W 0 9 x 279 chapter 8 10 r10 x15 r 2x3 10 10 15 15

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Unformatted text preview: t 1 + t2 x (b) b ck fk (x)dx = ck tan(π/4−2) 1 dt 1 + t2 (a) F (x) = x−3 ; increasing on [3, +∞), decreasing on (−∞, 3] x2 + 7 (b) 24. F (x) = 7 + 6x − x2 (7 − x)(1 + x) = ; concave up on (−1, 7), concave down on (−∞, −1) and (x2 + 7)2 (x2 + 7)2 (7, +∞) (c) x−3 = 0 when x = 3, which is a relative minimum, and hence the absolute minimum, x2 + 7 by the first derivative test. F (x) = (d) F(x) 3 2 1 -10 10 20 x 1 1 + (−1/x2 ) = 0 so F is constant on (0, +∞). 2 1+x 1 + (1/x)2 25. F (x) = 26. (−3, 3) because f is continuous there and 1 is in (−3, 3) 27. (a) The domain is (−∞, +∞); F (x) is 0 if x = 1, positive if x > 1, and negative if x < 1, because the integrand is positive, so the sign of the integral depends on the orientation (forwards or backwards). (b) The domain is [−2, 2]; F (x) is 0 if x = −1, positive if −1 < x ≤ 2, and negative if −2 ≤ x < −1; same reasons as in part (a). 253 Chapter 7 28. The left endpoint of the top boundary is ((b − a)/2, h) and the right endpoint of the top boundary is ((b + a)/2, h) so x < (b − a)/2 2hx/(b − a), h, (b − a)/2 < x < (b + a)/2 f (x) = 2h(x − b)/(a − b), x > (a + b)/2 The area of the trapezoid is given by (b−a)/2 0 2hx dx + b−a (b−a)/2 b hdx + (b+a)/2 2h(x − b) dx = (b − a)h/4 + ah + (b − a)h/4 = h(a + b)/2. a−b √ 2000e−t/48 + 500 sin(πt/12) dt = 96000(1 − 1/ e) ≈ 37, 773.06 24 29. (b+a)/2 (a) 0 (b) (c) 1 8−0 8 2000e−t/48 + 500 sin(πt/12)dt = 1125/π + 12000(1 − e−1/6 ) ≈ 2, 200.32 0 2300 0 2000 (d) 30. 31. 8 maximum rate is 2285.32 kW/h at t = 4.8861 wave = 1 52 − 26 52 (t/7)dt = 39/7; t∗ /7 = 39/7, t∗ = 39 26 no, since the velocity curve is not a straight line (b) 25 < t < 40 (c) 3.54 ft/s (d) 141.5 ft (e) no since the velocity is positive and the acceleration is never negative (f ) need the position at any one given time (e.g. s0 ) (a) x = aekt + be−kt , dx/dt = akekt − bke−kt , d2 x/dt2 = ak 2 ekt + bk 2 e−kt = k 2 (aekt + be−kt ) = k 2 x (b) 32. (a) 2 At t = 0, v = ak − bk = (a − b)k = v0 so k = v0 /(a − b) and a = k 2 x = v0 x/(a − b)2 . 33. u = 5 + 2 sin 3x, du = 6 cos 3xdx; 34. u=3+ 35. u = ax3 + b, du = 3ax2 dx; √ 1 x, du = √ dx; 2x 36. u = ax2 , du = 2axdx; 1 2a 1√ 1 1 √ du = u1/2 + C = 5 + 2 sin 3x + C 3 3 6u √ √ 4 4 2 udu = u3/2 + C = (3 + x)3/2 + C 3 3 1 1 1 +C =− 2 3 +C du = − 3au2 3au 3a x + 3ab sec2 udu = 1 1 tan u + C = tan(ax2 ) + C 2a 2a 37. ln(ex ) + ln(e−x ) = ln(ex e−x ) = ln 1 = 0 so [ln(ex ) + ln(e−x )]dx = C Supplementary Exercises 7 − 38. 254 −1 1 3 1 −+4 3u3 u 4u = 389/192 −2 2 39. u = ln x, du = (1/x)dx; 1 1 40. 2 1 du = ln u u = ln 2 1 √ e−x/2 dx = 2(1 − 1/ e) 0 41. 42. u = e−2x , du = −2e−2x dx; − 1/4 1 2 (1 + cos u)du = 1 sin 1 − sin 1 4 1 1 sin3 πx 3π =0 0 b 43. 31 + 82 With b = 1.618034, area = (x + x2 − x3 )dx = 1.007514. 0 44. (a) (b) 45. (a) (b) 2 2 12 x sin 3x − sin 3x + x cos 3x − 0.251607 3 27 9 4 −6 f (x) = 4 + x2 + √ 4 + x2 f (x) = 7 14 k − k − k 2 + = 0 to get k = 2.073948. 4 4 1 1 1 Solve − cos 2k + k 3 + = 3 to get k = 1.837992. 2 3 2 Solve x 46. 47. x t dt, F (x) = √ , so F is increasing on [1, 3]; Fmax = F (3) ≈ 1.152082854 2 + t3 2 + x3 −1 and Fmin = F (1) ≈ −0.07649493141 F (x) = (a) √ (b) 0.7651976866 y (c) J0 (x) = 0 if x = 2.404826 1 y = J0(x) 0.5 x 12345678 -0.5 1 48. (a) x2 dx = A= 0 n (b) k =1 k−1 n n n (c) k =1 49. k =1 k n 2 1 = 1/3 0 1 1 =3 n n k−1 n lim n→+∞ 2 13 x 3 1 1 =3 n n 2 n n k2 − 2 k =1 n k+ k =1 1= k =1 1 n3 n(n + 1) n(n + 1)(2n + 1) −2 +n , 6 2 2 1 1 == n 6 3 n 1 n(n + 1)(2n + 1) and lim k= 3 n→+∞ n 6 n 2 k =1 100,000 100, 000/(ln 100, 000) ≈ 8,686; 2 k =1 k n 2 2 1 1 == n 6 3 1 dt ≈ 9,629, so the integral is better ln t 255 Chapter 7 CHAPTER 7 HORIZON MODULE 1. 2. 3. 4. vx (0) = 35 cos α, so from Equation (1), x(t) = (35 cos α)t; vy (0) = 35 sin α, so from Equation (2), y (t) = (35 sin α)t − 4.9t2 . dx(t) dy (t) = 35 cos α, vy (t) = = 35 sin α − 9.8t dt dt (b) vy (t) = 35 sin α − 9.8t, vy (t) = 0 when t = 35 sin α/9.8; y = vy (0)t − 4.9t2 = (35 sin α)(35 sin α)/9.8 − 4.9((35 sin α)/9.8)2 = 62.5 sin2 α, so ymax = 62.5 sin2 α. (a) vx (t) = 0.004 2 t = x/(35 cos α) so y = (35 sin α)(x/(35 cos α)) − 4.9(x/(35 cos α))2 = (tan α)x − x; cos2 α the trajectory is a parabola because y is a quadratic function of x. 15◦ no 25◦ yes 35◦ no 45◦ no 55◦ no 65◦ yes 75◦ no 85◦ no 65 0 120 0 5. y (t) = (35 sin α s)t − 4.9t2 = 0 when t = 35 sin α/4.9, at which time x = (35 cos α)(35 sin α/4.9) = 125 sin 2α; this is the maximum value of x, so R = 125 sin 2α m. 6. (a) R = 95 when sin 2α = 95/125 = 0.76, α = 0.4316565575, 1.139139769 rad ≈ 24.73◦ , 65.27◦ . (b) y (t) < 50 is required; but y (1.139) ≈ 51.56 m, so his height would be 56.56 m. 7. 0.4019 < α < 0.4636 (radians), or 23.03◦ < α < 26.57◦ CHAPTER 8 Applications of the Definite Integral in Geometry, Science, and Engineering EXERCISE SET 8.1 2 2 1. A = −1 4 2. A = (x2 + 1 − x)dx = (x3 /3 + x − x2 /2) = 9/2 −...
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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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