MAT 213 Chapter 13 HW Solutions - Solutions Section 13.1 1 � 2 � 3 � 4 � 5 � 6 � 7 8 9 � � � 10 11 12 13 14 = = 6 = 6 1 6 5 = = 5 ⇒ � 1

# MAT 213 Chapter 13 HW Solutions - Solutions Section...

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Unformatted text preview: Solutions Section 13.1 1. ∫ 2. ∫ 3. ∫ 4. ∫ 5. ∫ 6. ∫ 7. 8. 9. ∫ ∫ ∫ 10. 11. 12. 13. 14. = = 6 = 6 +1 6 5 = + ; = 5 ⇒ + ∫ +1 6 +1 8 7 = + ; = 7 ⇒ + ∫ +1 8 ∫ 1 = 6 + (−5) = (−5) 1 = −5 + ∫ = +1 2 = + ; = 1 ⇒ + ∫ +1 2 2 (−) = − = − + ∫ 2 ( 2 − ) = ( + 3) = (1 + ) = ∫ ∫ ∫ ∫ ∫ = ∫ ∫ ∫ (4 − ) = = 2 − ∫ ∫ + 1 + ∫ ∫ 3 = 3 2 − + 3 2 2 4 + + 2 4 = + 4 − = ∫ 2 + 2 = 4 − 2 + 2 +1 −4 −4 −5 = + ; = −5 ⇒ + =− + ∫ +1 4 (−4) +1 −6 −6 −7 = + ; = −7 ⇒ + =− + ∫ +1 6 (−6) ( 2.3 + −1.3) = ( −0.2 − 0.2) = ∫ ∫ 2.3 + ∫ −0.2 − −1.3 = ∫ 0.2 = 3.3 −0.3 − + 3.3 0.3 0.8 1.2 − + 0.8 1.2 Solutions Section 13.1 15. 16. ∫ ∫ ( 2 − −1) = = 3 √ = ∫ 18. ∫ 19. ∫ 21. 22. 23. 24. 25. 26. 27. 28. ∫ ∫ −1 = 3 − ln || + 3 ∫ ∫ 1 + 2ln || + 5/4 4 5/4 + = + 5/4 5 1/4 = 4/3 3 4/3 + = + 4/3 4 1/3 = (3 4 − 2 −2 + −5 + 4) = 3 = 20. ∫ 2 − ( −2 + 2 −1) = − −1 + 2ln || + = − 4 √ 17. 3 5 −4 + 2 −1 − + 4 + 5 4 (4 7 − −3 + 1) = 4 ∫ ∫ 7 − 4 − 2 ∫ ∫ −3 + −2 + ∫ ∫ 1 = −5 + ∫ 4 4 8 −2 8 −2 − ++ = + ++ 8 2 2 −2 2 1 2 = (2 −1 + (1/4)) = 2ln || + + + = 2ln || + 2/8 + ∫ ( 4) ∫ 4 2 2 2 2 −1 1 3 = (2 −2 + (1/4) 2) = + + + = −2 −1 + 3/12 + ) ∫ (2 ∫ 4 4 3 −1 1 2 1 −1 −2 2 1 = ( −1 + 2 −2 − −3) = ln || + 2 + − − + = ln ||− + + ∫ ( 2 3) ∫ 2 2 −1 −2 3 1 1 −4 −6 1 1 = (3 −1 − −5 + −7) = 3ln ||− − + + + = 3ln || + − + 4 ∫ ( 5 7) ∫ −4 −6 4 6 6 ∫ (3 0.1 − 4.3 − 4.1) = 3 1.1 5.3 − − 4.1 + 1.1 5.3 2.1 3.1 − 2.3 = − 2.3 + ( ) ∫ 2 6.2 3 ∫ ( 0.1 1 ( 1.1 − − 4 1.1 ) = ∫ (3 −0.1 − 4 −1.1) = 3 0.9 −0.1 0.9 40 −4 + = + + 0.9 0.3 0.1 −0.1 1 −0.1 10 = ( −1.1 − −1) = − ln || + = − − ln || + ∫ ) −0.1 0.1 Solutions Section 13.1 29. 1.2 3 5.1 2 −0.2 5.1 − = (5.1 − 1.2 −1 + 3 −1.2) = + − 1.2ln || + 3 + ∫( ∫ 2 −0.2 1.2 ) = 2.55 2 − 1.2ln ||− 30. 1 1.2 0.1 2.2 3.2 + = (3.2 + −0.9 + 1.2/3) = 3.2 + + + + ∫( ∫ 3 ) 0.1 3(2.2) 0.9 = 3.2 + 10 0.1 + 31. 32. 33. ∫ ∫ 36. 37. || || − /8 + = −4 + − /8 + 6 (2)(3) 6.1 0.5 1 0.5 1.5 6.1 −0.5 + 0.5 − = 6.1 + − = + − + 0.5 ) ) ∫ ( ∫( 6 6 0.5 6 · 1.5 1.5 − + 9 4.2 0.4 1 0.6 1.4 4.2 −0.4 + 0.4 − 2 = 4.2 + − 2 = + − 2 + ) ) ∫ ( 0.4 ∫( 3 3 0.6 3 · 1.4 ∫ ∫ 1.4 − 2 + 4.2 (2 − 3 ) = 2 3 − + ln 2 ln 3 (1.1 + 2 ) = 1.1 2 + + ln 1.1 ln 2 100(1.1 ) 2 || 2|| 2 100(1.1 ) − = 100(1.1 ) − || = − · + ∫( ∫( 3 ) 3 ) 3 2 ln(1.1) = 38. 5|| + + 2 4 (−4 + ||/3 − 1/8) = −4 + = 7 0.6 + 35. 2.2 + 6.6 (2 + 5|| + 1/4) = 2 + = 12.2 0.5 + 34. 15 + 0.2 100(1.1 ) ln(1.1) − || + 3 1, 000(0.9 ) 4 || 4|| 4 1, 000(0.9 ) + = 1, 000(0.9 ) + || = + · + ∫( ∫( 5 ) 5 ) 5 2 ln(0.9) = 1, 000(0.9 ) ln(0.9) + 2|| + 5 39. First multiply inside the integral: Solutions Section 13.1 3 3 3 3 −5 3 3 −5 −2 4 − = 2 − −6 = − · + = + + . ( ) ∫ ∫( 2 3 2 −5 3 10 2 4 ) 2 3 9 9 −1 9 −1 = 6 + −2 = 6 + · + + = 6 − + . ( 4 5 6 ) ) ∫( 5 5 −1 5 40. First multiply inside the integral: ∫ 3 4 +2 2 1 2 = = = ( −2 + 2 −3) + + ∫ 3 ∫ (3 3) ∫ (2 3) ∫ 41. First decompose the integrand into a sum of two fractions: = − −1 − −2 + = − 1 1 − + . 2 2 − 2 2 2 2 2 = = − = ( − 2 −1) = − − 2ln || + . ∫ ∫( ∫( ∫ ) ) 2 42. First decompose the integrand into a sum of two fractions: 43. ′() = , so () = ∫ = 02 + = 1 ⇒ = 1. 2 Substituting gives So, () = 2 + . To find the value of we use the given information that (0) = 1. 2 2 + 1. 2 1 1 44. ′() = , so () = = ln || + . To find the value of we use the given information that (1) = 1. ∫ Substituting gives ln 1 + = 1 ⇒ = 1. So, () = ln || + 1. 45. ′() = − 1, so () = (0) = 0. Substituting gives ∫ ( − 1) = − + . To find the value of we use the given information that 0 − 0 + = 0 ⇒ 1 + = 0 ⇒ = −1. So, () = − − 1. 46. ′() = 2 + 1, so () = ∫ that (1) = −1. Substituting gives (2 + 1) = 2 + + . To find the value of we use the given information 2 1 + 1 + = −1 ⇒ 2 + 1 + = −1 ⇒ = −2( + 1). So, () = 2 + − 2( + 1). Solutions Section 13.1 Applications 47. ′() = 5 − 2 ⇒ () = 5− = 5 − + ∫( 10,000 10,000 ) 20,000 (0) = 20,000, so 0 − 0 + = 20,000, giving = 20,000. Thus, () = 5 − 48. ′() = 10 + 2 + 20,000. 20,000 2 2 3 ⇒ () = 10 + = 10 + + ∫( 100,000 100,000 ) 300,000 (0) = 100,000, so 0 + 0 + = 100,000, giving = 100,000. Thus, () = 10 + 3 + 100,000. 300,000 49. ′() = 5 + 2 + 1 1 ⇒ () = 5 + 2 + = 5 + 2 + ln + ∫( ) (Note that > 0, so there is no need to write ln ||.) (1) = 1,000, so 5 + 1 + ln 1 + = 1,000, giving = 994. Thus, () = 5 + 2 + ln + 994. 50. ′() = 10 + + 1 1 2 ⇒ () = 10 + + = 10 + − −1 + 2 2 ∫( 2 ) (100) = 10,000, so 10(100) + Thus, () = 10 + 100 2 1 − + = 10,000, giving = 4,000.01. 2 100 2 − −1 + 4,000.01. 2 51. a. We are given ′() = () = −6.6 2 + 18 + 220, so () = ∫ (−6.6 2 + 18 + 220) = −2.2 3 + 9 2 + 220 + . To find we use the fact that (0) = 360. Substituting gives so 360 = −2.2(0) 3 + 9(0) 2 + 220(0) + = , () = −2.2 3 + 9 2 + 220 + 360. b. Midway through 2014 ( = 4.5), the membership was (4.5) = −2.2(4.5) 3 + 9(4.5) 2 + 220(4.5) + 360 = 1,331.775 ≈ 1,332 million members. 52. a. We are given ′() = 1.1 2 − 2.6 + 2.3, so () = ∫ (1.1 2 − 2.6 + 2.3) = 1.1 3/3 − 1.3 2 + 2.3 + . To find we use the fact that (0) = 0. Substituting gives So 0 = 1.1(0) 3/3 − 1.3(0) 2 + 2.3(0) + = . Solutions Section 13.1 () = 1.1 3/3 − 1.3 2 + 2.3. b. By the start of 2014 ( = 8.5), the total number of hours uploaded was (8.5) = 1.1(8.5) 3/3 − 1.3(8.5) 2 + 2.3(8.5) ≈ 151 million hours of video. 53. The rate of change of () is \$1,200 per year; ′() = 1,200. So, () = ∫ 1,200 = 1,200 + . To find we use the fact that in 2000 ( = 0) the median household income was \$42,000. Thus, (0) = 42,000: 1,200(0) + = 42,000 ⇒ = 42,000, so () = 42,000 + 1,200. 2005 income: (5) = 42,000 + 1,200(5) = \$48,000 54. The rate of change of () is \$1,500 per year; ′() = 1,500. So, () = ∫ 1,500 = 1,500 + . To find we use the fact that in 2000 ( = 0) the mean household income was \$57,000. Thus, (0) = 57,000: 1,500(0) + = 57,000 ⇒ = 57,000, so () = 57,000 + 1,500. 2006 income: (6) = 57,000 + 1,500(6) = \$66,000 55. We are given ′() = () = 0.08 2 − 0.26 + 8.8, so () = ∫ (0.08 2 − 0.26 + 8.8) = 0.08 3/3 − 0.13 2 + 8.8 + . Since () represents sales since 2007, it follows that (0) = 0. (Sales since the start of 2007 are zero at the start of 2007.) 0.08(0) 3/3 − 0.13(0) 2 + 8.8(0) + = , so = 0 and () = 0.08 3/3 − 0.13 2 + 8.8. At the start of 2014 ( = 7), the total sales were (7) = 0.08(7) 3/3 − 0.13(7) 2 + 8.8(7) ≈ 64 billion gallons (to the nearest billion gallons). 56. We are given ′() = () = 0.25 2 − + 29, so () = ∫ (0.25 2 − + 29) = 0.25 3/3 − 2/2 + 29 + . Since () represents sales since 2007, it follows that (0) = 0. (Sales since the start of 2007 are zero at the start of 2007.) 0 = 0.25(0) 3/3 − (0) 2/2 + 29(0) + = , so = 0 and () = 0.25 3/3 − 2/2 + 29. At the end of 2012 (the start of 2013: = 6), the total sales were (6) = 0.25(6) 3/3 − (6) 2/2 + 29(6) = 174 gallons. Solutions Section 13.1 57. a. = (100 − 65)/10 = 3.5, so ′() = 3.5 + 65 billion dollars per year b. () = ∫ (3.5 + 65) = 3.5 2 + 65 + 2 (0) = 700 = , so () = 1.75 2 + 65 + 700 billion dollars 58. a. = (190 − 100)/10 = 9, so ′() = 9 + 100 billion dollars per year b. () = ∫ (9 + 100) = 9 2 + 100 + 2 (0) = 1, 300 = , so () = 4.5 2 + 100 + 1, 300 billion dollars 59. a. From the quick example referred to in the hint, the rate of change (velocity) of the percentage is given by () = ∫ () = ∫ −0.4 = −0.4 + . To find , use the fact that at time = 0, = 1 percentage point per year, so (0) = 1, giving = 1: () = −0.4 + 1 points per year. b. From the quick example referred to in the hint, the value of the percentage is given by () = ∫ () = ∫ (−0.4 + 1) = −0.2 2 + + . To find , use the fact that at time = 0, = 13 percentage points, so (0) = 13, giving = 13: () = −0.2 2 + + 13. In 2008, = 1, so (1) = −0.2(1) 2 + 1 + 13 = 13.8%. 60. a. From the quick example referred to in the hint, the rate of change (velocity) of the value is given by () = ∫ () = ∫ −20 = −20 + . To find , use the fact that at time = 0, = 40 billion dollars per year, so (0) = 40, giving = 40: () = −20 + 40 billion dollars per year. b. From the quick example referred to in the hint, the value of outstanding debt is given by () = ∫ () = ∫ (−20 + 40) = −10 2 + 40 + . To find , use the fact that at time = 0, = 1,300, so (0) = 1,300, giving = 1,300: () = −10 2 + 40 + 1,300. In 2009, = 1, so (1) = −10(1) 2 + 40(1) + 1,300 = \$1,330 billion. 61. () = 2 + 1 a. () = ∫ () = ∫ b. 0 + 0 + = 1 so = ( 2 + 1) = 3 ++1 3 3 ++ 3 Solutions Section 13.1 62. () = 3 + a. () = ∫ () = ∫ (3 + ) = 3 + b. 3 0 + 0 + = 3, = 0 so = 3 + 63. () = −32 () = ∫ () = ∫ 2 2 2 + 2 (−32) = −32 + (0) = 0 is given, so 0 = 0 + ⇒ = 0, and so () = −32. After 10 seconds, (10) = −32(10) = −320, so the stone is traveling 320 ft/sec downward. 64. () = −32 () = ∫ () = ∫ (−32) = −32 + (0) = 10 is given, so 10 = 0 + ⇒ = 10, and so () = −32 + 10. After 10 seconds, (10) = −32(10) + 10 = −310, so the stone is traveling 310 ft/sec downward. 65. a. () = ∫ () = ∫ (−32) = −32 + At time = 0, = 16 ft/sec, so 16 = −32(0) + ⇒ = 16, giving () = −32 + 16. b. () = ∫ () = ∫ (−32 + 16) = −16 2 + 16 + At time = 0, = 185 ft, so 185 = −16(0) 2 + 16(0) + ⇒ = 185, giving () = −16 2 + 16 + 185. It reaches its zenith when () = 0 −32 + 16 = 0 ⇒ = 16/32 = 0.5 sec. Its height at that moment is (0.5) = −16(0.5)2 + 16(0.5) + 185 = 189 feet, 4 feet above the top of the tower. 66. a. () = ∫ () = ∫ (−32) = −32 + At time = 0, = 24 ft/sec, so 24 = −32(0) + ⇒ = 24, giving () = −32 + 24. b. () = ∫ () = ∫ (−32 + 24) = −16 2 + 24 + At time = 0, = 185 ft, so 185 = −16(0) 2 + 24(0) + ⇒ = 185, giving () = −16 2 + 24 + 185. It reaches its zenith when () = 0 Solutions Section 13.1 −32 + 24 = 0 ⇒ = 24/32 = 0.75 sec. Its height at that moment is (0.5) = −16(0.75)2 + 24(0.75) + 185 = 194 feet, 9 feet above the top of the tower. 67. a. The ground speed is = Air speed + Tailwind speed = 500 + (25 + 50) = 525 + 50. Thus, = 525 + 50, where is the total distance traveled. Thus the total distance traveled is = ∫ [525 + 50] = 525 + 25 2 + . To obtain , use the information that, at time = 0, = 0 as well (zero distance was traveled at time = 0). Thus, 0 = 525(0) + 25(0) 2 + ⇒ = 0, giving = 525 + 25 2. b. At the end of a 1,800-mile trip, = 1,800 and so, by (a) 1,800 = 525 + 25 2. To obtain we solve the quadratic 25 2 + 525 − 1,800 = 0. Dividing by 25 gives 2 + 21 − 72 = 0 ⇒ ( + 24)( − 3) = 0. So, rejecting the negative solution gives = 3 hours. c. The negative solution indicates that, at time = −24, the tailwind would have been large and negative, causing the plane to be moving backward through that position 24 hours prior to departure and arrive at the starting point of the flight at time 0! 68. a. The ground speed is = Air speed − Headwind speed = 500 − (25 + 50) = 475 − 50. Thus, = 475 − 50, where is the total distance traveled. Thus the total distance traveled is = ∫ [475 − 50] = 475 − 25 2 + . To obtain , use the information that, at time = 0, = 0 as well (zero distance was traveled at time = 0). Thus, 0 = 475(0) −25(0) 2 + ⇒ = 0, giving = 475 − 25 2. b. For a 1,500-mile trip, = 1,500 and so, by (a) 1,500 = 475 − 25 2. To obtain we solve the quadratic 25 2 − 475 + 1,500 = 0. Dividing by 25 gives 2 − 19 + 60 = 0 ⇒ ( − 4)( − 15) = 0. So, rejecting the larger solution gives = 4 hours. c. Had the flight continued until time = 15, the headwinds would have become so powerful (they were increasing by 50 mph each hour) that the plane would be flying backward by then and would arrive at the destination a second time at = 15 (11 hours later)! Solutions Section 13.1 69. From the formulas at the end of the section in the textbook, () = −32 + 0 . The projectile has zero velocity when () = 0, so that 0 = −32 + 0 . Solving for gives = 0 /32. This is when it reaches its highest point. () = −16 2 + 0 + 0 . 70. From the formulas at the end of the section in the textbook, If we take the starting height as 0, 0 = 0, () = −16 2 + 0 . and so From Exercise 69, the projectile reaches its highest point at = 0 /32; its height then is (0 /32) = −16(0 /32) 2 + 0 (0 /32) = 02/64 feet. 71. By Exercise 70, the ball reaches a maximum height of 02/64 feet. Thus, 02/64 = 20 ⇒ 0 = (1,280) 1/2 ≈ 35.78 ft/sec. 72. By Exercise 70, the ball reaches a maximum height of 02/64 feet. Thus, 02/64 = 40 ⇒ 0 = (2,560) 1/2 ≈ 51 ft/sec. 73. a. By Exercise 70, the chalk reaches a maximum height of 02/64 feet. Thus, 02/64 = 100 ⇒ 0 = (6,400) 1/2 = 80 ft/sec. b. () = −16 2 + 0 + 0 0 = 0, and so () = −16 2 + 100. If we take the starting height as 0, () = 100 − 16 2 + 100 = 100 The chalk strikes the ceiling when 16 2 − 100 + 100 = 0 ⇒ 4(4 − 5)( − 5) = 0 ⇒ = 1.25 or 5. We take the first solution, which is the first time it strikes the ceiling. Now, () = −32 + 0 = −32 + 100, so the velocity when it strikes the ceiling is (1.25) = 60 ft/sec. c. Start with (0) = 100 and (0) = −60. So, () = −32 + 0 = −32 − 60. () = −16 2 + 0 + 0 = −16 2 − 60 + 100. We have Now we find when () = 0: −16 2 − 60 + 100 = 0 ⇒ − 4(4 − 5)( + 5) = 0 ⇒ = 1.25 or −5. We take the positive solution and say that it takes 1.25 seconds to hit the ground. Solutions Section 13.1 74. a. 02/64 = (16,000) 2/64 = 4,000,000 feet b. 0 /32 = 16,000/32 = 500 seconds c. () = −16 2 + 0 = −16 2 + 16,000 The projectile hits the ground when (0) = 0: −16 2 + 16,000 = 0 ⇒ − 16( − 1000) = 0 ⇒ = 0 or 1,000. The solution = 0 is the starting point, so we are interested in the other solution, = 1,000. () = −32 + 0 = −32 + 16,000 so (1,000) = −16,000. Thus, the projectile is traveling at a speed of 16,000 ft/sec when it hits the ground. 75. Let 0 be the speed at which Prof. Strong throws and let 0 be the speed at which Prof. Weak throws. We have 02/64 = 202/64, so solving for 0 gives 0 = 0 √2. Thus, Prof. Strong throws it √2 ≈ 1.414 times as fast as Prof. Weak. 76. Let 0 be the speed at which Prof. Strong throws and let 0 be the speed at which Prof. Weak throws. We have 302/64 = 02/64, so solving for 0 gives 0 = 0 √3. Thus, Prof. Weak throws it √3 ≈ 1.732 times as fast as Prof. Strong. Communication and reasoning exercises 77. The term indefinite refers to the arbitrary constant term in the indefinite integral; we do not obtain a definite value for , and hence the integral is "not definite." 78. an antiderivative 79. Constant; since the derivative of a linear function is constant, linear functions are antiderivatives of constant functions. 80. The zero function; since the derivative of a constant function is zero, constant functions are antiderivatives of the zero function. 81. No; there are infinitely many antiderivatives of a given function, each pair of them differing by a constant. Knowing the value of the function at a specific point suffices. 82. Yes; if () is an antiderivative of (), then its derivative is (), so that we can obtain the exact function by taking the derivative of (). 83. They differ by a constant, () − () = Constant. 84. If () is one antiderivative of (), then every other antiderivative has the form () + for some constant . 85. antiderivative, marginal 86. velocity, acceleration Solutions Section 13.1 87. Up to a constant, ∫ 88. Up to a constant, ∫ () represents the total cost of manufacturing items. The units of product of the units of () and the units of . ∫ () are the () represents the total volume of rocket fuel burned since liftoff. 89. The indefinite integral of the constant 1 is not zero; it is (+ constant). Thus, the correct answer is ∫ (3 + 1) = 3 2/2 + + . 90. The indefinite integral of 11 is not zero; it is 11 2/2 (+ constant). Thus, the correct answer is ∫ (3 2 − 11) = 3 − 11 2/2 + . 91. There should be no integral sign ( ) in the answer. The correct way to write the calculation is ∫ ∫ (12 5 − 4) = 2 6 − 2 2 + . 92. The "" indicates that the variable of integration is , and not . Correct answer: 5 + 93. The integral of a constant times a function is the constant times the integral of the function, not the integral of the constant times the integral of the function. (In general, the integral of a product is not the product of the integrals.) The correct way to write the calculation is: ∫ 4( − 2) = 4 ∫ ( − 2) = 4( − 2) + . 94. The integral of 2 is 2 /(ln 2) and not 2 + 1/( + 1). Correct answer: 2 /(ln 2) − + 95. It is the integral of 1/ that equals ln || + , not 1/ itself. Correct answer: 1 = ln || + ∫ 96. 1/ 3 can be written as −3, and its integral is therefore − −2/2 + . (The logarithm comes in only when you are integrating to the −1 power.) 97. ∫ (() + ()) is, by definition, an antiderivative of () + (). Let () be an antiderivative of (), and let () be an antiderivative of (). Then, since the derivative of () + () is () + () (by the rule for sums of derivatives), this means that () + () is an antiderivative of () + (). In symbols, ∫ (() + ()) = () + () + = ∫ () + ∫ (), the sum of the indefinite integrals. Solutions Section 13.1 98. ∫ (()) is, by definition, an antiderivative of (). Let () be an antiderivative of (). Then, since the derivative of () is () (by the rule for constant multiples of derivatives), this means that () is an antiderivative of (). In symbols, indefinite integral. ∫ (()) = () + = ∫ (), the constant multiple of the 99. Answers will vary. ∫ ∫ · 1 = · ∫ ∫ = 2/2 + , whereas 1 = 2 + · ( + ), (2 ) which is not the same as 2/2 + , no matter what values we choose for the constants , , and . 100. Answers will vary. ∫ (/1) = ∫ = 2/2 + , whereas ( )/( 1) = ( 2/2 + )/( + ), ∫ ∫ which is not the same as 2/2 + , no matter what values we choose for the constants , , and . 101. If you take the derivative of the indefinite integral of (), you obtain () back. On the other hand, if you take the indefinite integral of the derivative of (), you obtain () + . 102. The only possible antiderivatives of − 1 have the form 2/2 − + for some constant . Thus, the "classified" Martian function is () = 2/2 − + , and so the Institute of Alien Mathematics is not being truthful. Solutions Section 13.2 1. 4. 5. ∫ =2 ∫ ∫ (3 − 5) 3 = ∫ 3 · (3 − 5) 4 1 1 4 4 = + = + = + 3 3 4 12 12 1 3 = 2 + 5 = 3. ∫ =3 = 2. = 3 − 5 (2 + 5) −2 = ∫ −2 · (2 + 5) 1 1 −1 −1 = + =− + =− 2 2 −1 2 2 1 2 (3 − 5) 3 = 4 (3 − 5) 4 1 (3 − 5) + = + 3 4 12 (2 + 5) −2 = = − = −1 −1 (2 + 5) −1 1 (2 + 5) + =− + 2 2 −1 ∫ − = − = − + = − − + ∫ = − 6. = 2 1 = 2 ∫ /2 = 2 + = 2 /2 + = 2 7. ∫ 8. ∫ − = /2 = 1 − + = − − + −1 1 /2 + = 2 /2 + 1/2 −1 + Solutions Section 13.2 9. 2( + 1) = 3( − 1) 2 =3 ∫ ∫ ( + 1) 2 1 1 + = ( + 1) + 2 2 ∫ = 3 + 1 2 ( + 1) ( + 1) = = 3( − 1) 2 = 12. = = ( − 1) 3 = 11. ∫ = 2( + 1) = 10. = ( + 1) 2 3 ( − 1) 2 ( − 1) = ∫ ( − 1) 2 3 1 1 + = ( − 1) + 3 3 (3 + 1) 5 = ∫ 5 1 = ∫ 2 2( + 1) 3( − 1) 2 = 1 ∫3 (3 + 1) 6 1 1 6 6 = + = + = + 3 3 6 18 18 1 3 = − 1 = −1 (− − 1) 8 (− − 1) 7 = − 7 = − + =− + ∫ ∫ 8 8 8 = − 13. = 3 − 4 =3 = 14. ∫ 7.2√3 − 4 = ∫ 7.2√ 1 3/2 = 2.4 1/2 = 2.4 + = 1.6(3 − 4) 3/2 + ∫ 3 3/2 1 3 = −3 + 4 = −3 1 = − 3 1 4.4 4.4 (−3 + 4) 4.4 (−3 + 4) = − 4.4 = − + =− + ∫ ∫ 3 3 3 Solutions Section 13.2 15. = 0.6 + 2 = 0.6 = 16. 1 1.2 (0.6 + 2) = − 1.2 = 2 + = 2 (0.6 + 2) + ∫ ∫ 0.6 1 0.6 = −3 + 4 = −3 1 3/2 8.1√−3 + 4 = − 8.1√ = −2.7 + = −1.8(−3 + 4) 3/2 + ∫ ∫ 3 3/2 1 = − 3 17. = 3 2 + 3 = 6 = 18. = − 2 − 1 = −2 20. ∫ 3 (3 2+3) 4 1 1 4 = 3 = + = + 6 6∫ 12 24 (− 2−1) 4 1 1 4 (− 2−1) 3 = − 3 = − 3 = − + =− + ∫ ∫ 2 2∫ 8 8 1 2 = 3 2 − 1 = 6 = ∫ (3 2+3) 3 = 1 6 = − 19. ∫ 2√3 2 − 1 = ∫ 2√ 1 1 3/2 2 = 1/2 = + = (3 2−1) 3/2 + 6 3∫ 9/2 9 1 6 = − 2 + 1 = −2 = − 1 2 1 3 3√− 2 + 1 = − 3√ = − 1/2 ∫ ∫ 2 2∫ =− 3 3/2 · + = −(− 2+1) 3/2 + 2 3/2 Solutions Section 13.2 21. = 2 + 1 = 2 = 22. = 1 + 3 = 3 2 = 23. = 8 = 12 2 1 12 2 = 3.1 − 2 = 3.1 = 26. 1 8 = 4 3 + 1 = 25. 1 3 2 = 4 2 − 1 = 24. 1 2 1 3.1 = 1.2 − 3 = 1.2 = 1 1.2 ∫ ( 2+1) 1.3 = 1 1 · = −1.3 ∫ 1.3 2 2∫ ( 2+1) −0.3 −0.3 =− + =− + 0.6 0.6 2 ∫ (1 + 3) 1.4 =− ∫ = = 2 1 1 · = −1.4 ∫ 1.4 3 2 3∫ (1 + 3) −0.4 −0.4 + =− + 1.2 1.2 |4 2 − 1| = ∫ || (4 2 − 1)|4 2 − 1| + 16 ∫ = ∫ 2|4 3 + 1| = ∫ 1 1 1 || = || = + 8 8∫ 8 2 2|| 1 1 = || 12 ∫ 12 2 (4 3 + 1)|4 3 + 1| 1 || + = + 12 2 24 (1 + 9.3 3.1 − 2) = ∫ + ∫ = + 3 + = + 3 3.1 − 2 + ∫ (3.2 − 4 1.2 − 3) = 3.2 − = 3.2 − 10 1.2 − 3 + 3 ∫ 9.3 3.1 − 2 = + 4 ∫ 9.3 1 4 = 3.2 − + 1.2 1.2 1 3.1 Solutions Section 13.2 27. = −2 = − 28. = 4 ∫ 2 1 1 1 1 = = + = 2 − 1 + 4 4∫ 4 4 2 1 1 ( + 1) −( + 2) = − ( + 1) =...
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