Answers A-75
85. I=I
e–
.
x; x≈117 ft 86. P(t)=P
e–at 87. (A) Q=5e–
.
t
(B) 12 minutes (C) After 15 minutes
Exercises 12.4
1. 80 in.2 2. 180 cm2 3. 36 m2 4. 90 ft2 5. No, p–271 m
16.
u(x)
v(x)
74. L4=5,520 …L100
N(t)
dt …6,180 =R4
Exercises 12.5
1. 500 2. 200 3. 28 4. 8 5. 48 6. 62,500 7. 4.
p≈14.14
8. 12.5p≈39.27 9. (A) F(15) –F(10) =375 (B) F9(x)
F9(x) 5 9
20. 2
1
3– ln
2≈0.4055 23. 2e3–2≈38.171
37. 56=15,625 38. 510 39.
n 4 ≈1.38
40.
n
3≈1.099
41. 20
e0.2
–e–0.
≈13.550 42. 32.637 43.
1
7
2 11–e–12≈0.316 46.
2 1e–12≈0.859 47. 0 48. 0
(B)
500
Ave f(x) 5 250
50. (A) Average g(x)=12
the reverse is true.
is decreasing on [1, 4], L
overestimates the area and R
underestimates the
area. 21. In both figures, the error bound for L3 and R3 is 7.
24. S4=–1,
2
25. S4=–1,194 26. S
=10 27. S
=–33.01
32. 10.667 33. 8.533 34. 2.132 35. 4.949 36. 7.083 37. –10.
7
43. 15 44. 63 45. 58.5 46. 10.5 47. –54 48. 16.5 49. 248
3 51. 0 52. 0 53. –183 54. 13.5 55. False 56. True
57. False 58. True 59. False 60. False 61. L1
=286,100 ft
; error
2.63. Geometrically, the definite integral over the interval [2, 5] is the sum of
67. Decreasing; L4≈–9.308 68. Decreasing; L4≈–5.245
73. L
=2,580, R
=3,900; error bound for L3 and R3 is 1,320