answers tutor 8
Upcoming SlideShare
Loading in...5
×
 

answers tutor 8

on

  • 788 views

please study

please study

Statistics

Views

Total Views
788
Views on SlideShare
788
Embed Views
0

Actions

Likes
0
Downloads
31
Comments
0

0 Embeds 0

No embeds

Accessibility

Categories

Upload Details

Uploaded via as Adobe PDF

Usage Rights

© All Rights Reserved

Report content

Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

Cancel
  • Full Name Full Name Comment goes here.
    Are you sure you want to
    Your message goes here
    Processing…
Post Comment
Edit your comment

answers tutor 8 answers tutor 8 Document Transcript

  • 7.5 5.0 2.5Chapter 4 0.0 −2 −1 0 1 2 x −2.5Integration 4. sin x, sin x + 2, sin x − 5 2 x −3 −2 −1 0 1 2 34.1 Antiderivatives 0 −2 x4 x4 x4 1. , + 3, −2 4 4 4 −4 20 −6 15 3 5 3 2 5. (3x4 − 3x)dx = x − x +c 10 5 2 1 4 5 6. (x3 − 2)dx = x − 2x + c 4 √ 1 x−3 −3 −2 −1 1 2 3 7. 3 x− 4 dx = 2x3/2 + +c x 3 1 8. 2x−2 + √ dx x x4 x2 x4 x2 x4 x2 = −2x−1 + 2x1/2 + c 2. − , − − 1, − +4 4 2 4 2 4 2 x1/3 − 3 6 9. dx = (x−1/3 − 3x−2/3 )dx x2/3 5 3 = x2/3 − 9x1/3 + c 4 2 x + 2x3/4 3 10. dx = (x−1/4 + 2x−1/2 )dx x5/4 2 4 = x3/4 + 4x1/2 + c 1 3 0 −2 −1 0 1 2 11. (2 sin x + cos x)dx = −2 cos x + sin x + c x −1 12. (3 cos x − sin x)dx = 3 sin x + cos x + c 3. ex , ex + 1, ex − 3 13. 2 sec x tan xdx = 2 sec x + c 240
  • 4.1. ANTIDERIVATIVES 241 4 d14. √ dx = 4 arcsin x + c 30. ln |sin x · 2| 1 − x2 dx 1 d = (sin x · 2)15. 5 sec2 xdx = 5 tan x + c sin x · 2 dx 2 cos x = = cot x 2 sin x 4 cos x16. dx = −4 csc x + c sin2 x 31. (a) N/A (b) By Power Formula,17. (3ex − 2)dx = 3ex − 2x + c √ 2 ( x3 + 4)dx = x5/2 + 4x + c. 518. (4x − 2ex )dx = 2x2 − 2ex + c 32. (a) By Power Formula, 3x2 − 419. (3 cos x − 1/x)dx = 3 sin x − ln |x| + c dx = (3 − 4x−2 )dx x2 = 3x + 4x−1 + c20. (2x−1 + sin x)dx = 2 ln |x| − cos x + c (b) N/A 33. (a) N/A 4x21. dx = 2 ln |x2 + 4| + c (b) By Reversing derivative formula, x2 + 4 sec2 xdx = tan x + c 3 322. dx = tan−1 x + c 4x2 + 4 4 34. (a) By Power Formula, 1 1 cos x − 1 dx = − − x + c23. dx = ln | sin x| + c x 2 x sin x (b) N/A24. (2 cos x − ex )dx = 2 sin x − ex + c 35. Finding the antiderivative, x2 f (x) = 3ex + + c. ex 225. dx = ln | ex + 3| + c ex + 3 Since f (0) = 4, ex + 3 we have 4 = f (0) = 3 + c.26. dx = (1 + 3e−x )dx Therefore, ex = x − 3e−x + c x2 f (x) = 3ex + + 1. 227. x1/4 (x5/4 − 4)dx = (x3/2 − 4x1/4 )dx 36. Finding the antiderivative, 2 5/2 16 5/4 f (x) = 4 sin x + c. = x − x +c Since f (0) = 3, 5 5 we have 3 = f (0) = c. Therefore,28. x2/3 (x−4/3 − 3)dx = (x−2/3 − 3x2/3 )dx f (x) = 4 sin x + 3. 9 = 3x1/3 − x5/3 + c 37. Finding the antiderivative 5 f (x) = 4x3 + 2ex + c1 . d Since, f (0) = 2.29. ln |sec x + tan x| dx We have 2 = f (0) = 2 + c1 1 d = (sec x + tan x) and therefore sec x + tan x dx f (x) = 4x3 + 2ex . 2 sec x tan x + sec x = Finding the antiderivative, sec x + tan x f (x) = x4 + 2ex + c2 . sec x (tan x + sec x) = Since f (0) = 3, sec x + tan x = sec x We have 3 = f (0) = 2 + c2 Therefore,
  • 242 CHAPTER 4. INTEGRATION f (x) = x4 + 2ex + 1. 1 f (x) = −3 sin x + x4 + c1 x + c2 . 3 42. Taking antiderivatives, 38. Finding the antiderivative, f (x) = x1/2 − 2 cos x f (x) = 5x4 + e2x + c1 . 2 Since f (0) = −3, f (x) = x3/2 − 2 sin x + c1 3 we have −3 = f (0) = 1 + c1 4 5/2 f (x) = x + 2 cos x + c1 x + c2 . Therefore, 15 f (x) = 5x4 + e2x − 4. 43. Taking antiderivatives, Finding the antiderivative, f (x) = 4 − 2/x3 e2x f (x) = x5 + − 4x + c2 . f (x) = 4x + x−2 + c1 2 f (x) = 2x2 − x−1 + c1 x + c2 Since f (0) = 2, 1 2 c1 We have 2 = f (0) = + c2 f (x) = x3 − ln |x| + x2 + c2 x + c3 2 3 2 Therefore, 44. Taking antiderivatives, e2x 3 f (x) = sin x − ex f (x) = x5 + − 4x + . 2 2 f (x) = − cos x − ex + c1 f (x) = − sin x − ex + c1 x + c2 c1 39. Taking antiderivatives, f (x) = cos x − ex + x2 + c2 x + c3 f (t) = 2t + t2 + c1 2 t3 45. Position is the antiderivative of velocity, f (t) = t2 + + c1 t + c2 3 s(t) = 3t − 6t2 + c. Since f (0) = 2, Since s(0) = 3, we have c = 3. Thus, we have 2 = f (0) = c2 s(t) = 3t − 6t2 + 3. Therefore, t3 46. Position is the antiderivative of velocity, f (t) = t2 + + c1 t + 2. 3 s(t) = −3e−t − 2t + c. Since f (3) = 2, Since s(0) = 0, we have −3 + c = 0 and there- we have fore c = 3. Thus, 2 = f (3) = 9 + 9 + 3c1 + 2 s(t) = −3e−t − 2t + 3. − 6 = c1 Therefore, 47. First we find velocity, which is the antideriva- t3 tive of acceleration, f (t) = + t2 − 6t + 2. 3 v(t) = −3 cos t + c1 . Since v(0) = 0 we have −3 + c1 = 0, c1 = 3 and 40. Taking antiderivatives, v(t) = −3 cos t + 3. f (t) = 4t + 3t2 + c1 Position is the antiderivative of velocity, f (t) = 2t2 + t3 + c1 t + c2 s(t) = −3 sin t + 3t + c2 . Since f (1) = 3, Since s(0) = 4, we have c2 = 4. Thus, we have 3 = f (1) = 2 + 1 + c1 + c2 s(t) = −3 sin t + 3t + 4. Therefore, c1 + c2 = 0 48. First we find velocity, which is the antideriva- Since f (−1) = −2, tive of acceleration, we have −2 = f (−1) = 2 − 1 − c1 + c2 1 v(t) = t3 + t + c1 . Therefore, −c1 + c2 = −3. 3 So, c1 = 2 and c2 = − 3 3 2 Since v(0) = 4 we have c1 = 4 and Hence, 1 3 3 v(t) = t3 + t + 4. f (t) = t3 + 2t2 + t − . 3 2 2 Position is the antiderivative of velocity, 1 4 1 2 41. Taking antiderivatives, s(t) = t + t + 4t + c2 . 12 2 f (x) = 3 sin x + 4x2 Since s(0) = 0, we have c2 = 0. Thus, 4 1 4 1 2 f (x) = −3 cos x + x3 + c1 s(t) = t + t + 4t. 3 12 2 View slide
  • 4.1. ANTIDERIVATIVES 24349. (a) There are many correct answers, but any 15 correct answer will be a vertical shift of these answers. 10 10.0 5 7.5 0 5.0 y −3 −2 −1 0 1 2 3 x −5 2.5 0.0 −10 −4.0 −3.2 −2.4 −1.6 −0.8 0.0 0.8 1.6 2.4 3.2 x −2.5 51. We start by taking antiderivatives: −5.0 f (x) = x2 /2 − x + c1 f (x) = x3 /6 − x2 /2 + c1 x + c2 . Now, we use the data that we are given. We (b) There are many correct answers, but any know that f (1) = 2 and f (1) = 3, which gives correct answer will be a vertical shift of us these answers. 3 = f (1) = 1/2 − 1 + c1 , and 8.8 1 = f (1) = 1/6 − 1/2 + c1 + c2 . 8.0 Therefore c1 = 7/2 and c2 = −13/6 and the 7.2 function is 6.4 x3 x2 7x 13 f (x) = − + − . 5.6 6 2 2 6 4.8 4.0 52. We start by taking antiderivatives: 3.2 f (x) = 3x2 + 4x + c1 2.4 f (x) = x3 + 2x2 + c1 x + c2 . Now, we use the data that we are given. We −3 −2 −1 0 1 2 3 x know that f (−1) = 1 and f (−1) = 2, which gives us 2 = f (−1) = −1 + c1 , and 1 = f (−1) = 1 − c1 + c2 .50. (a) There are many correct answers, but any Therefore c1 = 3 and c2 = 3 and the function correct answer will be a vertical shift of is these answers. f (x) = x3 + 2x2 + 3x − 3. 14 d 12 53. sin x2 = 2x cos x2 dx 10 Therefore, 8 y 6 2x cos x2 dx = sin x2 + c 4 2 d 9 54. (x3 + 2)3/2 = x2 (x3 + 2)1/2 −4 −2 0 0 2 dx 2 x −2 Therefore, −4 2 x2 x3 + 2dx = (x3 + 2)3/2 + c 9 (b) There are many correct answers, but any d 55. x2 sin 2x = 2(x sin 2x + x2 cos 2x) correct answer will be a vertical shift of dx these answers. Therefore, View slide
  • 244 CHAPTER 4. INTEGRATION 1 36 x2 x sin 2x + x2 cos 2x dx = 3− = 2 9 12 − 9x 2 3x − 4 1 2 For 33(a): Almost the same as in Exercise 59, = x sin 2x + c 2 example 1.11 (b). 1 x−1 d x2 2xe3x − 3x2 e3x For 34(b): ln +c 56. = 2 x+1 dx e 3x e6x Verify: Therefore, d 1 x−1 ln 2xe3x − 3x2 e3x x2 dx 2 x + 1 6x dx = 3x + c 1 x + 1 (x + 1) − (x − 1) e e = · · 2 x−1 (x + 1)2 x cos(x2 ) 1 57. dx = sin(x2 ) + c = 2 sin(x2 ) x −1 61. Use a CAS to find antiderivatives and verify by d √ √ 1 computing the derivatives: 58. 2 x sin x = 2 x cos x + √ sin x dx x √ 1 3 1 3 2 x cos x + √ sin x dx (a) x2 e−x dx = − e−x + c x 3 √ Verify: = 2 x sin x + c d 1 3 − e−x dx 3 59. Use a CAS to find antiderivatives and verify by 1 3 = − e−x · (−3x2 ) computing the derivatives: 3 3 For 11.1(b): = x2 e−x 1 sec xdx = ln | sec x + tan x| + c (b) dx = ln |x − 1| − ln |x| + c Verify: x2 − x Verify: d (ln |x − 1| − ln |x|) d dx ln | sec x + tan x| 1 1 x − (x − 1) dx = − = sec x tan x + sec2 x x−1 x x(x − 1) = = sec x 1 1 sec x + tan x = = 2 For 11.1(f): x(x − 1) x −x sin 2x x cos 2x x sin 2xdx = − +c (c) sec xdx = ln | sec x + tan x| + c 4 2 Verify: Verify: d sin 2x x cos 2x d − [ln | sec x + tan x|] dx 4 2 dx 2 cos 2x cos 2x − 2x sin 2x sec x tan x + sec2 x = − = 4 2 sec x + tan x = x sin 2x sec x(sec x + tan x) = = sec x sec x + tan x 60. Use a CAS to find antiderivatives and verify by computing the derivatives: 62. Use a CAS to find antiderivatives and verify For 31(a): The answer is too complicated to be by computing the derivatives: presented here. √ 1 √ 2 3 − 3x x 1 For 32(b): 3x + 3 ln √ +c (a) dx = arctan x2 + c 9 2 3 + 3x x4 + 1 2 Verify: Verify: √ d 1 d 1 √ 2 3 − 3x arctan x2 3x + 3 ln √ dx 2 dx 9 2 3 + 3x √ 1 1 x 1 2 3 + 3x = · 4 · 2x = 4 = 3+ √ · 2 x +1 x +1 9 2 3 − 3x √ √ (b) 3x sin 2xdx −3(2 3 + 3x) − 3(2 3 − 3x) √ 3 3x (2 3 + 3x)2 = sin 2x − cos 2x + c 4 2
  • 4.1. ANTIDERIVATIVES 245 Verify: 67. The key is to find the velocity and position d 3 3x functions. We start with constant acceleration sin 2x − cos 2x dx 4 2 a, a constant. Then, v(t) = at + v0 where v0 3 3 is the initial velocity. The initial velocity is 30 = cos 2x − cos 2x + 3x sin 2x 2 2 miles per hour, but since our time is in seconds, = 3x sin 2x it is probably best to work in feet per second (30mph = 44ft/s). v(t) = at + 44. (c) ln xdx = x ln x − x + c We know that the car accelerates to 50 mph Verify: (50mph = 73ft/s) in 4 seconds, so v(4) = 73. d 29 (x ln x − x) = ln x + 1 − 1 Therefore, a · 4 + 44 = 73 and a = ft/s dx 4 = ln x So, 29 −1 v(t) = t + 44 and63. √ dx = cos−1 (x) + c1 4 1 − x2 29 2 s(t) = t + 44t + s0 −1 8 √ dx = − sin−1 (x) + c2 where s0 is the initial position. We can assume 1 − x2 the the starting position is s0 = 0. Therefore, 29 2 cos−1 x + c1 = − sin−1 x + c2 Then, s(t) = t + 44t and the distance 8 Therefore, traveled by the car during the 4 seconds is sin−1 x + cos−1 x = constant s(4) = 234 feet. To find the value of the constant, let x be any convenient value. 68. The key is to find the velocity and position Suppose x = 0; then sin−1 0 = 0 and cos−1 0 = functions. We start with constant acceleration π/2, so a, a constant. Then, v(t) = at + v0 where v0 π is the initial velocity. The initial velocity is 60 sin−1 x + cos−1 x = 2 miles per hour, but since our time is in seconds, it is probably best to work in feet per second64. To derive these formulas, all that needs to be (60mph = 88ft/s). v(t) = at + 88. done is to take the derivatives to see that the We know that the car comes to rest in 3 sec- integrals are correct: d onds, so v(3) = 0. (tan x) = sec2 x Therefore, dx d a(3) + 88 = 0 and a = −88/3ft/s (the accelera- (sec x) = sec x tan x tion should be negative since the car is actually dx decelerating.65. To derive these formulas, all that needs to be So, done is to take the derivatives to see that the 88 integrals are correct: v(t) = − t + 88 and 3 d x 44 (e ) = ex s(t) = − t2 + 88t + s0 where s0 is the initial dx 3 d position. We can assume the the starting po- −e−x = e−x dx sition is s0 = 0. 44 1 1 1 Then, s(t) = − t2 + 88t and the stopping66. (a) dx = dx 3 kx k x distance is s(3) = 132 feet. 1 = ln |x| + c1 69. To estimate the acceleration over each inter- k val, we estimate v (t) by computing the slope 1 1 k (b) dx = dx of the tangent lines. For example, for the in- kx k kx terval [0, 0.5]: 1 = ln |kx| + c2 v(0.5) − v(0) k a≈ = −31.6 m/s2 . 0.5 − 0 Because Notice, acceleration should be negative since 1 1 ln |kx| = (ln |k| + ln |x|) the object is falling. k k 1 1 1 To estimate the distance traveled over the in- = ln |x| + ln |k| = ln |x| + c terval, we estimate the velocity and multiply k k k The two antiderivatives are both correct. by the time (distance is rate times time). For
  • 246 CHAPTER 4. INTEGRATION an estimate for the velocity, we will use the Time Speed Dist average of the velocities at the endpoints. For 0 70 0 example, for the interval [0, 0.5], the time inter- 0.5 69.55 34.89 val is 0.5 and the velocity is −11.9. Therefore 1.0 70.3 69.85 the position changed is (−11.9)(0.5) = −5.95 1.5 70.35 105.01 meters. The distance traveled will be 5.95 me- 2.0 70.65 104.26 ters (distance should be positive). Interval Accel Dist 72. To estimate the speed over the interval, we first [0.0, 0.5] −31.6 5.95 approximate the acceleration over the interval [0.5, 1.0] −2 12.925 by averaging the acceleration at the endpoint [1.0, 1.5] −11.6 17.4 of the interval. Then, the velocity will be the [1.5, 2.0] −3.6 19.3 acceleration times the length of time. the slope of the tangent lines. For example, for the in- terval [0.0, 0.5] the average acceleration is −0.8 and v(0.5) = 20+(−0.8)(.5) = 19.6. Of course, speed is the absolute value of the velocity. 70. To estimate the acceleration over each inter- And, the distance traveled is the average speed val, we estimate v (t) by computing the slope times the length of time. For the time t = 0.5, of the tangent lines. For example, for the in- 20 + 19.6 terval [0, 1.0]: the distance would be × 0.5 = 9.9 2 v(1.0) − v(0) meters. a≈ = −9.8 m/s2 . Time Speed Dist 1.0 − 0 Notice, acceleration should be negative since 0 20 0 the object is falling. 0.5 19.6 9.9 To estimate the distance traveled over the in- 1.0 17.925 19.281 terval, we estimate the velocity and multiply 1.5 16.5 27.888 by the time (distance is rate times time). For 2.0 16.125 34.044 an estimate for the velocity, we will use the av- erage of the velocities at the endpoints. For example, for the interval [0, 1.0], the time in- 4.2 Sums And Sigma Notation terval is 1.0 and the velocity is −4.9. Therefore the position changed is (−4.9)(1.0) = −4.9 me- 1. The given sum is the sum of twice the ters. The distance traveled will be 4.9 meters squares of the integers from 1 to 14. (distance should be positive). 14 2 2 2 2 Interval Accel Dist 2(1) + 2(2) + 2(3) + . . . + 2(14) = 2i2 i=1 [0.0, 1.0] −9.8 4.9 [1.0, 2.0] −8.8 14.2 2. The given sum is the sum of squares [2.0, 3.0] −6.3 21.75 roots of √the integers from 1 to 14. √ √ √ [3.0, 4.0] −3.6 26.7 2 − 1 + 3 − 1 + 4 − 1 + . . . + 15 − 1 √ √ √ √ √ = 1 + 2 + 3 + ... + 13 + 14 14 √ = i 71. To estimate the speed over the interval, we i=1 first approximate the acceleration over the in- terval by averaging the acceleration at the end- 50 (50)(51)(101) point of the interval. Then, the velocity will be 3. (a) i2 = = 42, 925 6 the acceleration times the length of time. The i=1 slope of the tangent lines. For example, for the 50 2 2 50(51) interval [0, 0.5] the average acceleration is −0.9 (b) i = = 1, 625, 625 and v(0.5) = 70 + (−0.9)(0.5) = 69.55. i=1 2 And, the distance traveled is the speed times 10 the length of time. For the time t = 0.5, the √ 70 + 69.55 4. (a) i distance would be ×0.5 ≈ 34.89 me- i=1 2 √ √ √ √ √ ters. =1+ 2+ 3+ 4+ 5+ 6
  • 4.2. SUMS AND SIGMA NOTATION 247 √ √ √ √ + 7 + 8 + 9 + 10 = 338, 350 − 15, 150 + 200 = 323, 400 ≈ 22.47 10 140 10(11) √ (b) i= = 55 14. n2 + 2n − 4 i=1 2 n=1 140 140 140 6 = n2 + 2 n− 4 5. 3i2 = 3 + 12 + 27 + 48 + 75 + 108 n=1 n=1 n=1 i=1 (140)(141)(281) 140(141) = 273 = +2 − 4 (140) 6 2 7 = 943, 670 6. i2 + i = 12 + 20 + 30 + 42 + 56 i=3 30 = 160 2 15. (i − 3) + i − 3 10 i=3 7. (4i + 2) 30 30 2 i=6 = (i − 3) + (i − 3) = (4(6) + 2) + (4(7) + 2) + (4(8) + 2) i=3 i=3 + (4(9) + 2) + (4(10) + 2) 27 27 = 26 + 30 + 34 + 38 + 42 = n2 + n (substitute i − 3 = n) = 170 n=0 n=0 27 27 8 2 =0+ n2 + 0 + n 8. (i + 2) n=1 n=1 i=6 27 (28) (55) 27 (28) = (62 + 2) + (72 + 2) + (82 + 2) = + = 7308 6 2 = 38 + 51 + 66 = 155 70 70 20 20 9. (3i − 1) = 3 · i − 70 16. (i − 3) (i + 3) = i2 − 9 i=1 i=1 i=4 i=4 70(71) 20 20 =3· − 70 = 7, 385 2 = i2 − 9 1 45 45 45 i=4 i=410. (3i − 4) = 3 i−4 1 20 3 20 i=1 i=1 i=1 = i2 − i2 −9 1 45(46) i=1 i=1 i=4 =3 − 4(45) = 2925 2 20 (21) (41) = − 1 − 4 − 9 − 9 (17) 40 40 611. (4 − i2 ) = 160 − i2 = 2703 i=1 i=1 (40)(41)(81) n = 160 − 6 17. k2 − 3 = 160 − 22, 140 = −21, 980 k=3 n n 50 50 5012. (8 − i) = 8 1− i = k2 + (−3) k=3 k=3 i=1 i=1 i=1 n 2 50(51) = 8(50) − = −875 = k2 − k2 2 k=1 k=1 100 n 213. n2 − 3n + 2 + (−3) − (−3) n=1 k=1 k=1 100 100 100 n (n + 1) (2n + 1) = −1−4 = n2 − 3 n+ 2 6 n=1 n=1 n=1 + (−3) n − (−3) (2) (100)(101)(201) 100(101) n (n + 1) (2n + 1) = −3 + 200 = − 5 − 3n + 6 6 2 6
  • 248 CHAPTER 4. INTEGRATION n (n + 1) (2n + 1) = ((2.05)3 + 4)(0.1) + . . . = − 3n + 1 6 + ((2.95)3 + 4)(0.1) n = (202.4375)(0.1) 18. k2 + 5 = 20.24375 k=0 n 2 n n 1 i i 23. +2 = k2 + 5 n n n i=1 k=0 k=0 n n n n 1 i2 i =0+ k2 + 5 + 5 = +2 k=1 k=1 n i=1 n2 i=1 n n (n + 1) (2n + 1) n n = + 5 + 5n 1 1 2 6 = i2 + i n n2 i=1 n i=1 n 19. f (xi )∆x 1 1 n(n + 1)(2n + 1) = i=1 n n2 6 5 = (x2 + 4xi ) · 0.2 i 2 n(n + 1) + i=1 n 2 = (0.22 + 4(0.2))(0.2) + . . . n(n + 1)(2n + 1) n(n + 1) + (12 + 4)(0.2) = + 6n3 n2 = (0.84)(0.2) + (1.76)(0.2) n 2 + (2.76)(0.2) + (3.84)(0.2) 1 i i lim +2 + (5)(0.2) n→∞ n n n i=1 = 2.84 n n(n + 1)(2n + 1) n(n + 1) = lim + 20. f (xi )∆x n→∞ 6n3 n2 i=1 2 4 5 = +1= = (3xi + 5) · 0.4 6 3 i=1 n 2 = (3(0.4) + 5)(0.4) + . . . 1 i i 24. −5 + (3(2) + 5)(0.4) i=1 n n n = (6.2)(0.4) + (7.4)(0.4) n n + (8.6)(0.4) + (9.8)(0.4) 1 i2 i = −5 + (11)(0.4) n i=1 n2 i=1 n = 17.2 n n n 1 1 5 = i2 − i 21. f (xi )∆x n n2 i=1 n i=1 i=1 10 1 1 n(n + 1)(2n + 1) = = (4x2 − 2) · 0.1 i n n2 6 i=1 = (4(2.1)2 − 2)(0.1) + . . . 5 n(n + 1) − + (4(3)2 − 2)(0.1) n 2 = (15.64)(0.1) + (17.36)(0.1) n(n + 1)(2n + 1) 5n(n + 1) + (19.16)(0.1) + (21.04)(0.1) = − 6n3 2n2 + (23)(0.1) + (25.04)(0.1) 2 −13n − 12n + 1 + (27.16)(0.1) + (29.36)(0.1) = 6n2 + (31.64)(0.1) + (34)(0.1) n 2 1 i i = 24.34 lim −5 n n→∞ i=1 n n n 22. f (xi )∆x −13n2 − 12n + 1 i=1 = lim 10 n→∞ 6n2 = (x3 + 4) · 0.1 13 12 1 = lim − − + 2 i=1 n→∞ 6 6n 6n
  • 4.2. SUMS AND SIGMA NOTATION 249 13 n =− n2 (n + 1)2 6 i3 = i=1 4 n 1 2i 2 2i is true for all integers n ≥ 1.25. 4 − For n = 1, we have n n n 1 i=1 12 (1 + 1)2 n n i3 = 1 = , 1 i2 i i=1 4 = 16 −2 n i=1 n2 i=1 n as desired. So the proposition is true for n = 1. n n 1 16 2 Next, assume that = i2 − i k n n2 n k 2 (k + 1)2 i=1 i=1 i3 = , 1 16 n(n + 1)(2n + 1) i=1 4 = for some integer k ≥ 1. n n2 6 In this case, we have by the induction assump- 2 n(n + 1) tion that for n = k + 1, − n 2 n k+1 k i3 = i3 = i3 + (k + 1)3 16n(n + 1)(2n + 1) n(n + 1) = − i=1 i=1 i=1 6n3 n2 k 2 (k + 1)2 n 2 = + (k + 1)3 1 2i 2i 4 lim 4 − k 2 (k + 1)2 + 4(k + 1)3 n→∞ i=1 n n n = 4 16n(n + 1)(2n + 1) n(n + 1) (k + 1)2 (k 2 + 4k + 4) = lim − = n→∞ 6n3 n2 4 (k + 1)2 (k + 2)2 16 13 = = −1= 4 3 3 n2 (n + 1)2 = n 2 4 1 2i i as desired.26. +4 i=1 n n n 28. Want to prove that n n n 1 4i2 i n2 (n + 1)2 (2n2 + 2n − 1) = +4 i5 = n i=1 n2 i=1 n 12 i=1 n n is true for all integers n ≥ 1. 1 4 4 For n = 1, we have = i2 + i n n2 i=1 n i=1 1 12 (1 + 1)2 (2 + 2 − 1) i3 = 1 = , 1 4 n(n + 1)(2n + 1) i=1 12 = n n2 6 as desired. So the proposition is true for n = 1. 4 n(n + 1) + Next, assume that n 2 k k 2 (k + 1)2 (2k 2 + 2k − 1) 4n(n + 1)(2n + 1) 4n(n + 1) i5 = , = + i=1 12 6n3 2n2 for some integer k ≥ 1. 2 10n + 12n + 2 In this case, we have by the induction assump- = 3n2 tion that for n = k + 1, n 2 n k+1 k 1 2i i lim +4 i5 = i5 = i5 + (k + 1)5 n→∞ i=1 n n n i=1 i=1 i=1 k 2 (k + 1)2 (2k 2 + 2k − 1) 10n + 12n + 2 2 = + (k + 1)5 = lim 12 n→∞ 3n2 10 12 2 10 k 2 (k + 1)2 (2k 2 + 2k − 1) + 12(k + 1)5 = lim + + = = n→∞ 3 3n 3n2 3 12 (k + 1)2 [k 2 (2k 2 + 2k − 1) + 12(k + 1)3 ]27. Want to prove that = 12
  • 250 CHAPTER 4. INTEGRATION (k + 1)2 [2k 4 + 14k 3 + 35k 2 + 36k + 12] a − ar = 34. When n = 0, a = . 12 1−r Assume the formula holds for n = k − 1, which (k + 1) (k + 4k + 4)(2k 2 + 6k + 3) 2 2 gives = 12 a − ark a + ar + · · · ark−1 = . n2 (n + 1)2 (2n2 + 2n − 1) 1−r = Then for n = k, 12 as desired. we have a + ar + · · · ark 10 = a + ar + · · · ark−1 + ark a − ark 29. (i3 − 3i + 1) = + ark 1−r i=1 10 10 a − ark + ark (1 − r) = = i3 − 3 i + 10 1−r i=1 i=1 a − ark + ark − ark+1 = 100(11)2 10(11) 1−r = −3 + 10 a − ark+1 4 2 = = 2, 870 1−r a − arn+1 = 20 1−r 30. (i3 + 2i) as desired. i=1 20 20 n = i3 + 2 i 6 35. e6i/n i=1 i=1 i=1 n 400(21)2 20(21) n = +2 = 44, 520 6 4 2 = e6i/n n i=1 100 31. (i5 − 2i2 ) 6 e6/n − e6 = i=1 n 1 − e6/n 100 100 = 5 i −2 i2 6 1 − e6 = −1 i=1 i=1 n 1 − e6/n (100 )(1012 )[2(1002 ) + 2(100) − 1] 2 6 1 − e6 6 = = − 12 n 1 − e6/n n 6 100(101)(201) Now lim = 0, and −2 x→∞ n 6 = 171, 707, 655, 800 6 1 − e6 lim x→∞ n 1 − e6/n 100 1/n 32. (2i5 + 2i + 1) = 6(1 − e6 ) lim x→∞ 1 − e6/n i=1 100 100 1 = 6(1 − e6 ) lim =2 i5 + 2 i + 100 x→∞ −6e6/n i=1 i=1 = e6 − 1. 2 2 2 n (100 )(101 )[2(100 ) + 2(100) − 1] 6 =2 Thus lim e6i/n = e6 − 1. 12 x→∞ n i=1 100(101) +2· + 100 2 n = 343, 416, 675, 200 2 36. e(2i)/n i=1 n n n n 33. (cai + dbi ) = cai + dbi 2 e2/n − e2 = i=1 i=1 i=1 n 1 − e2/n n n =c ai + d bi 2 1 − e2 = −1 i=1 i=1 n 1 − e2/n
  • 4.3. AREA 251 2 1 − e2 2 Notice that ∆x = 0.25. = − n 1 − e2/n n A4 = [f (0.125) + f (0.375) + f (0.625) 2 + f (0.875)](0.25) Now lim = 0, and x→∞ n = [(0.125)2 + 1 + (0.375)2 + 1 2 1 − e2 + (0.625)2 + 1 + (0.875)2 + 1](0.25) lim x→∞ n 1 − e2/n = 1.38125. 1/n = 2(1 − e2 ) lim 2 x→∞ 1 − e2/n 1 = 2(1 − e2 ) lim 1.5 x→∞ −2e2/n = e2 − 1. 1 n 2 Thus lim e2i/n = e2 − 1. x→∞ i=1 n 0.537. Distance 0 = 50(2) + 60(1) + 70(1/2) + 60(3) 0 0.2 0.4 0.6 x 0.8 1 1.2 = 375 miles.38. Distance = 50(1) + 40(1) + 60(1/2) + 55(3) = 285 miles. (b) Evaluation points: 0.25, 0.75, 1.25, 1.75.39. On the time interval [0, 0.25], the estimated ve- Notice that ∆x = 0.5. 120 + 116 A4 = [f (0.25) + f (0.75) + f (1.25) locity is the average velocity = 118 2 + f (1.75)](0.5) feet per second. = [(0.25)2 + 1 + (0.75)2 + 1 + (1.25)2 We estimate the distance traveled during the + 1 + (1.75)2 + 1](0.5) time interval [0, 0.25] to be = 4.625. (118)(0.25 − 0) = 29.5 feet. Altogether, the distance traveled is estimated 7 as 6 = (236/2)(0.25) + (229/2)(0.25) + (223/2)(0.25) + (218/2)(0.25) 5 + (214/2)(0.25) + (210/2)(0.25) 4 + (207/2)(0.25) + (205/2)(0.25) 3 = 217.75 feet. 240. On the time interval [0, 0.5], the estimated ve- 1 10 + 14.9 0 locity is the average velocity = 12.45 -0.5 0 0.5 1 1.5 2 2.5 2 x meters per second. We estimate the distance fallen during the time interval [0, 0.5] to be (12.45)(0.5 − 0) = 6.225 meters. Altogether, the distance fallen (estimated) = (12.45)(0.5) + (17.35)(0.5) + (22.25)(0.5) + (27.15)(0.5) + (32.05)(0.5) + (36.95)(0.5) + (41.85)(0.5) + (46.75)(0.5) 2. (a) Evaluation points: = 118.4 meters. 1.125, 1.375, 1.625, 1.875. Notice that ∆x = 0.25. A4 = [f (1.125) + f (1.375) + f (1.625)4.3 Area + f (1.875)](0.25) = [(1.125)3 − 1 + (1.375)3 − 1 1. (a) Evaluation points: + (1.625)3 − 1 + (1.875)3 − 1](0.25) 0.125, 0.375, 0.625, 0.875. = 2.7265625.
  • 252 CHAPTER 4. INTEGRATION 11π/16, 13π/16, 15π/16. 7 Notice that ∆x = π/8. 6 A4 = [f (π/16) + f (3π/16) + f (5π/16) 5 + f (7π/16) + f (9π/16) + f (11π/16) 4 + f (13π/16) + f (15π/16)](π/8) = [sin(π/16) + sin(3π/16) + sin(5π/16) 3 + sin(7π/16) + sin(9π/16) 2 + sin(11π/16) + sin(13π/16) 1 + sin(15π/16)](π/8) = 2.0129. 0 1 1.2 1.4 1.6 1.8 2 x 1 (b) Evaluation points: 0.8 1.25, 1.75, 2.25, 2.75. Notice that ∆x = 0.5. 0.6 A4 = [f (1.25) + f (1.75) + f (2.25) + f (2.75)](0.5) 0.4 = [(1.25)3 − 1 + (1.75)3 − 1 + (2.25)3 − 1 + (2.75)3 − 1](0.5) 0.2 = 17.75. 0 0 0.5 1 1.5 2 2.5 3 x 30 25 20 15 4. (a) Evaluation points: −0.75, −0.25, 0.25, 0.75. 10 Notice that ∆x = 0.5. 5 A4 = [f (−0.75) + f (−0.25) + f (0.25) 0 + f (0.75)](0.5) 1 1.5 2 2.5 3 x = [4 − (−0.75)2 + 4 − (−0.25)2 + 4 − (0.25)2 + 4 − (0.75)2 ](0.5) 3. (a) Evaluation points: = 7.375. π/8, 3π/8, 5π/8, 7π/8. 4 Notice that ∆x = π/4. A4 = [f (π/8) + f (3π/8) + f (5π/8) 3 + f (7π/8)](π/4) = [sin(π/8) + sin(3π/8) + sin(5π/8) 2 + sin(7π/8)](π/4) = 2.05234. 1 1 0 0.8 -1 -0.5 0 0.5 1 x 0.6 0.4 (b) Evaluation points: −2.75, −2.25, −1.75, −1.25. 0.2 Notice that ∆x = 0.5. 0 A4 = [f (−2.75) + f (−2.25) + f (−1.75) 0 0.5 1 1.5 x 2 2.5 3 + f (−1.25)](0.5) = [4 − (−2.75)2 + 4 − (−2.25)2 + 4 (b) Evaluation points: − (−1.75)2 + 4 − (−1.25)2 ](0.5) π/16, 3π/16, 5π/16, 7π/16, 9π/16, = −0.625.
  • 4.3. AREA 253 15 A16 = ∆x f (ci ) 2 i=0 15 2 x 1 i 1 -3 -2.5 -2 -1.5 -1 = + + 1 ≈ 4.6640 0 8 i=0 8 16 -2 (c) There are 16 rectangles and the evalua- tion points are given by ci = i∆x + ∆x -4 where i is from 0 to 15. 15 -6 A16 = ∆x f (ci ) i=0 15 2 1 i 1 = + + 1 ≈ 4.9219 5. (a) There are 16 rectangles and the evalua- 8 i=0 8 8 tion points are given by ci = i∆x where i is from 0 to 15. 7. (a) There are 16 rectangles and the evalua- 15 tion points are the left endpoints which A16 = ∆x f (ci ) are given by i=0 15 2 ci = 1 + i∆x where i is from 0 to 15. 1 i 15 = + 1 ≈ 1.3027 16 16 A16 = ∆x f (ci ) i=0 i=0 (b) There are 16 rectangles and the evalua- 3 15 3i ∆x = 1+ + 2 ≈ 6.2663 tion points are given by ci = i∆x + 16 16 2 i=0 where i is from 0 to 15. 15 (b) There are 16 rectangles and the evalua- A16 = ∆x f (ci ) tion points are the midpoints which are i=0 given by 15 2 ∆x 1 i 1 ci = 1 + i∆x + where i is from 0 to = + +1 2 16 i=0 16 32 15. 15 ≈ 1.3330 A16 = ∆x f (ci ) (c) There are 16 rectangles and the evalua- i=0 15 tion points are given by ci = i∆x + ∆x 3 3i 3 = 1+ + +2 where i is from 0 to 15. 16 16 32 15 i=0 A16 = ∆x f (ci ) ≈ 6.3340 i=0 15 2 (c) There are 16 rectangles and the evalua- 1 i 1 = + +1 tion points are the right endpoints which 16 i=0 16 16 are given by ≈ 1.3652 ci = 1 + i∆x where i is from 1 to 16. 16 A16 = ∆x f (ci ) 6. (a) There are 16 rectangles and the evalua- i=1 tion points are given by ci = i∆x where i 16 3 3i is from 0 to 15. = 1+ + 2 ≈ 6.4009 15 16 i=1 16 A16 = ∆x f (ci ) i=0 15 2 8. (a) There are 16 rectangles and the evalua- 1 i tion points are the left endpoints which = + 1 ≈ 4.4219 8 i=0 8 are given by ci = −1 + i∆x − ∆x (b) There are 16 rectangles and the evalua- where i is from 1 to 16. ∆x 16 tion points are given by ci = i∆x + 2 A16 = ∆x f (ci ) where i is from 0 to 15. i=1
  • 254 CHAPTER 4. INTEGRATION 16 1 i 1 given by ci = −1 + i∆x − ∆x where i is = e−2(−1+ 8 − 8 ) ≈ 4.0991 from 1 to 100. 8 i=1 100 (b) There are 16 rectangles and the evalua- A100 = ∆x f (ci ) tion points are the midpoints which are i=1 100 3 given by 2 2i 2 ∆x = −1 + − −1 ci = −1 + i∆x − 100 i=1 100 100 2 where i is from 1 to 16. ≈ −2.02 16 A16 = ∆x f (ci ) (b) There are 100 rectangles and the evalua- i=1 tion points are midpoints which are given 16 ∆x 1 i 1 by ci = −1 + i∆x − where i is from 1 = e−2(−1+ 8 − 16 ) ≈ 3.6174 2 8 to 100. i=1 100 (c) There are 16 rectangles and the evalua- A100 = ∆x f (ci ) tion points are the right endpoints which i=1 100 3 are given by 2 2i 1 ci = −1 + i∆x where i is from 1 to 16. = −1 + − −1 100 i=1 100 100 16 A16 = ∆x f (ci ) = −2 i=1 16 (c) There are 100 rectangles and the evalua- 1 i tion points are right endpoints which are = e−2(−1+ 8 ) ≈ 3.1924 8 i=1 given by ci = −1 + i∆x where i is from 1 to 100. 9. (a) There are 50 rectangles and the evalua- 100 tion points are given by ci = i∆x where i A100 = ∆x f (ci ) is from 0 to 49. i=1 50 100 3 2 2i A50 = ∆x f (ci ) = −1 + − 1 ≈ −1.98 100 i=1 100 i=0 50 π πi 1 = cos ≈ 1.0156 11. (a) ∆x = . We will use right endpoints as 100 i=0 100 n i (b) There are 50 rectangles and the evalua- evaluation points, xi = . n ∆x n tion points are given by ci = + i∆x An = f (xi )∆x 2 where i is from 0 to 49. i=1 50 n 2 n 1 i 1 A50 = ∆x f (ci ) = +1 = 3 i2 + 1 i=0 n i=1 n n i=1 50 1 n(n + 1)(2n + 1) π π πi = 3 +1 = cos + n 6 100 i=0 200 100 ≈ 1.00004 8n2 + 3n + 1 = 6n2 (c) There are 50 rectangles and the evalua- Now to compute the exact area, we take tion points are given by ci = ∆x + i∆x the limit as n → ∞: where i is from 0 to 49. 8n2 + 3n + 1 50 A = lim An = lim n→∞ n→∞ 6n2 A50 = ∆x f (ci ) 8 3 1 4 i=0 = lim + + = 50 n→∞ 6 6n 6n2 3 π π πi 2 = cos + (b) ∆x = . We will use right endpoints as 100 i=0 100 100 n 2i ≈ 0.9842 evaluation points, xi = . n n 10. (a) There are 100 rectangles and the evalu- An = f (xi )∆x ation points are left endpoints which are i=1
  • 4.3. AREA 255 2 n 2i 2 8 32 = +1 =4+4+ = n n 3 3 i=1 n 2 1 2 2i 12. (a) ∆x = . We will use right endpoints as = +1 n n n i=1 i 2 n 2i 2 2 n evaluation points, xi = . = + 1 n n n n n i=1 i=1 An = f (xi )∆x n 8 i=1 = 3 i2 + 2 n 2 n i=1 1 i i = +3 8 n (n + 1) (2n + 1) n i=1 n n = 3 +2 n 6 n n 8 (n + 1) (2n + 1) 1 3 = 2 +2 = i2 + i n 6 n3 i=1 n2 i=1 4 = 2 2n2 + 3n + 1 + 2 1 n(n + 1)(2n + 1) 3n = 14n2 + 12n + 4 n3 6 = 3n2 3 n(n + 1) Now, to compute the exact area, we take + n2 2 the limit as n → ∞ : A = lim An 11n2 + 12n + 1 n→∞ = 6n2 14n2 + 12n + 4 Now to compute the exact area, we take = lim n→∞ 3n2 the limit as n → ∞: 14 = A = lim An 3 n→∞ 11n2 + 12n + 1 2 = lim (c) ∆x = We will use right endpoints as n→∞ 6n2 n 11 12 1 11 2i = lim + + = evaluation points,xi = 1 + . n→∞ 6 6n 6n2 6 n 2 n (b) ∆x = . We will use right endpoints as n An = f (xi ) ∆x 2i i=1 evalution points, xi = . n n 2 = xi 2 + 1 n i=1 n An = f (xi )∆x n 2 i=1 2 2i n = 1+ +1 2 2i 2 2i n i=1 n = +3 n n n n 2 4i 4i2 i=1 n n = 2+ + 2 8 12 n i=1 n n = 3 i2 + 2 i n n n i=1 n i=1 8 8 =4+ 2 i+ 3 i2 8 n (n + 1) (2n + 1) n i=1 n i=1 = 3 n 6 8 n (n + 1) 12 n (n + 1) =4+ 2 + 2 n 2 n 2 8 n (n + 1) (2n + 1) + 3 8n2 + 12n + 4 6n + 6 n 6 = 2 + 4n + 4 8n2 + 12n + 4 3n n =4+ + Now, to compute the exact area, we take n 3n2 Now, to compute the exact area, we take the limit as n → ∞ : A = lim An n→∞ the limit as n → ∞: 8n2 + 12n + 4 6n + 6 A = lim An = lim + n→∞ n→∞ 3n2 n 4n + 4 8n2 + 12n + 4 8 26 = lim 4+ + = +6= n→∞ n 3n2 3 3
  • 256 CHAPTER 4. INTEGRATION 2 2 (c) ∆x = . We will use right endpoints as (b) ∆x = . We will use right endpoints as n n 2i 2i evalution points, xi = 1 + . evalution points, xi = −1 + . n n n n An = f (xi ) ∆x An = f (xi )∆x i=1 n i=1 2 n 2 = 2xi 2 + 1 = 2 xi + 3xi i=1 n i=1 n n 2 n 2 2i 2 2i 2 2i = 2 −1 + +1 = 1+ +3 1+ n i=1 n n n n n i=1 n 2 8i 8i2 2 10i 4i2 = 3− + 2 = 4+ + 2 n i=1 n n n n n n n i=1 16 16 20 8 n n =6− i+ 3 i2 =8+ 2 i+ 3 i2 n2 i=1 n i=1 n i=1 n i=1 16 n (n + 1) 20 n (n + 1) =6− 2 =8+ 2 n 2 n 2 16 n (n + 1) (2n + 1) 8 n (n + 1) (2n + 1) + 3 + 3 n 6 n 6 8n + 8 16n2 + 24n + 8 10 4 = 6− + = 8+ (n + 1) + 2 2n2 + 3n + 1 n 3n2 n 3n Now, to compute the exact area, we take Now, to compute the exact area, we take the limit as n → ∞: the limit as n → ∞: A = lim An A = lim An n→∞ n→∞ = lim = lim n→∞ n→∞ 8n + 8 16n2 + 24n + 8 10 4 6− + 8+ (n + 1) + 2 2n2 + 3n + 1 n 3n2 n 3n 16 10 8 62 =6−8+ = = 8 + 10 + = 3 3 3 3 2 1 (c) ∆x = . We will use right endpoints as 13. (a) ∆x = . We will use right endpoints as n n 2i i evaluation points, xi = 1 + . evalution points, xi = . n n n n An = f (xi )∆x An = f (xi )∆x i=1 i=1 n 2 n 2 2 2i 1 i = 2 1+ +1 = 2 +1 n i=1 n n i=1 n n 2 n 2 8i2 8i = 3 i2 + 1 = 2 + +3 n i=1 n i=1 n n 2 n (n + 1) (2n + 1) n n = 3 +1 16 16 n 6 = i2 + i+6 n3 i=1 n2 i=1 5n2 + n + 1 = 16 n(n + 1)(2n + 1) 3n2 = Now, to compute the exact area, we take n3 6 the limit as n → ∞ : 16 n(n + 1) A = lim An + +6 n→∞ n2 2 2 5n + n + 1 5 = lim = . 16n(n + 1)(2n + 1) n→∞ 3n2 3 = 6n3
  • 4.3. AREA 257 16n(n + 1) n + +6 2 18i 16i2 2n2 = 5− + 2 n i=1 n n Now to compute the exact area, we take n n n the limit as n → ∞: 10 36 32 = 1− i+ i2 A = lim An n i=1 n2 i=1 n3 i=1 n→∞ 16n(n + 1)(2n + 1) 36 n (n + 1) = lim = 10 − n→∞ 6n3 n2 2 16n(n + 1) 32 n (n + 1) (2n + 1) + +6 + 3 2n2 n 6 18 16 32 16 58 = 10 − (n + 1) + 2 2n2 + 3n + 1 = lim + +6 = n 3n n→∞ 6 2 3 8 2 16 = − + 2 3 n 3n 114. (a) ∆x = . We will use right endpoints as Now, to compute the exact area, we take n i the limit as n → ∞ : evalution points, xi = . A = lim An n n→∞ n n 1 An = f (xi )∆x = 4xi 2 − xi 8 2 16 n = lim − + 2 i=1 i=1 n→∞ 3 n 3n n 2 8 1 i i = = 4 − 3 n i=1 n n n 2 1 4i2 i (c) ∆x = . We will use right endpoints as = − n n n2 n 2i i=1 evaluation points xi = 1 + . n n n 4 i2 1 i n n = − 2 n n2 n n An = f (xi )∆x = 4xi 2 − xi i=1 i=1 i=1 i=1 n 4 n (n + 1) (2n + 1) n 2 = 3 2 2i 2i n 6 = 4 1+ − 1+ 1 n (n + 1) n i=1 n n − 2 n 2 n 2 1 2 14i 16i2 = 2 2n2 + 3n + 1 − (n + 1) = 3+ + 2 3n 2n n i=1 n n 5 3 2 = + + 6 n 28 n 32 n 6 2n 3n2 = 1+ 2 i+ 3 i2 Now, to compute the exact area, we take n n n i=1 i=1 i=1 the limit as n → ∞: A = lim An 28 n (n + 1) n→∞ =6+ n2 2 5 3 2 32 n (n + 1) (2n + 1) = lim + + + 3 n→∞ 6 2n 3n2 n 6 5 14 16 = = 6+ (n + 1) + 2 2n2 + 3n + 1 6 n 3n 92 30 16 = + + 2 2 3 n 3n (b) ∆x = . We will use right endpoints as n Now, to compute the exact area, we take 2i evalution points, xi = −1 + . the limit as n → ∞: n n n 2 A = lim An An = f (xi )∆x = 4xi 2 − xi n→∞ i=1 i=1 n 92 30 16 = lim + + 2 2 n 2i 2 2i n→∞ 3 n 3n = 4 −1 + − −1 + 92 n n n = i=1 3
  • 258 CHAPTER 4. INTEGRATION 15. 20. Let L, M , and R be the values of the Riemann n Left Midpoint Right sums with left endpoints, midpoints and right Endpoint Endpoint endpoints. Let A be the area under the curve. 10 10.56 10.56 10.56 Then: L < A < M < R. 50 10.662 10.669 10.662 100 10.6656 10.6672 10.6656 250 500 10.6666 10.6667 10.6666 1000 10.6667 10.6667 10.6667 200 5000 10.6667 10.6667 10.6667 150 16. 100 n Left Midpoint Right Endpoint Endpoint 50 10 0.91940 1.00103 1.07648 0 50 0.98421 1.00004 1.01563 1 1.5 2 x 2.5 3 100 0.99213 1.00001 1.00783 500 0.99843 1.00000 1.00157 21. Let L, M , and R be the values of the Riemann 1000 0.99921 1.00000 1.00079 sums with left endpoints, midpoints and right 5000 0.99984 1.00000 1.00016 endpoints. Let A be the area under the curve. 17. Then: R < A < M < L. n Left Midpoint Right Endpoint Endpoint 0.12 10 15.48000 17.96000 20.68000 0.1 50 17.4832 17.9984 18.5232 0.08 100 17.7408 17.9996 18.2608 500 17.9480 17.9999 18.0520 0.06 1000 17.9740 17.9999 18.0260 0.04 5000 17.9948 17.9999 18.0052 0.02 18. n Left Midpoint Right 0 2 2.5 3 3.5 4 Endpoint Endpoint x 10 −2.20000 −2 −1.80000 22. Let L, M , and R be the values of the Riemann 50 −2.04000 −2 −1.96000 sums with left endpoints, midpoints and right 100 −2.02000 −2 −1.98000 endpoints. Let A be the area under the curve. 500 −2.00400 −2 −1.99600 Then: R < A < M < L. 1000 −2.00200 −2 −1.99800 5000 −2.00040 −2 −1.99960 250 19. Let L, M , and R be the values of the Riemann sums with left endpoints, midpoints and right 200 endpoints. Let A be the area under the curve. 150 Then: L < M < A < R. 100 1000 50 800 0 1 1.5 2 2.5 3 x 600 400 23. There are many possible answers here. One possibility is to use x = 1/6 on [0, 0.5] and √ 200 x = 23/6 on [0.5, 1]. 0 2 2.5 3 3.5 4 24. There are many possible answers here. One x possibility is to use x = 1/4 on [0, 0.5] and
  • 4.3. AREA 259 x = 25/36 on [0.5, 1]. 2 28. Consider interval[0, 2] , then ∆x = . n25. (a) We subdivide the interval [a, b] into n Use mid points as evaluation points, xi = 2(i−1) equal subintervals. If you are located at n + 2i n a + (b − a)/n (the first right endpoint), . 2 then each step of distance ∆x takes you to a new right endpoint. To arrive at the i-    n 2(i−1) + 2i 2 th right endpoint, you have to take (i − 1) A = lim  n n   steps to the right of distance ∆x. There- n→∞ i=1 2 n fore, n 2i − 2 + 2i 2 ci = a + (b − a)/n + (i − 1)∆x = a + i∆x. = lim n→∞ i=1 2n n (b) We subdivide the interval [a, b] into n Hence, equal subintervals. The first evaluation n 1 √ 2 point is a + ∆x/2. From this evaluation A = lim √ 2i − 1 . n→∞ n n point, each step of distance ∆x takes you i=1 to a new evaluation point. To arrive at Assume the i-th evaluation point, you have to take i = k + 1. n−1 (i − 1) steps to the right of distance ∆x. 1 2 A= √ 2 (k + 1) − 1 Therefore, n n k=0 ci = a + ∆x/2 + (i − 1)∆x n 1 √ 2 = a + (i − 1/2) ∆x, for i = 1, . . . , n. = √ 2k + 1 n n k=126. (a) We subdivide the interval [a, b] into n hence, n equal subintervals. If you are located at a 1 √ 2 A1 = √ 2k + 1 . (the first left endpoint), then each step of n n k=1 distance ∆x takes you to a new left end- point. To arrive at the i-th left endpoint, 4 2 you have to take (i − 1) steps to the right 2 i 29. U4 = of distance ∆x. Therefore, 4 i=1 2 ci = a + (i − 1)∆x. 4 1 1 2 = i2 = 1 + 22 + 32 + 42 (b) We subdivide the interval [a, b] into n 8 8 i=1 equal subintervals. The first evaluation 4 2 point is a + ∆x/3. From this evaluation 30 2 i−1 = = 3.75 L4 = point, each step of distance ∆x takes you 8 4 2 i=1 to a new evaluation point. To arrive at 4 the i-th evaluation point, you have to take 1 1 2 = i2 = 0 + 12 + 22 + 32 (i − 1) steps to the right of distance ∆x. 8 i=1 8 Therefore, 14 ci = a + ∆x/3 + (i − 1)∆x = = 1.75 8 = a + (i − 2/3) ∆x, for i = 1, . . . , n. 30. The function f (x) = x2 is symmetric on the 227. Consider interval [2, 4] , then ∆x = . two intervals [−2, 0] and [0, 2], so the upper n sum U8 is just double the value of U4 as cal- Use right endpoints as evaluation points, 2i culated in Exercise 35, and the same is for L8 . xi = 2 + . The answers are n n U8 = 2 · 3.75 = 7.5, L8 = 2 · 1.75 = 3.5. 2i 2 A = lim 2+ n 2 n→∞ i=1 n n 2 2i n 31. (a) Un = √ i 2 n n i=1 = lim 2 1+ 3 n n→∞ n n 2 i=1 = i2 Hence, n n i=1 √ i 2 A2 = lim 2 1+ . 3 n→∞ n n 2 n(n + 1)(2n + 1) i=1 = n 6
  • 260 CHAPTER 4. INTEGRATION 4 n(n + 1)(2n + 1) 24 (n − 1)2 n2 2 = = + (n) 3 n3 n4 4 n 4 1 1 4(n − 1)2 = 1+ 2+ = +2 3 n n n2 4 8 lim Un = (2) = 4(n2 − 2n + 1) n→∞ 3 3 = +2 n2 n 2 2 2(i − 1) 8 4 (b) Ln = =6− + 2 n i=1 n n n lim Ln = 6 3 n n→∞ 2 2 = (i − 1) 33. Here, f (x) = a2 − x2 and interval is [−a, a]. n i=1 2a 3 n−1 Hence ∆x = . 2 n = i2 Use right endpoints as evaluation points, n i=1 2ai xi = −a + . 3 n 2 (n − 1)(n)(2n − 1) n = n 6 An = f (xi )∆x i=1 4 (n − 1)(n)(2n − 1) n = 3 n3 = a2 − xi 2 ∆x 4 1 1 i=1 = 1− 2− n 2 3 n n 2ia 2a = a2 − −a + 4 8 n n lim Ln = (2) = i=1 n→∞ 3 3 n 4ia2 4i2 a2 2a = − i=1 n n2 n n 3 2 2 3 n n 32. (a) Un = 0+ i +1 8a 8a3 n n = i− i2 i=1 n2 i=1 n3 i=1 n 2 2i 3 8a3 n (n + 1) = +1 = 2 n n n 2 i=1 8a3 n (n + 1) (2n + 1) 2 4 n n − 3 = 3 i + 1 n 6 n 4a3 4a3 i=1 i=1 = (n + 1) − 3 2n2 + 3n + 1 n 3n 24 n2 (n + 1)2 2 = 4 + (n) Now, to compute the exact area, we take the n 4 n limit as n → ∞: 4(n + 1)2 A = lim An = +2 n→∞ n2 4a3 4a3 4(n2 + 2n + 1) = lim (n + 1) − 3 2n2 + 3n + 1 = +2 n→∞ n 3n n2 8 4 3 8 4 = 4− a3 = a =6+ + 2 3 3 n n 2 2 lim Un = 6 = (2a) a n→∞ 3 n−1 3 2 2 34. Here,f (x) = ax2 and interval is [0, b]. (b) Ln = 0+ i +1 n i=0 n b Hence ∆x = . n−1 3 n 2 2i Use right endpoints as evaluation points, xi = = +1 bi n i=0 n . n 4 n−1 n n 2 = i3 + 1 An = f (xi )∆x n i=0 i=1 i=1
  • 4.4. THE DEFINITE INTEGRAL 261 n = (0.0 + 0.4 + 0.6 + 0.8 + 1.2 + 1.4 + 1.2 + = axi 2 ∆x 1.4)(0.2) = 1.40 i=1 n 2 Right endpoints: bi b R8 = [f (1.2)+f (1.4)+f (1.6)+f (1.8)+f (2.0)+ = a i=1 n n f (2.2) + f (2.4) + f (2.6)](0.2) 3 n ab = (0.4 + 0.6 + 0.8 + 1.2 + 1.4 + 1.2 + 1.4 + = i2 1.0)(0.2) = 1.60 n3 i=1 ab3 n (n + 1) (2n + 1) = 3 39. A ≈ (0.2 − 0.1)(0.002) + (0.3 − 0.2)(0.004) + n 6 (0.4 − 0.3)(0.008) + (0.5 − 0.4)(0.014) + ab3 2 (0.6 − 0.5)(0.026) + (0.7 − 0.6)(0.048) + = 2 2n + 3n + 1 6n (0.8 − 0.7)(0.085) + (0.9 − 0.8)(0.144) + Now, to compute the exact area, we take the (0.95 − 0.9)(0.265) + (0.98 − 0.95)(0.398) + limit as n → ∞ : (0.99 − 0.98)(0.568) + (1 − 0.99)(0.736) + 1/2 · A = lim An n→∞ [(0.1 − 0)(0.002) ab3 +(0.2−0.1)(0.004−0.002)+(0.3−0.2)(0.008− = lim 2n2 + 3n + 1 n→∞ 6n2 0.004) + (0.4 − 0.3)(0.014 − 0.008) + (0.5 − 2ab3 ab3 1 0.4)(0.026 − 0.014) + (0.6 − 0.5)(0.048 − = = = b ab2 6 3 3 0.026) + (0.7 − 0.6)(0.085 − 0.048) + (0.8 −35. Using left hand endpoints: 0.7)(0.144 − 0.085) + (0.9 − 0.8)(0.265 − L8 = [f (0.0)+f (0.1)+f (0.2)+f (0.3)+f (0.4)+ 0.144) + (0.95 − 0.9)(0.398 − 0.265) + (0.98 − f (0.5) + f (0.6) + f (0.7)](0.1) 0.95)(0.568 − 0.398) + (0.99 − 0.98)(0.736 − = (2.0 + 2.4 + 2.6 + 2.7 + 2.6 + 2.4 + 2.0 + 0.568) (1 − 0.99)(1 − 0.736)] 1.4)(0.1) = 1.81 ≈ 0.092615 The Lorentz curve looks like: Right endpoints: 1 R8 = [f (0.1)+f (0.2)+f (0.3)+f (0.4)+f (0.5)+ f (0.6) + f (0.7) + f (0.8)](0.2) 0.8 = (2.4 + 2.6 + 2.7 + 2.6 + 2.4 + 2.0 + 1.4 + 0.6 0.6)(0.1) = 1.6736. Using left hand endpoints: 0.4 L8 = [f ().0)+f (0.2)+f (0.4)+f (0.6)+f (0.8)+ 0.2 f (1.0) + f (1.2) + f (1.4)](0.2) = (2.0 + 2.2 + 1.6 + 1.4 + 1.6 + 2.0 + 2.2 + 0 2.4)(0.2) = 3.08 0.2 0.4 0.6 0.8 1 Right endpoints: R8 = [f (0.2)+f (0.4)+f (0.6)+f (0.8)+f (1.0)+ 40. Obviously G = A1 /A2 is greater or equal to f (1.2) + f (1.4) + f (1.6)](0.2) 0. From the above figure we see that the = (2.2 + 1.6 + 1.4 + 1.6 + 2.0 + 2.2 + 2.4 + Lorentz curve is below the diagonal line y = x 2.0)(0.2) = 3.08 on the interval [0, 1], hence the area A1 ≤37. Using left hand endpoints: the area A2 . Furthermore, A2 = the area of L8 = [f (1.0)+f (1.1)+f (1.2)+f (1.3)+f (1.4)+ the triangle formed by the points (0, 0), (1, 0) f (1.5) + f (1.6) + f (1.7)](0.1) and (1, 1), hence equal to 1/2. Now G = = (1.8 + 1.4 + 1.1 + 0.7 + 1.2 + 1.4 + 1.82 + A1 /A2 = 2A1 . Using the date in Exercise 33, 2.4)(0.1) = 1.182 G ≈ 2 · 0.092615 = 0.185230. Right endpoints: R8 = [f (1.1)+f (1.2)+f (1.3)+f (1.4)+f (1.5)+ f (1.6) + f (1.7) + f (1.8)](0.1) 4.4 The Definite Integral = (1.4 + 1.1 + 0.7 + 1.2 + 1.4 + 1.82 + 2.4 + 2.6)(0.1) = 1.262 1. We know that 3 n x3 + x dx ≈ c3 + ci ∆x i38. Using left hand endpoints: 0 i=1 L8 = [f (1.0)+f (1.2)+f (1.4)+f (1.6)+f (1.8)+ xi + xi−1 3i f (2.0) + f (2.2) + f (2.4)](0.2) Where ci = , xi = , n = 6. 2 n
  • 262 CHAPTER 4. INTEGRATION 3i 3(i−1) 6 + 6 (2i − 1) 4. We know that Here ci = = . 2 n 2 2 2 4 e−x dx ≈ e−ci ∆x. n 3 −2 c3 + ci . i i=1 n xi + xi−1 4i i=1 Where ci = , xi = −2 + , n = 6. 6 3 2 n (2i − 1) (2i − 1) 1 Here, = + . 64 4 2 −2 + 4i + −2 + 4(i−1) i=1 6 6 2i − 7 1 1 27 3 125 5 343 ci = = . = + + + + + + 2 3 64 4 64 4 64 4 64 n 2 4 6 2 4 7 729 9 1331 11 1 e−ci = e−ci + + + + + . n 6 4 64 4 64 4 2 i=1 i=1 3 = e−25/9 + e−1 + e−1/9 ⇒ x3 + x dx ≈ 24.47 2 0 +e−1/9 + e−1 + e−25/9 . 2. We know that 3 n 4 3 = e−25/9 + e−1 + e−1/9 . x2 + 1dx ≈ c2 + 1∆x i 3 0 2 i=1 2 xi + 3i xi−1 ⇒ e−x dx ≈ 1.7665 Where ci = , xi = , n = 6. −2 2 n 3i + 3(i−1) (2i − 1) 5. Notice that the graph of y = x2 is above the Here ci = 6 6 = . 3 2 4 x-axis. So, 1 x2 dx is the area of the region n 3 bounded by y = x2 and the x-axis, between c2 + 1 i i=1 n x = 1 and x = 3.   6 2 2i − 1 1 6. Notice that the graph of y = ex is above the =  + 1. 1 4 2 x-axis. So, 0 ex dx is the area of the region i=1 √ √ √ bounded by y = ex , and the x-axis, between 17 5 41 65 x = 0 and x = 1. = + + + 4 4 4 4 √ √ 7. Notice that the graph of y = x2 − 2 is below √ 97 137 1 the x−axis for |x| ≤ 2 above the, x−axis for + + . √ 4 4 2 |x| ≥ 2. 3 Also, ⇒ x2 + 1dx ≈ 5.64 2 0 x2 − 2 dx 0 √ 3. We know that 2 2 π n sin x2 dx ≈ sin c2 ∆x. = x2 − 2 dx + √ x2 − 2 dx. i 0 0 2 i=1 2 xi + xi−1 iπ So, 0 x2 − 2 dx is the additon of the ar- Where ci = , xi = , n = 6. 2 n eas of the regions bounded by y = √2 − 2and x πi π(i−1) 6 + 6 (2i − 1) π the x−axis, between x = 0 and x = 2√ (which Hereci = = . is below the x−axis) and between x = 2 and 2 12 n π x = 2 (which is above the x−axis) sin c2 i i=1 n 6 2 (2i − 1) π π 8. Notice that the graph of y = x3 − 3x2 + 2x = sin . is below the x-axis, for 1 ≤ x ≤ 2 and x ≤ 0 i=1 12 6 2 2 and above the x-axis, for all other values of x. π 2 3π 5π = sin + sin + sin Also, 12 12 12 2 2 2 2 x3 − 3x2 + 2x dx 7π 9π 11π π 0 + sin + sin + sin . 1 12 12 12 6 = x3 − 3x2 + 2x dx π 0 ⇒ sin x2 dx ≈ 0.8685 2 0 + x3 − 3x2 + 2x dx 1
  • 4.4. THE DEFINITE INTEGRAL 263 8 n(n + 1)(2n + 1) 2 = 3 2 n3 6 So, x − 3x + 2x dx is the additon of 0 4(n + 1)(2n + 1) the areas of the regions bounded by = y = x3 − 3x2 + 2x and the x-axis between 3n2 To compute the value of the integral, we take x = 0 and x = 1 (which is above the x-axis) the limit as n → ∞, and between x = 1 and x = 2 (which is below 2 the x-axis). x2 dx = lim Rn 0 n→∞ 1 4(n + 1)(2n + 1) 8 9. For n rectangles, ∆x = , xi = i∆x. = lim 2 = n n→∞ 3n 3 n 12. For n rectangles, Rn = f (xi )∆x 3 3i i=1 ∆x = , xi = i∆x = . n n n n n 1 i 2 n = 2xi ∆x = 2 = i Rn = f (xi )∆x i=1 n i=1 n n2 i=1 i=1 2 n(n + 1) (n + 1) n n 2 = = 3 3i n2 2 n = (x2 i + 1)∆x = 2 +1 To compute the value of the integral, we take i=1 n i=1 n the limit as n → ∞, n 1 3 18i2 (n + 1) = +1 2xdx = lim Rn = lim =1 n n2 0 n→∞ n→∞ n i=1 n n 1 54 310. For n rectangles, ∆x = , xi = 1 + i∆x. = i2 + 1 n n3 i=1 n i=1 n Rn = f (xi ) ∆x 54 n(n + 1)(2n + 1) 3 = + n n i=1 n n3 6 n 1 i = 2xi ∆x = 2 1+ 9(n + 1)(2n + 1) n n = +3 i=1 i=1 n2 n n To compute the value of the integral, we take 2 2 = 1+ i the limit as n → ∞, n i=1 n2 i=1 3 (x2 + 1)dx = lim Rn 2 2 n(n + 1) 0 n→∞ = (n) + 2 n n 2 9(n + 1)(2n + 1) = lim +3 (n + 1) n→∞ n2 =2+ n = 9 + 3 = 12 To compute the value of the integral, we take the limit as n → ∞, 2 13. For n rectangles, ∆x = , 2 (n + 1) n 2xdx = lim Rn = lim 2 + 2i 1 n→∞ n→∞ n xi = 1 + i∆x = 1 + n n =2+1=3 Rn = f (xi ) ∆x i=1 n11. For n rectangles, 2 2i = (x2 − 3)∆x i ∆x = , xi = i∆x = . i=1 n n n n 2 Rn = f (xi )∆x 2 2i = 1+ −3 i=1 n i=1 n n n 2 2 2i n = (x2 )∆x = i 8i 8i2 4 i=1 n i=1 n = + 3 − n 2 n n n n i=1 2 2 4i 8 2 = = 3 i 8n(n + 1) 8n(n + 1)(2n + 1) n n2 n = + −4 i=1 i=1 2n2 6n3
  • 264 CHAPTER 4. INTEGRATION 2 To compute the value of the integral, we take −(x2 − 4)dx the limit as n → ∞, −2 3 2 (x − 3)dx = lim Rn 1 n→∞ 18. Notice that the graph of y = x2 − 4x is below 8 16 8 the x-axis between x = 0 and x = 4. Since we = + −4= are asked for area and the area in question is 2 6 3 below the x-axis, we have to be a bit careful. 14. For n rectangles, 4 4 4i −(x2 − 4x)dx ∆x = , xi = −2 + i∆x = −2 + 0 n n n n π Rn = f (xi )∆x = (x2 − 1)∆x i 19. sin xdx i=1 i=1 0 n 2 0 π/4 4 4i 20. − sin xdx + sin xdx = −2 + −1 n i=1 n −π/2 0 n 21. The total distance is the total area under the 4 16i 16i2 = 3− + 2 curve whereas the total displacement is the n i=1 n n signed area under the curve. In this case, from n n n t = 0 to t = 4, the function is always positive 12 64 64 = 1− i+ i2 so the total distance is equal to the total dis- n i=1 n2 i=1 n3 i=1 placement. This means we want to compute 4 12 64 n(n + 1) the definite integral 40(1 − e−2t )dt. We = n− 2 n n 2 0 compute various right hand sums for different 64 n(n + 1)(2n + 1) values of n: + n3 6 n Rn 32(n + 1) 32(n + 1)(2n + 1) = 12 − + 10 146.9489200 n 3n2 To compute the value of the integral, we take 20 143.7394984 the limit as n → ∞, 50 141.5635684 2 100 140.7957790 (x2 − 1)dx = lim Rn 500 140.1662293 −2 n→∞ 1000 140.0865751 32(n + 1) = lim 12 − It looks like these are converging to about 140. n→∞ n So, the total distance traveled is approximately 32(n + 1)(2n + 1) 140 and the final position is + s(b) ≈ s(0) + 140 = 0 + 140 = 140. 3n2 64 4 22. The total distance is the total area under the = 12 − 32 + = 3 3 curve whereas the total displacement is the 15. Notice that the graph of y = 4 − x2 is above signed area under the curve. In this case, from the x-axis between x = −2 and x = 2: t = 0 to t = 4, the function is always posi- 2 tive so the total distance is equal to the total (4 − x2 )dx displacement. This means we want to com- −2 4 pute the definite integral 0 30e−t/4 dt. We 16. Notice that the graph of y = 4x − x2 is above compute various right hand sums for different the x-axis between x = 0 and x = 4: values of n: 4 n Rn (4x − x2 )dx 0 10 72.12494524 20 73.97390774 17. Notice that the graph of y = x2 − 4 is below 50 75.09845086 the x-axis between x = −2 and x = 2. Since 100 75.47582684 we are asked for area and the area in question 500 75.77863788 is below the x-axis, we have to be a bit careful. 1000 75.81654616
  • 4.4. THE DEFINITE INTEGRAL 265 n It looks like these are converging to about 75.8. 1 i2 2i = lim + So, the total distance traveled is approximately n→∞ n n2 n i=1 75.8 and the final position is s(b) ≈ s(0) + 75.8 = −1 + 75.8 = 74.8. n(n + 1)(2n + 1) 2n(n + 1) = lim + n→∞ 6n3 n2 423. f (x)dx 2 7 = +2= 0 6 3 1 4 = f (x)dx + f (x)dx 1 3 0 1 4 1 27. fave = (x2 − 1)dx 3−1 1 = 2xdx + 4dx n 2 0 1 1 2i 1 = lim 1+ −1 n→∞ n n 2xdx is the area of a triangle with base i=1 0 1 and height 2 and therefore has area = 1 4i 4i2 1 = lim + 2 n→∞ n n n 2 (1)(2) = 1. 4 4dx is the area of a rectangle with base 3 4n(n + 1) 4n(n + 1)(2n + 1) = lim + 1 n→∞ 2n2 6n3 and height 4 and therefore has area = (3)(4) = 4 10 12. =2+ = Therefore 3 3 4 f (x)dx = 1 + 12 = 13 1 1 0 28. fave = (2x − 2x2 )dx 1−0 0 4 n 2 1 i i24. f (x)dx = lim 2 −2 0 n→∞ i=1 n n n 2 4 n = f (x)dx + f (x)dx 1 2i 2i2 0 2 = lim + 2 2 4 n→∞ i=1 n n n = 2dx + 3xdx 0 2 2n(n + 1) 2n(n + 1)(2n + 1) 2 = lim + 2dx is the area of a square with base 2 and n→∞ 2n2 6n3 0 2 5 height 2 (it is, after all, a square) and therefore =1+ = has area = 4. 3 3 4 3xdx is a trapezoid with height 3 and bases 29. The function f (x) = 3 cos x2 is decreasing on 2 [π/3, π/2]. Therefore, on this interval, the 6 and 12 and therefore has area (using the for- maximum occurs at the left endpoint and is mula in the front of the text) 1 f (π/3) = 3 cos(π 2 /9). The minimum occurs at area = (6 + 12)(2) = 18. 2 the right endpoint and is f (π/2) = 3 cos(π 2 /4). Therefore Using these to estimate the value of the inte- 4 gral gives the following inequality: f (x)dx = 4 + 18 = 22 π/2 0 π π2 · (3 cos ) ≤ 3 cos x2 dx 4 6 4 π/3 1 π π225. fave = (2x + 1)dx ≤ · (3 cos ) 4 0 6 9 n 1 4 8i π/2 = lim +1 4 n→∞ n n −1.23 ≤ 3 cos x2 dx ≤ 0.72 i=1 π/3 8n(n + 1) = lim +1 2 n→∞ 2n2 30. The function f (x) = e−x is decreasing on =4+1=5 [0, 1/2]. Therefore, on this interval, the maxi- mum occurs at the left endpoint and is f (0) = 1 1. The minimum occurs at the right endpoint 126. fave = (x2 + 2x)dx and is f (1/2) = e−1/4 . Using these to estimate 1 0
  • 266 CHAPTER 4. INTEGRATION the value of the integral gives the following in- equality: We are interested in the value that is in the √ 1 −1/4 1/2 2 1 3−2 3 (e )≤ e−x dx ≤ (1) interval [−1, 1], so c = . 2 0 2 3 1/2 2 3 3 2 0.3894 ≤ e−x dx ≤ 0.5 35. (a) f (x)dx + f (x)dx = f (x)dx 0 0 2 0 √ 3 3 2 31. The function f (x) = 2x2 + 1 is increasing (b) f (x)dx − f (x)dx = f (x)dx on [0, 2]. Therefore, on this interval, the maxi- 0 2 0 mum occurs at the right endpoint and is f (2) = 2 1 1 3. The minimum occurs at the left endpoint 36. (a) f (x)dx + f (x)dx = f (x)dx and is f (0) = 1. Using these to estimate the 0 2 0 value of the integral gives the following inequal- 2 3 3 ity: (b) f (x)dx + f (x)dx = f (x)dx 2 −1 2 −1 (2)(1) ≤ 2x2 + 1dx ≤ (2)(3) 3 0 2 37. (a) (f (x) + g (x)) dx 2≤ 2x2 + 1dx ≤ 6 1 3 3 0 = f (x) dx + g (x) dx 3 1 1 32. The function f (x) = 3 is decreasing = 3 + (−2) = 1 x +2 on [−1, 1]. Therefore, on this interval, the 3 maximum occurs at the left endpoint and is (b) (2f (x) − g (x)) dx f (−1) = 3. The minimum occurs at the right 1 endpoint and is f (1) = 1. Using these to esti- 3 3 mate the value of the integral gives the follow- =2 f (x) dx − g (x) dx 1 1 ing inequality: = 2 (3) − (−2) = 8 1 3 (2)(1) ≤ dx ≤ (2)(3) −1 x3 + 2 3 1 3 38. (a) (f (x) − g (x)) dx 2≤ 3+2 dx ≤ 6 1 −1 x 3 3 = f (x) dx − g (x) dx 33. We are looking for a value c, such that 1 1 2 1 = 3 − (−2) = 5 f (c) = 3x2 dx 2−0 0 2 3 Since 3x2 dx = 8, we want to find c so that (b) (4g (x) − 3f (x)) dx 0 1 2 3 3 f (c) = 4 or, 3c = 4 Solving this equation using the quadratic for- =4 g (x) dx − 3 f (x) dx 1 1 2 = 4 (−2) − 3 (3) = −17 mula gives c = ± √ 3 We are interested in the value that is in the 2 39. (a) interval [0, 2], so c = √ . 3 3 34. We are looking for a value c, such that 1 1 f (c) = (x2 − 2x)dx 2 1 − (−1) −1 1 2 Since (x2 − 2x)dx = , we want to find c 1 −1 3 1 2 1 so that f (c) = or, c − 2c = 3 3 1 2 3 Solving this equation using the quadratic for- √ 3±2 3 mula gives c = 3
  • 4.4. THE DEFINITE INTEGRAL 267 (b) (b) Notice that x2 sin x is a continuous func- tion for all values of x and for 1 ≤ x ≤ 2, x2 sin 1 ≤ x2 sin x ≤ x2 . 12 On using theorem 4.3,we get 2 2 8 sin 1 x2 dx ≤ x2 sin xdx 1 1 2 ≤ x2 dx 4 1 2 2 2 x3 x3 sin 1 ≤ x2 sin xdx ≤ 3 1 1 3 1 2 1 2 3 4 7 2 7 sin 1 ≤ x sin xdx ≤ 3 1 3 240. (a) (c) Let us evaluate x2 sin xdx 1 2 n 2 1 using x sin xdx ≈ c2 sin ci ∆x i 1 i=1 and n = 6 xi + xi−1 i Where ci = , xi = 1 + , 2 6 2 + 6 + (i−1) i 6 Here ci = 2 (2i + 11) = 12 n 1 0 1 c2 sin ci i i=1 n 2 2 (b) 13 13 15 15 = sin + sin 8 12 12 12 12 2 2 17 17 19 19 + sin + sin 6 12 12 12 12 2 2 21 21 23 23 1 + sin + sin . 4 12 12 12 12 6 2 Therefore, x2 sin xdx ≈ 2.2465 2 1 2 (cos 1 − cos 2) ≤ x2 sin xdx -2 0 2 1 ≤ 4 (cos 1 − cos 2) ⇒ 0.9564 ≤ 2.2465 ≤ 3.825741. (a) Notice that x2 sin x is a continuous func- and 2 7 7 tion for all values of x and sin 1 ≤ x2 sin xdx ≤ for 1 ≤ x ≤ 2, 3 1 3 ⇒ 1.9634 ≤ 2.2465 ≤ 2.3333 sin x ≤ x2 sin x ≤ 4 sin x. The second inequality gives a range which On using theorem 4.3,we get 2 2 is more closer to the value of the integral. sin xdx ≤ x2 sin xdx Therefore, part (b) is more useful than 1 1 2 part (a). ≤ 4 sin xdx √ 1 2 42. Notice that x2 e− x is a continuous function (cos 1 − cos 2) ≤ x2 sin xdx for all values of x ≥ 0. 1 For 1 ≤ x ≤ 2, ≤ 4 (cos 1 − cos 2) √ √ e− 2 ≤ e− x ≤ e−1
  • 268 CHAPTER 4. INTEGRATION √ √ 0 Therefore x2 e− 2 ≤ x2 e− x ≤ x2 e−1 1 2 1 52. 9 − x2 dx = πr = π32 −3 4 4 Thus, on using theorem 4.3. 2 √ 2 √ 2 9π x2 e− 2 dx ≤ x e 2 − x dx ≤ x2 e−1 dx = 1 1 1 4 3 2 2 2 √ 2x √x3 e− ≤ x e 2 − x dx ≤ e−1 53. (a) Given limit 3 1 1 3 1 1 π nπ 2 lim sin + .... + sin 7 −√2 √ 2 − x 7 −1 n→∞ n n n e ≤ x e dx ≤ e n 3 1 3 1 iπ 2 √ = lim sin n→∞ n n 0.5672 ≤ x2 e− x dx ≤ 0.8583 i=1 1 We know that n b lim f (ci )∆x = a f (x)dx 43. This is just a restatement of the Integral Mean x→∞ i=1 Value Theorem. b−a Where ci = a + i∆x and ∆x = a+b n 44. Let c = . By definition, On comparision,we get 2 i 1 b n ci = , ∆x = and f (x)dx = lim f (ci )∆x. n n a n→∞ f (x) = sin(πx) ⇒ a = 0, b = 1 i=1 We can choose n to be always even, so that Therefore n 1 n = 2m, and 1 iπ b n lim sin = sin(πx)dx n→∞ n n 0 f (x)dx = lim f (ci )∆x i=1 a n→∞ i=1 m n (b) Given limit = lim f (ci )∆x + lim f (ci )∆x n+1 n+2 2n m→∞ m→∞ = lim + + ... + 2 i=1 i=m+1 n→∞ n2 n2 n c b n 1 n+i = f (x)dx + f (x)dx = lim a c n→∞ n n i=1 We know that 45. Between x = 0 and x = 2, the area below the n b x-axis is much less than the area above the x- lim f (ci )∆x = f (x)dx 2 x→∞ axis. Therefore 0 f (x)dx > 0 i=1 a b−a Where ci = a + i∆x and ∆x = 46. Between x = 0 and x = 2, the area above the n x-axis is much greater than the area below the On comparision,we get 2 x-axis. Therefore 0 f (x)dx > 0 i 1 ci = , ∆x = and f (x) = 1 + x n n 47. Between x = 0 and x = 2, the area below the ⇒ a = 0, b = 1 x-axis is slightly greater than the area above Therefore, n 1 2 the x-axis. Therefore 0 f (x)dx < 0 1 n+i lim ∆x = (1 + x)dx n→∞ n n 0 i=1 48. Between x = 0 and x = 2, the area below the x-axis is much greater than the area above the (c) Given limit 2 1 2 n x-axis. Therefore 0 f (x)dx < 0 f n +f n + ... + f n lim n→∞ n 2 1 1 n 49. 3xdx = bh = (2)(6) = 6 1 i 0 2 2 = lim f n→∞ n n i=1 4 1 1 We know that 50. 2xdx = (a + b)h = (2 + 8)(3) 1 2 2 n b = 15 lim f (ci )∆x = f (x)dx x→∞ a i=1 2 1 2 1 b−a 51. 4 − x2 = πr = π 22 = π Where ci = a + i∆x and ∆x = 0 4 4 n
  • 4.4. THE DEFINITE INTEGRAL 269 On comparision,we get per month), the total number of births from i 1 time t = 0 to t = 12 is given by the integral ci = and ∆x = n 12 n ⇒ a = 0, b = 1 b(t)dt. Therefore, 0 n 1 Similarly, the total number of deaths from time 1 i t = 0 to t = 12 is given by the integral lim f = f (x)dx 12 n→∞ n n 0 i=1 a(t)dt. 0 b 1 Of course, the net change in population is the54. f (x)dx = v number of birth minus the number of deaths: b−a a b Population Change f (x)dx = v(b − a) = Births − Deaths a and 12 12 c 1 = b(t)dt − a(t)dt f (x)dx = w 0 0 c−b b c 12 f (x)dx = w(c − b) = [b(t) − a(t)]dt. b 0 The average value of f over [a, c] is By graphing b(t) and a(t) we see that their c 1 graphs intersect 9 times, at f (x)dx c−a a t ≈ 38.5, 40.1, 44.4, 46.9, 50.2, 53.6, b c 56.1, 60.5, 61.9. 1 This tells us that we have b(t) > a(t) on the = f (x)dx + f (x)dx c−a a b intervals 1 (0, 38.5), (40.1, 44.4), (46.9, 50.2), = [v(b − a) + w(c − b)] (53.6, 56.1), (60.5, 61.9). c−a v(b − a) + w(c − b) The maximum population will occur when t = = 50.2. c−a55. Since b(t) represents the birthrate (in births 404 per month), the total number of births from 402 time t = 0 to t = 12 is given by the integral 400 12 0 b(t) dt. 398 Similarly, the total number of deaths from time t = 0 to t = 12 is given by the integral 396 12 0 a(t) dt. 394 Of course, the net change in population is the 392 number of birth minus the number of deaths: 390 Population Change 0 10 20 30 t 40 50 60 70 = Births − Deaths 12 12 57. From P V = 10 we get P (V ) = 10/V . By = b(t) dt − a(t) dt definition, 0 0 4 4 10 12 P (V ) dV = dV = [b(t) − a(t)] dt. 2 2 V n 0 2 10 Next we solve the inequality = · n 2 + 2i 410 − 0.3t > 390 + 0.2t i=1 n 20 > 0.5t then t < 40 months . An estimate of the value of this integral is Therefore b(t) > a(t) when t < 40 months. setting n = 100, and then the integral ≈ 6.93 The population is increasing when the birth rate is greater than the death rate, which is 58. The average temperature over the year is 12 during the first 40 month. After 40 months, 1 π 64 − 24 cos t dt. If you look at the the population is decreasing. The population- 12 0 6 would reach a maximum at t = 40 months. graphs T (t) and f (t) = 64 you should be able to see that the area under T (t) and f (t) be-56. Since b(t) represents the birthrate (in births tween t = 0 to t = 12 are equal. This means
  • 270 CHAPTER 4. INTEGRATION that the average temperature is 64. 61. Delivery is completed in time Q/p, and since in that time Qr/p items are shipped, the in- ventory when delivery is completed is 80 Qr r Q− =Q 1− . p p 60 The inventory at any time is given by   (p − r)t for t ∈ 0, Q 40 p g(t) =  Q − rt for t ∈ Q , Q p r 20 The graph of g has two linear pieces. The av- erage value of g over the interval [0, Q/r] is the 0 0 2 4 6 8 10 12 area under the graph (which is the area of a t triangle of base Q/r and height Q(1 − r/p)) di- vided by the length of the interval (which is the 59. Since r is the rate at which items are shipped, base of the triangle). Thus the average value rt is the number of items shipped between time of the function is (1/2)bh divided by b, which 0 and time t. Therefore, Q − rt is the num- is ber of items remaining in inventory at time t. (1/2)h = (1/2)Q(1 − r/p). Since Q − rt = 0 when t = Q/r, the formula This time the total cost is is valid for 0 ≤ t ≤ Q/r. The average value of D Q r f (Q) = c0 + cc (1 − ) f (t) = Q − rt on the time interval [0, Q/r] is Q 2 p Q/r r 1 c0 D cc (1 − p ) f (t)dt f (Q) = − + Q/r − 0 0 Q2 2 r Q/r c0 D cc r = (Q − rt)dt f (Q) = 0 gives 2 = (1 − ) Q 0 Q 2 p Q/r r 1 = Qt − rt2 2c0 D Q 2 Q= . 0 cc (1 − r/p) r Q2 r Q2 The order size to minimize the total cost is = − Q r 2 r2 2c0 D Q= . r Q2 Q cc (1 − r/p) = = . Q 2r 2 62. Use the result from Exercise 60, 2c0 D D Q Q= 60. f (Q) = c0 + cc cc Q 2 c0 D cc 2(50, 000)(4000) f (Q) = − 2 + = ≈ 324.44. Q 2 3800 Setting f (Q) = 0 gives Since this quantity already takes advantage of c0 D cc largest possible discount, the order size that = Q2 2 minimizes the total cost is about 324.44 items. 2c0 D Q = . This is the right answer of Q 63. The maximum of cc F (t) = 9 − 108 (t − 0.0003)2 minimizing the total cost f (Q), since when the occurs when 108 (t − 0.0003)2 reaches its mini- value of Q is very small, the value of D/Q mum, that is, when t = 0.0003. At that time will get very big, and when the value of Q F (0.0003) = 9 thousand pounds. is very small, the value of Q/2 will get very We estimate the value of big. This means that the function f (Q) is de- 0.0006 creasing on the interval [0, 2c0 D/cc ] and in- [9 − 108 (t − 0.0003)2 ] dt using midpoint creasing on the interval [ 2c0 D/cc , ∞]. When 0 sum and n = 20, and get m∆v ≈ 0.00360 thou- Q = 2c0 D/cc , sand pound-seconds, so ∆v ≈ 360 ft per sec- 2c0 D D c0 D cc Q ond. c0 = = cc = cc . Q 2c0 D 2 2 cc 64. The impulse-momentum equation of Prob-
  • 4.5. THE FUNDAMENTAL THEOREM OF CALCULUS 271 π/2 lem 65 gives 5∆v π/2 0.4 10. 3 csc x cot xdx = (−3 csc x) = (1000 − 25, 000(t − 0.2)2 ) dt π/4 π/4 0 √ 0.4 = −3 + 3 2 = (−25000t2 + 10000t) dt 0 π/4 π/4 Using a midpoint sum and n = 20 gives an 11. (sec t tan t) dt = sec t approximation for this integral of 267.0. This 0 means 5∆v ≈ 267 and ∆v ≈ 53.4 m/s √ 0 = 2−1 π/4 π/44.5 The Fundamental Theorem 12. 2 sec tdt = tan t =1 of Calculus 0 0 2 2 1/2 1/2 1. (2x − 3)dx = x2 − 3x = −2 3 13. √ dx = 3sin−1 x 0 0 0 1 − x2 0 3 x3 3 π π 2. 2 x − 2 dx = − 2x =3 =3 −0 = 3 6 2 0 0 1 1 1 1 4 3 x4 14. dx = 4 arctan x = 2π 3. x + 2x dx = + x2 =0 1 + x2 −1 4 −1 −1 −1 2 4 t−3 4. x3 + 3x − 1 dx 15. dt 0 1 t 2 4 x4 3x2 4 = − −x = −4 = 1 − 3t −1 dt = (t − 3 ln |t|) 4 2 0 1 1 4 = 3 − 3 ln 4 √ 3 5. x x+ dx 4 4 1 x t3 16 16. t (t − 2) dt = − t2 = 2 5/2 4 0 3 0 3 = x + 3 log x 5 1 t 2 t 2 2 17. ex/2 dx = (ex ) = et − 1 = · 32 + 3 log 4 − .1 − 3 log 1 0 0 5 5 62 t = + 3 log 4 18. sin2 x + cos2 x dx 5 0 2 2 t t 2 2 6. 4x − dx = 2x2 + =5 = 1dx = (x) =t 1 x2 x 1 0 0 1 1 6e−3x 19. The graph of y = 4 − x2 is above the x-axis 7. 6e−3x + 4 dx = + 4x 0 −3 0 over the interval [−2, 2]. 2 2 2 2 x3 32 =− 3 +4+2−0=− 3 +6 4 − x2 dx = 4x − = e e −2 3 −2 3 2 e2x − 2e3x 8. dx 20. The graph of y = x2 − 4x is below the x-axis 0 e3x 2 2 over the interval [0, 4]. = e−x − 2 dx = −e−x − 2x 4 x3 4 32 0 0 − x2 − 4x dx = − + 2x2 = 1 0 3 3 =− 2 −3 0 e π π 21. The graph of y = x2 is above the x-axis over 9. (2 sin x − cos x)dx = −2 cos x − sin x the interval [0, 2] . 2 2 π/2 π/2 x3 8 =3 x2 dx = = 0 3 0 3
  • 272 CHAPTER 4. INTEGRATION 0 sin x 22. The graph of y = x3 is above the x-axis over 32. f (x) == t2 + 4 dt + t2 + 4 dt the interval 3x 0 [0, 3] . 3x sin x 3 3 =− t2 + 4 dt + t2 + 4 dt x4 81 x3 dx = = 0 d 0 0 4 0 4 f (x) = − 9x2 + 4 (3x) dx 23. The graph of y = sin x is above the x-axis over d + sin2 x + 4 (sin x) the interval [0, π] . dx 2 2 π π = −27x − 12 + sin x cos x + 4 cos x sin xdx = − cos x =2 0 0 33. s (t) = 40t + cos t + c, s (0) = 0 + cos 0 + c = 2 24. The graph of y = sin x is below the x-axis over so therefore c = 1 and s (t) = 40t + cos t + 1. the interval − π , 0 and above the x-axis over 2 the interval 0, π . Hence we need to compute 4 34. s (t) = 10et + c, two seperate integrals and add them together: s (0) = 10 + c = 2 0 π/4 so therefore c = −8 and s (t) = 10e−t − 8. − sin xdx + sin xdx −π/2 0 t2 1 1 35. v (t) = 4t − + c1 , =1+ 1− √ =2− √ . 2 2 2 v (0) = c1 = 8 25. f (x) = x2 − 3x + 2 t2 so therefore c1 = 8 and v (t) = 4t − + 8. 2 t3 26. f (x) = x2 − 3x − 4 s (t) = 2t2 − + 8t + c2 , 6 s (0) = c2 = 0 2 2 d 27. f (x) = e−(x ) + 1 x2 t3 dx so therefore c2 = 0 and s (t) = 2t2 − + 8t. 4 = 2x e−x + 1 6 t3 36. v (t) = 16t − + c1 , 28. f (x) = − sec x 3 v (0) = c1 = 0 0 2−x so therefore c1 = 0 and 29. f (x) = sin t2 dt + sin t2 dt ex 0 t3 v (t) = 16t − . d x 3 f (x) = − sin e2x (e ) t4 dx 2 s (t) = 8t − + c2 , 2 d 12 + sin (2 − x) (2 − x) s (0) = c2 = 30 dx x 2x 2 = −e sin e − sin (2 − x) t4 so therefore c2 = 30 and s (t) = 8t2 − + 30. 0 xex 12 2t 30. f (x) = e dt + e2t dt 37. Let w (t) be the number of gallons in the tank 2−x 0 at time t. d f (x) = −e2(2−x) (2 − x) (a) The water level decreases if w (t) = dx x d f (t) < 0 i.e. if f (t) = 10 sin t < 0, for + e2(xe ) (xex ) dx x which π < t < 2π. = e4−2x + e2xe (xex + ex ) Alternatively, the water level increases if w (t) = f (t) > 0 i.e. if f (t) = 10 sin t > 0 x3 0, for which 0 < t < π. 31. f (x) = sin (2t) dt + sin (2t) dt x2 0 (b) Now,we start with d f (x) = − sin 2x2 x2 w (t) = 10 sin t dx π π d + sin 2x3 x3 Therefore, w (t) dt = 10 sin tdt dx 0 π 0 = −2x sin 2x2 + 3x2 sin 2x3 w (π) − w (0) = − 10 cos t|0
  • 4.5. THE FUNDAMENTAL THEOREM OF CALCULUS 273 2 n But w (0) = 100. 2 2i 43. x2 + 1dx = lim +1 Therefore, 0 n→∞ n n i=1 w (π) − 100 = −10 (−1 − 1) = 20 Estimating using n = 20, we get the Riemann ⇒ w (π) = 120. sum ≈ 2.96. Therefore the tank will have 120 gallons 2 √ 2 √ at t = π. 2 44. x + 1 dx = x + 2 x + 1 dx 0 038. Let w (t) be the number of thousand gallons in 2 2 √ the pond at time t. x 4 3 8 2 = + x2 + x =4+ . 2 3 0 3 (a) The water level decreases if w (t) = 2 f (t) < 0 i.e. if f (t) = 4t − t2 < 0, for 4 x2 n 3 1 + (3i/n) which 4 < t ≤ 6. 45. dx = lim 1 x2 + 4 n→∞ i=1 n (3i/n)2 + 4 Alternatively, the water level increases if w (t) = f (t) > 0 i.e. if f (t) = 4t−t2 > 0, Estimating using n = 20, we get the Riemann for which 0 < t < 4. sum ≈ 1.71. 4 (b) Now, we start with w (t) = 4t−t2 , There- 4x2 + 4 4 4 fore 46. 1 dx = 1 1+ dx = x − 4x−1 6 6 x2 x2 1 =6 w (t) dt = 4t − t2 dt 0 0 π/4 t3 6 sin x w (6) − w (0) = 2 2t − 47. dx 3 0 cos2 x 0 π/4 But w (0) = 40. π/4 √ = tan x sec xdx = sec x = 2−1 Therefore, 0 0 w (6) − 40 = 72 − 72 = 0 π/4 π/4 ⇒ w (6) = 40. tan x 48. dx = sin x cos xdx Therefore the pond has 40,000 gallons at 0 sec2 x 0 π/4 t=6. π/4 1 1 1 = sin 2xdx = − cos 2x = √ 0 2 4 439. y (x) = sin x2 + π 2 . 0 At the point in question, y (0) = 0 and y (0) = 49. From the graph of f (x), sin π = 0. Therefore, the tangent line has slope 3 2 1 0 and passes through the point (0, 0). The f (x) dx < f (x) dx < f (x) dx. equation of this line is y = 0. 0 0 0 The function increases if g (x) = f (x) > 040. y (x) = ln x2 + 2x + 2 . i.e. when x < 1 or x > 3. Thus, the function At the point in question, y (−1) = 0 and g (x) is increasing in the intervals (−∞, 1) and y (−1) = ln 1 = 0. Therefore, the tangent (3, ∞). The function g (x) has critical points line has slope 0 and passes through the point at g (x) = 0. i.e. when x = 1 or x = 3. There- (−1, 0). The equation of this line is y = 0. fore the critical points of g(x) are x = 1 and41. y (x) = cos πx3 . x = 3. At the point in question, y (2) = 0 and y (2) = 1 3 2 cos 8π = 1. Therefore, the tangent line has 50. f (x) dx < f (x) dx < f (x) dx. 0 0 0 slope 1 and passes through the point (2, 0). The equation of this line is y = x − 2. The function increases if g (x) = f (x) > 0 i.e. when 0 < x < 2 or x > 4. Thus, the function 242. y (x) = e−x +1 . g (x) is increasing in the intervals (0, 2) and At the point in question, y (0) = 0 and y (0) = (4, ∞). The function g (x) has critical points e. Therefore, the tangent line has slope e and at g (x) = 0 i.e.when x = 0, x = 2 and x = 4. passes through the point (0, 0). The equation Therefore the critical points of g(x) are x = 0, of this line is y = ex. x = 2 and x = 4.
  • 274 CHAPTER 4. INTEGRATION 51. If you look at the graph of 1/x2 , it is obvious (b) is improper since the point x = 3 lies in that there is positive area between the curve 1 the interval [0, 4], and is not defined and the x-axis over the interval [−1, 1]. In ad- (x − 3)2 dition to this, there is a vertical asymptote in at x = 3. The integral in part (c) is improper the interval that we are integrating over which since the point x = π/2 lies in the interval should alert us to a possible problem. [0, 2], and sec x is not defined at x = π/2. The problem is that 1/x2 is not continuous 1 3 on [−1, 1] (the discontinuity occurs at x = 0) 55. fave = x2 − 1 dx 3−1 1 and that continuity is one of the conditions in 3 1 x3 10 the Fundamental Theorem of Calculus, Part = −x = I(Theorem 4.1). 2 3 1 3 1 1 50 56. fave = 2x − 2x2 dx 1−0 0 1 40 2x3 1 = x2 − = 3 0 3 30 π/2 y 1 20 57. fave = cos xdx π/2 − 0 0 2 π/2 2 10 = (sin x)|0 = π π 2 0 1 ex dx -1 -0.5 0 0.5 1 x 58. fave = 2−0 0 2 52. If you look at the graph of sec2 x, it is obvious 1 x = (e ) that there is positive area between the curve 2 0 and the x-axis over the interval [0, π]. In ad- 1 2 = e −1 dition to this, there is a vertical asymptote in 2 the interval that we are integrating over which 59. (a) Using the Fundamental Theorem of Cal- should alert us to a possible problem. The culus, it follows that an antiderivative of problem is that sec2 x is not continuous on [0, π] 2 x 2 and that continuity is one of the conditions in e−x is e−t dt where a is a constant. a the Fundamental Theorem of Calculus, Part I (b) Using the Fundamental Theorem of Cal- (Theorem 4.1). culus, it follows that an antiderivative of x 10 sin x2 + 1 is sin t2 + 1dt where a a 8 is a constant. y 6 60. It may be observed that f is piecewise contin- uous over its domain. 4 For 0 < x ≤ 4, x x 2 g(x) = f (t) dt = t2 + 1 dt 0 0 x t3 x3 -1 0 0 1 2 3 4 = +t = +x x 3 0 3 Now, for x > 4 x 53. The integrals in parts (a) and (c) are improper, g(x) = f (t) dt 0 because the integrands have asymptotes at one 4 x of the limits of integration. The Fundamental = f (t) dt + f (t) dt 0 4 Theorem of Calculus applies to the integral in 4 x part (b). = t2 + 1 dt + t3 − t dt 0 4 4 x 54. The Fundamental Theorem of Calculus applies t3 4 t t2 to the integral in part (a). The integral in part = +t + − 3 0 4 2 4
  • 4.5. THE FUNDAMENTAL THEOREM OF CALCULUS 275 43 x4 x2 44 42 1 = +4 + − − + f (1) = t2 − 3t + 2 dt 3 4 2 4 2 0 x4 x2 92 t3 3t2 1 5 = − − = − + 2t = 4 2 3 3 2 6 x3 0 +x for 0 < x ≤ 4 2 g (x) = x4 3 x2 92 f (2) = 2 t − 3t + 2 dt 4 − 2 − 3 for 4 < x 0 Consider 2 g (4 + h) − g (4) t3 3t2 2 g (4) = lim = − + 2t = h→0 h 3 2 0 3 4+h 4 Hence f (x) has a local maximum at the 1 = lim f (t) dt − f (t) dt 5 h→0 h 0 0 point 1, and local minimum at the point 6 4+h 1 2 = lim f (t) dt. 2, . h→0 h 4 3 The Right Hand Limit: x u 1 4+h 62. g (x) = f (t) dt du lim+ f (t) dt 0 0 h→0 h 4 x 1 4+h 3 g (x) = f (t) dt = lim t − t dt 0 h→0+ h 4 g (x) = f (x) 4+h 1 t4 t2 A zero of f corresponds to a zero of the second = lim − h→0+ h 4 2 4 derivative of g (possibly an inflection point of 1 (4 + h) 4 (4 + h) 2 44 42 g). = lim+ − − + h→0 h 4 2 4 2 63. When a < 2 or a > 2, f is continuous. Using 1 h4 47h2 the Fundamental Theorem of Calculus, = lim + 4h3 − + 60h h→0+ h 4 2 lim F (x) − F (a) x→a h3 47h = lim [F (x) − F (a)] = lim+ + 4h2 − + 60 = 60. x→a h→0 4 2 x a Now, the Left Hand Limit: = lim f (t) dt − f (t) dt x→a 0 0 1 4+h x lim− f (t) dt = lim f (t) dt = 0 h→0 h 4 x→a a 1 4+h 2 When a = 2, = lim t + 1 dt x x h→0− h 4 4+h lim f (t) dt = lim tdt 1 t3 x→a− a x x→2− 2 = lim +t t2 2 x 22 h→0− h 3 4 = lim = lim− − =0 3 x→2− 2 x→2 2 2 1 (4 + h) 43 2 x = lim− +4+h− −4 h→0 h 3 3 and lim+ f (t) dt x→a a 1 h3 + 12h2 + 48h + 64 64 x = lim +h− = lim+ (t + 1) dt h→0+ h 3 3 x→2 2 h2 t 2 x = lim+ + 4h + 17 = 17. = lim+ +t h→0 3 x→2 2 2 Therefore, g (4) doesn’t exist though f (4) x2 22 = lim +x− −2 exists. Therefore g (x) = f (x) is not true for x→2+ 2 2 all x ≥ 0. =0 Thus, for all value of a, lim F (x) − F (a) = 0 x→a61. f (x) = x2 − 3x + 2. lim F (x) = F (a) x→a Setting f (x) = 0, we get (x − 1) (x − 2) = 0 Thus, F is continuous for all x. However, F (2) which implies x = 1, 2. does not exist, which is shown as follows: > 0 when t < 1 or t > 2 F (2 + h) − F (2) f (x) = F (2) = lim < 0 when 1 < t < 2 h→0 h
  • 276 CHAPTER 4. INTEGRATION 1 2+h 2 lim gn (x) = lim f (xn ) = lim f (t) dt − f (t) dt n→∞ n→∞ h→0 h 0 0 = lim f (1) = f (1) . n→∞ 2+h 1 1 Thus the integral 0 gn (x) dx represents the = lim f (t) dt h→0 h 2 net area between the graph of f (xn ) and the We will show that this limit does not exist by x-axis. As n approaches ∞, showing that the left and right limits are dif- f (0) when 0 ≤ x < 1 ferent. The right limit is f (xn ) → f (1) when x = 1 1 2+h 1 lim+ f (t) dt Thus the integral 0 gn (x) dx approaches the h→0 h 2 area of the shape of a rectangle with length 1 2+h 1 and width f (0) (possibly negative), which = lim+ (t + 1) dt 1 h→0 h 2 means lim 0 gn (x) dx = f (0). 2+h n→∞ 1 t2 = lim +t x h→0+ h 2 2 2 65. [f (t) − g (t)] dt 1 (2 + h) 22 0 = lim +2+h− −2 x h→0+ h 2 2 = [55 + 10 cos t − (50 + 2t)] dt 1 h2 + 4h + 4 0 x = lim +2+h−4 h→0+ h 2 = [5 + 10 cos t − 2t] dt 2 0 1 h x = lim+ + 3h = 5t + sin t − t2 0 h→0 h 2 1 h = 5x + sin x − x2 = lim+ +3 =3 Since we are integrating the difference in h→0 h 2 The left limit is speeds, the integral represents the distance 1 2+h that Katie is ahead at time x. Of course, if lim− f (t) dt this value is negative, is means that Michael is h→0 h 2 1 2+h really ahead. = lim tdt h→0− h 2 Q 2+h 1 t2 66. (a) CS = D (q) dq − P Q = lim− 0 h→0 h 2 2 Q 1 (2 + h) 2 22 = 150q − 2q − 3q 2 dq − P Q = lim− − 0 Q h→0 h 2 2 = 150q − q 2 − q 3 0 − P Q 2 1 h + 4h + 4 = 150Q − Q2 − Q3 = lim− −2 h→0 h 2 − 150 − 2Q − 3Q2 Q 1 h = Q2 + 2Q3 . = lim− +2 =2 h→0 h 2 When Q = 4, Thus, F (2) does not exist. This result does CS = 16 + 2 (64) = 144 dollors not contradict the Fundamental Theorem of When Q = 6, CS = 36 + 2 (216) = Calculus, because in this situation, f (x) is not 468 dollors continuous, and thus The Fundamental Theo- The consumer surplus is higher for Q = 6 rem of Calculus does not apply. than that for Q = 4. Q 64. When x = 0, (b) CS = D (q) dq − P Q lim gn (x) = lim f (xn ) Q 0 n→∞ n→∞ = lim f (0) = f (0) = 40e−0.05q dq − P Q n→∞ 0 When 0 < x < 1, Q = −800e−0.05q 0 − P Q limn = 0, and then n→x = −800e−0.05Q + 800 − 40e−0.05Q lim gn (x) = lim f (xn ) = −840e−0.05Q + 800. n→∞ n→∞ When Q = 10, CS = −840e−0.5 + 800 ≈ =f lim xn = f (0) n→∞ 290.5 dollors = lim f (0) = f (0) When Q = 20, CS = −840e−1 + 800 ≈ n→∞ When x = 1, 491.0 dollors
  • 4.6. INTEGRATION BY SUBSTITUTION 277 The consumer surplus is higher for Q = −2/3 1 x3 (x4 + 1) dx = u−2/3 du 20 than that for Q = 10. 4 3 1/3 3 1/3 = u + c = (x4 + 1) + c.67. The next shipment must arrive when the in- 4 4 ventory is zero. This occurs at time T : f (t) = √ √ 1 Q−r t √ 3. Let u = x + 2 and then du = x−1/2 dx and f√ ) = 0 = Q − r T (T √ 2 3 ( x + 2) r T =Q √ dx = 2 u3 du Q2 x T = 2 2 1 √ 4 r = u4 + c = ( x + 2) + c. The average value of f on [0, T] is 4 2 1 T 4. Let u = sin xand then du = cos xdx and f (t) dt T 0 sin x cos xdx = udu 1 T = Q − rt1/2 dt u2 sin2 x T 0 = +c= + c. T 2 2 1 2 = Qt − rt3/2 T 3 0 5. Let u = x4 + 3 and then du = 4x3 dx and 1 2 3/2 1 = QT − rT x3 x4 + 3dx = u1/2 du T 3 4 2 √ 1 1 3/2 =Q− r T = u3/2 + c = (x4 + 3) + c. 3 6 6 2 Q =Q− r 6. Let u = 1 + 10x, and then du = 10dx and 3 r Q √ 1 √ = 1 + 10xdx = udu 3 10 1 1 3/2 = u1/2 du = u +c D 10 1568. The total annual cost f (Q) = c0 + cc A = 1 Q = 3/2 (1 + 10x) + c. D Q 15 c0 + cc Q 3 D 1 7. Let u = cos x and then du = − sin xdx and f (Q) = −c0 + cc sin x du Q2 3 √ dx = − √ cos x √ √ u f (Q) = 0 = −2 u + c = −2 cos x + c. gives that Q = 3c0cD . c This value of Q minimizes the total cost, since 8. Let u = sin x and then du = cos xdx and   > 0 when Q < 3c0 D sin3 x cos xdx = u3 du cc f (Q)  < 0 when Q > 3c0 D u4 sin4 x cc = +c= + c. 4 4 3c0 D When Q = , cc 9. Let u = t3 and then du = 3t2 dt and D D Q 1 c0 = c0 = cc = cc A t2 cos t3 dt = cos udu Q 3c0 D/cc 3 3 1 1 = sin u + c = sin t3 + c 3 34.6 Integration By Substitu- 10. Let u = cos t + 3 and then du = − sin tdt and 3/4 tion sin t(cos t + 3) dt = − u3/4 du 4 4 7/4 1. Let u = x3 + 2 and then du = 3x2 dx and = − u7/4 + c = − (cos t + 3) + c. 7 7 1 x2 x3 + 2dx = u−1/2 du 3 11. Let u = x2 + 1 and then du = 2xdx and 2 2 2 1 u 1 = u3/2 + c = (x3 + 2)u3/2 + c. xex +1 dx = e du = eu + c 9 9 2 2 1 2 2. Let u = x4 + 1and then du = 4x3 dx and = ex +1 + c 2
  • 278 CHAPTER 4. INTEGRATION 3 12. Let u = ex + 4 and then du = ex dx and (sin−1 x) √ √ 2 √ dx = u3 du ex ex + 4dx = udu = u3/2 + c 1 − x2 3 4 1 x 3/2 u4 (sin−1 x) = (e + 4) + c = +c= +c 2 4 4 √ 1 22. Let u = x2 and then du = 2xdx and 13. Let u = x and then du = 2√x dx and x 1 1 √ √ dx = √ du e x √ 1 − x4 2 1 − u2 √ dx = 2 eu du = 2eu + c = 2e x +c 1 1 x = sin−1 u + c = sin−1 x2 + c 2 2 1 1 14. Let u = and then du = − 2 dx and 23. (a) Let u = x2 and then du = 2xdx and x x cos x 1 x 1 1 √ dx = √ du dx = − cos udu = − sin u + c 1 − x4 2 1 − u2 x2 1 1 1 = − sin + c = sin−1 u + c = sin−1 x2 + c x 2 2 (b) Let u = 1 − x4 and then du = −4x3 dx 1 15. Let u = ln x and then du = x dx and and √ ln x √ 2 x3 1 dx = udu = u3/2 + c dx = − u−1/2 du x 3 (1 − x 4 )1/2 4 2 3/2 1 1 1/2 = (ln x) + c 3 = − u1/2 + c = − (1 − x4 ) + c 2 2 16. Let u = tan x and then du = sec2 xdx and 24. (a) Let u = x3 and then du = 3x2 dx and 1 Let u = ln x and then du = x dx and x2 1 1 √ 6 dx = du sec2 x tan xdx = u1/2 du 1+x 3 1 + u2 1 1 2 2 √ 3/2 = tan−1 u + c = tan−1 x3 + c = u3/2 + c = ( tan x) + c 3 3 3 3 (b) Let u = 1 + u6 and then du = 6x5 dx and √ 17. Let t = u + 1 and then x5 1 1 1 1 dx = du dt = u−1/2 du = √ du and 1 + x6 6 u 2 2 u 1 1 = ln |u| + c = ln |1 + x6 | + c 1 1 6 6 √ √ du = 2 dt = 2 ln |t| + c u (√ u + 1) t√ 1+x 25. (a) dx = 2 ln u + 1 + c = 2 ln u + 1 + c 1 + x2 1 x 18. Let u = v 2 + 4 and then du = 2vdv and = dx + dx 1 + x2 1 + x2 v 1 1 1 x 2+4 dv = du = ln |u| + c = tan−1 x + c1 + dx v 2 u 2 1 + x2 1 1 2 Let u = 1 + x and then du = 2xdx. = ln v + 4 + c = ln v 2 + 4 + c 2 2 2 1 1 = tan−1 x + c1 + du 1 2 u 19. Let u = ln x + 1 and then du = dx and 1 x = tan−1 x + c1 + ln |u| + c2 4 2 2 dx = 4 u−2 du 1 x(ln x + 1) = tan−1 x + ln 1 + x2 + c −1 2 = −4u−1 + c = −4(ln x + 1) + c −1 1 = tan x + ln 1 + x2 + c 2 20. Let u = cos 2x and then du = −2 sin 2xdx and 1+x 1+x 1 1 (b) dx = dx tan 2xdx = − du 1−x 2 (1 − x) (1 + x) 2 u 1 1 1 = − ln |u| + c = − ln | cos 2x| + c = dx 2 2 1−x Let u = 1 − x and then du = −dx. 1 1 21. Let u = sin−1 x and then du = √ dx and =− du = − ln |u| + c 1 − x2 u Let u = cos 2x and then du = −2 sin 2xdx and = − ln |1 − x| + c
  • 4.6. INTEGRATION BY SUBSTITUTION 27926. (a) Let u = x3/2 and then 33. Let u = t2 + 1 and then du = 2tdt, 3 3√ du = x1/2 dx = xdx and u (−1) = 2 = u (1) and √2 2 1 t 1 2 3 x 1 dt = u−1/2 du = 0 3 dx = 2 du −1 (t2 + 1)1/2 2 1+x 1 + u2 2 = 2tan−1 u + c = 2tan−1 x3/2 + c 34. Let u = t3 and then du = 3t2 dt, 5 (b) Let u = x 2 and then u (0) = 0, u (2) = 8 and 2 8 5 3 5 √ 3 1 8 u 1 du = x 2 dx = x xdx and t2 et dt = e du = eu √2 2 0 3 0 3 0 x x 2 1 1 8 5 dx = du = e −1 1+x 5 1 + u2 3 2 2 = tan−1 u + c = tan−1 x5/2 + c 35. Let u = ex and then du = ex dx, 5 5 u(0) = 1, u(2) = e2 and27. Let u = t + 7 and then du = dt, t = u − 7 and e2 2t + 3 2 (u − 7) + 3 2 ex e 1 2 dt = du dx = du = tan−1 u t+7 u 2x 1 + u2 0 1+e 1 11 1 = 2− du = 2u − 11 ln |u| + c π u = tan−1 e2 − 4 = 2 (t + 7) − 11 ln |t + 7| + c 36. Let u = 1 + ex and then du = ex dx,28. Let u = t + 3 and then du = dt and u (0) = 2, u (2) = 1 + e2 and 2 t2 (u − 3) 1/3 dt = du 2 1+e2 1+e2 (t + 3) u1/3 ex 1 dx = du = ln (u) = u5/3 − 6u2/3 + 9u−1/3 du 0 1 + ex 2 u 2 2 1 + e2 3 18 18 = ln 1 + e − ln (2) = ln = u8/3 − u5/3 + u2/3 + c 2 8 5 2 3 8/3 18 5/3 18 2/3 37. Let u = sin x and then du = cos xdx = (t + 3) − (t + 3) + (t + 3) + c √ 8 5 2 u(π/4) = 1/ 2, u(π/2) = 1 and √ 2 129. Let u = 1 + x and then (u2 − 1) = x, π/2 1 1 2 2(u − 1)(2u)du = dx and cot xdx = √ du = ln |u| π/4 1/ 2 u √ 1 4u(u2 − 1) √ 1/ 2 √ dx = du = ln 2 1+ x u u3 1 = 4 (u2 − 1)du = 4( − u) + c 38. Let u = ln x and then du = dx, u(1) = 0, 3 x 4 √ 3/2 √ 1/2 u(e) = 1 and = (1 + x) − 4(1 + x) + c e ln x 1 u2 1 1 3 dx = udu = =30. Let u = x2 and then du = 2xdx and 1 x 0 2 0 2 dx du/2 4 4 √ = √ x−1 x x 4−1 u u2 − 1 39. √ dx = (x1/2 − x−1/2 )dx 1 −1 1 −1 2 1 x 1 = sec u + c = sec x + c 4 2 2 2 = ( x3/2 − 2x1/2 ) 331. Let u = x2 + 1 and then u = 2xdx, u(0) = 1, 16 2 1 8 u(2) = 5 and = ( − 4) − ( − 2) = 2 3 3 3 1 5√ x x2 + 1dx = udu 40. Let u = x2 + 1 and then du = 2xdx and 2 1 1 0 5 x 1 2 −1/2 1 2 1 √ 5√ 1 dx = u du = . u3/2 = ( 125 − 1) = 5− 0 (x2 + 1) 1/2 2 1 2 3 1 3 3 3 2 √ = (u1/2 ) = 2 − 132. Let u = πx2 and then du = 2πxdx and 1 3 9π 9π π 1 x sin(πx2 )dx = sin udu = (sin u) = 41. (a) sin x2 dx ≈ .77 using midpoint evalu- 1 2π π π 0 0 ation with n ≥ 40.
  • 280 CHAPTER 4. INTEGRATION a (b) Let u = x2 and then du = 2xdx and f (x) 2 π 1 π −a x sin x2 dx = sin udu 0 a 0 2 0 =− f (−u)du + f (x)dx π2 −a 0 1 a a = (− cos u) 2 0 = f (−u)du + f (x)dx 1 1 0 0 = − cos π 2 + 2 2 If f is even, then f (−u) = f (u), and so ≈ 0.95134 a f (x)dx 42. (a) Let u = x2 and then du = 2xdx, −a a a 1 u(−1) = 1, u(1) = 1 and 2 xe−x dx = = f (u)du + f (x)dx 0 0 −1 a a 1 1 −u = f (x)dx + f (x)dx e du = 0 0 0 2 1 a 1 2 =2 f (x)dx (b) e−x ≈ 1.4937 using midpoint evalua- 0 −1 If f is odd, then f (−u) = −f (u), and so tion with n ≥ 50. a 2 2 f (x)dx 4x −a 43. (a) 2 dx ≈ 1.414 using right end- a a (x2 + 1) 0 =− f (u)du + f (x)dx point evaluation with n ≥ 50. 0 0 a a (b) Let u = x2 + 1 and then du = 2xdx, =− f (x)dx + f (x)dx x2 = u − 1 and 0 0 2 4x3 5 u−1 =0 dx = 2. 2 du 2 + 1)2 u 0 (x 1 50. First, let u = x − T, then for any a, 5 a+T a −1 −2 = (2u − 2u )du f (x)dx = f (u + T )du 1 T 0 5 8 a a = (2 ln |u| + 2u−1 ) 1 = 2 ln 5 − = f (u)du = f (x)dx 5 0 0 π/4 If we let a = T, then we get 44. (a) sec xdx ≈ .88 using midpoint evalu- T 2T 0 ation with n ≥ 10. f (x)dx = f (x)dx. a T π/4 π/4 If we let a = 2T, then we get 2 2T 3T (b) sec xdx = (tan x) = 1. 0 f (x)dx = f (x)dx 0 0 T 4 and then 1 T 2T 45. f (u)du. f (x)dx = f (x)dx 2 0 0 T 2T T 8 1 = f (x)dx − f (x)dx 46. f (u)du. 3 1 0 3T 0 2T 1 = f (x)dx− f (x)dx 47. f (u)du. T 3T T 0 √ = f (x)dx 4 2 f ( x) 2T 48. √ dx = 2 f (u)du. It is straight forward to see that for any integer 0 x 0 a i, 49. f (x)dx T (i+1)T −a f (x)dx = f (x)dx 0 a 0 iT = f (x)dx + f (x)dx −a 0 Now suppose 0 ≤ a ≤ T , then Let u = −x and du = −dx in the first integral. T a+T Then f (x)dx − dx 0 a
  • 4.6. INTEGRATION BY SUBSTITUTION 281 a a+T 0 f (a − u) = f (x)dx − f (x)dx =− du 0 T a f (a − u) + f (u) a T a+T f (a − u) = du So f (x)dx = dx 0 f (a − u) + f (u) a 0 a f (a − x) I= dx Now suppose a is any number. Then a must lie 0 f (a − x) + f (x) in some interval [iT, (i + 1)T ] for some interger The last equation follows from the previ- i. Use the similar method as in above, we shall ous one because u and x are dummy vari- get ables of integration. Now note that (i+1)T a+T f (x) f (x)dx = f (x)dx f (x) + f (a − x) iT a f (x) + f (a − x) − f (a − x) (i+1)T T = And since f (x)dx = f (x)dx, f (x) + f (a − x) iT 0 f (a − x) T a+T =1− we get f (x)dx = f (x)dx f (a − x) + f (x) 0 a Thus, a f (x) dx51. (a) Let u = 10 − x, so that du = −dx. Then, 0 f (x) + f (a − x) 10 √ a x f (a − x) I= √ √ dx = 1− dx 0 x + √ 10 − x 0 f (a − x) + f (x) a a x=10 10 − u f (a − x) =− √ √ du = 1dx − dx x=0 10 − u + u 0 0 f (a − x) + f (x) u=0 √ 2I = a 10 − u =− √ √ du I = a/2 u=10 u=10 √ −u+ u 10 10 − u = √ √ du u=0 10 − u + u √ 52. (a) Let u = 6 − x, so that du = −dx. x=10 10 − x Then, I= √ √ dx 4 x=0 10 − x + x sin2 (9 − x) I= dx The last equation follows from the previ- 2 2 2 sin (9 − x) + sin (x + 3) ous one because u and x are dummy vari- 2 sin2 (u + 3) ables of integration. Now note that =− 2 2 du √ 4 sin (u + 3) + sin (9 − u) x 4 √ √ sin2 (u + 3) x + 10 − x = du √ √ √ 2 2 2 sin (u + 3) + sin (9 − u) x + 10 − x − 10 − x 4 = √ √ sin2 (x + 3) √ + 10 − x x = 2 2 dx 2 sin (x + 3) + sin (9 − x) 10 − x 4 =1− √ √ sin2 (9 − x) x + 10 − x = 1− dx 2 sin2 (x + 3) + sin2 (9 − x) Thus, √ 4 10 x I= 1dx − I √ √ dx 2 0 x + 10 √ x − 2I = 2 10 10 − x I=1 = 1− √ √ dx 0 x + 10 − x √ 10 10 10 − x (b) Let u = 6 − x, so that du = −dx. = 1dx − √ √ dx 0 0 x + 10 − x Then, 4 10 f (9 − x) I= 1dx − I I= dx 0 2 f (9 − x) + f (x + 3) 2 2I = 10 f (u + 3) =− du I=5 4 f (u + 3) + f (9 − u) 4 (b) Let u = a − x, so that f (u + 3) = du du = −dx Then, 2 f (u + 3) + f (9 − u) a 4 f (x) f (x + 3) I= dx = dx 0 f (x) + f (a − x) 2 f (x + 3) + f (9 − x)
  • 282 CHAPTER 4. INTEGRATION 4 f (9 − x) u3 = 1− dx =6 du 2 f (x + 3) + f (9 − x) u+1 4 1 I= 1dx − I =6 u2 − u + 1 − du 2 u+1 2I = 2 u3 u2 I=1 =6 − + u − ln |u + 1| + c 3 2 53. Let 6 − u = x + 4; that is, let u = 2 − x, so that = 2x1/2 − 3x1/3 + 6x1/6 du = −dx. = −6 ln |x1/6 + 1| + c Then, (c) Let u = x1/q , then q du = x(1−q)/q dx, and 2 f (x + 4) 1 I= dx I= dx 0 f (x + 4) + f (6 − x) x(p+1)/q + xp/q 0 (1−q)/q f (6 − u) x dx =− du = (p+2−q)/q + x(p+1−q)/q dx 2 f (6 − u) + f (u + 4) x 2 f (6 − u) 1 = du =q du up+2−q + up+1−q 0 f (6 − u) + f (u + 4) q−1−p 2 u f (6 − x) =q du = dx u+1 0 f (6 − x) + f (x + 4) The rest of the calculation will depend on 2 f (6 − x) + f (x + 4) − f (x + 4) the values of p and q. = dx 0 f (6 − x) + f (x + 4) √ 2 f (x + 4) 55. First let u = ln x, so that du = = 1− dx 1 0 f (6 − x) + f (x + 4) x−1/2 (1/2)x−1/2 dx, so that 2du = dx. Then, 2 x I= 1dx − I 1 1 0 √ dx = 2 du 2I = 2 x ln x u I=1 = 2 ln |u| + c √ = 2 ln | ln x| + c 54. (a) Let u = x1/6 , so that du = 1 x−5/6 dx. 6 Then, Now use the substitution u = ln x, so that 1 1 du = x dx. Then, I= dx x5/6 + x2/3 1 1 x−5/6 dx √ dx = dx = x ln x x ln(x1/2 ) 1 + x−1/6 1 6 du = dx = 1 x 1 ln x 1+ u 2 6u 1 = du =2 du u+1 u Let v = u+1, then dv = du and u = v −1. = 2 ln |u| + c1 6u = 2 ln | ln x| + c1 Then, I = du u+1 6(v − 1) The two results differ by a constant, and = dv so are √ equivalent, as can be seen as follows: v 6 2 ln | ln x| = 2 ln | ln(x1/2 )| = 6− dv 1 v = 2 ln ln x = 6v − 6 ln |v| + c 2 1 = 6(u + 1) − 6 ln |u + 1| + c = 2 ln + ln | ln x| = 6(x1/6 + 1) − 6 ln |x1/6 + 1| + c 2 1 (b) Let u = x1/6 , so that du = (1/6)x−5/6 dx, = 2 ln + 2 ln | ln x| 2 which means 6u5 du = dx. = 2 ln | ln x| + constant Thus, 1 56. The area of the region bounded by the curve √ √ dx x+ 3x y = πx − x2 and x-axis, where 0 ≤ x ≤ 1 is 1 u5 =6 du πx − x2 dx u3 + u2 0
  • 4.6. INTEGRATION BY SUBSTITUTION 283 1 x2 x3 58. The problem is that it is not true on entire in- = π − 2 3 0 terval [0, π] that cos x = 1 − sin2 x. This is π 1 only true on the interval 0, π . To make this = − . 2 2 3 substitution correctly, one must break up the The area of the region bounded by the curve integral: π y = π cos x − cos2 x sin x and x-axis, where π/2 cos x(cos x)dx π 0≤x≤ 2 is 0 π cos x − cos2 x sin xdx. 0 π/2 π Let u = cos x and then du = − sin xdx. = cos x(cos x)dx + cos x(cos x)dx π 0 π/2 u (0) = 1, u = 0. x=π/2 2 0 = 1 − u2 du = −πu + u2 du x=0 x=π 1 0 − 1 − u2 du u2 u3 x=π/2 = −π + x=π/2 2 3 1 u sin−1 u π 1 = + = − 2 2 x=0 2 3 x=π u sin−1 u Thus, the areas are equal. − + 2 2 x=π/2 −1 x=π/2 2.5 sin x sin (sin x) = + 2 2 x=0 x=π 2.0 sin x sin−1 (sin x) − + 1.5 2 2 x=π/2 1 π 1 π = + −0−0+ + 1.0 2 4 2 4 π =1+ 0.5 2 0.0 59. Let u = 1/x, so that du = −1/x2 dx, which 0.0 0.25 0.5 0.75 1.0 means that −1/u2 du = dx. Then, 1 1 2.0 1 1/u2 1.8 2 dx = − 2 du 0 x +1 1/a 1/u + 1 1.6 1/a 1/a 1 1 1.4 = du = dx 1 1 + u2 1 1 + x2 1.2 1.0 The last equation follows from the previous one 0.8 because u and x are dummy variables of inte- 0.6 gration. Thus, 1 1/a 0.4 tan−1 x a = tan−1 x 1 1 0.2 tan−1 1 − tan−1 a = tan−1 − tan−1 1 a 0.0 1 tan−1 a + tan−1 = 2tan−1 157. The point is that if we let u = x4 , then we get a 1 π x = ±u1/4 , and so we need to pay attention to tan−1 a + tan−1 = a 2 the sign of u and x. A safe way is to solve the original indefinite integral in terms of x, and 60. If u = 1/x, then du = −dx/x2 and then solve the definite integral using boundary 1 √ dx points in terms of x. |x| x2 − 1 1 x=1 1 4x4 dx = u1/4 du = √ 2 x2 − 1 dx −2 x=−2 x 4 5/4 4 x=1 x=1 1 = u = x5 =− √ du 5 5 x=−2 1 − u2 x=−2 4 5 4 132 = −sin−1 u + c 5 = 5 1 − (−2) = (1 − (−32)) = 5 5 = −sin−1 1/x + c
  • 284 CHAPTER 4. INTEGRATION On the other hand, 1 √ dx = sec−1 x + c1 0.4 |x| x2 − 1 So −sin−1 1/x = sec−1 x + c2 . 0.3 Let x = 1, we get sin−1 1 = sec−1 1 + c2 π 0.2 = 0 + c2 2 π 0.1 c2 = 2 0 0 0.5 1 1.5 2 x 2 √ x 4 − x2 dx 61. x = −2 63. V (t) = Vp sin(2πf t)V 2 (t) 2 √ 4 − x2 dx −2 = Vp2 sin2 (2πf t) Examine the denominator of x, the graph of √ 1 1 4 − x2 , which is indeed a semicircle, is sym- = Vp2 − cos (4πf t) metric over the two intervals [−2, 0] and [0, 2], 2 2 while multiplying by x changes the symmetry Vp2 = (1 − cos (4πf t)) into anti-symmetry. In other words, 2 0 2 1/f x 4 − x2 dx = − x 4 − x2 dx rms = f V 2 (t) dt −2 0 0 so that 1/f 2 Vp2 x 4 − x2 dx = f (1 − cos (4πf t)) dt −2 0 2 0 2 √ 1/f = x 4 − x2 dx + x 4 − x2 dx = 0 Vp f sin (4πf t) = √ t− −2 0 2 4πf Hence x = 0. √ 0 2 √ Vp f 1 Vp Now the integral −2 4 − x2 dx is the area of = √ =√ 2 f 2 a semicircle with radius 2, thus its value is (1/2) π22 = 2π. Then 2 √ 2 2 −2 4 − x2 dx 64. f 2 (t)dt y= 2 √ −2 2 −2 4 − x2 dx −1 1 2 2 4−x 2 dx = 1dt + t2 dt + 1dt −2 −2 −1 1 = 2 8 2.2π =1+ +1= 0 2 −2 4 − x2 dx + 0 4−x 2 dx 3 3 = 4π 1 2 2 2 rms = f (t) dt 2 4 − x2 dx 0 4 1 = 4π 1 8 2 2 4 − x2 dx = = = 0 4 3 3 2π 2 1 x3 8 = 4x − = 2π 3 0 3π 1 0.5 62. These animals are likely to be found 0.7 miles 0 from the pond. Let u = −x2 , then du = -2 -1 0 1 t 2 −2xdx, u(0) = 0, u(2) = −4 and -0.5 2 2 1 −4 u xe−x dx = − e du 0 2 0 1 − e−4 -1 1 = − e−4 − 1 = 2 2
  • 4.7. NUMERICAL INTEGRATION 2854.7 Numerical Integration 1 = (1 + 5 + 4 + 13 + 5) 6 14 1. Midpoint Rule: = 1 3 x2 + 1 dx 0 3. Midpoint Rule: 1 1 3 5 7 31 ≈ f +f +f +f dx 4 8 8 8 8 85 1 x = 3−1 5 7 9 11 64 ≈ f +f +f +f 4 4 4 4 4 Trapezoidal Rule: 1 4 4 4 4 1 = + + + x2 + 1 dx 2 5 7 9 11 0 3776 = 1−0 1 1 3 3465 ≈ f (0) + 2f + 2f + 2f 2 (4) 4 2 4 Trapezoidal Rule: +f (1)] 3 43 1 = dx 32 1 x Simpson’s Rule: 3−1 3 5 ≈ f (1) + 2f + 2f (2) + 2f 1 2 (4) 2 2 x2 + 1 dx +f (3)] 0 1 4 4 1 1−0 1 1 3 = 1+ +1+ + = f (0) + 4f + 2f + 4f 4 3 5 3 3 (4) 4 2 4 67 +f (1)] = 60 4 = Simpson’s Rule: 3 31 2. Midpoint Rule: dx 1 x 2 x2 + 1 dx 3−1 3 5 = f (1) + 4f + 2f (2) + 4f 0 3 (4) 2 2 1 1 3 5 7 +f (3)] ≈ f +f +f +f 2 4 4 4 4 1 8 8 1 1 17 25 41 65 = 1+ +1+ + = + + + 6 3 5 3 2 16 16 16 16 11 37 = = 10 8 Trapezoidal Rule: 4. Midpoint Rule: 2 x2 + 1 dx 1 0 2x − x2 dx 1 1 3 −1 ≈ f (0) + 2f + 2f (1) + 2f 1 3 1 1 3 4 2 2 ≈ f − +f − +f +f +f (2)] 2 4 4 4 4 1 5 13 1 33 9 7 15 = 1+ +4+ +5 = − − + + 4 2 2 2 16 16 16 16 19 −5 = = 4 8 Simpson’s Rule: Trapezoidal Rule: 2 1 x2 + 1 dx 2x − x2 dx 0 −1 1 1 3 1 1 1 = f (0) + 4f + 2f (1) + 4f ≈ f (−1) + 2f − + 2f (0) + 2f 6 2 2 4 2 2 +f (2)] +f (1)]
  • 286 CHAPTER 4. INTEGRATION 1 5 3 8. Midpoint Rule: e2 − 7.322986 = 7.389056 − = −3 − +0+ +1 4 2 2 7.322986 3 = 0.06607 =− 4 Trapezoidal Rule: e2 − 7.52161 = 7.389056 − Simpson’s Rule: 7.52161 1 = −0.132554 2x − x2 dx Simpson’s Rule: e2 − 7.391210 = 7.389056 − −1 7.391210 1 1 1 ≈ f (−1) + 4f − + 2f (0) + 4f = −0.002154 6 2 2 Hence, the approximation using Simpson’s +f (1)] 1 Rule is too small and the Approximation us- = (−3 − 5 + 0 + 3 + 1) ing Trapezoidal Rule is too large. 6 2 π =− 9. cos x2 dx 3 0 5. Midpoint Rule: n M idpoint T rapezoidal Simpson ln 4 − 1.366162 = 1.386294 − 1.366162 10 0.5538 0.5889 0.5660 = 0.020132 20 0.5629 0.5713 0.5655 50 0.5652 0.566 0.5657 Trapezoidal Rule: π ln 4 − 1.428091 = 1.386294 − 1.428091 4 = −0.041797 10. sin πx2 dx 0 Simpson’s Rule: n M idpoint T rapezoidal Simpson 10 0.386939 0.385578 0.386476 ln 4 − 1.391621 = 1.386294 − 1.391621 20 0.386600 0.386259 0.386485 = −0.005327 50 0.386504 0.386450 0.386486 Hence, the approximation using Simpson’s Rule is too small and the Approximation us- 2 2 ing Trapezoidal Rule is too large. 11. e−x dx 0 n M idpoint T rapezoidal Simpson 6. Midpoint Rule: ln 8 − 1.987287 = 2.079442 − 1.987287 10 0.88220 0.88184 0.88207 = 0.092155 20 0.88211 0.88202 0.88208 Trapezoidal Rule: 50 0.88209 0.88207 0.88208 ln 8 − 2.289628 = 2.079442 − 2.289628 3 2 = −0.210186 12. e−x dx Simpson’s Rule: 0 ln 8 − 2.137327 = 2.079442 − 2.137327 n M idpoint T rapezoidal Simpson = −0.057885 10 0.886210 0.886202 0.886207 Hence, the approximation using Simpson’s 20 0.886208 0.886206 0.886207 Rule is too small and the Approximation us- 50 0.886207 0.886207 0.886207 ing Trapezoidal Rule is too large. π 13. ecos x dx 7. Midpoint Rule: 0 sin 1 − 0.843666 = 0.841471 − 0.843666 n M idpoint T rapezoidal Simpson = −0.002195 10 3.9775 3.9775 3.9775 Trapezoidal Rule: sin 1 − 0.837084 = 20 3.9775 3.9775 3.9775 0.841471 − 0.837084 50 3.9775 3.9775 3.9775 = 0.004387 Simpson’s Rule: 1 √ 3 14. x2 + 1dx sin 1 − 0.841489 = 0.841471 − 0.841489 0 = −0.000018 n M idpoint T rapezoidal Simpson Hence, the approximation using Simpson’s 10 3.333017 3.336997 3.334337 Rule is too small and the Approximation us- 20 3.334012 3.335007 3.334344 ing Trapezoidal Rule is too large. 50 3.334291 3.334450 3.334344
  • 4.7. NUMERICAL INTEGRATION 28715. The exact value of this integral is n Simpson ESn 1 10 0 0 1 5x4 dx = x5 0 =1−0=1 20 0 0 0 n M idpoint EMn 40 0 0 10 1.00832 8.3 × 10−3 80 0 0 20 1.00208 2.1 × 10−3 18. The exact value of this integral is 40 1.00052 5.2 × 10−3 π 4 80 1.00013 1.3 × 10−3 1 cos xdx = √ n T rapezoidal ETn 2 0 10 0.98335 1.6 × 10−2 n M idpoint EMn 20 0.99583 4.1 × 10−3 10 0.707289 1.8 × 10−4 40 0.99869 1.0 × 10−3 20 0.707152 4.5 × 10−5 80 0.99974 2.6 × 10−4 40 0.707118 1.1 × 10−5 n Simpson ESn 80 0.707110 2.8 × 10−6 10 1.000066 6.6 × 10−5 n T rapezoidal ETn 20 1.0000041 4.2 × 10−6 10 0.706743 3.6 × 10−4 40 1.00000026 2.6 × 10−7 20 0.707016 9.1 × 10−5 80 1.00000016 1.6 × 10−8 40 0.707084 2.3 × 10−5 80 0.707101 5.7 × 10−616. The exact value of this integral is 2 1 n Simpson ESn dx = ln 2 10 0.7071087 1.5 × 10−7 1 x n M idpoint EMn 20 0.7071068 9.5 × 10−9 10 0.692835 3.1 × 10−4 40 0.7071068 6 × 10−10 20 0.693069 7.8 × 10−5 80 0.7071068 6 × 10−10 40 0.693128 2.0 × 10−5 19. If you double the error in the Midpoint Rule is 80 0.693142 4.9 × 10−6 divided by 4, the error in the Trapezoidal Rule n T rapezoidal ETn is divided by 4 and the error in the Simpson’s 10 0.693771 6.2 × 10−4 Rule is divided by 16. 20 0.693303 1.6 × 10−4 40 0.693186 3.9 × 10−5 20. If you halve the interval length b − a the error 80 0.693157 9.8 × 10−6 in the Midpoint Rule is divided by 8, the error in the Trapezoidal Rule is divided by 8 and the n Simpson ESn error in the Simpson’s Rule is divided by 32. 10 0.693150 3.1 × 10−6 20 0.693147 1.9 × 10−7 21. Trapezoidal Rule: 40 0.693147 1.2 × 10−8 2 80 0.693147 8.0 × 10−10 f (x) dx 0 2−017. The exact value of this integral is ≈ [f (0) + 2f (0.25) + 2f (0.5) 2 (8) π π + 2f (0.75) + 2f (1) + 2f (1.25) + 2f (1.5) cos xdx = sin x|0 = 0 0 + 2f (1.75) + f (2)] n M idpoint EMn 1 = [4.0 + 9.2 + 10.4 + 9.6 + 10 + 9.2 + 8.8 10 0 0 8 + 7.6 + 4.0] 20 0 0 = 9.1 40 0 0 Simpson’s Rule: 80 0 0 2 f (x) dx n T rapezoidal ETn 0 10 0 0 2−0 ≈ [f (0) + 4f (0.25) + 2f (0.5) 20 0 0 3 (8) 40 0 0 + 4f (0.75) + 2f (1) + 4f (1.25) + 2f (1.5) 80 0 0 +4f (1.75) + f (2)]
  • 288 CHAPTER 4. INTEGRATION 1 1 = [4.0 + 18.4 + 10.4 + 19.2 + 10.0 |ET4 | ≤ 1 ≈ 0.005 12 12 · 42 +18.4 + 8.8 + 15.2 + 4.0] 1 |EM4 | ≤ 1 ≈ 0.003 ≈ 9.033 24 · 42 1 |ES4 | ≤ 1 ≈ 2.17 × 10−5 22. Trapezoidal Rule: 180 · 44 2 (b − a)3 1 f (x) dx (b) Midpoint: |En |K = 0 24n2 24n2 0.25 ≈ [f (0) + 2f (0.25) + 2f (0.5) 1 2 We want ≤ 107 +2f (0.75) + 2f (1) + 2f (1.25) + 2f (1.5) 24n2 +2f (1.75) + f (2)] 24n2 ≥ 107 0.25 = [(1.0) + 2(0.6) + 2(0.2) + 2(−0.2) 107 2 n2 ≥ + 2(−0.4) + 2(0.4) + 2(0.8) 24 + 2(1.2) + (2.0)] 107 = 1.025. n≥ ≈ 645.5 Simpson’s Rule: 24 2 So need n ≥ 646. f (x) dx (b − a)3 1 0 Trapezoid: |ETn |K 2 = 0.25 12n 12n2 ≈ [f (0) + 4f (0.25) + 2f (0.5) 3 107 + 4f (0.75) + 2f (1) + 4f (1.25) + 2f (1.5) We want n2 ≥ 12 +4f (1.75) + f (2)] 107 0.25 n≥ ≈ 912.87 = [(1.0) + 4 (0.6) + 2 (0.2) + 4 (−0.2) 12 3 +2 (−0.4) + 4 (0.4) + 2 (0.8) + 4 (1.2) + (2.0)] n ≥ 913 ≈ 1.016667 (b − a)5 1 Simpson: |ESn |L 4 = 1 2 24 180n 180n4 23. (a) f (x) = , f (x) = 3 , f (4) (x) = 5 . 1 x x x ≤ 10−7 Then K = 2, L = 24. Hence according to 180n4 Theorems 9.1 and 9.2, 180n4 ≥ 107 3 (4 − 1) 107 |ET4 | ≤ 2 ≈ 0.281 n4 ≥ 12 · 42 180 3 (4 − 1) |EM4 | ≤ 2 ≈ 0.141 107 4 24 · 42 n≥ ≈ 15.4 (4 − 1) 5 180 |ES4 | ≤ 24 ≈ 0.127 So need n ≥ 16. 180 · 42 (b) Using Theorems 9.1 and 9.2, and the cal- 25. (a) f (x) = ln x. Hence, f (x) = x and 1 culation in Example 9.10, we find the 1 f (x) = − x2 . Therefore |f (x)| ≤ 1. following lower bounds for the number The error using Trapezoidal Rule is of steps needed to guarantee accuracy of (2 − 1) 3 10−7 in Exercise 5: |E (Tn )| ≤ 1 ≤ 10−6 12n2 2 · 33 1 Midpoint: ≈ 4745 |E (Tn )| ≤ ≤ 10−6 24 · 10−7 12n2 Solving for n, 2 · 33 1 6 Trapezoidal: ≈ 6709 |E (Tn )| ≤ 10 ≤ n2 14 · 10−7 12 1 6 4 24 · 35 n≥ 10 Simpson’s: ≈ 135 12 180 · 10−7 ≈ 288.67 24. (a) f (x) = cos x, f (x) = − cos x, 1 f (4) (x) = cos x.Then K = L = 1. (b) f (x) = ln x. Hence, f (x) = x , f (x) = 1 Hence according to − x2 . Therefore |f (x)| ≤ 1. Theorems 9.1 and 9.2, The error using Midpoint Rule is
  • 4.7. NUMERICAL INTEGRATION 289 3 (2 − 1) (c) f (x) = x ln x. Hence, f (x) = 1 + ln x, |E (Mn )| ≤ 1 ≤ 10−6 1 1 24n2 f (x) = , f (x) = − 2 1 x x |E (Mn )| ≤ ≤ 10−6 24n2 2 Solving for n, and f (4) (x) = 3 . 1 6 x |E (Mn )| ≤ 10 ≤ n2 Therefore f (4) (x) ≤ 2. 24 1 6 The error using Simpson’s Rule is n≥ 10 4 24 (4 − 1) ≈ 204.12 |E (Sn )| ≤ 2 4 ≤ 10−6 180n 9 1 |E (Sn )| ≤ ≤ 10−6 (c) f (x) = ln x. Hence, f (x) = , 10n4 x Solving for n, 1 2 9 6 f (x) = − 2 , f (x) = 3 and f (4) (x) = |E (Sn )| ≤ 10 ≤ n4 x x 10 6 9 6 − 4 . Therefore f (4) (x) ≤ 6. n≥ 4 10 x 10 The error using Simpson’s Rule is ≈ 30.8 4 (2 − 1) |E (Sn )| ≤ 6 ≤ 10−6 180n4 2 2 1 27. (a) f (x) = ex . Hence, f (x) = 2xex , |E (Sn )| ≤ ≤ 10−6 2 30n4 f (x) = 2ex 2x2 + 1 . Therefore, Solving for n, |f (x)| ≤ 6e ≈ 16.3097. 1 6 The error using Trapezoidal Rule is |E (Sn )| ≤ 10 ≤ n4 3 30 (1 − 0) 1 6 |E (Tn )| ≤ 16.3097 2 ≤ 10−6 n≥ 4 10 12n 30 16.3097 ≈ 13.5 |E (Tn )| ≤ ≤ 10−6 12n2 Solving for n,26. (a) f (x) = x ln x. Hence, f (x) = 1 + ln x and 16.3097 6 |E (Tn )| ≤ 10 ≤ n2 1 12 f (x) = . Therefore |f (x)| ≤ 1. 16.3097 6 x n≥ 10 3 (4 − 1) 12 |E (Tn )| ≤ 1 ≤ 10−6 ≈ 1165. 12n2 27 |E (Tn )| ≤ ≤ 10−6 2 2 12n2 (b) f (x) = ex . Hence, f (x) = 2xex , Solving for n, 2 27 6 f (x) = 2ex 2x2 + 1 . Therefore, |E (Tn )| ≤ 10 ≤ n2 |f (x)| ≤ 6e ≈ 16.3097. 12 27 6 The error using Trapezoidal Rule is 3 n≥ 10 (1 − 0) 12 |E (Mn )| ≤ 16.3097 ≤ 10−6 = 1500. 24n2 16.3097 (b) f (x) = x ln x. Hence, f (x) = 1 + ln x, |E (Mn )| ≤ ≤ 10−6 24n2 1 Solving for n, f (x) = . Therefore |f (x)| ≤ 1. 16.3097 6 x |E (Mn )| ≤ 10 ≤ n2 The error using Trapezoidal Rule is 24 3 (4 − 1) 16.3097 6 |E (Mn )| ≤ 1 ≤ 10−6 n≥ 10 24n2 24 27 ≈ 824.36 |E (Mn )| ≤ ≤ 10−6 24n2 Solving for n, 2 27 6 (c) f (x) = ex . Hence, |E (Mn )| ≤ 10 ≤ n2 2 24 f (x) = 2xex , 27 6 2 f (x) = 2ex 2x2 + 1 , n≥ 10 24 2 ≈ 1060.66 f (x) = 4ex 2x3 + 3x 2 f (4) (x) = 4ex 4x4 + 12x2 + 3 .
  • 290 CHAPTER 4. INTEGRATION Therefore, |f (x)| ≤ 76e ≈ 206.5823. f (4) (x) ≤ 6e2 ≈ 31.82 The error using Simpson’s Rule is The error using Simpson’s Rule is 4 (1 − 0) (2 − 1) 4 |E (Sn )| ≤ 206.5823 ≤ 10−6 |E (Sn )| ≤ 31.82 ≤ 10−6 180n4 180n4 206.5823 31.82 |E (Sn )| ≤ ≤ 10−6 |E (Sn )| ≤ ≤ 10−6 180n2 180n2 Solving for n, Solving for n, 206.5823 6 31.82 6 |E (Sn )| ≤ 10 ≤ n2 |E (Sn )| ≤ 10 ≤ n2 180 180 4 206.5823 n≥ 106 4 31.82 180 n≥ 106 180 ≈ 32.7307. ≈ 20.50486515 28. (a) f (x) = xex Hence, 29. We use K = 60, L = 120 f (x) = ex (x + 1) n EMn Error Bound f (x) = ex (x + 2) 10 8.3 × 10−3 2.5 × 10−2 Therefore, |f (x)| ≤ 4e2 ≈ 21.21 n ETn Error Bound The error using Midpoint Rule is 10 1.6 × 10−2 5 × 10−2 3 (2 − 1) n ESn Error Bound |E (Mn )| ≤ 21.21 ≤ 10−6 24n2 10 7.0 × 10−5 6.6 × 10−3 21.21 |E (Mn )| ≤ ≤ 10−6 24n2 30. We use K = L = 1. Solving for n, n EMn Error Bound 2402.0293 6 |E (Mn )| ≤ 10 ≤ n2 10 0 1.3 × 10−2 24 21.21 6 n ETn Error Bound n≥ 10 24 10 0 2.6 × 10−2 ≈ 940.0797838 n ESn Error Bound 10 0 1.7 × 10−4 (b) f (x) = xex Hence, 31. (a) Left Endpoints: f (x) = ex (x + 1) 2 f (x) = ex (x + 2) f (x)dx 0 Therefore, 2−0 ≈ [f (0) + f (.5) + f (1) |f (x)| ≤ 4e2 ≈ 21.21 4 The error using Trapezoidal Rule is +f (1.5)] 3 (2 − 1) |E (Tn )| ≤ 21.21 ≤ 10−6 1 = (1 + .25 + 0 + .25) 12n2 2 21.21 |E (Tn )| ≤ ≤ 10−6 = .75 12n2 Solving for n, (b) Midpoint Rule: 21.21 6 2 |E (Tn )| ≤ 10 ≤ n2 12 f (x)dx 21.21 6 0 n≥ 10 2−0 12 ≈ [f (.25) + f (.75) ≈ 1329.473580 4 +f (1.25) + f (1.75)] (c) f (x) = xex 1 = (.65 + .15 + .15 + .65) Hence, 2 f (x) = ex (x + 1) , f (x) = ex (x + 2) = .7 f (x) = ex (x + 3) (c) Trapezoidal Rule: f (4) (x) = ex (x + 4) 2 Therefore, f (x)dx 0
  • 4.7. NUMERICAL INTEGRATION 291 2−0 (b) Trapezoidal Rule: ≈ [f (0) + 2f (.5) + 2f (1) 2(4) b Tn > f (x)dx +2f (1.5) + f (2)] a 1 = (1 + .5 + 0 + .5 + 1) (c) Simpson’s Rule: 4 Not enough information. = .75 2 34. (a) Midpoint Rule: b (d) Simpson’s rule: f (x) dx Mn < f (x)dx 0 a 2 = [f (0) + 4f (0.5) + 2f (1) (b) Trapezoidal Rule: 12 +4f (1.5) + f (2)] b 1 Tn > f (x)dx = [1 + 4(0.25) + 2(0) + 4(0.25) + 1] a 6 1 (c) Simpson’s Rule: = [4] 6 b = 0.66666 Sn ≥ f (x)dx a32. (a) Left Endpoints: 2 f (x)dx 35. (a) Midpoint Rule: b 0 1 Mn > f (x)dx ≈ (f (0) + f (.5) + f (1) + f (1.5)) a 2 1 (b) Trapezoidal Rule: = (0.5 + 0.8 + 0.5 + 0.1) b 2 Tn < f (x)dx = 0.95 a (b) Midpoint Rule: 2 (c) Simpson’s Rule: f (x) dx Not enough information. 0 1 b ≈ (0.7 + 0.8 + 0.4 + 0.2) 36. (a) Midpoint Rule: Mn > f (x)dx 2 a = 1.05 b (c) Trapezoidal Rule: (b) Trapezoidal Rule: Tn < f (x)dx 2 a f (x)dx b 0 1 (c) Simpson’s Rule: Sn ≤ f (x)dx ≈ [0.5 + 2(0.8) + 2(0.5) + 2(0.1) a 4 b + 0.5] 37. (a) Midpoint Rule: Mn < f (x)dx = 0.95 a 2 b (d) Simpson’s rule: f (x) dx (b) Trapezoidal Rule: Tn > f (x)dx a 0 2−0 b = [f (0) + 4f (0.5) + 2f (1) (c) Simpson’s Rule: Sn = f (x)dx 12 +4f (1.5) + f (2)] a 1 = [0.5 + 4(0.9) + 2(0.5) + 4(0.1) + 0.5] b 6 38. (a) Midpoint Rule: Mn = f (x)dx 1 a = [0.5 + 3.6 + 1 + 0.4 + 0.5] 6 b =1 (b) Trapezoidal Rule: Tn = f (x)dx a33. (a) Midpoint Rule: b b Mn < f (x)dx (c) Simpson’s Rule: Sn = f (x)dx a a
  • 292 CHAPTER 4. INTEGRATION 39. 1 1 1 2 (RL + RR ) −√ + √ =0 n−1 n 3 3 = f (xi ) + f (xi ) i=0 i=1 1 n−1 n−1 2 = f (x0 ) + f (xi ) + f (xi ) + f (xn ) (b) x2 dx = 3 i=1 i=1 −1 n−1 2 3 = f (x0 ) + 2 f (xi ) + f (xn ) = Tn 1 1 2 i=1 −√ + √ = 3 3 3 1 40. y (c) x3 dx = 0 2 −1 3 3 1 1 −√ + √ =0 3 3 1 44. Simpson’s Rule with n = 2 : 1 πx π cos dx 2 −1 0 x 2 −1 0.5 1 ≈ f (−1) + 4f + f (1) 6 3 1√ 1 −π −π π 41. I1 = 1 − x2 dx is one fourth of the area of = π cos + 4π cos + π cos 0 3 2 6 2 1 π √ 2π π = 0+2 3+0 = √ a circle with radius 1, so 1 − x2 dx = 3 3 4 ≈ 3.6276 0 1 1 1 Gaussian quadrature: I2 = dx = arctan x|0 1 1 + x2 πx 0 π cos dx π 2 = arctan 1 − arctan 0 = −1 √ 4 1 n Sn ( 1 − x2 ) Sn ( 1+x2 ) −1 1 ≈f √ +f √ 4 0.65652 0.78539 3 3 π π 8 0.66307 0.78539 = π cos − √ + π cos √ 2 3 2 3 1 ≈ 3.87164 The second integral dx provides a 1 + x2 better algorithm for estimating π. sin x h 45. Simpson’s Rule is not applicable because x 42. Ax2 + Bx + c dx sin x is not defined at x = 0. L = lim x→0 x −h cos x h = lim = cos 0 = 1 A 3 B 2 x→0 1 = x + x + cx sin x 3 2 −h The two functions f (x) and differ only 2 3 x π π sin x = Ah + 2Ch 3 at one point,so f (x) dx = dx We can h 0 0 x = 2Ah2 + 6C now apply Simpson’s Rule with n = 2 : 3 π h f (x) dx = [f (−h) + 4f (0) + f (h)] 3 0 π 4 sin π sin π 1 ≈ 1+ π + 6 2 π 43. (a) xdx = 0 π 1 8 = + −1 2 3 3π
  • 4.7. NUMERICAL INTEGRATION 293 π ≈ · 1.18 49. f (x) + f (1 − x) 2 2 x2 (1 − x) = 2 + 2x − 2x + 1 2(1 − x)2 − 2(1 − x) + 1 sin x46. The function is not defined at x = 0, and x2 x = 2 it is symmetric across the y-axis. We define a 2x − 2x + 1 2 new function (1 − x) + sin x/x if x = 0 2 (1 − 2x + x2 ) − 2 + 2x + 1 f (x) = x2 (1 − x) 2 1 if x = 0 = 2 + 2 π/2 sin x 2x − 2x + 1 2x − 2x + 1 over the interval [0, π/2], and dx = 2 x2 (1 − x) −π/2 x = + 2 2 π/2 x2 + (x − 1) (1 − x) + x2 2 f (x)dx 2 2 x + (1 − x) 0 = 2 Use Simpson’s Rule on n = 2: x2 + (1 − x) π/2 =1 f (x)dx By Trapezoidal Rule, 0 √ 1 2 π 1 ≈ 1+ 2 + f (x) dx 12 π/4 π/2 0 π ≈ · 15.22 (1 − 0) 2 = [f (x0 ) + 2f (x1 ) Hence 2n + 2f (x2 ) + ... + 2f (xn−1 ) + f (xn )] π/2 sin x π (1 − 0) 1 dx ≈ · 30.44 = f (0) + 2f ( ) −π/2 x 2 2n n 2 n−1 + 2f ( ) + ... + 2f ( ) + f (1)47. Let I be the exact integral. Then we have n n as f (x) + f (1 − x) = 1, Tn − I ≈ −2(Mn − I) we have, Tn − I ≈ 2I − 2Mn f (0) + f (1) = 1, Tn + 2Mn ≈ 3I 1 n−1 Tn 2 f( ) + f( )=1 + Mn ≈ I n n 3 3 2 n−2 f( ) + f( )=148. The text does not say this, but we want to n n . show that . 1 + 2 Mn = S2n . 3 Tn 3 n−1 1 In this case, we have data points: f( ) + f( ) = 1 n n x0 , x1 , x2 , x3 , ..., x2n . Adding the above n equations, we get The midpoint rule will use the points: 1 n−1 x1 , x3 , ..., x2n−1 . f (0) + 2f ( ) + .. + 2f ( ) + f (1) = n n n The trapezoidal rule will use the points: Hence, x0 x2 , ..., x2n . 1 1 2 1 1 Tn + Mn f (x) dx = (n) = 3 3 2n 2 1 b−a 0 = [f (x0 ) + 2f (x2 ) + 2f (x4 ) 3 2n n + ... + 2f (x2n−2 ) + f (x2n )] 2 b−a 50. xn dx + × [f (x1 ) + f (x3 ) 0 3 n + f (x5 ) + ... + f (x2n−1 ) + f (x2n )] n−0 = [f0 + 2f1 + 2f2 + ... + 2fn−1 + fn ] b−a 2n = [f (x0 ) + 4f (x1 ) + 2f (x2 ) 1 2n = [f (0) + 2f (1) + 2f (2) + 2f (3) + ......... + 4f (x3 ) + 2f (x4 ) + ... + 2f (x2n−2 ) 2 + 2f (n − 1) + f (n)] + 4f (x2n−1 ) + f (x2n )] 1 n = S2n = [(nn ) + 2(1n + 2n + 3n + ....... + (n − 1) ] 2
  • 294 CHAPTER 4. INTEGRATION Now n n 1.5 n xn+1 x dx = 1.25 n+1 0 0 1.0 nn+1 = 0.75 n+1 0.5 The sum of the areas of the trapezoids is 0.25 greater than the area defined by the curve 0.0 nn+1 nn over the interval 0 to n. < + 1n + 0 1 2 3 4 5 6 n n+1 2 x 2n + 3n + ... + (n − 1) nn+1 nn n + < 1n + 2n + +... + (n − 1) + nn 8.2 dx n+1 2 3. ln 8.2 = 2nn+1 + nn+1 + nn 1 x < 1n + 2n + ......... + nn 2(n + 1) 1.5 3nn+1 + nn < 1n + 2n + 3n + .......... + nn 1.25 2(n + 1) (3n + 1) n 1.0 n < 1n + 2n + 3n + ......... + nn 2(n + 1) 0.75 0.5 0.25 0.0 0 2 4 6 8 x 24 dx4.8 The Natural 4. ln 24 = 1 x Logarithm As An 1.5 Integral 1.25 1.0 0.75 4 4 dx 0.5 1. ln 4 = ln 4 − ln 1 = ln x|1 = 1 x 0.25 1.5 0.0 0 5 10 15 20 25 30 1.25 x 1.0 0.75 4 dx 5. ln 4 = 0.5 1 x 0.25 3 1 1 1 1 1 ≈ +4 +2 +4 + 0.0 12 1 1.75 1.5 3.25 4 0 1 2 3 4 5 ≈ 1.3868 x 5 dx 6. ln 5 = 1 x 4 1 1 1 1 1 ≈ +4 +2 +4 + 12 1 2 3 4 5 ≈ 1.6108 5 7. (a) Simpson’s Rule with n = 32 : dx 4 dx 2. ln 5 = ln 4 = 1 ≈ 1.386296874 1 x x
  • 4.8. THE NATURAL LOGARITHM AS AN INTEGRAL 295 √ (b) Simpson’s Rule with n = 64 : 20. y = 4 x 4 dx On taking natural logarithm. ln 4 = 1 ≈ 1.386294521 x √ √ ln y = ln 4 x = x ln 4 8. (a) Simpson’s Rule with n = 32 : 1 dy d √ 4 dx = x ln 4 ln 4 = 1 ≈ 1.609445754 y dx dx x d √ (b) Simpson’s Rule with n = 64 : = (ln 4) x dx 4 dx 1 ln 4 = 1 ≈ 1.609438416 = (ln 4) √ x 2 x 7 dy (ln 4) 9. ln 2 =y √ 2 dx √ 2 x10. ln 2 dy 4 x (ln 4) = √ √ dx 2 x 32 · 3 111. ln = 2 ln 3 1 9 21. dx = ln |ln x| + c x ln x 1 1 9 ·9 112. ln = −5 ln 3 22. √ dx = ln sin−1 x + c 3 1 − x2 sin−1 x 1 1 2 −113. √ x +1 . 2 .2x 23. Let u = x2 , du = 2xdx x2 + 1 2 2 1 3x 2 x3x dx = 3u du = +c 5x4 sin x cos x + x5 cos2 x − x5 sin x 2 2 ln 314. x5 sin x cos x 24. Let u = 2x , du = 2x (ln 2)dx 5 x + 1 4x 3 5 x +1 −x 4 5x 4 115. · 2x sin (2x ) dx = sin (u) du x4 (x5 + 1) 2 ln 2 x − cos (2 ) −1/2 = +c x5 + 1 1 x3 ln 216. · · x3 2 x5 + 1 2 −2 3x x + 1 − x3 5x4 2 5 25. Let u = , du = dx · x x2 2 (x5 + 1) e2/x 1 dx = − eu du x2 2 d 1 ln x2 + 1 1 117. = − u + c = − e2/x + c dx 2 ln 7 2e 2 1 d 3 = ln x2 + 1 26. Let u = ln x3 , du = dx 2 ln 7 dx x 1 x sin ln x3 1 = dx = sin udu ln 7 x2 + 1 x 3 1 d x ln 2 ln 2 d = − cos u + c18. = (x) = log10 2 3 dx ln 10 ln 10 dx 1 = − cos ln x3 + c 319. Let y = 3sin x 1 On taking natural logarithm. x2 27. dx ln y = ln 3 sin x = sin x ln 3 0 x3−4 1 dy d d 1 1 = (sin x ln 3) = ln 3 (sin x) = ln x3 − 4 0 y dx dx dx 3 1 1 1 3 1 dy = ln 3 − ln 4 = ln = (ln 3) cos x 3 3 3 4 y dx dy 1 ex − e−x = y (ln 3) cos x 28. dx dx x −x dy 0 e +e 1 = 3sin x (ln 3) cos x = ln ex + e−x 0 dx
  • 296 CHAPTER 4. INTEGRATION = ln e + e−1 − ln 2 5 e + e−1 = ln 4 2 3 y 2 1 1 sin x 29. tan xdx = dx 1 0 0 cos x 1 = − ln |cos x||0 0 = − ln |cos 1| − ln |cos 0| 0 1 2 3 x 4 5 6 = − ln (cos 1) From the graph, it may be observed that the dx 1 30. Let u = ln x, du = area bounded by y = ; the x-axis between x x ln x u2 the ordinates x = 1 and x = n is lesser than dx = udx = +c the shaded area which is the sum of areas of x 2 (ln x) 2 the (n − 1) rectangles having width 1 unit and = +c height f (i) 2 2 2 2 Thus from the graph, ln x (ln x) dx = n n−1 1 x 2 1 1 dx < (f (i) × 1) ln2 2 ln2 1 ln2 2 1 x i=1 = − = 2 2 2 ln(n) < f (1) + f (2) + f (3) + ... .. + f (n − 1) 1 1 or ln(n) < 1 + + ..... + a 1 1 2 n−1 31. ln = ln a · = ln a + ln b b b Hence proved. We know that, = ln a − ln b lim ln(n) = ∞ n→∞ 1 1 lim 1+ + ..... + ≥ lim ln(n) 32. Consider x = 2−n , where n is any integer for n→∞ 2 n−1 n→∞ =∞ x > 0. 34. We know that by definition, n On taking natural logarithm. 1 ln(n) = dx 1 x ln x = ln 2−n which is the area bounded by the curve ⇒ ln x = −n ln 2 1 Now x → 0, 2−n → 0 ⇒ n → ∞ y = , the positive x-axis between the ord- x ⇒ lim (ln x) = lim (−n ln 2) nates x = 1 and x = n. x→0+ n→∞ 1 = − (ln 2) lim (n). Let y = f (x) = . n→∞ x But, ln 2 ≈ 0.6931 and lim n = ∞ n→∞ 5 ⇒ lim (ln x) = −∞. x→0+ 4 3 n 1 y 33. We know that by definition, ln(n) = dx 2 1 x 1 1 which is the area bounded by the curve y = , 0 x 0 1 2 3 4 5 the positive x-axis between the ordinates x = 1 x 1 and x = n. Let y = f (x) = . x
  • 4.8. THE NATURAL LOGARITHM AS AN INTEGRAL 297 Let us consider (n − 1) rectangles, having for some x in (−h, 0) ¯ width 1 unit and height f (i + 1) where i = e−h − 1 =x ¯ 1, 2, 3, ........, n − 1. Thus from the graph, −h n 1 n−1 as h → 0 , −h → 0− , x → 0, then + ¯ dx > (f (i + 1) × 1) e−h − 1 1 x i=1 lim =0 h→0+ −h ln(n) > f (2) + f (3) + ..... + f (n) 1 1 1 1 38. f (x) = ln x, then f (x) = x and f (1) = 1. or ln(n) > + ..... + . 2 3 n On the other hand35. Since the domain of the function y = ln x ln x − ln a f (a) = lim 1 x→a x−a is (0, ∞) , f (x) = > 0 for x > 0. So f ln x − ln 1 x f (1) = lim =1 is increasing throughout the domain. Simi- x→1 x−1 1 ln x larly, f (x) = − 2 < 0 for x > 0. There- lim =1 x x→1 x − 1 fore, the graph is concave down everywhere, ln x the graph of the function y = ln x is as below. Thus the reciprocal of has the same x−1 3 limit, x−1 2 lim =1 x→1 ln x 1 eh − 1 Substituting x = eh , lim =1 0 h→0 h 0.0 0.5 1.0 1.5 2.0 2.5 3.0 −1 39. (a) Given that, y = ln(x+1) by using a linear approximation. −2 f (x) ≈ f (x0 ) + f (x0 ) (x − x0 ) −3 For small value of x, f (x) ≈ f (0) + f (0) (x − 0)36. Proof of (ii) ln(1 + x) ≈ 0 + 1 · (x − 0) ln(1 + x) ≈ x. By using the rules of logarithm we have, er (b) By using area under the curve. ln s = ln (er ) − ln (es ) Area the rectangle e = r ln e − s ln e = r − s = ln er−s = f (1) · x = x Since ln x is one to one, it follows that 1+x 1 1+x Also, dt = ln t|1 er t = er−s . 1 es = ln(1 + x) − ln(1) Proof of (iii) = ln(1 + x). By using the rules of logarithm we have, As x approaches to zero, we get: t ln(1 + x) ≈ x ln (er ) = t ln (er ) = rt ln e = ln ert Since ln x is one to one, it follows that 40. f (x) = ln x − 1 t 1 (er ) = ert . f (x) = x eh x0 = 3 1 eh − 137. h = ln eh = dx = , f (x0 ) ln 3 − 1 1 x x¯ x1 = x0 − =3− 1 for some x in (0, h) ¯ f (x0 ) 3 eh − 1 = 6 − 3 ln 3 ≈ 2.704163133 =x¯ f (x1 ) h x2 = x1 − ≈ 2.718245098 + as h → 0 , x → 0, then ¯ f (x1 ) f (x2 ) eh − 1 x3 = x2 − ≈ 2.718281827 lim+ =0 f (x2 ) h→0 h e ≈ 2.718282183 e−h −h 1 e−h − 1 Three steps are needed to start at x0 = 3 and − h = ln e = dx = , obtain five digits of accuracy. 1 x x ¯
  • 298 CHAPTER 4. INTEGRATION 1 <0 if x < e−1/2 41. f (x) = Since s (x) 1 + e−x >0 if x > e−1/2 2.0 The value x = e−1/2 maximizes the transmis- 1.5 sion speed. n 1.0 1 44. ln lim 1+ n→∞ n 0.5 n 1 0.0 = lim ln 1 + −3 −2 −1 0 1 2 3 n→∞ n −0.5 1 = lim n ln 1 + −1.0 n→∞ n ln(1 + 1/n) = lim n→∞ 1/n Using lim e−x = 0 we get x→∞ −1/n2 1 = lim lim = 1. n→∞ −1/n2 (1 + 1/n) x→∞ 1 + e−x Using lim e−x = ∞ we get = lim 1 x→−∞ n→∞ 1 + 1/n 1 lim = 0. =1 x→∞ 1 + e−x The function f (x) is increasing over (−∞, ∞) and when x = 0, Ch. 4 Review Exercises 1 1 f (0) = = . 4 3 1+1 2 1. (4x2 − 3) dx = x − 3x + c 0 if x < 0 3 So g(x) = 1 if x ≥ 0 x2 1 2. (x − 3x5 ) dx = − x6 + c The threshold value for g(x) to switch is x = 0. 2 2 One way of modifying the function to move 4 the threshold to x = 4 is to let f (x) = 3. dx = 4 ln |x| + c 1 x . 1 + e−(x−4) 4 4 4. dx = − + c 1 42. 1 − (9/10) 0 ≈ 0.65132 x2 x 1 − (19/20)2 0 ≈ 0.64151 1 1 − (9/10)1 0 > 1 − (19/20)2 0 5. 2 sin 4x dx = − cos 4x + c 2 The probability of winning is lower. When taking the limit as n → ∞, 6. 3 sec2 x dx = 3 tan x + c n n−1 lim 1 − x2 1 n→∞ n 7. (x − e4x ) dx = − e4x + c 2 4 n n−1 √ = 1 − lim n→∞ n 8. 3 x dx = 2x3/2 + c n −1 = 1 − lim 1+ x2 + 4 n→∞ n 9. dx = (x + 4x−1 ) dx x = 1 − e−1 x2 = + 4 ln |x| + c 2 43. s(x) = x2 ln(1/x) s (x) = 2x ln 1/x + x2 · x · (−1/x2 ) x 1 10. 2+4 dx = ln(x2 + 4) + c = 2x ln(1/x) − x = x(2 ln(1/x) − 1) x 2 s (x) = 0 gives x = 0 (which is impossible) or 11. ex (1 − e−x ) dx = (ex − 1) dx ln(1/x) = 1/2, x = e−1/2 . = ex − x + c
  • CHAPTER 4 REVIEW EXERCISES 299 112. ex (1 + ex )2 dx 22. f (x) = e−2x dx = − e−2x + c 2 1 = (ex + 2e2x + e3x ) dx f (0) = − + c = 3 2 1 3x 7 = ex + e2x + e +c c= 3 2 1 7 f (x) = − e−2x +13. Let u = x2 + 4, then du = 2x dx and 2 2 x x2 + 4 dx 23. s(t) = (−32t + 10) dt 1 1 = −16t2 + 10t + c = u1/2 du = u3/2 + c 2 3 s(0) = c = 2 1 2 3/2 s(t) = −16t2 + 10t + 2 = (x + 4) + c 3 24. v(t) = 6 dt = 6t + c114. x(x2 + 4) dx = (x3 + 4x) dx v(0) = c1 = 10 x4 v(t) = 6t + 10 = + 2x2 + c 4 s(t) = (6t + 10) dt = 3t2 + 10t + c215. Let u = x3 , du = 3x2 dx s(0) = c2 = 0 s(t) = 3t2 + 10t 6x2 cos x3 dx = 2 cos u du 6 3 = 2 sin u + c = 2 sin x + c 25. (i2 + 3i) i=116. Let u = x2 , du = 2x dx = (12 + 3 · 1) + (22 + 3 · 2) + (32 + 3 · 3) 4x sec x2 tan x2 dx + (42 + 3 · 4) + (52 + 3 · 5) + (62 + 3 · 6) = 4 + 10 + 18 + 28 + 40 + 54 =2 sec u tan u du = 154 12 = 2 sec u + c = 2 sec x2 + c 26. i2 = 65017. Let u = 1/x, du = −1/x2 dx i=1 e1/x 100 dx = − eu du 27. (i2 − 1) x2 = −eu + c = −e1/x + c i=1 100 100 = i2 − 118. Let u = ln x, du = dx/x i=1 i=1 ln x dx = u du 100(101)(201) x = − 100 2 6 u (ln x)2 = +c= +c = 338, 250 2 2 100 sin x 28. (i2 + 2i)19. tan x dx = dx cos x i=1 = − ln | cos x| + c 100 100 2 = i +2· i20. Let u = 3x + 1, du = 3 dx i=1 i=1 √ 1 100(101)(201) 3x + 1dx = u1/2 du = + 100(101) 3 6 1 2 2 = 348, 450 = · u3/2 + c = (3x + 1)3/2 + c 3 3 9 n 1 29. (i2 − i)21. f (x) = (3x2 + 1) dx = x3 + x + c n3 i=1 n n f (0) = c = 2 1 = 3 i2 − · i f (x) = x3 + x + 2 n i=1 i=1
  • 300 CHAPTER 4. INTEGRATION 1 n(n + 1)(2n + 1) n(n + 1) + f (.6) + f (.8) + f (1) + f (1.2) = − n3 6 2 + f (1.4)) 1 (n + 1)(2n + 1) n + 1 = (1 + 1.4 + 1.6 + 2 + 2.2 + 2.4 = − 5 6n2 2n2 + 2 + 1.6) n 1 = 2.84 lim (i2 − i) n→∞ n3 i=1 (b) Right-endpoints: (n + 1)(2n + 1) n + 1 1.6 = lim − f (x) dx n→∞ 6n2 2n2 0 2 1 1.6 − 0 = −0= ≈ (f (.2) + f (.4) + f (.6) 6 3 8 + f (.8) + f (1) + f (1.2) + f (1.4) 30. Evaluation points: 0.25, 0.75, 1.25, 1.75 n + f (1.6)) Riemann sum = ∆x f (ci ) 1 = (1.4 + 1.6 + 2 + 2.2 + 2.4 i=1 5 4 + 2 + 1.6 + 1.4) 2 = (c2 − 2ci ) = 2.92 4 i=1 i 1 (c) Trapezoidal Rule: = (0.252 − 2 · 0.25) + (0.752 − 2 · 0.75) 1.6 2 f (x) dx +(1.252 − 2 · 1.25) + (1.752 − 2 · 1.75) 0 = −2.75 1.6 − 0 ≈ [f (0) + 2f (.2) + 2f (.4) 2(8) 0.4 + 2f (.6) + 2f (.8) + 2f (1) + 2f (1.2) + 2f (1.4) + f (1.6)] x 0 0.5 1 1.5 2 = 2.88 0 (d) Simpson’s Rule: 1.6 -0.4 f (x) dx 0 1.6 − 0 -0.8 ≈ [f (0) + 4f (.2) + 2f (.4) 3(8) + 4f (.6) + 2f (.8) + 4f (1) + 2f (1.2) + 4f (1.4) + f (1.6)] 8 2 ≈ 2.907 31. Riemann sum = c2 = 2.65625 i 8 i=1 36. 8 (a) Left-endpoints: 2 32. Riemann sum = c2 = 0.6875 i 4.2 8 i=1 f (x) dx 1 8 ≈ (0.4)[f (1.0) + f (1.4) + f (1.8) 3 + f (2.2) + f (2.6) + f (3.0) 33. Riemann sum = c2 ≈ 4.668 i 8 i=1 + f (3.4) + f (3.8)] = (0.4)(4.0 + 3.4 + 3.6 + 3.0 8 1 + 2.6 + 2.4 + 3.0 + 3.6) 34. Riemann sum = c2 ≈ 0.6724 i = 10.24 8 i=1 (b) Right-endpoints: 35. 4.2 f (x) dx (a) Left-endpoints: 1 1.6 ≈ (0.4)[f (1.4) + f (1.8) + f (2.2) f (x) dx + f (2.6) + f (3.0) + f (3.4) 0 1.6 − 0 + f (3.8) + f (4.2)] ≈ (f (0) + f (.2) + f (.4) = (0.4)(3.4 + 3.6 + 3.0 + 2.6 8
  • CHAPTER 4 REVIEW EXERCISES 301 + 2.4 + 3.0 + 3.6 + 3.4) (n + 1)(2n + 1) = lim = 10.00 n→∞ 3n2(c) Trapezoidal Rule: 2 4.2 = 3 f (x) dx 1 40. We will compute the area An of n rectangles 0.4 2 ≈ [f (1.0) + 2f (1.4) + 2f (1.8) using right endpoints. In this case ∆x = 2 n + 2f (2.2) + 2f (2.6) + 2f (3.0) 2i and xi = + 2f (3.4) + 2f (3.8) + f (4.2)] n n n = (0.2)[4.0 + 2(3.4) + 2(3.6) 2 2i An = f (xi )∆x = f + 2(3.0) + 2(2.6) + 2(2.4) i=1 n i=1 n + 2(3.0) + 2(3.6) + 3.4] n 2 = 10.12 2 2i = +1 n i=1 n(d) Simpson’s Rule: 4.2 n n 8 2 f (x) dx = i2 + 1 1 n3 i=1 n i=1 0.4 ≈ [f (1.0) + 4f (1.4) + 2f (1.8) 8 n(n + 1)(2n + 1) 2 3 = + n + 4f (2.2) + 2f (2.6) + 4f (3.0) n3 6 n + 2f (3.4) + 4f (3.8) + f (4.2)] 4(n + 1)(2n + 1) 0.4 = +2 = [4.0 + 4(3.4) + 2(3.6) 3n2 3 Now, to find the integral, we take the limit: 2 + 4(3.0) + 2(2.6) + 4(2.4) + 2(3.0) + 4(3.6) + 3.4] (x2 + 1) dx = lim An 0 n→∞ ≈ 10.05333 4(n + 1)(2n + 1) = lim +237. See Example 7.10. n→∞ 3n2 Simpson’s Rule is expected to be most accu- 8 14 = +2= rate. 3 3 338. In this situation, the Midpoint Rule will be less 41. Area = (3x − x2 ) dx than the actual integral. The Trapezoid Rule 0 will be an overestimate. 3x2 x3 3 9 = − = 2 3 0 239. We will compute the area An of n rectangles 1 using right endpoints. In this case ∆x = n and 42. Area 1 xi = ni = (x3 − 3x2 + 2x) dx n n 0 1 i 2 An = f (xi )∆x = f − (x3 − 3x2 + 2x) dx i=1 n i=1 n 1 n 2 1 1 1 1 i = + = = 2· 4 4 2 n n i=1 43. The velocity is always positive, so distance n traveled is equal to change in position. 2 = i2 2 n3 i=1 Dist = (40 − 10t) dt 1 2 n(n + 1)(2n + 1) 2 = = (40t − 5t2 ) = 25 n3 6 1 (n + 1)(2n + 1) 44. The velocity is always positive, so distance = traveled is equal to change in position. 3n2 2 2 Now, to find the integral, we take the limit: Dist = 20e−t/2 dt = (−40e−t/2 ) 1 0 0 x2 dx = lim An = 40(−e−1 + 40) ≈ 25.2848 0 n→∞
  • 302 CHAPTER 4. INTEGRATION 2 π 1 e2 − 1 45. fave = ex dx = ≈ 3.19 58. cos(x/2) dx 2 0 2 −π π 1 4 8 = (2 sin(x/2)) =4 −π 46. fave = (4x − x2 ) dx = 4 0 3 59. f (x) = sin x2 − 2 2 3 2 x 4 47. (x2 − 2) dx = − 2x =− 0 3 0 3 60. f (x) = (x2 )2 + 1 · 2x 1 x4 1 61. 48. (x3 − 2x) dx = − x2 =0 −1 4 −1 a) Midpoint Rule: π/2 π/2 1 1 49. sin 2x dx = − cos 2x =1 x2 + 4 dx 0 2 0 0 1−0 1 3 π/4 π/4 ≈ f +f 50. 2 sec x dx = tan x =1 4 8 8 0 0 5 7 10 +f +f 51. (1 − e −t/4 ) dt 8 8 0 10 ≈ 2.079 = t + 4e−t/4 = 6 + 4e−5/2 0 b) Trapezoidal Rule: 1 1 52. te −t2 dt x2 + 4 dx 0 0 1 1 1 1−0 1 = − e−t 2 = − (e−1 − 1) ≈ f (0) + 2f 2 0 2 2(4) 4 2 2 1 3 x 1 +2f + 2f 53. 2+1 dx = ln |x2 + 1| 2 4 0 x 2 0 ln 5 +f (1)] = ≈ 2.083 2 2 ln x ln2 x 2 ln2 2 c) Simpson’s Rule: 54. dx = = 1 1 x 2 1 2 x2 + 4 dx 2 0 55. x x2 + 4 dx 1−0 1 ≈ f (0) + 4f 0 3(4) 4 1 2 2 = · · (x2 + 4)3/2 1 3 2 3 0 +2f + 4f + f (1) √ 2 4 16 2 − 8 ≈ 2.080 = 3 2 62. 56. x(x2 + 1) dx 0 a) Midpoint Rule: 1 2 2 2 = (x + 1)2 =6 e−x 2 /4 dx 4 0 0 1 1 2 x ≈ [f (0.25) + f (0.75) 57. (ex − 2)2 dx = (e2x−4e +4 ) dx 4 0 0 + f (1.25) + f (1.75)] 1 2x 2 = e − 4ex + 4x ≈ 1.497494 2 0 2 b) Trapezoidal Rule: e 1 2 = − 4e + 4 − −4 2 2 2 e−x /4 dx 0 e2 15 2 = − 4e + ≈ [f (0) + 2f (.5) + 2f (1) 2 2 8
  • CHAPTER 4 REVIEW EXERCISES 303 t t + 2f (1.5) + f (2)] e 2 −e− 2 2 t t ≈ 1.485968 2u e 2 +e− 2 similarly, = 2 1 + u2 t e 2 −e− 2 t c) Simpson’s Rule: 1+ t t e 2 +e− 2 2 t t t t e−x 2 /4 dx 2 e 2 − e− 2 e2 + e −2 0 = 2 2 2 t e 2 + e− 2 t t + e 2 − e− 2 t ≈ [f (0) + 4f (.5) + 2f (1) 12 2 (et − e−t ) + 4f (1.5) + f (2)] = = sinh t 4 ≈ 1.493711 1 (a) dt63. sinh t + cosh t n Midpoint Trapezoid Simpson’s 1 = (1+u2 du 20 2.08041 2.08055 2.08046 + (1−u2 ) 2u (1−u2 ) ) 40 2.08045 2.08048 2.08046 (Put:u = tanh(t/2)) 1 − u264. = 2 du n Midpoint Trapezoid Simpson’s (1 + u) 1−u 20 1.493802 1.493342 1.493648 = du 1+u 40 1.493687 1.493572 1.493648 2 = − 1 du t 1+u t sinh 2 = 2 ln (1 + u) − u65. Consider u = tanh 2 = t cosh 2 = 2 ln (1 + tanh(t/2)) − tanh(t/2) t t e 2 −e− 2 t t sinh t + cosh t 2 e 2 − e− 2 (b) dt = t t = 1 + cosh t e 2 +e− 2 t e 2 + e− 2 t 2 2u (1+u2 ) 2 (1−u2 ) + (1−u2 ) t e 2 −e− 2 t = (1+u2 du 1 − u2 1− t t e 2 +e− 2 1 + (1−u2 )) therefore = 2 2 1 + u2 t e 2 −e− 2 t (1 + u) 1+ t t = du e 2 +e− 2 2 t t 2 t t 2 3 e 2 + e− 2 − e2 − e −2 1 (1 + u) = = 2 2 2 3 t t t t e 2 + e− 2 + e 2 − e− 2 3 (1 + tanh(t/2)) t 2 (e + e ) −t = = = cosh t , 6 4