# N n exercise set 112 1 an1 an 1 1 1 0 for n

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Unformatted text preview: x x +x+1 x(x2 −x − 1 dx = − +x+1 x+1 dx = − (x + 1/2)2 + 3/4 x2 u + 1/2 du, u2 + 3/4 u = x + 1/2 √ 1 1 = − ln(u2 + 3/4) − √ tan−1 2u/ 3 + C1 2 3 1 1 2x + 1 dx = ln |x| − ln(x2 + x + 1) − √ tan−1 √ +C x(x2 + x + 1) 2 3 3 so √ 25. u = 2 0 2u2 du = 2 2+4 u 26. u = √ 3 2 0 27. u = x − 4, x = u2 + 4, dx = 2u du, 1 0 29. 30. √ 3 lim →+∞ 9 1− 2 u +9 0 −1 du = 2u − 6 tan 1 2(x2 + 1) 1 1− 0 1 2+1 u 34. =4−π 0 0 3 =6− π 2 2u du, −1 2u du, u2 + 1 1 0 1 4 π 2 1 1 1 + = + 1) 2(a2 + 1) 2(a2 + 1) − = lim a =2− 0 π 1 b tan−1 = ab a 2ab →+∞ →+∞ √ 2( 2 1 1 1 du = sin−1 u + C = sin−1 (x4 ) + C . 2 4 4 1−u (cos32 x sin30 x − cos30 x sin32 x)dx = x− 3 u2 = lim 1 bx tan−1 ab a x2 u 3 du = (2u − 2 tan−1 u) = 33. 2 √x 1 e +1−1 1 +C − du = ln |u − 1| − ln |u + 1| + C = ln √ x u−1 u+1 e +1+1 ex − 1, ex = u2 + 1, x = ln(u2 + 1), dx = 2u2 du = 2 2+1 u →+∞ 4 du = 2u − 4 tan−1 (u/2) u2 + 4 ex + 1, ex = u2 − 1, x = ln(u2 − 1), dx = 31. Let u = x4 to get 32. 0 u2 du = 2 u2 + 9 √ lim 1− x, x = u2 , dx = 2u du, 2 du = u2 − 1 28. u = 2 1 − 4dx = √ 2 1 1 dx = − x10 (1 + x−9 ) 9 cos30 x sin30 x(cos2 x − sin2 x)dx 1 230 sin30 2x cos 2x dx = √ √ ( x + 2 − x − 2)dx = sin31 2x +C 31(231 ) √ 2 [(x + 2)3/2 − (x − 2)3/2 ] + C 3 1 1 1 du = − ln |u| + C = − ln |1 + x−9 | + C u 9 9 Chapter 9 Supplementary Exercises 340 35. (a) (x + 4)(x − 5)(x2 + 1)2 ; (b) − B Cx + D Ex + F A + +2 +2 x+4 x−5 x +1 (x + 1)2 2 x−2 3 3 + − − x + 4 x − 5 x2 + 1 (x2 + 1)2 1 3 ln(x2 + 1) − 2 2 (c) −3 ln |x + 4| + 2 ln |x − 5| + 2 tan−1 x − +∞ 36. (a) Γ(1) = e−t dt = lim −e−t +∞ (b) Γ(x + 1) = x + tan−1 x +1 = lim (−e− + 1) = 1 →+∞ 0 x2 →+∞ 0 tx e−t dt; let u = tx , dv = e−t dt to get 0 +∞ 0 Γ(x + 1) = −tx e−t +∞ +x tx−1 e−t dt = −tx e−t 0 +∞ 0 + xΓ(x) tx = 0 (by multiple applications of L’Hˆpital’s rule) o t→+∞ t→+∞ et so Γ(x + 1) = xΓ(x) lim tx e−t = lim (c) Γ(2) = (1)Γ(1) = (1)(1) = 1, Γ(3) = 2Γ(2) = (2)(1) = 2, Γ(4) = 3Γ(3) = (3)(2) = 6 It appears that Γ(n) = (n − 1)! if n is a positive integer. +∞ 1 2 (e) Γ = 3 2 (d) Γ = +∞ t−1/2 e−t dt = 2 0 e−u du (with u = 2 √ √ √ t) = 2( π/2) = π 0 1 Γ 2 1 2 = 1√ π, Γ 2 5 2 = 3 Γ 2 3 2 = 3√ π 4 37. (a) t = − ln x, x = e−t , dx = −e−t dt, 1 0 (ln x)n dx = − 0 +∞ +∞ (−t)n e−t dt = (−1)n tn e−t dt = (−1)n Γ(n + 1) 0 (b) t = xn , x = t1/n , dx = (1/n)t1/n−1 dt, +∞ n 0 = = t1/n−1 e−t dt = (1/n)Γ(1/n) = Γ(1/n + 1) 0 cos θ − cos θ0 = 38. (a) +∞ e−x dx = (1/n) √ 1 2 T= 2 sin2 (θ0 /2) − sin2 (θ/2) = 2 k cos φ; k sin φ = sin(θ/2) so k cos φ dφ = 1 − k 2 sin2 φ dθ, thus dθ = 8L g π /2 0 √ 1 · 2k cos φ 1− 1 1 cos(θ/2) dθ = 2 2 2k cos φ 1 − k 2 sin2 φ 2k cos φ k2 2 2(k 2 − k 2 sin2 φ) = (b) If L = 1.5 ft and θ0 = (π/180)(20) = π/9, then √ 3 π/2 dφ ≈ 1.37 s. T= 20 1 − sin2 (π/18) sin2 φ 1 − sin2 (θ/2) dθ dφ and hence dφ = 4 sin φ 2k 2 cos2 φ L g π /2 0 1 1 − k 2 sin2 φ dφ 341 Chapter 9 CHAPTER 9 HORIZON MODULE 1. The depth of the cut equals the terrain elevation minus the track elevation. From Figure 2, the 1 cross sectional area of a cut of depth D meters is 10D + 2 · D2 = D2 + 10D square meters. 2 Distance from Terrain elevation Track elevation Depth of cut Cross-sectional area town A (m) (m) (m) (m) f (x) of cut (m2 ) 0 2000 4000 6000 8000 10,000 12,000 14,000 16,000 18,000 20,000 100 105 108 110 104 106 120 122 124 128 130 100 101 102 103 104 105 106 107 108 109 110 0 4 6 7 0 1 14 15 16 19 20 0 56 96 119 0 11 336 375 416 551 600 2000 The total volume of dirt to be excavated, in cubic meters, is f (x) dx. 0 By Simpson’s Rule, this is approximately 20,000 − 0 [0 + 4 · 56 + 2 · 96 + 4 · 119 + 2 · 0 + 4 · 11 + 2 · 336 + 4 · 375 + 2 · 416 + 4 · 551 + 600] 3 · 10 = 4,496,000 m3 . Excavation costs \$4 per m3 , so the ttal cost of the railroad from kA to M is about 4 · 4,496,000 = 17,984,000 dollars. 2. (a) Distance from Terrain elevation Track elevation Depth of cut Cross-sectional area town A (m) (m) (m) (m) f (x) of cut (m2 ) 20,000 20,100 20,200 20,300 20,400 20,500 20,600 20,700 20,800 20,900 21,000 130 135 139 142 145 147 148 146 143 139 133 110 109.8 109.6 109.4 109.2 109 108.8 108.6 108.4 108.2 108 20 25.2 29.4 32.6 35.8 38 39.2 37.4 34.6 30.8 25 300 887.04 1158.36 1388.76 1639.64 1824 1928.64 1772.76 1543.16 1256.64 875 21,000 The total volume of dirt to be excavated, in cubic meters, is f (x) dx. 20,000 By Simpson’s Rule this is approximately 21,000 − 20,000 [600 + 4 · 887.04 + 2 · 1158.36 + . . . + 4 · 1256.64 + 875] = 1,417,713.33 m3 . 3 · 10 The total cost of a trench from M to N is about 4 · 1,417,713.33 ≈ 5,670,853 dollars. Chapter 9 Horizon Module (b) 342 Distance from Terrain elevation Track elevation Depth of cut Cross-sectional area...
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