. Last Updated: November 14, 2016 Page 3 of 28 Calculus II Materials Math 252 (13768) (a) Sketch the region. (Be as careful as you can. You should know how to find intersection points and you should know

how to find the vertex of a parabola. Include these things in your sketch.)

(b) Set up AND solve a definite integral which computes the area of the region.

(c) Set up AND solve a definite integral which computes the volume of the solid obtained by revolving the region around

the line y = ?1.

(d) Set up AND solve a definite integral which computes the volume of the solid obtained by revolving the region around

the line x = 10.

Important Note: In parts (b), (c), and (d) you need to include both the setup and the computation. I want your

answers to look like the following:

Z

d

=

The expression on the left is the setup of the definite integral and the right side should be the number to which the integral

evaluates.

5. The graph below shows the region in the first quadrant between y = 0 and y = 3x3 ? x4 . Find the volume of the solid

obtained by revolving this region about the y-axis. (You should be able to find the roots of this polynomial easily.)

y x y = 3x3 ? x4 6. Consider the region in the plane which is between x = 0, x = ?, y = 0, and y = sin(x).

(a) Set up but do not solve a definite integral which computes the volume of the solid obtained by revolving this region

about the x-axis.

(b) Set up but do not solve a definite integral which computes the volume of the solid obtained by revolving the same

region about the y-axis.

(c) Solve the integral that you obtained in part (b). That is, compute the volume of the solid described in part (b).

5

7. (a) The figure below contains the graphs of y + 3 = x+2

and x + y = 1. Find the two points at which the graphs intersect.

Finding points means finding both coordinates and not just x-values. Each point should look like ( , ). (Ignore

the shading for now; that isn?t used in this part of the problem.)

y x (b) Consider only the shaded region (the region which is between the two graphs and below the x-axis). Set up but do

not solve a definite integral which computes the volume of the solid obtained by revolving this region about the line

y = 7.

(c) Set up but do not solve a definite integral which computes the volume of the solid obtained by revolving the same

region about the line x = 9. Last Updated: November 14, 2016 Page 4 of 28 Calculus II Materials Math 252 (13768) Homework Due November 4th

1. Each of the problems below describes an area. You are to set up (but NOT solve) a definite integral which evaluates to

the area described. For some of the problems you may need to use a sum of multiple definite integrals to describe the area.

REMEMBER: Do not solve the integrals. If your answers are numbers then you will receive no credit.

Your answers should be definite integrals.

(a) The area is the shaded region in the figure below.

y = ln(x) 2 1 e Z e2 2 ey dy Solution:

1 (b) The area enclosed by y = x2 ? 4x and y = 6 ? x2 . Z 3

(6 ? x2 ) ? (x2 ? 4x) dx Solution:

?1 (c) The graphs of y = sin(x) and y = cos(x) are shown below. The area is the shaded region. Note: There were no

numbers given in the figure. That was intentional; you should be able to work them out.

y y = sin(x) x y = cos(x) Z

Solution: ?

4 0 Last Updated: November 14, 2016 Z

(cos(x) ? sin(x)) dx + ?

2

?

4 (sin(x) ? cos(x)) dx Page 5 of 28 Calculus II Materials Math 252 (13768) 2. Consider the region (in the first quadrant) enclosed by y = 4

x, y = 4x, and y = 1. (a) Sketch the region. Try to be as accurate as possible and include as much detail as you can. If you?ve graphed the

equations properly then there should be three intersection points in the first quadrant. I?d like you to include all three

in your graph (including both coordinates).

Solution: The answer is the figure below:

y=

y 4

x y = 4x

(1, 4) (4, 1)

y=1

(0.25, 1)

x (b) Set up (but do NOT) solve a definite integral which evaluates to the area of the region. Z 4 Solution:

1

4

y ? y

4 dy (c) Solve the definite integral that you set up in the previous part. Solution: I?ll assume that you can do this. The answer is approximately 3.67. 3. Find the area enclosed by 2x + y 2 = 12 and x = 2y. Z 2 Solution:

?6 12?y 2

2 ? (2y) dy = 128

3 4. Consider the shaded region below.

y ?6 = x ? y 2 x y = 2x + 2 (a) Set up (but do not solve) a definite integral (or a sum of multiple definite integrals if necessary) which evaluates to

the area of the shaded region. (Note that you?re not given any scale on the x-axis and y-axis. You should be able to

find any values that you need.) Last Updated: November 14, 2016 Page 6 of 28 Calculus II Materials Math 252 (13768) Z 0 Solution:

?2 y?2

2

? y 2 ? 6 dy (b) Evaluate the integral that you set up in the previous part. Solution: Hopefully you can work these out by now. The integral evaluates to 19

3 . 5. Repeat parts (a) and (b) of the previous problem using the region shown below. (Part (b) can be tricky. I suggest that

you start by breaking up the integral into two seperate integrals and handling each on its own.)

y y = xe?x

2 y = xe?x x Solution: This one is a little bit more challenging so I?ll give a better explanation.

2

(a) To find the intersection points we need to solve xe?x = xe?x .

xe?x = xe?x 2 2 xe?x ? xe?x = 0

2

x e?x ? e?x = 0

2 At this point we have that x = 0 or e?x ? e?x = 0. The second equation is solved as follows:

2 e?x ? e?x = 0

2 e?x = e?x

2

ln e?x = ln e?x

?x = ?x2

x2 ? x = 0

x(x ? 1) = 0

Hence we have that x = 0 (which is a solution that we already found) or x = 1. The x-values of the two

intersection points are x = 0 and x = 1. Hence the area of the region is

Z 1

2

xe?x ? xe?x dx.

0 (b) First separate the integral as

Z 1 2 xe?x ? xe?x dx = 0 Z

0 1 2 xe?x dx ? Z 1 xe?x dx. 0 I?m going to solve these two integrals separately: Last Updated: November 14, 2016 Page 7 of 28 Calculus II Materials Math 252 (13768) ? To compute the first (left) integral we do a w-substitution with w = ?x2 so that dx =

1 Z

0 2 xe?x dx = ? 12 Z x=1 x=0 h

2 x=1

x=1

=

eu du = ? 12 [eu |x=0 = ? 21 e?x x=0 dw

?2x . e?1

2e ? To compute the second (right) integral we do integration by parts with u = x and dv = e?x dx so that

du = dx and v = ?e?x .

Z 1

Z 1 1 1 1

e?x dx = ?xe?x 0 + ?e?x 0 = ? 1e ? 0 + ? 1e + 1 = e?2

xe?x dx = ?xe?x 0 +

e

0 0 Putting all of these things together, we have

Z

Z 1

Z 1

?x2

?x2

?x

xe

dx ?

xe

? xe dx = xe?x dx = 0 0 0 1

e?1

2e ? e?2

e = 3?e

2e . Homework Due October 28th

Z ? 25x2 ? 4

dx. I suggest using substitution with x =

x

2

sec (x) ? 1. (In fact, you?re going to want to use it twice.) 1. Evaluate Solution:?

There were several

correct answers for this.

?

2

2 tan?1 ( 12 25x2 ? 4) + c and 25x2 ? 4 ? 2 cos?1 ( 5x

) + c.

46656x5

dx. I suggest using substitution with x =

36x2 + 1 Z

2. Evaluate 1

6 2

5 sec(u). You might need the identity tan2 (x) = The two most common were ? 25x2 ? 4 ? tan(u). You might also need the identity sec2 (x) = tan2 (x) + 1. Solution: 324x4 ? 18x2 + Z 1

2 ln |36x2 + 1| + c ? xe?x dx, if it converges. 3. Compute

1 Solution: 0.74

Z ? 2 xe?x dx, if it converges. 4. Compute

?? Solution: 0 Z 0 5. Compute

?? 1

dx, if it converges.

(x ? 3)2 Last Updated: November 14, 2016 Page 8 of 28 Calculus II Materials Math 252 (13768) Solution: 0.33 Z 0 6. Compute

?? 1

dx, if it converges.

x?3 Solution: divergent Z

7. Compute

0 5 1

dx, if it converges.

(x ? 5)2 Solution: divergent Z 5 8. (a) Compute 2x ? 3

p 0 (5 ? x)(x + 2) dx, if it converges. Solution: 6.32 Z 3 (b) Compute

?2 2x ? 3

p (5 ? x)(x + 2) dx, if it converges. Solution: ?6.32 Z 5 (c) Compute

?2 2x ? 3

p (5 x)(x + 2) dx, if it converges. Solution: 0 Z ? 9. Compute cos(x) dx, if it converges.

0 Solution: divergent Z

10. Compute

1 ? ln(x)

dx, if it converges.

x2 Solution: 1 Last Updated: November 14, 2016 Page 9 of 28 Calculus II Materials Math 252 (13768) Homework Due October 21st

Z

1. Evaluate sin5 (x) cos7 (x) dx. Solution: The (marginally) easier substitution is letting u = cos(x) so that dx =

Z du

? sin(x) . Z

Z

Z

sin5 (x) cos7 (x) dx = ? sin4 (x)u7 du = ? (1 ? cos2 (x))2 u7 du = ? (1 ? u2 )2 u7 du

Z

1 12

u + c = 15 cos10 (x) ? 81 cos8 (x) ?

= 2u9 ? u7 ? u11 du = 15 u10 ? 18 u8 ? 12 Z

2. Evaluate cos12 (x) + c sec3 (x)

dx.

csc5 (x) Solution: Let u = cos(x) so du =

Z 1

12 dx

? sin(x) . Z

Z

sin5 (x)

sin4 (x)

(1 ? u2 )2

dx

=

?

du

=

?

du

3

3

cos (x)

u

u3

Z

1

2

1

= ? 3 + ? u du = 2 + 2 ln |u| ? 12 u2 + c =

u

u

2u 3. (a) Find real numbers A and B such that

B= 1

2 sec2 (x) + 2 ln | cos(x)| ? 1

2 cos2 (x) + c 2x ? 16

A

B

=

+

. (Write your answer as ?A =

(x + 1)(x ? 5)

x+1

x?5 and .?) Solution: It?s hard to type an explanation of the hand-cover method, but that?s what you should use:

?18

2(?1) ? 16

=

A=X

=3

X

(x

+

1)(?1

?

5)

?6

X

X B= 2(5) ? 16

?6

=

= ?1

X

(5 + 1)X

(x

?

5)

6

X

X A = 3 and B = ?1. Z

(b) Evaluate 2x ? 16

dx.

(x + 1)(x ? 5) Z

Solution: Z

4. Evaluate 2x ? 16

dx =

(x + 1)(x ? 5) Z 3

1

?

dx = 3 ln |x + 1| ? ln |x ? 5| + c

x+1 x?5 t2 ? t + 3

dt.

2t ? 1 Last Updated: November 14, 2016 Page 10 of 28 Calculus II Materials Math 252 (13768) Solution: Let u = 2t ? 1 so that dt =

Z Z

5. Evaluate x2 t2 ? t + 3

dt =

2t ? 1

= 1

2 du and t = 21 u + 12 . ( 12 u + 21 )2 ? ( 12 u + 12 ) + 3

du =

u

2

11

1

16 (2t ? 1) + 8 ln |2t ? 1| + c 1

2 Z Z u

8 + 11

8u du = 1 2

16 u + 11

8 ln |u| + c x + 37

dx.

+ 4x ? 21 Solution: Using a partial fraction decomposition we have

Z

Z

Z

x + 37

x + 37

?3

4

dx

=

dx

=

+

dx = ?3 ln |x + 7| + 4 ln |x ? 3| + c.

2

x + 4x ? 21

(x + 7)(x ? 3)

x+7 x?3 Z

6. Evaluate 1

dx.

x2 ? x ? 2 Solution: Using a partial fraction decomposition we have

Z

Z

Z

1

1/3

?1/3

1

dx =

dx =

+

dx =

x2 ? x ? 2

(x ? 2)(x + 1)

x?2 x+1 Z

7. Evaluate Z Z Z Z Z 10. Evaluate ln |x + 1| + c. sin3 x cos19 (x) dx = 1

22 cos22 (x) ? 1

20 cos20 (x) + c 84

dx =

(x ? 3)(x + 1)(x + 4) Z 3

7

4

?

+

dx = 3 ln |x ? 3| ? 7 ln |x + 1| + 4 ln |x + 4| + c

x?3 x+1 x+4 3x2 ? 20x ? 18

dx Hint: Start by factoring an x out the denominator.

x3 + 5x2 ? 6x Solution: Z 1

3 84

dx.

(x ? 3)(x + 1)(x + 4) Solution: 9. Evaluate ln |x ? 2| ? sin3 (x) cos19 (x) dx. Solution: 8. Evaluate 1

3 3x2 ? 20x ? 18

dx =

x3 + 5x2 ? 6x Z 3

5

5

+

?

dx = 3 ln |x| + 5 ln |x + 6| ? 5 ln |x ? 1| + c

x x+6 x?1 9x + 8

dx.

(3x + 5)2 Last Updated: November 14, 2016 Page 11 of 28 Calculus II Materials Math 252 (13768) Z 9x + 8

dx =

(3x + 5)2 Solution: Z

11. Evaluate x2

dx =

x2 + 4x + 3 Solution: 12. Evaluate 3

7

7

dx = ln |3x + 5| +

?

+c

3x + 5 (3x + 5)2

9x + 15 x2

dx.

x2 + 4x + 3 Z Z Z Z

1? 9/2

1/2

+

dx = x ? 4.5 ln |x + 3| + 0.5 ln |x + 1| + c

x+3 x+1 12

dx. I suggest using a substitution with 2x ? 1 = 4u.

(2x ? 1)2 + 16 Z 12

dx = ? 32

(2x ? 1)2 + 16 Solution: Z

tan ?1 (u) du = ? 23 tan 8 9 ?1 1 ? 2x

4

+c Homework Due October 14th

1. The graph of y = P (x) is shown below.

y

4 3 2 1

x

1 2 3 4 5 6 7 10 11 12 13 14 ?1

?2 Z

(a) Approximate 4 P (x) dx using a Riemann sum with four samples and left endpoints.

0 Z 4 P (x) dx ? 2(1) + 1(1) + 4(1) + 1(1) = 8 Solution:

0 Z

(b) Use geometry to compute 14 P (x) dx.

4 Z 14 Solution:

4 Last Updated: November 14, 2016 P (x) dx = 12 (5)(3) ? 21 (4 + 2)(2) + 12 (1)(2) = 5

2 Page 12 of 28 Calculus II Materials Math 252 (13768) Z

(c) Define Q(x) = x P (t) dt. What is Q0 (7)? 0 Solution: Q0 (7) = P (7) = 2 Z

2. Compute x cos(x2 ) dx. I suggest using substitution with u = x2 . du

Solution: First, u = x2 so dx = 2x

. Then

Z

Z

x cos(x2 ) dx = x cos(u) Z

3. Compute Z

(7x + 15) Z 2.8 Z Z Z Z 1

2 cos(u) du = 1

2 sin(u) + c = 1

2 sin(x2 ) + c 1

7 Z

dx = du. Then

2.8 u 1

7

du = 1

7 1 3.8

u

3.8

+c= 5

133 (7x + 15)3.8 + c du

dx = 1

x (ln(x))2

dx =

x and dx = x du. Then

Z u2

(x du) =

x Z u2 du = 31 u3 + c = 13 (ln(x))3 + c ?

?

sin( x + 1)

?

dx. I suggest using substitution with u = x + 1.

x Solution: First, u = 6. Compute Z

= (ln(x))2

dx. I suggest using substitution with u = ln(x).

x Solution: First, u = ln(x) so 5. Compute (7x + 15)2.8 dx. I suggest using substitution with u = 7x + 15. Solution: First, u = 7x + 15 so dx = 4. Compute du

2x ? x + 1 so du

dx = 1

?

2 x ?

and dx = 2 x du. Then ?

Z

Z

?

sin( x + 1)

sin(u) ?

?

?

dx =

2 x du = 2 sin(u) du = ?2 cos(u) + c = ?2 cos( x + 1) + c

x

x sin?1 (x)

?

dx. I suggest using substitution with u = sin?1 (x).

1 ? x2 Last Updated: November 14, 2016 Page 13 of 28 Calculus II Materials Math 252 (13768) Solution: First, u = sin?1 (x) so

Z Z

7. Compute x2 p sin?1 (x)

?

dx =

1 ? x2 Z du

dx ? = ? 1

1?x2 and dx = 8. Compute 1 ? x2 du. Then p

Z

u

2

1 ? x du = u du = 12 u2 + c =

2

1?x 1

2 2

sin?1 (x) + c x3 + 8 dx. du

Solution: First, u = x3 + 8 so dx = 3x

2 . Then

Z

Z

p

?

du

x2 x3 + 8 dx = x2 u 3x

=

2 Z ? 1

3 Z u1/2 du = 92 u3/2 + c = 2

9 x3 + 8 3/2 +c ex

dx.

ex + 1 Solution: First, u = ex + 1 so dx = edu

x . Then

Z x

Z

Z

e

ex

du

dx

=

=

u?1 du = ln(u) + c = ln(ex + 1) + c

x

ex + 1

u e

Two notes:

R

? It is possible to use y = ex as your substitution, although you will then end up with

a second substitution to compute.

? Please note that ln(ex + 1) does not equal ln(ex ) + ln(1). Z

9. Compute ?

x. cos( ?x )

dx = ? ?1

x2 Z Z du which requires cos( ?x )

dx

x2 Solution: Let u = 10. Compute 1

u+1 Z cos(u) du = ? ?1 sin(u) + c = ? ?1 sin( ?x ) + c 10 sin(x) sec9 (x) dx Solution: Let u = cos(x).

Z

10 Z

11. Compute sin(x)

dx = ?10

cos9 (x) Z 1

1 ?8

du = ?10( ?8

u )+c=

u9 5

4 sec8 (x) + c x2 + 2

dx

x?1 Last Updated: November 14, 2016 Page 14 of 28 Calculus II Materials Math 252 (13768) Solution: Let u = x ? 1.

Z 2

Z

Z 2

x +2

(u + 1)2 + 2

u + 2u + 3

dx =

du =

du

x?1

u

u

Z

= u + 2 + u3 du = 21 u2 + 2u + 3 ln(u) + c = 12 x2 + x ? Z

12. Compute 3

2 + 3 ln(x ? 1) + c 2 x sin(x2 ) dx 1 Solution: Let u = x2 .

Z 2

x sin(x2 ) dx =

1 Z

13. Compute 1

2 Z x=2 sin u du =

x=1 1

2 x=2 [? cos(u)|x=1 = 1

2 x=2

? cos(x2 )x=1 ? 0.597 (6x ? 1)e7x dx. I suggest using integration by parts. Solution: Let u = 6x ? 1 and dv = e7x dx so that du = 6 dx and v = 71 e7x

Z

Z

7x 6

7x

6 7x

(6x ? 1)e7x dx = 6x?1

e

?

e7x dx = 6x?1

e ? 49

e +c=

7

7

7 Z

14. Compute x2 cos x

2 42x?13

49 e7x + c dx. I suggest using integration by parts twice. Solution: Let u = x2 and dv = cos( x2 ) dx. Then, let U = x and dV = sin( x2 ) dx.

Z

Z

Z

x2 cos x2 dx = 2x2 sin( x2 ) ? 4 x sin( x2 ) dx = 2x2 sin( x2 ) ? 4 ?2x cos( x2 ) + 2 cos( x2 ) dx

= 2x2 sin( x2 ) + 8x cos( x2 ) ? 16 sin( x2 ) + c

Z

15. Compute x2 ln(x) dx. I suggest using integration by parts. If you?re not sure what the variables should be, try each of the following two choices:

? u = x2 and dv = ln(x) dx

? u = ln(x) and dv = x2 dx

Only one of the two will work out nicely (but it?s useful to learn why assignments don?t work nicely sometimes). Solution: Let u = ln(x) and dv = x2 dx. Z x2 ln(x) dx = 31 x3 ln(x) ? 1

3 Z x2 dx = 13 x3 ln(x) ? 91 x3 + c Z

16. Compute ln(x) dx. I suggest doing this problem in a manner similar to how we did Last Updated: November 14, 2016 R tan?1 (x) dx in class. Page 15 of 28 Calculus II Materials Math 252 (13768) Solution: Let u = ln(x) and let dv = dx. Then du = x1 dx and v = x.

Z

Z

Z

x

ln(x) dx = x ln(x) ?

dx = x ln(x) ?

dx = x ln(x) ? x + c

x Z

17. Compute e2x sin( 31 x) dx. Note: We did a problem like this in class. This is very similar and uses the same method but the arithmetic is more challenging (the 2 and 1

3 really ramp things up). Make sure that you?re very careful. Solution: Start by performing integration by parts with u = e2x and dv = sin( 31 x) dx so that du = 2e2x dx and

v = ?3 cos( 13 x).

Z

Z

e2x sin( 13 x) dx = ?3e2x cos( 31 x) + 6 e2x cos( 13 x) dx

Now perform integration by parts a second time with U = e2x and dV = cos( 31 x) dx so that dU = 2e2x dx and

V = 3 sin( 31 x).

Z

Z

e2x sin( 13 x) dx = ?3e2x cos( 13 x) + 6 3e2x sin( 13 x) ? 6 e2x sin( 13 x) dx

Z

Z

e2x sin( 13 x) dx = ?3e2x cos( 13 x) + 18e2x sin( 13 x) ? 36 e2x sin( 31 x) dx

Z

By setting I = e2x sin( 13 x) dx we have I = ?3e2x cos( 13 x) + 18e2x sin( 31 x) ? 36I ? 37I = ?3e2x cos( 13 x) + 18e2x sin( 13 x) and finally

I= Z

18. Compute

3

3e2x

6 sin( 13 x) ? cos( 13 x) + c

37 6 t sin( 13 t) dt. Z

Solution:

3 6

t sin( 13 t) dt = 9 sin t

3 ? 3t cos t

3 6 = 12.96

3 Homework Due October 7th

Z

1. True or False: 2x dx = x2 Z

Solution: False. Last Updated: November 14, 2016 2x dx = x2 + c Page 16 of 28 Calculus II Materials Math 252 (13768) Z

2. True or False: ln(x) dx = x ln(x) ? x + c Solution: True. d

dx (x ln(x) ? x) = (x( x1 ) + (1) ln(x)) ? 1 ? ln(x) 3. Compute each of the following indefinite integrals.

Z

(a)

e dx

Z

x(1 ? x) dx (b)

Z

(c)

Z

(d)

Z

(e) ?

3 x dx 1

3x dx ?

?5 sin(x) + 12 x dx Solution:

Z

(a)

e dx = ex + c

Z

Z

(b)

x(1 ? x) dx = x ? x2 dx = 12 x2 ? 13 x3 + c

Z

Z

?

3

1

x dx = x1/3 dx = 4/3

(c)

x4/3 + c = 34 x4/3 + c

Z

Z

1

1

1

1

(d)

3x dx = 3

x dx = 3 ln |x| + c

Z

?

(e)

?5 sin(x) + 12 x dx = 5 cos(x) + 8x3/2 + c 4. Compute each of the following definite integrals.

Z 1

(a)

ex ? xe dx

0 Z 10 (b)

0 Z 10

1

x (c)

1 Z 5

dx

x2 + 1

dx 0.5 (d)...

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