Chapter E

Exercises E.1

  1. 1.

    Answers will vary.

  2. 2.

    “an”

  3. 3.

    Answers will vary.

  4. 4.

    opposite; opposite

  5. 5.

    Answers will vary.

  6. 6.

    velocity

  7. 7.

    velocity

  8. 8.

    F⁢(x)+G⁢(x)

  9. 9.

    3⁢x4/4+C

  10. 10.

    x9/9+C

  11. 11.

    10⁢x3/3−2⁢x+C

  12. 12.

    t+C

  13. 13.

    −1/(3⁢t)+C

  14. 14.

    −3/(t)+C

  15. 15.

    2⁢x+C

  16. 16.

    tan⁡θ+C

  17. 17.

    −cos⁡θ+C

  18. 18.

    sec⁡x−csc⁡x+C

  19. 19.

    5⁢eθ+C

  20. 20.

    et2+C

  21. 21.

    4/3⁢t3+6⁢t2+9⁢t+C

  22. 22.

    t6/6+t4/4−3⁢t2+C

  23. 23.

    x6/6+C

  24. 24.

    eπ⁢x+C

  25. 25.

    −x−3+C

  26. 26.

    43⁢x3+72⁢x−2+C

  27. 27.

    29⁢x9/2+C

  28. 28.

    27⁢x7/2−143⁢x3/2+C

  29. 29.

    5⁢x−29⁢x3+316⁢x4+C

  30. 30.

    17⁢u7−13⁢u6−14⁢u4+27⁢u+C

  31. 31.

    23⁢u3+92⁢u2+4⁢u+C

  32. 32.

    27⁢t7/2+65⁢t5/2+43⁢t3/2+C

  33. 33.

    2⁢x+x+23⁢x⁢x+C

  34. 34.

    x+C

  35. 35.

    θ+tan⁡θ+C

  36. 36.

    tan⁡t+sec⁡t+C

  37. 37.

    −cot⁡t−t+C

  38. 38.

    2⁢sin⁡x+C

  39. 39.

    8⁢u+4⁢u⁢u+C

  40. 40.

    −cos⁡t+C

  41. 41.

    6⁢t1/3+34⁢t4/3+C

  42. 42.

    49⁢x9/4+49⁢x9/5+C

  43. 43.

    −cos⁡x+3

  44. 44.

    5⁢ex+5

  45. 45.

    x4−x3+7

  46. 46.

    tan⁡x+4

  47. 47.

    7⁢x36−9⁢x2+403

  48. 48.

    5⁢ex−2⁢x

  49. 49.

    θ−sin⁡(θ)−π+4

  50. 50.

    3⁢x−2

  51. 51.

    x−2+1

  52. 52.

    2⁢x−4

  53. 53.
    • x>0

      1/x

      x<0

      1/x

      ln⁡|x|+C. Explanations will vary.

  54. 54.

    s⁢(t)=2⁢t3/2.

  55. 55.

    s⁢(t)=241.67−16⁢t2 ft, so s⁢(t)=0 at t=3.89sec.

  56. 56.

    2424xy

    Other antiderivatives are vertical shifts of this one.

  57. 57.

    24−224xy

    Other antiderivatives are vertical shifts of this one.

  58. 58.

    Use technology to verify

  59. 59.

    d⁡y=(2⁢x⁢ex⁢cos⁡x+x2⁢ex⁢cos⁡x−x2⁢ex⁢sin⁡x)⁢d⁡x

Exercises E.2

  1. 1.

    Answers will vary.

  2. 2.

    Answers will vary.

  3. 3.

    0

  4. 4.

    ∫02⁢(2⁢x+3)⁢d⁡x

  5. 5.

    • 3

      4

      3

      0

      −4

      9

  6. 6.

    • −4

      −5

      −3

      1

      −2

      10

  7. 7.

    • 4

      2

      4

      2

      1

      2

  8. 8.

    • −1/2

      0

      3/2

      3/2

      9/2

      15/2

  9. 9.

    • π

      π

      2⁢π

      10⁢π

  10. 10.

    • 15

      12

      0

      3⁢(b−a)

  11. 11.

    • −59

      −48

      −27

      −33

      70

      91

  12. 12.

    • 4/π

      −4/π

      0

      2/π

      4/π

      8/π

  13. 13.

    • 4

      4

      −4

      −2

      6

      2

  14. 14.

    • 40/3

      26/3

      8/3

      38/3

  15. 15.

    • 2ft/s

      2ft

      1.5ft

  16. 16.

    • 3ft/s

      9.5ft

      9.5ft

  17. 17.

    • 64ft/s

      64ft

      t=2

      t=2+7≈4.65 seconds

  18. 18.

    • 96ft/s

      6 seconds

      6 seconds

      Never; the maximum height is 208ft.

  19. 19.

    2

  20. 20.

    5

  21. 21.

    16

  22. 22.

    Answers can vary; one solution is a=−2, b=7

  23. 23.

    22

  24. 24.

    −7

  25. 25.

    0

  26. 26.

    Answers can vary; one solution is a=−11, b=18

  27. 27.

    This is a triangle with base b and height m⁢b.

  28. 28.
  29. 29.

    1/4⁢x4−2/3⁢x3+7/2⁢x2−9⁢x+C

  30. 30.

    −cos⁡x−sin⁡x+tan⁡x+C

  31. 31.

    3/4⁢t4/3−1/t+2t/ln⁡2+C

  32. 32.

    ln⁡|x|+csc⁡x+C

Exercises E.3

  1. 1.

    limits

  2. 2.

    14

  3. 3.

    Rectangles.

  4. 4.

    T

  5. 5.

    22+32+42=29

  6. 6.

    −6−2+2+6+10=10

  7. 7.

    0−1+0+1+0=0

  8. 8.

    5+5+5+5+5+5+5+5+5+5=50

  9. 9.

    1+1/2+1/3+1/4+1/5=137/60

  10. 10.

    −1+2−3+4−5+6=3

  11. 11.

    1/2+1/6+1/12+1/20=4/5

  12. 12.

    1+1+1+1+1+1=6

  13. 13.

    Answers may vary; ∑i=153⁢i

  14. 14.

    Answers may vary; ∑i=08(i2−1)

  15. 15.

    Answers may vary; ∑i=14ii+1

  16. 16.

    Answers may vary; ∑i=04(−1)i⁢ei

  17. 17.

    50

  18. 18.

    325

  19. 19.

    1045

  20. 20.

    28,650

  21. 21.

    −8525

  22. 22.

    2050

  23. 23.

    5050

  24. 24.

    2870

  25. 25.

    155

  26. 26.

    91,225

  27. 27.

    24

  28. 28.

    11,700

  29. 29.

    ∫0πsin⁡x1+x⁢d⁡x

  30. 30.

    ∫25x⁢1+x3⁢d⁡x

  31. 31.

    ∫275⁢x3−4⁢x+7⁢d⁡x

  32. 32.

    ∫13xx2+4⁢d⁡x

  33. 33.

    limn→∞[3n⁢∑i=1n4−2⁢(2+3⁢in)]

  34. 34.

    limn→∞2n⁢∑i=1n[(−2+2⁢in)2+3⁢(−2+2⁢in)]

  35. 35.

    limn→∞πn⁢∑i=1nsin3⁡(−π/2+π⁢i/n)2+cos⁡(−π/2+π⁢i/n)

  36. 36.

    limn→∞2n⁢∑i=1ne2⁢i/n

  37. 37.

    19

  38. 38.

    59/8

  39. 39.

    π/3+π/(2⁢3)≈1.954

  40. 40.

    8.16986

  41. 41.

    0.388584

  42. 42.

    496/315≈1.5746

  43. 43.

    • Exact expressions will vary; (1+n)24⁢n2.

      121/400, 10201/40000, 1002001/4000000

      1/4

  44. 44.

    • Exact expressions will vary; 2+4/n2.

      51/25, 5001/2500, 500001/250000

      2

  45. 45.

    • 8.

      8, 8, 8

      8

  46. 46.

    • Exact expressions will vary; 20/3−96/(3⁢n)+64/(3⁢n2).

      92/25, 3968/625, 103667/15625

      20/3

  47. 47.

    • Exact expressions will vary; 100−200/n.

      80, 98, 499/5

      100

  48. 48.

    • Exact expressions will vary; −(1−1/n2)/12.

      −33/400, −3333/40000, −333333/4000000

      −1/12

  49. 49.

    • Exact expressions will vary; 80.5.

      72.25

      62.5

  50. 50.

    • (5⁢ s)⁢((0+6+14+23+30+36)⁢ mph)=545⁢mi shr×1⁢ hr3600⁢ s×5280⁢ ft1⁢ mi=799⁢ ft

      (5⁢ s)⁢((6+14+23+30+36+40)⁢ mph)=585⁢mi shr×1⁢ hr3600⁢ s×5280⁢ ft1⁢ mi=858⁢ ft

  51. 51.

    ∫abk⋅f⁢(x)⁢d⁡x =limn→∞∑i=1nk⋅f⁢(ci)⁢Δ⁢x T5.3.2.2
    =limn→∞k⋅∑i=1nk⋅f⁢(ci)⁢Δ⁢x T5.3.1.3
    =k⋅limn→∞∑i=1nk⋅f⁢(ci)⁢Δ⁢x T1.3.1.4
    =k⁢∫abf⁢(x)⁢d⁡x T5.3.2.2
  52. 52.

    Let f and M be as given.

    ∫abf⁢(x)⁢d⁡x =limn→∞∑i=1nf⁢(ci)⁢Δ⁢x T5.3.2.2
    ≤limn→∞∑i=1nM⁢Δ⁢x
    =∫abM⁢d⁡x T5.3.2.2
    =M⁢(b−a)
  53. 53.

    F⁢(x)=5⁢tan⁡x+4

  54. 54.

    F⁢(x)=7⁢ln⁡|x|+14

  55. 55.

    G⁢(t)=4/6⁢t6−5/4⁢t4+8⁢t+9

  56. 56.

    G⁢(t)=5⋅et+900

  57. 57.

    G⁢(t)=sin⁡t−cos⁡t−78

  58. 58.

    F⁢(x)=2⁢x−π

Exercises E.4

  1. 1.

    Answers will vary.

  2. 2.

    0

  3. 3.

    T

  4. 4.

    Answers will vary.

  5. 5.

    20

  6. 6.

    28/3

  7. 7.

    0

  8. 8.

    1

  9. 9.

    1

  10. 10.

    1

  11. 11.

    23/2

  12. 12.

    −4

  13. 13.

    e3−e

  14. 14.

    16/3

  15. 15.

    4

  16. 16.

    45/4

  17. 17.

    ln⁡2

  18. 18.

    1/2

  19. 19.

    1/4

  20. 20.

    1/101

  21. 21.

    15

  22. 22.

    2−2/3

  23. 23.

    2

  24. 24.

    72

  25. 25.

    632

  26. 26.

    883

  27. 27.

    6⁢π7

  28. 28.

    0

  29. 29.

    2

  30. 30.

    1

  31. 31.

    36

  32. 32.

    12

  33. 33.

    694

  34. 34.

    89

  35. 35.

    Explanations will vary. A sketch will help.

  36. 36.

    ∫aa+2⁢πsin⁡t⁢d⁡t=cos⁡(a+2⁢π)−cos⁡(a). Since cosine is periodic with period 2⁢π, cos⁡(a+2⁢π)=cos⁡(a), and hence the integral is 0.

  37. 37.

    c=2/3

  38. 38.

    c=±2/3

  39. 39.

    c=ln⁡(e−1)≈0.54

  40. 40.

    c=64/9≈7.1

  41. 41.

    2/π

  42. 42.

    2/p⁢i

  43. 43.

    2

  44. 44.

    16/3

  45. 45.

    16

  46. 46.

    1/(e−1)

  47. 47.

    (a) −300ft; (b) 312.5ft

  48. 48.

    (a) 400ft; (b) 850ft

  49. 49.

    (a) −1ft; (b) 3ft

  50. 50.

    (a) 128/5ft; (b) same

  51. 51.

    −64ft/s

  52. 52.

    50ft/s

  53. 53.

    2ft/s

  54. 54.

    0ft/s

  55. 55.

    F′⁢(x)=(3⁢x2+1)⁢1x3+x

  56. 56.

    F′⁢(x)=−3⁢x11

  57. 57.

    F′⁢(x)=2⁢x⁢(x2+2)−(x+2)

  58. 58.

    F′⁢(x)=ex⁢sin⁡(ex)−1x⁢sin⁡(ln⁡x)

  59. 59.

    F′⁢(x)=ln⁡x+4x2+7

  60. 60.

    F′⁢(x)=[cos3⁡(sin⁡x)+3⁢tan3⁡(sin⁡x)]⁢cos⁡x

  61. 61.

    F′⁢(x)=−15⁢x2⁢cos⁡(5⁢x3)+525⁢x6+e5⁢x3

  62. 62.

    F′⁢(x)=−2⁢tan⁡x⁢sec2⁡x⁢[ln⁡(tan2⁡x)+etan4⁡x−7]

  63. 63.

    • x 0 1 2 3 4 5 6
      g(x) 0 .5 0 -.5 0 1.5 4

      g⁢(7)≈5.7

      min at x=3; max at x=7

      Approximately
      246246xy

  64. 64.

    • This is a consequence of Theorem 5.4.1.

      The derivative of the left is g′⁢(x⁢y)⁢y=1x⁢y⁢y. The derivative of the right is g′⁢(x)=1x. Theorem 5.1.1 tells us that the left and right therefore differ by a constant. Looking at x=1 and g⁢(1)=0 tells us that this difference is 0.

      The derivative of the left is g′⁢(xr)⁢r⁢xr−1=1xr⁢r⁢xr−1. The derivative of the right is r⁢g′⁢(x)=r⁢1x. Theorem 5.1.1 tells us that the left and right therefore differ by a constant. Looking at x=1 and g⁢(1)=0 tells us that this difference is 0.

  65. 65.

    • bn=4/n⁢π for odd n and bn=0 for even n

      answers will vary

Exercises E.5

  1. 1.

    Chain Rule.

  2. 2.

    T

  3. 3.

    18⁢(x3−5)8+C

  4. 4.

    14⁢(x2−5⁢x+7)4+C

  5. 5.

    118⁢(x2+1)9+C

  6. 6.

    13⁢(3⁢x2+7⁢x−1)6+C

  7. 7.

    12⁢ln⁡|2⁢x+7|+C

  8. 8.

    2⁢x+3+C

  9. 9.

    23⁢(x+3)3/2−6⁢(x+3)1/2+C=23⁢(x−6)⁢x+3+C

  10. 10.

    221⁢x3/2⁢(3⁢x2−7)+C

  11. 11.

    2⁢ex+C

  12. 12.

    2⁢x5+15+C

  13. 13.

    −12⁢x2−1x+C

  14. 14.

    ln2⁡(x)2+C

  15. 15.

    sin3⁡(x)3+C

  16. 16.

    −cos4⁡(x)4+C

  17. 17.

    −16⁢sin⁡(3−6⁢x)+C

  18. 18.

    −tan⁡(4−x)+C

  19. 19.

    12⁢ln⁡|sec⁡(2⁢x)+tan⁡(2⁢x)|+C

  20. 20.

    sin⁡(x2)2+C

  21. 21.

    tan⁡(x)−x+C

  22. 22.

    The key is to rewrite cot⁡x as cos⁡x/sin⁡x, and let u=sin⁡x.

  23. 23.

    The key is to multiply csc⁡x by 1 in the form (csc⁡x+cot⁡x)/(csc⁡x+cot⁡x).

  24. 24.

    13⁢e3⁢x−1+C

  25. 25.

    ex33+C

  26. 26.

    12⁢e(x−1)2+C

  27. 27.

    x−e−x+C

  28. 28.

    ln⁡(ex+1)+C

  29. 29.

    e−3⁢x3−e−x+C

  30. 30.

    12⁢ln2⁡(x)+C

  31. 31.

    (ln⁡x)33+C

  32. 32.

    32⁢ln2⁡(x)+C

  33. 33.

    12⁢ln⁡|ln⁡(x2)|+C

  34. 34.

    x22+3⁢x+ln⁡|x|+C

  35. 35.

    x33+x22+x+ln⁡|x|+C

  36. 36.

    −13⁢(x3+3)+C

  37. 37.

    145⁢(5⁢x3+5⁢x2+2)9+C

  38. 38.

    −1−x2+C

  39. 39.

    −13⁢cot⁡(x3+1)+C

  40. 40.

    −23⁢cos32⁡(x)+C

  41. 41.

    −15⁢cos⁡(5⁢x+1)+C

  42. 42.

    ln⁡|x−5|+C

  43. 43.

    73⁢ln⁡|3⁢x+2|+C

  44. 44.

    ln⁡|x2+7⁢x+3|+C

  45. 45.

    3⁢ln⁡|3⁢x2+9⁢x+7|+C

  46. 46.

    3⁢x2−2⁢x−6+C

  47. 47.

    x2−6⁢x+8+C

  48. 48.

    2⁢sin⁡x+C

  49. 49.

    12⁢sec2⁡θ+C or 12⁢tan2⁡θ+C

  50. 50.

    110⁢(2⁢x+3)5/2−12⁢(2⁢x+3)3/2+C

  51. 51.

    −12⁢(x2+1)+14⁢(x2+1)2+C

  52. 52.

    12⁢(x2+1)2−2⁢(x2+1)+ln⁡(x2+1)+C

  53. 53.

    111⁢(x3+2)11−25⁢(x3+2)10+49⁢(x3+2)9+C

  54. 54.

    −3⁢cos⁡(x3)+C

  55. 55.

    23⁢sin6⁡(x4)+C

  56. 56.

    23⁢sin⁡(x3/2+1)+C

  57. 57.

    −ln⁡2

  58. 58.

    352/15

  59. 59.

    2/3

  60. 60.

    1/5

  61. 61.

    (1−e)/2

  62. 62.

    e−1

  63. 63.

    0

  64. 64.

    ln⁡(41+e)

  65. 65.

    23

  66. 66.

    112

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