# 2x x2 mxdx 28 the line through 0 0 and 56 12

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Unformatted text preview: −4.9t(t − 10) = 0, t = 10 s (d) v (10) = −9.8(10) + 49 = −49 m/s (e) s(t) = 0 when the projectile hits the ground, −4.9t2 + 49t + 150 = 0 when (use the quadratic formula) t ≈ 12.46 s (f ) v (12.46) = −9.8(12.46) + 49 ≈ −73.1, the speed at impact is about 73.1 m/s 46. take s = 0 at the water level and let h be the height of the bridge, then s = h and v = 0 when t = 0 so s(t) = −16t2 + h (a) s = 0 when t = 4 thus −16(4)2 + h = 0, h = 256 ft (b) First, ﬁnd how long it takes for the stone to hit the water (ﬁnd t for s = 0) : −16t2 + h = 0, √ t = h/4. Next, ﬁnd how long it takes the sound to travel to the bridge: this time is h/1080 because the speed is constant at 1080 ft/s. Finally, use the fact that the total of these two √ √ √ h h + = 4, h + 270 h = 4320, h + 270 h − 4320 = 0, and by the times must be 4 s: 1080 4 √ √ −270 ± (270)2 + 4(4320) , reject the negative value to get h ≈ 15.15, quadratic formula h = 2 h ≈ 229.5 ft. 47. g = 9.8/6 = 4.9/3 m/s2 , so v = −(4.9/3)t, s = −(4.9/6)t2 + 5,s = 0 when t = v = −(4.9/3) 30/4.9 ≈ −4.04, so the speed of the module upon landing is 4.04 m/s 48. s(t) = − 1 gt2 + v0 t; s = 1000 when v = 0, so 0 = v = −gt + v0 , t = v0 /g , 2 √ 2 2 1000 = s(v0 /g ) = − 1 g (v0 /g )2 + v0 (v0 /g ) = 1 v0 /g , so v0 = 2000g , v0 = 2000g . 2 2 √ The initial velocity on the Earth would have to be 6 times faster than that on the Moon. 30/4.9 and Exercise Set 7.7 49. 51. 52. 53. fave = fave = fave = fave = 54. fave = 55. (a) 240 3 1 3−1 3x dx = 1 π 1 π−0 sin x dx = − 0 π 1 π−0 0 e 1 =6 50. fave = 1 1 cos x π −1 2 −1 x2 dx = 13 x 9 2 =1 −1 π = 2/π 0 π =0 0 ex dx = 1 5 − e−1 (5 − e−1 ) = ln 5 + 1 1 + ln 5 √ (b) (x∗ )2 = 4/3, x∗ = ±2/ 3, √ but only 2/ 3 is in [0, 2] 2 1 2−0 1 2 − (−1) 1 1 1 dx = (ln e − ln 1) = x 1−e e−1 ln 5 1 ln 5 − (−1) (c) 3 1 sin x π cos x dx = 1 = e−1 fave = 32 x 4 x2 dx = 4/3 0 y 4 2 3 56. (a) fave = (c) x 2 4 1 4−0 2x dx = 4 (b) 2x∗ = 4, x∗ = 2 0 y 8 4 2 57. (a) vave = x 4 4 1 4−1 (3t3 + 2)dt = 1 263 1 789 = 34 4 (b) 58. vave = 100 − 7 s(4) − s(1) = = 31 4−1 3 (a) aave = 1 5−0 5 (t + 1)dt = 7/2 0 v (π/4) − v (0) = (b) aave = π/4 − 0 59. √ √ 2/2 − 1 = (2 2 − 4)/π π/4 time to ﬁll tank = (volume of tank)/(rate of ﬁlling) = [π (3)2 5]/(1) = 45π , weight of water in tank at time t = (62.4) (rate of ﬁlling)(time) = 62.4t, 1 weightave = 45π 45π 62.4t dt = 1404π lb 0 241 Chapter 7 If x is the distance from the cooler end, then the temperature is T (x) = (15 + 1.5x)◦ C, and 10 1 (15 + 1.5x)dx = 22.5◦ C Tave = 10 − 0 0 By the Mean-Value Theorem for Integrals there exists x∗ in [0, 10] such that 10 1 (15 + 1.5x)dx = 22.5, 15 + 1.5x∗ = 22.5, x∗ = 5 f (x∗ ) = 10 − 0 0 (a) amount of water = (rate of ﬂow)(time) = 4t gal, total amount = 4(30) = 120 gal (b) 61. (a) (b) 60. amount of water = 60 (4 + t/10)dt = 420 gal 0 120 (c) amount of water = (10 + √ √ t)dt = 1200 + 160 30 ≈ 2076.36 gal 0 62. (a) The maximum value of R occurs at 4:30 P.M. when t = 0. 60 (b) 100(1 − 0.0001t2 )dt = 5280 cars 0 b 63. b [f (x) − fave ] dx = (a) a b f (x)dx − a b f (x)dx − fave (b − a) = 0 fave dx = a a b because fave (b − a) = f (x)dx a b (b) no, because if a c= b [f (x) − c]dx = 0 then f (x)dx − c(b − a) = 0 so a b 1 b−a f (x)dx = fave is the only value a EXERCISE SET 7.8 3 1. u7 du (a) (b) 1 2. (a) − 4 1 2 u1/2 du 7 1 −1 −π (b) u = 2x + 1, (d) udu 1 u2 du 4. u = 4x − 2, 5. 0 sin u du 2 eu du (d) 0 3. π 1 1 2 (c) 1 π (c) 3 1 2 u4 du = 1 6 1 4 u = 1 − 2x, − 6. u = 4 − 3x, − u3 du = 2 1 2 1 3 1 3 14 u 16 3 = 121/5, or 1 6 = 80, or 2 1 u3 du = − u4 8 −2 1 15 u 10 u8 du = − 1 3 19 u 27 1 2 = 19, or − 1 1 = 121/5 0 2 = 80 1 1 = 10, or − (1 − 2x)4 8 −2 (u − 3)u1/2 du 3 1 (2x + 1)5 10 1 (4x − 2)4 16 4 0 = 10 −1 1 (4 − 3x)9 27 2 = 19 1 −3 (u + 5)u20 du Exercise Set 7.8 242 9 7. u = 1 + x, 9 (u − 1)u1/2 du = 1 8. 1 4 u = 4 − x, 4 (u3/2 − 4u1/2 )du = 9 −5 π /4 sin u du = −8 cos u 11. √ = 8 − 4 2, or − 8 cos(x/2) 0 π /2 2 3 u = 3x, 0 = −506/15 9 π /2 = 2/3, or 0 2 sin 3x 3 π /6 = 2/3 0 13 1 +4= when x = − ln 3, 3 3 7 7 1 = ln(7) − ln(13/3) = ln(21/13) du = ln u u = eln 3 + 4 = 3 + 4 = 7 when x = ln 3, 13/3 u 13/3 u = ex + 4, du = ex dx, u = e− ln 3 + 4 = 12. u = 3 − 4ex , du = −4ex dx, u = −1 when x = 0, u = −17 when x = ln 5 − 13. 14. 15. −17 1 4 −1 5 1 3 1 u du = − u2 8 11 25 π (5)2 = π 34 12 11 π (4)2 = 2π 24 0 − 0 1 2 1 1 − u2 du = 1 sin πxdx = − 0 3 cos 2x dx = A= 0 −2 (x + 5) 0 3 sin 2x 2 = −(x + 5) π /8 A= 0 1 dx =− (3x + 1)2 3(3x + 1) 1 fave = 4−0 4 −2x e 0 √ = 3 2/4 0 =− 3 1 21. 11 · [π (1)2 ] = π/8 24 1 = − (−1 − 1) = 2/π π −1 3 20. 1 7 7 19. 1 − u2 du = 0 1 cos πx π π /8 18. 1 1 2 36 − u2 du = π (6)2 /2 = 18π −6 17. = −36 −1 16 − u2 du = 4 1 2 −17 25 − u2 du =...
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## This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.

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