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) = 0.01xe0.1x 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) = 0.01xe0.1x 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\u0018 X\u0018 X 16th 11th Homework Due November \u0018 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 \u0002 \u0003 (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 \u0010 \u0011 4 y y 4 \u0001 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 \u0010 12y 2 2 \u0011 (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 y2 2 \u0001 \u0001 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 = xex 2 y = xex 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 xex = xex . xex = xex 2 2 xex xex = 0 \u0010 \u0011 2 x ex ex = 0 2 At this point we have that x = 0 or ex ex = 0. The second equation is solved as follows: 2 ex ex = 0 2 ex = ex \u0010 \u0011 \u0001 2 ln ex = ln ex 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 xex xex dx. 0 (b) First separate the integral as Z 1 2 xex xex dx = 0 Z 0 1 2 xex dx Z 1 xex 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 xex dx = 12 Z x=1 x=0 h 2 x=1 x=1 = eu du = 12 [eu |x=0 = 21 ex x=0 dw 2x . e1 2e To compute the second (right) integral we do integration by parts with u = x and dv = ex dx so that du = dx and v = ex . Z 1 Z 1 1 1 \u0002 1 \u0002 \u0002 \u0001 \u0001 ex dx = xex 0 + ex 0 = 1e 0 + 1e + 1 = e2 xex dx = xex 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 = xex dx = 0 0 0 1 e1 2e e2 e = 3e 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 tan1 ( 12 25x2 4) + c and 25x2 4 2 cos1 ( 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 xex dx, if it converges. 3. Compute 1 Solution: 0.74 Z 2 xex 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. x3 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 x5 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 \u0018 =3 \u0018 X \u0018 (x + 1)(1 5) 6 X \u0018 X B= 2(5) 16 6 = = 1 \u0018 \u0018 X \u0018 (5 + 1)X (x 5) 6 X \u0018 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 x5 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 x3 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) x2 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 x3 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 x1 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 \u0012 1 1 2x 4 \u0013 +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 \u0001 du = 1 7 \u0012 1 3.8 u 3.8 \u0013 +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 \u0001 (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 \u0001 sin( x + 1) sin(u) dx = 2 x du = 2 sin(u) du = 2 cos(u) + c = 2 cos( x + 1) + c x x sin1 (x) dx. I suggest using substitution with u = sin1 (x). 1 x2 Last Updated: November 14, 2016 Page 13 of 28 Calculus II Materials Math 252 (13768) Solution: First, u = sin1 (x) so Z Z 7. Compute x2 p sin1 (x) dx = 1 x2 Z du dx = 1 1x2 and dx = 8. Compute 1 x2 du. Then \u0010p \u0011 Z u 2 1 x du = u du = 12 u2 + c = 2 1x 1 2 \u00012 sin1 (x) + c x3 + 8 dx. du Solution: First, u = x3 + 8 so dx = 3x 2 . Then Z Z p \u0001 du x2 x3 + 8 dx = x2 u 3x = 2 Z 1 3 Z u1/2 du = 92 u3/2 + c = 2 9 x3 + 8 \u00013/2 +c ex dx. ex + 1 Solution: First, u = ex + 1 so dx = edu x . Then Z x Z Z \u0001 e ex du dx = = u1 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 x1 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 x1 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 \u0002 cos(x2 )\fx=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 \u0001 7x 6 \u0001 7x 6 7x (6x 1)e7x dx = 6x1 e e7x dx = 6x1 e 49 e +c= 7 7 7 Z 14. Compute x2 cos x 2 \u0001 42x13 49 \u0001 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. \u0014 \u0015 Z Z Z \u0001 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 tan1 (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). \u0014 \u0015 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 \u0001 3e2x 6 sin( 13 x) cos( 13 x) + c 37 6 t sin( 13 t) dt. Z Solution: 3 6 \u0002 t sin( 13 t) dt = 9 sin t 3 \u0001 3t cos t 3 \u0001\f6 = 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) 0.5 1 dx 1 x2 Solution: Z 1 1 h 1 (a) ex xe dx = ex e+1 xe+1 1.45 0 Z0 10 Z 10 10 \u0002 5 1 (b) dx = 5 dx = 5 tan1 (x)\f0 7.36 2 2 x +1 x +1 0 0 Last Updated: November 14, 2016 Page 17 of 28 Calculus II Materials Z Math 252 (13768) 10 1 x (c) Z1 10 dx = [ln(x)|1 2.30 0.5 (d) 0.5 0.5 \u0002 1 dx = sin1 (x)\f0.5 1.05 1 x2 5. The graph of y = sin(x) is shown below. y y = sin(x) x 2 (a) Set up a definite integral which would compute the area of the shaded region. Your answer to this part of the problem Z d . isn't a number, but rather the definite integral itself; it should look like (b) Use the evaluation theorem to compute the definite integral that you set up in part (a). Solution: Z (a) sin(x) dx Z0 (b) sin(x) dx = [ cos(x)|0 = ( cos()) ( cos(0)) = 2 0 6. The graph of y = P (t) is shown below. y 3 2 1 t 1 1 2 3 4 5 6 7 8 9 10 11 12 1 2 Z 11 (a) Compute P (t) dt. 0 Z 11 |P (t)| dt. (b) Compute 0 Solution: (a) This value is computed by a signed area. Areas that lie above the x-axis are counted as positive and areas Last Updated: November 14, 2016 Page 18 of 28 Calculus II Materials Math 252 (13768) that lie below the x-axis are counted as negative (as demonstrated by the figure below). The signed area is Z 11 1 1 1 1 1 P (t) dt = 2 (1)(3) + (2)(3) + 2 (1)(3) + 2 (3)(2) (3)(1) + 2 (3)(1) + 2 (1)(2) = 6.5 0 | {z } | {z } areas above the x-axis areas below the x-axis y t (b) This value is the un-signed area. That is, the area below the x-axis and the area above the x-axis are added together (both counted positively). Z 11 1 1 1 1 1 |P (t)| dt = 2 (1)(3) + (2)(3) + 2 (1)(3) + 2 (3)(2) + (3)(1) + 2 (3)(1) + 2 (1)(2) = 17.5 0 | | {z } {z } areas above the x-axis areas below the x-axis Z Thinking of this computation another way, the graph of y = |P (t)| is shown below and 11 |P (t)| dt is the 0 signed gray area; since everything in this graph is above the x-axis, all of the areas count positively. y t 7. A ball was launched straight up (exactly vertical) into the air. It started at ground level and reached it's maximum height after three seconds. The speed of the ball t seconds after it was launched was v(t) = 19.6t + 58.8 meters per seconds. What was the maximum height that the ball reached? (Keep in mind that speed is the derivative of position.) Solution: From the time the ball was launched (t = 0) to the time the ball reaches it's peak (t = 3), the change in height is Z 3 Z 3 3 \u0002 v(t) dt = (19.6t + 58.8) dt = 9.8t2 + 58.8t\f0 = (9.8(3)2 + 58.8(3)) (9.8(0)2 + 58.8(0)) = 88.2 m. 0 0 2x + 4 8. Let f (x) = 1 x + 3 Z if x < 1 19 3 Compute if x 1 19 Solution: Z Last Updated: November 14, 2016 f (x) dx using whatever method you want. 2 1 f (x) dx = 2 19 Z . Z f (x) dx + 2 19 Z 1 f (x) dx = 1 Z (2x + 4) dx + 2 1 19 ( 31 x + 19 3 ) dx = 9 + 54 = 63 Page 19 of 28 Calculus II Materials Math 252 (13768) 9. After school Jane needs to run an errand. She leaves school, rides her bike to the bank, and then rides home and all three locations are on the same road which runs north and south. Her position is defined by a function s as follows: t hours after she leaves school she is s(t) miles north of her school. Her velocity (in miles per hour) is given by 16 12 3 t if 0 t 0.75 s0 (t) = 0 24 5 if 0.75 < t < 1.25 . t 19 if 1.25 t 2.5 (a) Is she going north or south at t = 2? (b) How far had she gone in the first fifteen minutes of her trip? (c) She reaches her house 2.5 hours after leaving school. Is her house to the north or the south of her school and how far north or south? (d) What was the total distance that she biked over her whole trip? Hint: Parts (c) and (d) will both involve an indefinite integral. Once you've set them up, I suggest splitting them into three Z 0.75 Z 1.25 Z 2.5 seperate integrals that look like dt + dt + dt. 0 0.75 1.25 Solution: (a) She is going south because s0 (2) = 24 5 (2) 19 < 0. Z 0.25 Z 0.25 (b) s0 (t) dt = (12 16 3 t) dt 2.83 mi. 0 0 (c) The net change in her position over her trip is given by Z 2.5 Z 0 0.75 0 s (t) dt = Z 1.25 0 s (t) dt + 0 0 Z 0.75 (12 0 2.5 s0 (t) dt s (t) dt + 0.75 = Z 16 3 t) dt Z 1.25 + 1.25 Z 2.5 0 dt + 0.75 1.25 ( 24 5 t 19) dt = 7.5 + 0 + (12.5) = 5. It follows that her house is five miles south of her school. (d) The total distance that she traveled over her trip is given by Z 2.5 |s0 (t)| dt = 0 Z 0.75 |s0 (t)| dt + 0 Z 1.25 |s0 (t)| dt + 0.75 0.75 (12 = Z 0 16 3 t) dt Z 1.25 + 0.75 Z 2.5 |s0 (t)| dt 1.25 Z 2.5 0 dt + 1.25 = 7.5 + 0 + | 12.5| = 20. t 19) dt ( 24 5 Hence her whole trip was 20 mi. 10. Let f (x) = 3 3. Compute x Z 8 |f (x)| dx. 2 Solution: The only root of y = f (x) lies outside of [2, 8] and f (x) < 0 whenever 2 x 8. Using the evaluation theorem, Z 8 3 3 dx = 13.84 x 2 Z 8 and hence |f (x)| dx = 13.84. 2 Last Updated: November 14, 2016 Page 20 of 28 Calculus II Materials Math 252 (13768) 11. A company sells subscriptions to a service online. They currently charge $6 per subscription and they make a revenue of $3300 each month but the market research department is thinking of adjusting the pricing scheme. The company finds that if they charge x dollars for each subscription then they will make a revenue of R(x) dollars each month. The marginal + 750 . Find an equation for R(x). (Hint: What is R(6)? ) revenue function is given by R0 (x) = 500 x x Solution: First, R(x) = 1000 x + 750 ln(x) + c. Since R(6) = 3300, 3300 = 1000 6 + 750 ln(6) + c and c = 493.31. Hence R(x) = 1000 x + 750 ln(x) 493.31. 12. A small electronics company sells a flashlight and is examining their profit from sales of this flashlight. If they sell x flashlights each month then P (x) is their approximate monthly profit from sales of the flashlights. You are given that the marginal profit function is P 0 (x) = 12.5 .05x. The company also approximates that P (0) = 150. (a) Find an equation for P (x). (b) They currently sell 240 flashlights each month but the company is considering making efforts to increase sales figures to 270 flashlights each month. If that were to happen, by how much would the company's profit increase or decrease? Solution: (a) Since P is an antiderivative of P 0 (obviously), P (x) = 0.025x2 + 12.5x + c for some constant, c. Since P (0) = 150 we have 150 = 0.025(0)2 +12.5(0)+c so that c = 150. Thus P (x) = 0.025x2 +12.5x150. (b) The answer to this problem is found through P (270) P (240). There are two ways to determine that value. The first is by using part (a) to determine both P (270) and P (240). The other is by computing Z 270 P 0 (x) dx. In either case, we find that P (270) P (240) = 7.5. This tells us that if sales are increased 240 from 240 flashlights per month to 270 flashlights per month then monthly profit will decrease by $7.50. Homework Due September 30th 1. Differentiate the function g(x) = e(2 sin(x)) . Solution: g 0 (x) = 2 cos(x)e(2 sin(x)) 2. When the market opened today the price of a share of a particular stock was $5 and t hours later it was P (t) dollars where P (t) = 12 x ln(x). At what rate was the price of a share of this stock increasing or decreasing five hours after the market opened? Make sure to include units. Solution: P 0 (t) = 12 + 1.305 dollars per hour. 1 2 ln(x) and P 0 (5) = 1 2 + 1 2 ln(5) 1.305. The price of the stock is increasing at a rate of 3. Let f (x) = x + ln(x). Compute a Riemann Sum for f on the interval [1, 4] using. . . (a) . . . three samples and midpoints. Last Updated: November 14, 2016 Page 21 of 28 Calculus II Materials Math 252 (13768) Solution: (1.5 + ln(1.5)) (1) + (2.5 + ln(2.5)) (1) + (3.5 + ln(3.5)) (1) 10.07 (b) . . . three samples and left endpoints. Solution: (1 + ln(1)) (1) + (2 + ln(2)) (1) + (3 + ln(3)) (1) 7.79 (c) . . . six samples and right endpoints. Solution: (1.5 + ln(1.5)) (0.5) + (2 + ln(2)) (0.5) + (2.5 + ln(2.5)) (0.5) + (3 + ln(3)) (0.5) + (3.5 + ln(3.5)) (0.5) + (4 + ln(4)) (0.5) 11.13 Z 5 4. In this problem we're going to be approximating A = 2 sin 0 below for convenience: 2x \u0001 \u0001 + 4 dx. The graph of y = 2 sin 2x \u0001 + 4 is provided y 6 y = 2 sin 2x \u0001 +4 5 4 3 2 1 x 2 1 1 2 3 4 5 6 7 8 (a) Sketch the region whose area is A. Solution: The answer to this question is the figure shown below (the actual region is drawn in red): y 6 5 4 3 2 1 x 2 1 1 2 3 4 5 6 7 8 (b) Find an approximation of A by computing a Riemann sum with five samples and right endpoints. Last Updated: November 14, 2016 Page 22 of 28 Calculus II Materials Math 252 (13768) Solution: The area is approximately 1(6) + 1(4) + 1(2) + 1(4) + 1(6) = 22. (bonus) Redo the sketch from part (a). In addition, sketch the rectangles that you used in part (b) to approximate A. Solution: The answer to this question is the figure shown below (the rectangles are drawn in gray): y 6 5 4 3 2 1 x 1 2 3 4 5 (d) Find a different approximation of A using rectangles. This time, use a Riemann sum with five samples and left endpoints. Solution: The area is approximately 1(4) + 1(6) + 1(4) + 1(2) + 1(4) = 20. 5 Z 5. Use a Riemann sum to approximate ln(x) dx. Use four samples and use right endpoints. 3 Solution: 1 2 ln(3.5) + 1 2 1 2 ln(4) + Z 6. Use a Riemann sum to approximate ln(4.5) + 1 2 ln(5) 2.88 5 ln(x) dx. Use four samples and use midpoints. 3 Solution: 1 2 ln(3.25) + 1 2 ln(3.75) + 1 2 ln(4.25) + 1 2 ln(4.75) 2.75 7. Compute the following definite integrals geometrically. Z 8 (a) 3 dx 2 Z 8 3 dx (b) 2 Z 5 (c) 2 1 2 (x 1) dx Last Updated: November 14, 2016 Page 23 of 28 Calculus II Materials Z Math 252 (13768) 5 (d) 2 1 2 |x 1| dx (If you don't know what the graph of y = 12 |x 1| looks like then use technology to find out.) Solution: Each of these is found by drawing a picture of the region and using geometry. (a) 3(8 2) = 18 (b) [3(8 2)] = 18 (c) [ 21 (1.5)(3)] + 12 (2)(4) = 1.75 (d) 12 (1.5)(3) + 12 (2)(4) = 6.25 8. Suppose that f and g are functions and assume the following: Z Z 0 Z 1 f (x) dx = 6 f (x) dx = 4 5 5 9 Z 0 g(x) dx = 3 f (x) dx = 7 5 1 Compute each of the definite integrals below. Z 9 (a) f (x) dx 5 Z 1 (b) f (x) dx 0 Z 0 (c) 12f (x) dx 5 Z 0 (d) f (x) + g(x) dx 5 Z 0 f (x) 3g(x) dx (e) 5 Solution: Z 9 Z 1 Z 9 (a) f (x) dx = f (x) dx + f (x) dx = 4 7 = 3 5 1 Z5 Z Z Z 1 0 1 1 (b) f (x) dx + f (x) dx = f (x) dx 6 + f (x) = 4 0 5 0 Z5 Z 0 0 12f (x) dx = 12 f (x) dx = 12(6) = 72 (c) 5 Z5 Z Z 0 0 0 (d) f (x) + g(x) dx = f (x) dx + g(x) dx = 6 + (3) = 3 5 5 Z5 Z Z 0 0 0 (e) f (x) 3g(x) dx = f (x) dx 3 g(x) dx = (6) 3(3) = 15 5 Last Updated: November 14, 2016 5 Z 1 f (x) = 4 6 = 2 0 5 Page 24 of 28 Calculus II Materials Math 252 (13768) Quiz and Exam Solutions Week 2 Quiz Z 10 1. The graph of y = P (x) is shown below. Compute P (x) dx. 0 y 4 3 2 1 x 1 2 3 4 5 6 7 8 9 10 1 2 Solution: 2.5 Z 6 2. Approximate 2 12 dx using a Riemann sum with four samples and left endpoints. x1 Solution: 25 Z 3. Compute (x3 4)2 dx. x4 Solution: Z 4. Compute 1 3 8 x 5. Compute 2 8 ln |x| sec2 (x) + ex + Solution: Z 1 3 3x 1 3 16 3 3 x +c 7 dx. 1 + x2 tan(x) + ex + 7 tan1 (x) + c 20 dx using the evaluation theorem. x Solution: 2.27 Last Updated: November 14, 2016 Page 25 of 28 Calculus II Materials Math 252 (13768) 6. A train in Arizona leaves its station on a track running north and south and t hours later it is s(t) mile's north of the Mexican border. The train's velocity is s0 (t) = 50t(t 1)(t 3) = 150t 200t2 + 50t3 and the train travels for a total of three hours. (a) At the end of the train's trip, is it north or south of the station and by how far? (You must do this problem without the information in part (c).) 3 Z s0 (t) dt = Solution: 3 Z 0 south of the station. 150t 200t2 + 50t3 dt = 112.5. At the end of the trip, the train is 112.5 mi 0 (b) What is the total distance that the train travels over the course of the three hour trip? Z 3 Solution: |s0 (t)| dt = 1 Z 0 150t 200t2 + 50t3 dt 0 Z 1 3 150t 200t2 + 50t3 dt = 125 6 + 400 3 154.17 mi. (c) When the train finally stops (at the end of the three hour trip) it is 20 mi north of the Mexican border. Find an equation for s(t). 25 4 200 3 25 4 3 2 Solution: s(t) = 75t2 200 3 t + 2 t + c and s(3) = 20 so c = 132.5. Thus s(t) = 75t 3 t + 2 t + 132.5 Week 4 Quiz Z (a) Evaluate 17x + 33 dx. 2x2 5x 3 Solution: 12 ln |3 x| Z (b) Evaluate 7 2 ln |2x + 1| + c 4x + 1 dx. 2x2 + x 21 Solution: 2 2x2 + x 21 + c Z (c) Evaluate (3x2 + 1) ln(x) dx. Solution: (x3 + x) ln(x) 13 x3 x + c Last Updated: November 14, 2016 Page 26 of 28 Calculus II Materials Math 252 (13768) Z (d) Evaluate sin(ln(x)) dx. x Solution: cos(ln(x)) + c Z 5 xe3x dx. (e) Evaluate 0 5 \u0014 1 3x Solution: e (3x + 1) = 0.11 9 0 Z (f) Evaluate 0 5 3x + 2 dx. x2 + 2x + 1 \u0014 Solution: 5 1 + 3 ln |x + 1| = 4.54 x+1 0 Last Updated: November 14, 2016 Page 27 of 28 Calculus II Materials Math 252 (13768) Some Review Problems Here are some problems from the book that are appropriate to what we've been doing in class. This is not a perfect review; it is just some problems to practice if you're looking for problems. Keep in mind that the book does not word things the same way that I word things and not all the problems fit the mold of the class. If you find a problem that doesn't seem appropriate then you should skip it. Conversely, these problems don't cover every single thing I could ask. I don't want to hear you say \"that didn't look like anything in the review.\" Section 5.1: 1-6, 11-13 Section 5.2: 1-12, 21-25 (use the evaluation theorem for the integrals), 31-44, 47-49 Section 5.3: 1-30 (the \"Table of Indefinite Integrals\" in section 5.3 of the book might be helpful for these), 31, 32, 37-40, 43-48, 51-56, 59-68 Section 5.4: 2-4, 7-18, 19-21, 23-25, 30, 31 Section 5.5: 1-6, 7-36, 51-57, 62-64, 66-68 Section 5.6: 1-24 (some of these are exceptionally hard but I'm not going to go through and check all of them), 25-30 (these are all exceptionally hard), 31-34 (don't worry about illustrating), 43, 44 Section 5.7: 1-4, 7-10, 21-23 Section 5.10: 1, 2, 5-34 (some of these might not be appropriate because they are too difficult to antidifferentiate; if you see one whose antiderivative you can't find then just skip it) Last Updated: November 14, 2016 Page 28 of 28