Exercícios de cálculo integral, Manuais, Projetos, Pesquisas de Eletrônica

Baixe Exercícios de cálculo integral e outras Manuais, Projetos, Pesquisas em PDF para Eletrônica, somente na Docsity! 223 CHAPTER 6 Integration EXERCISE SET 6.1 1. Endpoints 0, 1 n , 2 n , . . . , n− 1 n , 1; using right endpoints, An = [√ 1 n + √ 2 n + · · ·+ √ n− 1 n + 1 ] 1 n n 2 5 10 50 100 An 0.853553 0.749739 0.710509 0.676095 0.671463 2. Endpoints 0, 1 n , 2 n , . . . , n− 1 n , 1; using right endpoints, An = [ n n+ 1 + n n+ 2 + n n+ 3 + · · ·+ n 2n− 1 + 1 2 ] 1 n n 2 5 10 50 100 An 0.583333 0.645635 0.668771 0.688172 0.690653 3. Endpoints 0, π n , 2π n , . . . , (n− 1)π n , π; using right endpoints, An = [sin(π/n) + sin(2π/n) + · · ·+ sin(π(n− 1)/n) + sinπ] π n n 2 5 10 50 100 An 1.57080 1.93376 1.98352 1.99935 1.99984 4. Endpoints 0, π 2n , 2π 2n , . . . , (n− 1)π 2n , π 2 ; using right endpoints, An = [cos(π/2n) + cos(2π/2n) + · · ·+ cos((n− 1)π/2n) + cos(π/2)] π 2n n 2 5 10 50 100 An 0.555359 0.834683 0.919405 0.984204 0.992120 5. Endpoints 1, n+ 1 n , n+ 2 n , . . . , 2n− 1 n , 2; using right endpoints, An = [ n n+ 1 + n n+ 2 + · · ·+ n 2n− 1 + 1 2 ] 1 n n 2 5 10 50 100 An 0.583333 0.645635 0.668771 0.688172 0.690653 6. Endpoints −π 2 ,−π 2 + π n ,−π 2 + 2π n , . . . ,−π 2 + (n− 1)π n , π 2 ; using right endpoints, An = [ cos ( −π 2 + π n ) + cos ( −π 2 + 2π n ) + · · ·+ cos ( −π 2 + (n− 1)π n ) + cos (π 2 )] π n n 2 5 10 50 100 An 1.99985 1.93376 1.98352 1.99936 1.99985 224 Chapter 6 7. Endpoints 0, 1 n , 2 n , . . . , n− 1 n , 1; using right endpoints, An =   √ 1− ( 1 n )2 + √ 1− ( 2 n )2 + · · ·+ √ 1− ( n− 1 n )2 + 0   1 n n 2 5 10 50 100 An 0.433013 0.659262 0.726130 0.774567 0.780106 8. Endpoints −1,−1 + 2 n ,−1 + 4 n , . . . ,−1 + 2(n− 1) n , 1; using right endpoints, An =   √ 1− ( n− 2 n )2 + √ 1− ( n− 4 n )2 + · · ·+ √ 1− ( n− 2 n )2 + 0   2 n n 2 5 10 50 100 An 1 1.423837 1.518524 1.566097 1.569136 9. 3(x− 1) 10. 5(x− 2) 11. x(x+ 2) 12. 3 2 (x− 1)2 13. (x+ 3)(x− 1) 14. 3 2 x(x− 2) 15. The area in Exercise 13 is always 3 less than the area in Exercise 11. The regions are identical except that the area in Exercise 11 has the extra trapezoid with vertices at (0, 0), (1, 0), (0, 2), (1, 4) (with area 3). 16. (a) The region in question is a trapezoid, and the area of a trapezoid is 1 2 (h1 + h2)w. (b) From Part (a), A′(x)= 1 2 [f(a) + f(x)] + (x− a)1 2 f ′(x) = 1 2 [f(a) + f(x)] + (x− a)1 2 f(x)− f(a) x− a = f(x) 17. B is also the area between the graph of f(x) = √ x and the interval [0, 1] on the y−axis, so A+B is the area of the square. 18. If the plane is rotated about the line y = x then A becomes B and vice versa. EXERCISE SET 6.2 1. (a) ∫ x√ 1 + x2 dx = √ 1 + x2 + C (b) ∫ (x+ 1)exdx = xex + C 2. (a) d dx (sinx− x cosx+ C) = cosx− cosx+ x sinx = x sinx (b) d dx ( x√ 1− x2 + C ) = √ 1− x2 + x2/ √ 1− x2 1− x2 = 1 (1− x2)3/2 3. d dx [√ x3 + 5 ] = 3x2 2 √ x3 + 5 so ∫ 3x2 2 √ x3 + 5 dx = √ x3 + 5 + C Exercise Set 6.2 227 40. f ′(x) = m = (x+ 1)2, so f(x) = ∫ (x+ 1)2dx = 1 3 (x+ 1)3 + C; f(−2) = 8 = 1 3 (−2 + 1)3 + C = −1 3 + C,= 8 + 1 3 = 25 3 , f(x) = 1 3 (x+ 1)3 + 25 3 41. (a) y(x) = ∫ x1/3dx = 3 4 x4/3 + C, y(1) = 3 4 + C = 2, C = 5 4 ; y(x) = 3 4 x4/3 + 5 4 (b) y(t) = ∫ (sin t+ 1) dt = − cos t+ t+ C, y (π 3 ) = −1 2 + π 3 + C = 1/2, C = 1− π 3 ; y(t) = − cos t+ t+ 1− π 3 (c) y(x) = ∫ (x1/2 + x−1/2)dx = 2 3 x3/2 + 2x1/2 + C, y(1) = 0 = 8 3 + C, C = −8 3 , y(x) = 2 3 x3/2 + 2x1/2 − 8 3 42. (a) y(x) = ∫ ( 1 8 x−3 ) dx = − 1 16 x−2 + C, y(1) = 0 = − 1 16 + C, C = 1 16 ; y(x) = − 1 16 x−2 + 1 16 (b) y(t) = ∫ (sec2 t − sin t) dt = tan t + cos t + C, y(π 4 ) = 1 = 1 + √ 2 2 + C, C = − √ 2 2 ; y(t) = tan t+ cos t− √ 2 2 (c) y(x) = ∫ x7/2dx = 2 9 x9/2 + C, y(0) = 0 = C, C = 0; y(x) = 2 9 x9/2 43. (a) y = ∫ 4ex dx = 4ex + C, 1 = y(0) = 4 + C,C = −3, y = 4ex − 3 (b) y(t) = ∫ t−1dt = ln |t|+ C, y(−1) = C = 5, C = 5; y(t) = ln |t|+ 5 44. (a) y = ∫ 3√ 1− t2 dt = 3 sin −1 t+ C, y (√ 3 2 ) = 0 = π + C,C = −π, y = 3 sin−1 t− π (b) dy dx = 1− 2 x2 + 1 , y = ∫ [ 1− 2 x2 + 1 ] dx = x− 2 tan−1 x+ C, y(1) = π 2 = 1− 2π 4 + C,C = π − 1, y = x− 2 tan−1 x+ π − 1 45. f ′(x) = 2 3 x3/2 + C1; f(x) = 4 15 x5/2 + C1x+ C2 46. f ′(x) = x2/2 + sinx+ C1, use f ′(0) = 2 to get C1 = 2 so f ′(x) = x2/2 + sinx+ 2, f(x) = x3/6− cosx+ 2x+ C2, use f(0) = 1 to get C2 = 2 so f(x) = x3/6− cosx+ 2x+ 2 47. dy/dx = 2x+ 1, y = ∫ (2x+ 1)dx = x2 + x+ C; y = 0 when x = −3 so (−3)2 + (−3) + C = 0, C = −6 thus y = x2 + x− 6 48. dy/dx = x2, y = ∫ x2dx = x3/3 + C; y = 2 when x = −1 so (−1)3/3 + C = 2, C = 7/3 thus y = x3/3 + 7/3 228 Chapter 6 49. dy/dx = ∫ 6xdx = 3x2 + C1. The slope of the tangent line is −3 so dy/dx = −3 when x = 1. Thus 3(1)2 +C1 = −3, C1 = −6 so dy/dx = 3x2 − 6, y = ∫ (3x2 − 6)dx = x3 − 6x+C2. If x = 1, then y = 5− 3(1) = 2 so (1)2 − 6(1) + C2 = 2, C2 = 7 thus y = x3 − 6x+ 7. 50. dT/dx = C1, T = C1x+ C2; T = 25 when x = 0 so C2 = 25, T = C1x+ 25. T = 85 when x = 50 so 50C1 + 25 = 85, C1 = 1.2, T = 1.2x+ 25 51. (a) F ′(x) = G′(x) = 3x+ 4 (b) F (0) = 16/6 = 8/3, G(0) = 0, so F (0)−G(0) = 8/3 (c) F (x) = (9x2 + 24x+ 16)/6 = 3x2/2 + 4x+ 8/3 = G(x) + 8/3 52. (a) F ′(x) = G′(x) = 10x/(x2 + 5)2 (b) F (0) = 0, G(0) = −1, so F (0)−G(0) = 1 (c) F (x) = x2 x2 + 5 = (x2 + 5)− 5 x2 + 5 = 1− 5 x2 + 5 = G(x) + 1 53. ∫ (sec2 x− 1)dx = tanx− x+ C 54. ∫ (csc2 x− 1)dx = − cotx− x+ C 55. (a) 1 2 ∫ (1− cosx)dx = 1 2 (x− sinx) + C (b) 1 2 ∫ (1 + cosx) dx = 1 2 (x+ sinx) + C 56. (a) F ′(x) = G′(x) = f(x), where f(x) = { 1, x > 0 −1, x < 0 (b) G(x)− F (x) = { 2, x > 0 3, x < 0 so G(x) = F (x) plus a constant (c) no, because (−∞, 0) ∪ (0,+∞) is not an interval 57. v = 1087 2 √ 273 ∫ T−1/2 dT = 1087√ 273 T 1/2+C, v(273) = 1087 = 1087+C so C = 0, v = 1087√ 273 T 1/2 ft/s EXERCISE SET 6.3 1. (a) ∫ u23du = u24/24 + C = (x2 + 1)24/24 + C (b) − ∫ u3du = −u4/4 + C = −(cos4 x)/4 + C (c) 2 ∫ sinu du = −2 cosu+ C = −2 cos√x+ C (d) 3 8 ∫ u−1/2du = 3 4 u1/2 + C = 3 4 √ 4x2 + 5 + C 2. (a) 1 4 ∫ sec2 u du = 1 4 tanu+ C = 1 4 tan(4x+ 1) + C (b) 1 4 ∫ u1/2du = 1 6 u3/2 + C = 1 6 (1 + 2y2)3/2 + C (c) 1 π ∫ u1/2du = 2 3π u3/2 + C = 2 3π sin3/2(πθ) + C (d) ∫ u4/5du = 5 9 u9/5 + C = 5 9 (x2 + 7x+ 3)9/5 + C Exercise Set 6.3 229 3. (a) − ∫ u du = −1 2 u2 + C = −1 2 cot2 x+ C (b) ∫ u9du = 1 10 u10 + C = 1 10 (1 + sin t)10 + C (c) 1 2 ∫ cosu du = 1 2 sinu+ C = 1 2 sin 2x+ C (d) 1 2 ∫ sec2 u du = 1 2 tanu+ C = 1 2 tanx2 + C 4. (a) ∫ (u− 1)2u1/2du= ∫ (u5/2 − 2u3/2 + u1/2)du = 2 7 u7/2 − 4 5 u5/2 + 2 3 u3/2 + C = 2 7 (1 + x)7/2 − 4 5 (1 + x)5/2 + 2 3 (1 + x)3/2 + C (b) ∫ csc2 u du = − cotu+ C = − cot(sinx) + C (c) ∫ sinu du = − cosu+ C = − cos(x− π) + C (d) ∫ du u2 = − 1 u + C = − 1 x5 + 1 + C 5. (a) ∫ 1 u du = ln |u|+ C = ln | lnx|+ C (b) −1 5 ∫ eu du = −1 5 eu + C = −1 5 e−5x + C (c) −1 3 ∫ 1 u du = −1 3 ln |u|+ C = −1 3 ln |1 + cos 3θ|+ C (d) ∫ du u = lnu+ C = ln(1 + ex) + C 6. (a) u = x3, 1 3 ∫ du 1 + u2 = 1 3 tan−1(x3) + C (b) u = lnx, ∫ 1√ 1− u2 du = sin −1(lnx) + C (c) u = 3x, ∫ 1 u √ u2 − 1du = sec −1(3x) + C (d) u = √ x, 2 ∫ du 1 + u2 = 2 tan−1 u+ C = 2 tan−1( √ x) + C 7. u = 2− x2, du = −2x dx; −1 2 ∫ u3du = −u4/8 + C = −(2− x2)4/8 + C 8. u = 3x− 1, du = 3dx; 1 3 ∫ u5du = 1 18 u6 + C = 1 18 (3x− 1)6 + C 9. u = 8x, du = 8dx; 1 8 ∫ cosu du = 1 8 sinu+ C = 1 8 sin 8x+ C 10. u = 3x, du = 3dx; 1 3 ∫ sinu du = −1 3 cosu+ C = −1 3 cos 3x+ C 232 Chapter 6 48. sec2 3θ = tan2 3θ + 1, u = 3θ, du = 3dθ∫ sec4 3θ dθ = 1 3 ∫ (tan2 u+ 1) sec2 u du = 1 9 tan3 u+ 1 3 tanu+ C = 1 9 tan3 3θ + 1 3 tan 3θ + C 49. ∫ ( 1 + 1 t ) dt = t+ ln |t|+ C 50. e2 ln x = eln x 2 = x2, x > 0, so ∫ e2 ln xdx = ∫ x2dx = 1 3 x3 + C 51. ln(ex) + ln(e−x) = ln(exe−x) = ln 1 = 0 so ∫ [ln(ex) + ln(e−x)]dx = C 52. ∫ cosx sinx dx; u = sinx, du = cosxdx; ∫ 1 u du = ln |u|+ C = ln | sinx|+ C 53. (a) sin−1(x/3) + C (b) (1/ √ 5) tan−1(x/ √ 5) + C (c) (1/ √ π) sec−1(x/ √ π) + C 54. (a) u = ex, ∫ 1 4 + u2 du = 1 2 tan−1(ex/2) + C (b) u = 2x, 1 2 ∫ 1√ 9− u2 du = 1 2 sin−1(2x/3) + C, (c) u = √ 5y, ∫ 1 u √ u2 − 3du = 1√ 3 sec−1( √ 5y/ √ 3) + C 55. u = a+ bx, du = bdx,∫ (a+ bx)ndx = 1 b ∫ undu = (a+ bx)n+1 b(n+ 1) + C 56. u = a+ bx, du = b dx, dx = 1 b du 1 b ∫ u1/ndu = n b(n+ 1) u(n+1)/n + C = n b(n+ 1) (a+ bx)(n+1)/n + C 57. u = sin(a+ bx), du = b cos(a+ bx)dx 1 b ∫ undu = 1 b(n+ 1) un+1 + C = 1 b(n+ 1) sinn+1(a+ bx) + C 59. (a) with u = sinx, du = cosx dx; ∫ u du = 1 2 u2 + C1 = 1 2 sin2 x+ C1; with u = cosx, du = − sinx dx; − ∫ u du = −1 2 u2 + C2 = − 1 2 cos2 x+ C2 (b) because they differ by a constant:( 1 2 sin2 x+ C1 ) − ( −1 2 cos2 x+ C2 ) = 1 2 (sin2 x+ cos2 x) + C1 − C2 = 1/2 + C1 − C2 60. (a) First method: ∫ (25x2 − 10x+ 1)dx = 25 3 x3 − 5x2 + x+ C1; second method: 1 5 ∫ u2du = 1 15 u3 + C2 = 1 15 (5x− 1)3 + C2 Exercise Set 6.3 233 (b) 1 15 (5x− 1)3 + C2 = 1 15 (125x3 − 75x2 + 15x− 1) + C2 = 25 3 x3 − 5x2 + x− 1 15 + C2; the answers differ by a constant. 61. y(x) = ∫ √ 3x+ 1dx = 2 9 (3x+ 1)3/2 + C, y(1) = 16 9 + C = 5, C = 29 9 so y(x) = 2 9 (3x+ 1)3/2 + 29 9 62. y(x) = ∫ (6− 5 sin 2x)dx = 6x+ 5 2 cos 2x+ C, y(0) = 5 2 + C = 3, C = 1 2 so y(x) = 6x+ 5 2 cos 2x+ 1 2 63. y(t) = ∫ 2e−t dt = −2e−t + C, y(1) = −2 e + C = 3− 2 e , C = 3; y(t) = −2e−t + 3 64. y = ∫ dx 100 + 4x2 , u = x/5, dx = 5 du, y = 1 20 ∫ du 1 + u2 = 1 20 tan−1 u+ C = 1 20 tan−1 (x 5 ) + C; y(−5) = 3π 80 = 1 20 ( −π 4 ) + C, C = π 20 , y = 1 20 tan−1 (x 5 ) + π 20 65. 5 0 -5 5 66. y x 4 -4 4 67. f ′(x) = m = √ 3x+ 1, f(x) = ∫ (3x+ 1)1/2dx = 2 9 (3x+ 1)3/2 + C f(0) = 1 = 2 9 + C, C = 7 9 , so f(x) = 2 9 (3x+ 1)3/2 + 7 9 68. p(t) = ∫ (4 + 0.15t)3/2dt = 8 3 (4 + 0.15t)5/2 + C; p(0) = 100,000 = 8 3 45/2 + C = 256 3 + C, C = 100,000− 256 3 ≈ 99,915, p(t) ≈ 8 3 (4+0.15t)5/2 +99,915, p(5) ≈ 8 3 (4.75)5/2 +99,915 ≈ 100,046 69. u = a sin θ, du = a cos θ dθ; ∫ du√ a2 − u2 = aθ + C = sin −1 u a + C 70. If u > 0 then u = a sec θ, du = a sec θ tan θ dθ, ∫ du u √ u2 − a2 = 1 a θ = 1 a sec−1 u a + C 234 Chapter 6 EXERCISE SET 6.4 1. (a) 1 + 8 + 27 = 36 (b) 5 + 8 + 11 + 14 + 17 = 55 (c) 20 + 12 + 6 + 2 + 0 + 0 = 40 (d) 1 + 1 + 1 + 1 + 1 + 1 = 6 (e) 1− 2 + 4− 8 + 16 = 11 (f) 0 + 0 + 0 + 0 + 0 + 0 = 0 2. (a) 1 + 0− 3 + 0 = −2 (b) 1− 1 + 1− 1 + 1− 1 = 0 (c) π2 + π2 + · · ·+ π2 = 14π2 (14 terms) (d) 24 + 25 + 26 = 112 (e) √ 1 + √ 2 + √ 3 + √ 4 + √ 5 + √ 6 (f) 1− 1 + 1− 1 + 1− 1 + 1− 1 + 1− 1 + 1 = 1 3. 10∑ k=1 k 4. 20∑ k=1 3k 5. 10∑ k=1 2k 6. 8∑ k=1 (2k − 1) 7. 6∑ k=1 (−1)k+1(2k − 1) 8. 5∑ k=1 (−1)k+1 1 k 9. (a) 50∑ k=1 2k (b) 50∑ k=1 (2k − 1) 10. (a) 5∑ k=1 (−1)k+1ak (b) 5∑ k=0 (−1)k+1bk (c) n∑ k=0 akx k (d) 5∑ k=0 a5−kbk 11. 1 2 (100)(100 + 1) = 5050 12. 7 100∑ k=1 k + 100∑ k=1 1 = 7 2 (100)(101) + 100 = 35,450 13. 1 6 (20)(21)(41) = 2870 14. 20∑ k=1 k2 − 3∑ k=1 k2 = 2870− 14 = 2856 15. 30∑ k=1 k(k2 − 4) = 30∑ k=1 (k3 − 4k) = 30∑ k=1 k3 − 4 30∑ k=1 k = 1 4 (30)2(31)2 − 4 · 1 2 (30)(31) = 214,365 16. 6∑ k=1 k − 6∑ k=1 k3 = 1 2 (6)(7)− 1 4 (6)2(7)2 = −420 17. n∑ k=1 3k n = 3 n n∑ k=1 k = 3 n · 1 2 n(n+ 1) = 3 2 (n+ 1) 18. n−1∑ k=1 k2 n = 1 n n−1∑ k=1 k2 = 1 n · 1 6 (n− 1)(n)(2n− 1) = 1 6 (n− 1)(2n− 1) 19. n−1∑ k=1 k3 n2 = 1 n2 n−1∑ k=1 k3 = 1 n2 · 1 4 (n− 1)2n2 = 1 4 (n− 1)2 20. n∑ k=1 ( 5 n − 2k n ) = 5 n n∑ k=1 1− 2 n n∑ k=1 k = 5 n (n)− 2 n · 1 2 n(n+ 1) = 4− n Exercise Set 6.4 237 39. ∆x = 3 n , x∗k = 0 + k 3 n ; f(x∗k)∆x = ( 9− 9k 2 n2 ) 3 n n∑ k=1 f(x∗k)∆x = n∑ k=1 ( 9− 9k 2 n2 ) 3 n = 27 n n∑ k=1 ( 1− k 2 n2 ) = 27− 27 n3 n∑ k=1 k2 A = lim n→+∞ [ 27− 27 n3 n∑ k=1 k2 ] = 27− 27 ( 1 3 ) = 18 40. ∆x = 3 n , x∗k = k 3 n f(x∗k)∆x = [ 4− 1 4 (x∗k) 2 ] ∆x = [ 4− 1 4 9k2 n2 ] 3 n = 12 n − 27k 2 4n3 n∑ k=1 f(x∗k)∆x= n∑ k=1 12 n − 27 4n3 n∑ k=1 k2 = 12− 27 4n3 · 1 6 n(n+ 1)(2n+ 1) = 12− 9 8 (n+ 1)(2n+ 1) n2 A = lim n→+∞ [ 12− 9 8 ( 1 + 1 n )( 2 + 1 n )] = 12− 9 8 (1)(2) = 39/4 41. ∆x = 4 n , x∗k = 2 + k 4 n f(x∗k)∆x = (x ∗ k) 3∆x = [ 2 + 4 n k ]3 4 n = 32 n [ 1 + 2 n k ]3 = 32 n [ 1 + 6 n k + 12 n2 k2 + 8 n3 k3 ] n∑ k=1 f(x∗k)∆x= 32 n [ n∑ k=1 1 + 6 n n∑ k=1 k + 12 n2 n∑ k=1 k2 + 8 n3 n∑ k=1 k3 ] = 32 n [ n+ 6 n · 1 2 n(n+ 1) + 12 n2 · 1 6 n(n+ 1)(2n+ 1) + 8 n3 · 1 4 n2(n+ 1)2 ] = 32 [ 1 + 3 n+ 1 n + 2 (n+ 1)(2n+ 1) n2 + 2 (n+ 1)2 n2 ] A= lim n→+∞ 32 [ 1 + 3 ( 1 + 1 n ) + 2 ( 1 + 1 n )( 2 + 1 n ) + 2 ( 1 + 1 n )2] = 32[1 + 3(1) + 2(1)(2) + 2(1)2] = 320 42. ∆x = 2 n , x∗k = −3 + k 2 n ; f(x∗k)∆x = [1− (x∗k)3]∆x= [ 1− ( −3 + 2 n k )3] 2 n = 2 n [ 28− 54 n k + 36 n2 k2 − 8 n3 k3 ] n∑ k=1 f(x∗k)∆x = 2 n [ 28n− 27(n+ 1) + 6(n+ 1)(2n+ 1) n − 2(n+ 1) 2 n ] A= lim n→+∞ 2 [ 28− 27 ( 1 + 1 n ) + 6 ( 1 + 1 n )( 2 + 1 n ) − 2 ( 1 + 1 n )2] = 2(28− 27 + 12− 2) = 22 238 Chapter 6 43. ∆x = 3 n , x∗k = 1 + (k − 1) 3 n f(x∗k)∆x = 1 2 x∗k∆x = 1 2 [ 1 + (k − 1) 3 n ] 3 n = 1 2 [ 3 n + (k − 1) 9 n2 ] n∑ k=1 f(x∗k)∆x = 1 2 [ n∑ k=1 3 n + 9 n2 n∑ k=1 (k − 1) ] = 1 2 [ 3 + 9 n2 · 1 2 (n− 1)n ] = 3 2 + 9 4 n− 1 n A = lim n→+∞ [ 3 2 + 9 4 ( 1− 1 n )] = 3 2 + 9 4 = 15 4 44. ∆x = 5 n , x∗k = 5 n (k − 1) f(x∗k)∆x = (5− x∗k)∆x = [ 5− 5 n (k − 1) ] 5 n = 25 n − 25 n2 (k − 1) n∑ k=1 f(x∗k)∆x = 25 n n∑ k=1 1− 25 n2 n∑ k=1 (k − 1) = 25− 25 2 n− 1 n A = lim n→+∞ [ 25− 25 2 ( 1− 1 n )] = 25− 25 2 = 25 2 45. ∆x = 3 n , x∗k = 0 + (k − 1) 3 n ; f(x∗k)∆x = (9− 9 (k − 1)2 n2 ) 3 n n∑ k=1 f(x∗k)∆x = n∑ k=1 [ 9− 9(k − 1) 2 n2 ] 3 n = 27 n n∑ k=1 ( 1− (k − 1) 2 n2 ) = 27− 27 n3 n∑ k=1 k2 + 54 n3 n∑ k=1 k − 27 n2 A = lim n→+∞ = 27− 27 ( 1 3 ) + 0 + 0 = 18 46. ∆x = 3 n , x∗k = (k − 1) 3 n f(x∗k)∆x = [ 4− 1 4 (x∗k) 2 ] ∆x = [ 4− 1 4 9(k − 1)2 n2 ] 3 n = 12 n − 27k 2 4n3 + 27k 2n3 − 27 4n3 n∑ k=1 f(x∗k)∆x= n∑ k=1 12 n − 27 4n3 n∑ k=1 k2 + 27 2n3 n∑ k=1 k − 27 4n3 n∑ k=1 1 = 12− 27 4n3 · 1 6 n(n+ 1)(2n+ 1) + 27 2n3 n(n+ 1) 2 − 27 4n2 = 12− 9 8 (n+ 1)(2n+ 1) n2 + 27 4n + 27 4n2 − 27 4n2 A = lim n→+∞ [ 12− 9 8 ( 1 + 1 n )( 2 + 1 n )] + 0 + 0− 0 = 12− 9 8 (1)(2) = 39/4 47. ∆x = 1 n , x∗k = 2k − 1 2n f(x∗k)∆x = (2k − 1)2 (2n)2 1 n = k2 n3 − k n3 + 1 4n3 n∑ k=1 f(x∗k)∆x = 1 n3 n∑ k=1 k2 − 1 n3 n∑ k=1 k + 1 4n3 n∑ k=1 1 Using Theorem 6.4.4, A = lim n→+∞ n∑ k=1 f(x∗k)∆x = 1 3 + 0 + 0 = 1 3 Exercise Set 6.4 239 48. ∆x = 2 n , x∗k = −1 + 2k − 1 n f(x∗k)∆x = ( −1 + 2k − 1 n )2 2 n = 8k2 n3 − 8k n3 + 2 n3 − 2 n n∑ k=1 f(x∗k)∆x = 8 n3 n∑ k=1 k2 − 8 n3 n∑ k=1 k + 2 n2 − 2 A = lim n→+∞ n∑ k=1 f(x∗k)∆x = 8 3 + 0 + 0− 2 = 2 3 49. ∆x = 2 n , x∗k = −1 + 2k n f(x∗k)∆x = ( −1 + 2k n ) 2 n = − 2 n + 4 k n2 n∑ k=1 f(x∗k)∆x = −2 + 4 n2 n∑ k=1 k = −2 + 4 n2 n(n+ 1) 2 = −2 + 2 + 2 n A = lim n→+∞ n∑ k=1 f(x∗k)∆x = 0 The area below the x-axis cancels the area above the x-axis. 50. ∆x = 3 n , x∗k = −1 + 3k n f(x∗k)∆x = ( −1 + 3k n ) 3 n = − 3 n + 9 n2 k n∑ k=1 f(x∗k)∆x = −3 + 9 n2 n(n+ 1) 2 A = lim n→+∞ n∑ k=1 f(x∗k)∆x = −3 + 9 2 + 0 = 3 2 The area below the x-axis cancels the area above the x-axis that lies to the right of the line x = 1; the remaining area is a trapezoid of width 1 and heights 1, 2, hence its area is 1 + 2 2 = 3 2 51. ∆x = 2 n , x∗k = 2k n f(x∗k) = [( 2k n )2 − 1 ] 2 n = 8k2 n3 − 2 n n∑ k=1 f(x∗k)∆x = 8 n3 n∑ k=1 k2 − 2 n n∑ k=1 1 = 8 n3 n(n+ 1)(2n+ 1) 6 − 2 A = lim n→+∞ n∑ k=1 f(x∗k)∆x = 16 6 − 2 = 2 3 52. ∆x = 2 n , x∗k = −1 + 2k n f(x∗k)∆x = ( −1 + 2k n )3 2 n = − 2 n + 12 k n2 − 24k 2 n3 + 16 k3 n4 242 Chapter 6 (c) 1 n3 n∑ k=1 k2 = 1 n3 n(n+ 1)(2n+ 1) 6 = 2 6 + 3 6n + 1 6n2 , so lim n→+∞ 1 n3 n∑ k=1 k2 = 1 3 (d) 1 n4 n∑ k=1 k3 = 1 n4 ( n(n+ 1) 2 )2 = 1 4 + 1 2n + 1 4n2 , so lim n→+∞ 1 n4 n∑ k=1 k3 = 1 4 EXERCISE SET 6.5 1. (a) (4/3)(1) + (5/2)(1) + (4)(2) = 71/6 (b) 2 2. (a) ( √ 2/2)(π/2) + (−1)(3π/4) + (0)(π/2) + ( √ 2/2)(π/4) = 3( √ 2− 2)π/8 (b) 3π/4 3. (a) (−9/4)(1) + (3)(2) + (63/16)(1) + (−5)(3) = −117/16 (b) 3 4. (a) (−8)(2) + (0)(1) + (0)(1) + (8)(2) = 0 (b) 2 5. ∫ 2 −1 x2 dx 6. ∫ 2 1 x3dx 7. ∫ 3 −3 4x(1− 3x)dx 8. ∫ π/2 0 sin2 x dx 9. (a) lim max ∆xk→0 n∑ k=1 2x∗k∆xk; a = 1, b = 2 (b) limmax ∆xk→0 n∑ k=1 x∗k x∗k + 1 ∆xk; a = 0, b = 1 10. (a) lim max ∆xk→0 n∑ k=1 √ x∗k∆xk, a = 1, b = 2 (b) lim max ∆xk→0 n∑ k=1 (1 + cosx∗k)∆xk, a = −π/2, b = π/2 11. (a) A = 1 2 (3)(3) = 9/2 3 x y A (b) −A = −1 2 (1)(1 + 2) = −3/2 -2 -1 x y A (c) −A1 +A2 = − 1 2 + 8 = 15/2 -1 4 x y A1 A2 (d) −A1 +A2 = 0 -5 5 x y A1 A2 Exercise Set 6.5 243 12. (a) A = 1 2 (1)(2) = 1 2 1 x y A (b) A = 1 2 (2)(3/2 + 1/2) = 2 -1 1 1 x y A (c) −A = −1 2 (1/2)(1) = −1/4 2 1 x y A (d) A1 −A2 = 1− 1/4 = 3/4 2 1 x y A2 A1 13. (a) A = 2(5) = 10 y x 1 2 5 A (b) 0; A1 = A2 by symmetry 6 c x y A1 A2 (c) A1 +A2 = 1 2 (5)(5/2) + 1 2 (1)(1/2) = 13/2 -1 5 2 x y 3 2 A1 A2 (d) 1 2 [π(1)2] = π/2 y x 1 -1 1 A 14. (a) A = (6)(5) = 30 -10 -5 6 x y A (b) −A1 +A2 = 0 because A1 = A2 by symmetry $ 4 x y A1 A2 (c) A1 +A2 = 1 2 (2)(2) + 1 2 (1)(1) = 5/2 2 2 x y A1 A2 (d) 1 4 π(2)2 = π y x 2 2 A 15. (a) 0.8 (b) −2.6 (c) −1.8 (d) −0.3 244 Chapter 6 16. (a) ∫ 1 0 f(x)dx = ∫ 1 0 2xdx = x2 ]1 0 = 1 (b) ∫ 1 −1 f(x)dx = ∫ 1 −1 2xdx = x2 ]1 −1 = 12 − (−1)2 = 0 (c) ∫ 10 1 f(x)dx = ∫ 10 1 2dx = 2x ]10 1 = 18 (d) ∫ 5 1/2 f(x)dx = ∫ 1 1/2 2xdx+ ∫ 5 1 2dx = x2 ]1 1/2 + 2x   5 1 = 12−(1/2)2+2·5−2·1 = 3/4+8 = 35/4 17. ∫ 2 −1 f(x)dx+ 2 ∫ 2 −1 g(x)dx = 5 + 2(−3) = −1 18. 3 ∫ 4 1 f(x)dx− ∫ 4 1 g(x)dx = 3(2)− 10 = −4 19. ∫ 5 1 f(x)dx = ∫ 5 0 f(x)dx− ∫ 1 0 f(x)dx = 1− (−2) = 3 20. ∫ −2 3 f(x)dx = − ∫ 3 −2 f(x)dx = − [∫ 1 −2 f(x)dx+ ∫ 3 1 f(x)dx ] = −(2− 6) = 4 21. (a) ∫ 1 0 xdx+ 2 ∫ 1 0 √ 1− x2dx = 1/2 + 2(π/4) = (1 + π)/2 (b) 4 ∫ 3 −1 dx− 5 ∫ 3 −1 xdx = 4 · 4− 5(−1/2 + (3 · 3)/2) = −4 22. (a) ∫ 0 −3 2dx+ ∫ 0 −3 √ 9− x2dx = 2 · 3 + (π(3)2)/4 = 6 + 9π/4 (b) ∫ 2 −2 dx− 3 ∫ 2 −2 |x|dx = 4 · 1− 3(2)(2 · 2)/2 = −8 23. (a) √ x > 0, 1− x < 0 on [2, 3] so the integral is negative (b) x2 > 0, 3− cosx > 0 for all x so the integral is positive 24. (a) x4 > 0, √ 3− x > 0 on [−3,−1] so the integral is positive (b) x3 − 9 < 0, |x|+ 1 > 0 on [−2, 2] so the integral is negative 25. ∫ 10 0 √ 25− (x− 5)2dx = π(5)2/2 = 25π/2 26. ∫ 3 0 √ 9− (x− 3)2dx = π(3)2/4 = 9π/4 27. ∫ 1 0 (3x+ 1)dx = 5/2 28. ∫ 2 −2 √ 4− x2dx = π(2)2/2 = 2π 29. (a) f is continuous on [−1, 1] so f is integrable there by Part (a) of Theorem 6.5.8 (b) |f(x)| ≤ 1 so f is bounded on [−1, 1], and f has one point of discontinuity, so by Part (b) of Theorem 6.5.8 f is integrable on [−1, 1] Exercise Set 6.6 247 15. − cos θ ]π/2 −π/2 = 0 16. tan θ ]π/4 0 = 1 17. sinx ]π/4 −π/4 = √ 2 18. ( 1 2 x2 − secx )]1 0 = 3/2− sec(1) 19. 5ex ]3 ln 2 = 5e3 − 5(2) = 5e3 − 10 20. (lnx)/2 ]1 1/2 = (ln 2)/2 21. sin−1 x ]1/√2 0 = sin−1(1/ √ 2)− sin−1 0 = π/4 22. tan−1 x ]1 −1 = tan−1 1− tan−1(−1) = π/4− (−π/4) = π/2 23. sec−1 x ]2 √ 2 = sec−1 2− sec−1 √ 2 = π/3− π/4 = π/12 24. − sec−1 x ]−2/√3 − √ 2 = − sec−1(−2/ √ 3) + sec−1(− √ 2) = −5π/6 + 3π/4 = −π/12 25. ( 6 √ t− 10 3 t3/2 + 2√ t )]4 1 = −55/3 26. ( 8 √ y + 4 3 y3/2 − 2 3y3/2 )]9 4 = 10819/324 27. ( 1 2 x2 − 2 cotx )]π/2 π/6 = π2/9 + 2 √ 3 28. ( a1/2x− 2 3 x3/2 )]4a a = −5 3 a3/2 29. (a) ∫ 3/2 0 (3− 2x)dx+ ∫ 2 3/2 (2x− 3)dx = (3x− x2) ]3/2 0 + (x2 − 3x) ]2 3/2 = 9/4 + 1/4 = 5/2 (b) ∫ π/2 0 cosx dx+ ∫ 3π/4 π/2 (− cosx)dx = sinx ]π/2 0 − sinx ]3π/4 π/2 = 2− √ 2/2 30. (a) ∫ 0 −1 √ 2− x dx+ ∫ 2 0 √ 2 + x dx = −2 3 (2− x)3/2 ]0 −1 + 2 3 (2 + x)3/2 ]2 0 = −2 3 (2 √ 2− 3 √ 3) + 2 3 (8− 2 √ 2) = 2 3 (8− 4 √ 2 + 3 √ 3) (b) ∫ π/6 0 (1/2− sinx) dx+ ∫ π/2 π/6 (sinx− 1/2) dx = (x/2 + cosx) ]π/6 0 − (cosx+ x/2) ]π/2 π/6 = (π/12 + √ 3/2)− 1− π/4 + ( √ 3/2 + π/12) = √ 3− π/12− 1 248 Chapter 6 31. (a) ∫ 0 −1 (1−ex)dx+ ∫ 1 0 (ex−1)dx = (x−ex) ]0 −1 +(ex−x) ]1 0 = −1−(−1−e−1)+e−1−1 = e+1/e−2 (b) ∫ 2 1 2− x x dx+ ∫ 4 2 x− 2 x dx = 2 lnx ]2 1 − 1 + 2− 2 lnx ]4 2 = 2 ln 2 + 1− 2 ln 4 + 2 ln 2 = 1 32. (a) The function f(x) = x2 − 1− 15 x2 + 1 is an even function and changes sign at x = 2, thus ∫ 3 −3 |f(x)| dx = 2 ∫ 3 0 |f(x)| dx = −2 ∫ 2 0 f(x) dx+ 2 ∫ 3 2 f(x) dx = 28 3 − 30 tan−1(3) + 60 tan−1(2) (b) ∫ √3/2 0 ∣∣∣∣ 1√1− x2 − √ 2 ∣∣∣∣ dx = − ∫ √2/2 0 [ 1√ 1− x2 − √ 2 ] dx+ ∫ √3/2 √ 2/2 [ 1√ 1− x2 − √ 2 ] dx = −2 sin−1 (√ 2 2 ) + sin−1 (√ 3 2 ) − √ 2 (√ 3 2 − √ 2 2 ) + 1 = −2π 4 + π 3 − √ 3√ 2 + 2 = 2− √ 3√ 2 − π 6 33. (a) 17/6 (b) F (x) =   1 2 x2, x ≤ 1 1 3 x3 + 1 6 , x > 1 34. (a) ∫ 1 0 √ x dx+ ∫ 4 1 1 x2 dx = 2 3 x3/2 ]1 0 − 1 x ]4 1 = 17/12 (b) F (x) =   2 3 x3/2, x < 1 − 1 x + 5 3 , x ≥ 1 35. 0.665867079; ∫ 3 1 1 x2 dx = − 1 x ]3 1 = 2/3 36. 1.000257067; ∫ π/2 0 sinx dx = − cosx ]π/2 0 = 1 37. 3.106017890; ∫ 1 −1 sec2 x dx = tanx ∣∣∣∣∣ 1 −1 = 2 tan 1 ≈ 3.114815450 38. 1.098242635; ∫ 3 1 1 x dx = lnx ]3 1 = ln 3 ≈ 1.098612289 39. A = ∫ 3 0 (x2 + 1)dx = ( 1 3 x3 + x )]3 0 = 12 40. A = ∫ 2 1 (−x2 + 3x− 2)dx = ( −1 3 x3 + 3 2 x2 − 2x )]2 1 = 1/6 Exercise Set 6.6 249 41. A = ∫ 2π/3 0 3 sinx dx = −3 cosx ]2π/3 0 = 9/2 42. A = − ∫ −1 −2 x3dx = −1 4 x4 ]−1 −2 = 15/4 43. (a) A = ∫ 0.8 0 1√ 1− x2 dx = sin −1 x ]0.8 0 = sin−1(0.8) (b) The calculator was in degree mode instead of radian mode; the correct answer is 0.93. 44. (a) the area is positive (b) ∫ 5 −2 ( 1 100 x3 − 1 20 x2 − 1 25 x+ 1 5 ) dx = ( 1 400 x4 − 1 60 x3 − 1 50 x2 + 1 5 x )]5 −2 = 343 1200 45. (a) the area between the curve and the x-axis breaks into equal parts, one above and one below the x-axis, so the integral is zero (b) ∫ 1 −1 x3dx = 1 4 x4 ]1 −1 = 1 4 (14 − (−1)4) = 0; ∫ π/2 −π/2 sinxdx = − cosx ]π/2 −π/2 = − cos(π/2) + cos(−π/2) = 0 + 0 = 0 (c) The area on the left side of the y-axis is equal to the area on the right side, so∫ a −a f(x)dx = 2 ∫ a 0 f(x)dx (d) ∫ 1 −1 x2dx = 1 3 x3 ]1 −1 = 1 3 (13 − (−1)3) = 2 3 = 2 ∫ 1 0 x2dx; ∫ π/2 −π/2 cosxdx = sinx ]π/2 −π/2 = sin(π/2)− sin(−π/2) = 1 + 1 = 2 = 2 ∫ π/2 0 cosxdx 46. The numerator is an odd function and the denominator is an even function, so the integrand is an odd function and the integral is zero. 47. (a) x3 + 1 (b) F (x) = ( 1 4 t4 + t )]x 1 = 1 4 x4 + x− 5 4 ; F ′(x) = x3 + 1 48. (a) cos 2x (b) F (x) = 1 2 sin 2t ]x π/4 = 1 2 sin 2x− 1 2 , F ′(x) = cos 2x 49. (a) sin √ x (b) ex 2 50. (a) 1 1 + √ x (b) lnx 51. − x cosx 52. |u| 53. F ′(x) = √ 3x2 + 1, F ′′(x) = 3x√ 3x2 + 1 (a) 0 (b) √ 13 (c) 6/ √ 13 54. F ′(x) = tan−1 x, F ′′(x) = 1 1 + x2 (a) 0 (b) π/3 (c) 1/4 252 Chapter 6 5. (a) v(t) = 20 + ∫ t 0 a(u)du; add areas of the small blocks to get v(4) ≈ 20 + 1.4 + 3.0 + 4.7 + 6.2 = 35.3 m/s (b) v(6) = v(4) + ∫ 6 4 a(u)du ≈ 35.3 + 7.5 + 8.6 = 51.4 m/s 6. a > 0 and therefore (Theorem 6.5.6(a)) v > 0, so the particle is always speeding up for 0 < t < 10 7. (a) s(t) = ∫ (t3 − 2t2 + 1)dt = 1 4 t4 − 2 3 t3 + t+ C, s(0) = 1 4 (0)4 − 2 3 (0)3 + 0 + C = 1, C = 1, s(t) = 1 4 t4 − 2 3 t3 + t+ 1 (b) v(t) = ∫ 4 cos 2t dt = 2 sin 2t+ C1, v(0) = 2 sin 0 + C1 = −1, C1 = −1, v(t) = 2 sin 2t− 1, s(t) = ∫ (2 sin 2t− 1)dt = − cos 2t− t+ C2, s(0) = − cos 0− 0 + C2 = −3, C2 = −2, s(t) = − cos 2t− t− 2 8. (a) s(t) = ∫ (1 + sin t)dt = t− cos t+C, s(0) = 0− cos 0 +C = −3, C = −2, s(t) = t− cos t− 2 (b) v(t) = ∫ (t2 − 3t+ 1)dt = 1 3 t3 − 3 2 t2 + t+ C1, v(0) = 1 3 (0)3 − 3 2 (0)2 + 0 + C1 = 0, C1 = 0, v(t) = 1 3 t3 − 3 2 t2 + t, s(t) = ∫ ( 1 3 t3 − 3 2 t2 + t ) dt = 1 12 t4 − 1 2 t3 + 1 2 t2 + C2, s(0) = 1 12 (0)4 − 1 2 (0)3 + 1 2 (0)2 + C2 = 0, C2 = 0, s(t) = 1 12 t4 − 1 2 t3 + 1 2 t2 9. (a) s(t) = ∫ (2t− 3)dt = t2 − 3t+ C, s(1) = (1)2 − 3(1) + C = 5, C = 7, s(t) = t2 − 3t+ 7 (b) v(t) = ∫ cos tdt = sin t+ C1, v(π/2) = 2 = 1 + C1, C1 = 1, v(t) = sin t+ 1, s(t) = ∫ (sin t+ 1)dt = − cos t+ t+ C2, s(π/2) = 0 = π/2 + C2, C2 = −π/2, s(t) = − cos t+ t− π/2 10. (a) s(t) = ∫ t2/3dt = 3 5 t5/3 + C, s(8) = 0 = 3 5 32 + C, C = −96 5 , s(t) = 3 5 t5/3 − 96 5 (b) v(t) = ∫ √ tdt = 2 3 t3/2 + C1, v(4) = 1 = 2 3 8 + C1, C1 = − 13 3 , v(t) = 2 3 t3/2 − 13 3 , s(t) = ∫ ( 2 3 t3/2 − 13 3 ) dt = 4 15 t5/2− 13 3 t+C2, s(4) = −5 = 4 15 32− 13 3 4+C2 = − 44 5 +C2, C2 = 19 5 , s(t) = 4 15 t5/2 − 13 3 t+ 19 5 11. (a) displacement = s(π/2)− s(0) = ∫ π/2 0 sin tdt = − cos t ]π/2 0 = 1 m distance = ∫ π/2 0 | sin t|dt = 1 m Exercise Set 6.7 253 (b) displacement = s(2π)− s(π/2) = ∫ 2π π/2 cos tdt = sin t ]2π π/2 = −1 m distance = ∫ 2π π/2 | cos t|dt = − ∫ 3π/2 π/2 cos tdt+ ∫ 2π 3π/2 cos tdt = 3 m 12. (a) displacement = s(6)− s(0) = ∫ 6 0 (2t− 4)dt = (t2 − 4t) ]6 0 = 12 m distance = ∫ 6 0 |2t−4|dt = ∫ 2 0 (4−2t)dt+ ∫ 6 2 (2t−4)dt = (4t−t2) ]2 0 +(t2−4t) ]6 2 = 20 m (b) displacement= ∫ 5 0 |t− 3|dt = ∫ 3 0 −(t− 3)dt+ ∫ 5 3 (t− 3)dt = 13/2 m distance= ∫ 5 0 |t− 3|dt = 13/2 m 13. (a) v(t)= t3 − 3t2 + 2t = t(t− 1)(t− 2) displacement= ∫ 3 0 (t3 − 3t2 + 2t)dt = 9/4 m distance= ∫ 3 0 |v(t)|dt = ∫ 1 0 v(t)dt+ ∫ 2 1 −v(t)dt+ ∫ 3 2 v(t)dt = 11/4 m (b) displacement= ∫ 3 0 ( √ t− 2)dt = 2 √ 3− 6 m distance= ∫ 3 0 |v(t)|dt = − ∫ 3 0 v(t)dt = 6− 2 √ 3 m 14. (a) displacement= ∫ 3 1 ( 1 2 − 1 t2 )dt = 1/3 m distance= ∫ 3 1 |v(t)|dt = − ∫ √2 1 v(t)dt + ∫ 3 √ 2 v(t)dt = 10/3− 2 √ 2 m (b) displacement= ∫ 9 4 3t−1/2dt = 6 m distance= ∫ 9 4 |v(t)|dt = ∫ 9 4 v(t)dt = 6 m 15. v(t)= −2t+ 3 displacement= ∫ 4 1 (−2t+ 3)dt = −6 m distance= ∫ 4 1 | − 2t+ 3|dt = ∫ 3/2 1 (−2t+ 3)dt+ ∫ 4 3/2 (2t− 3)dt = 13/2 m 16. v(t)= 1 2 t2 − 2t displacement= ∫ 5 1 ( 1 2 t2 − 2t ) dt = −10/3 m distance= ∫ 5 1 ∣∣∣∣12 t2 − 2t ∣∣∣∣ dt = ∫ 4 1 − ( 1 2 t2 − 2t ) dt+ ∫ 5 4 ( 1 2 t2 − 2t ) dt = 17/3 m 254 Chapter 6 17. v(t)= 2 5 √ 5t+ 1 + 8 5 displacement= ∫ 3 0 ( 2 5 √ 5t+ 1 + 8 5 ) dt = 4 75 (5t+ 1)3/2 + 8 5 t ]3 0 = 204/25 m distance= ∫ 3 0 |v(t)|dt = ∫ 3 0 v(t)dt = 204/25 m 18. v(t)= − cos t+ 2 displacement= ∫ π/2 π/4 (− cos t+ 2)dt = (π + √ 2− 2)/2 m distance= ∫ π/2 π/4 | − cos t+ 2|dt = ∫ π/2 π/4 (− cos t+ 2)dt = (π + √ 2− 2)/2 m 19. (a) s = ∫ sin 1 2 πt dt = − 2 π cos 1 2 πt+ C s = 0 when t = 0 which gives C = 2 π so s = − 2 π cos 1 2 πt+ 2 π . a = dv dt = π 2 cos 1 2 πt. When t = 1 : s = 2/π, v = 1, |v| = 1, a = 0. (b) v = −3 ∫ t dt = −3 2 t2 + C1, v = 0 when t = 0 which gives C1 = 0 so v = − 3 2 t2 s = −3 2 ∫ t2dt = −1 2 t3 + C2, s = 1 when t = 0 which gives C2 = 1 so s = − 1 2 t3 + 1. When t = 1 : s = 1/2, v = −3/2, |v| = 3/2, a = −3. 20. (a) negative, because v is decreasing (b) speeding up when av > 0, so 2 < t < 5; slowing down when 1 < t < 2 (c) negative, because the area between the graph of v(t) and the t-axis appears to be greater where v < 0 compared to where v > 0 21. A = A1 +A2 = ∫ 1 0 (1− x2)dx+ ∫ 3 1 (x2 − 1)dx = 2/3 + 20/3 = 22/3 22. A = A1 +A2 = ∫ π 0 sinxdx− ∫ 3π/2 π sinxdx = 2 + 1 = 3 23. A= A1 +A2 = ∫ 0 −1 [ 1− √ x+ 1 ] dx+ ∫ 1 0 [ √ x+ 1− 1] dx = ( x− 2 3 (x+ 1)3/2 )]0 −1 + ( 2 3 (x+ 1)3/2 − x )]1 0 = −2 3 + 1 + 4 √ 2 3 − 1− 2 3 = 4 √ 2− 1 3 24. A = A1 +A2 = ∫ 1 1/2 1− x2 x2 dx+ ∫ 2 1 x2 − 1 x2 dx = ( − 1 x − x ) ∣∣∣∣∣ 1 1/2 + ( x+ 1 x ) ∣∣∣∣∣ 2 1 = −2 + 2 + 1 2 + 2 + 1 2 − 2 = 1 Exercise Set 6.7 257 36. (a) From (9) t = v − v0 a ; from that and (8) s− s0 = v0 v − v0 a + 1 2 a (v − v0)2 a2 ; multiply through by a to get a(s− s0) = v0(v − v0) + 1 2 (v − v0)2 = (v − v0) [ v0 + 1 2 (v − v0) ] = 1 2 (v2 − v20). Thus a = v2 − v20 2(s− s0) . (b) Put the last result of Part (a) into the first equation of Part (a) to obtain t = v − v0 a = (v − v0) 2(s− s0) v2 − v20 = 2(s− s0) v + v0 . (c) From (9) v0 = v − at; use this in (8) to get s− s0 = (v − at)t+ 1 2 at2 = vt− 1 2 at2 This expression contains no v0 terms and so differs from (8). 37. (a) a = −1 mi/h/s = −22/15 ft/s2 (b) a = 30 km/h/min = 1/7200 km/s2 38. Take t = 0 when deceleration begins, then a = −10 so v = −10t + C1, but v = 88 when t = 0 which gives C1 = 88 thus v = −10t+ 88, t ≥ 0 (a) v = 45 mi/h = 66 ft/s, 66 = −10t+ 88, t = 2.2 s (b) v = 0 (the car is stopped) when t = 8.8 s s = ∫ v dt = ∫ (−10t + 88)dt = −5t2 + 88t + C2, and taking s = 0 when t = 0, C2 = 0 so s = −5t2 + 88t. At t = 8.8, s = 387.2. The car travels 387.2 ft before coming to a stop. 39. a = a0 ft/s2, v = a0t+ v0 = a0t+ 132 ft/s, s = a0t2/2 + 132t+ s0 = a0t2/2 + 132t ft; s = 200 ft when v = 88 ft/s. Solve 88 = a0t+ 132 and 200 = a0t2/2 + 132t to get a0 = − 121 5 when t = 20 11 , so s = −12.1t2 + 132t, v = −121 5 t+ 132. (a) a0 = − 121 5 ft/s2 (b) v = 55 mi/h = 242 3 ft/s when t = 70 33 s (c) v = 0 when t = 60 11 s 40. dv/dt = 3, v = 3t+C1, but v = v0 when t = 0 so C1 = v0, v = 3t+ v0. From ds/dt = v = 3t+ v0 we get s = 3t2/2 + v0t+ C2 and, with s = 0 when t = 0, C2 = 0 so s = 3t2/2 + v0t. s = 40 when t = 4 thus 40 = 3(4)2/2 + v0(4), v0 = 4 m/s 41. Suppose s = s0 = 0, v = v0 = 0 at t = t0 = 0; s = s1 = 120, v = v1 at t = t1; and s = s2, v = v2 = 12 at t = t2. From Exercise 36(a), 2.6= a = v21 − v20 2(s1 − s0) , v21 = 2as1 = 5.2(120) = 624. Applying the formula again, −1.5= a = v 2 2 − v21 2(s2 − s1) , v22 = v 2 1 − 3(s2 − s1), so s2 = s1 − (v22 − v21)/3 = 120− (144− 624)/3 = 280 m. 258 Chapter 6 42. a(t) = { 4, t < 2 0, t > 2 , so, with v0 = 0, v(t) = { 4t, t < 2 8, t > 2 and, since s0 = 0, s(t) = { 2t2, t < 2 8t− 8, t > 2 s = 100 when 8t− 8 = 100, t = 108/8 = 13.5 s 43. The truck’s velocity is vT = 50 and its position is sT = 50t+5000. The car’s acceleration is aC = 2, so vC = 2t, sC = t2 (initial position and initial velocity of the car are both zero). sT = sC when 50t+5000 = t2, t2−50t−5000 = (t+50)(t−100) = 0, t = 100 s and sC = sT = t2 = 10, 000 ft. 44. Let t = 0 correspond to the time when the leader is 100 m from the finish line; let s = 0 cor- respond to the finish line. Then vC = 12, sC = 12t − 115; aL = 0.5 for t > 0, vL = 0.5t + 8, sL = 0.25t2 + 8t − 100. sC = 0 at t = 115/12 ≈ 9.58 s, and sL = 0 at t = −16 + 4 √ 41 ≈ 9.61, so the challenger wins. 45. s = 0 and v = 112 when t = 0 so v(t) = −32t+ 112, s(t) = −16t2 + 112t (a) v(3) = 16 ft/s, v(5) = −48 ft/s (b) v = 0 when the projectile is at its maximum height so −32t + 112 = 0, t = 7/2 s, s(7/2) = −16(7/2)2 + 112(7/2) = 196 ft. (c) s = 0 when it reaches the ground so −16t2 + 112t = 0, −16t(t − 7) = 0, t = 0, 7 of which t = 7 is when it is at ground level on its way down. v(7) = −112, |v| = 112 ft/s. 46. s = 112 when t = 0 so s(t) = −16t2 + v0t+ 112. But s = 0 when t = 2 thus −16(2)2 + v0(2) + 112 = 0, v0 = −24 ft/s. 47. (a) s(t) = 0 when it hits the ground, s(t) = −16t2 + 16t = −16t(t− 1) = 0 when t = 1 s. (b) The projectile moves upward until it gets to its highest point where v(t) = 0, v(t) = −32t+ 16 = 0 when t = 1/2 s. 48. (a) s(t) = 0 when the rock hits the ground, s(t) = −16t2 + 555 = 0 when t = √ 555/4 s (b) v(t) = −32t, v( √ 555/4) = −8 √ 555, the speed at impact is 8 √ 555 ft/s 49. (a) s(t) = 0 when the package hits the ground, s(t) = −16t2 + 20t+ 200 = 0 when t = (5 + 5 √ 33)/8 s (b) v(t) = −32t+ 20, v[(5 + 5 √ 33)/8] = −20 √ 33, the speed at impact is 20 √ 33 ft/s 50. (a) s(t) = 0 when the stone hits the ground, s(t) = −16t2 − 96t+ 112 = −16(t2 + 6t− 7) = −16(t+ 7)(t− 1) = 0 when t = 1 s (b) v(t) = −32t− 96, v(1) = −128, the speed at impact is 128 ft/s 51. s(t) = −4.9t2 + 49t+ 150 and v(t) = −9.8t+ 49 (a) the projectile reaches its maximum height when v(t) = 0, −9.8t+ 49 = 0, t = 5 s (b) s(5) = −4.9(5)2 + 49(5) + 150 = 272.5 m (c) the projectile reaches its starting point when s(t) = 150, −4.9t2 + 49t+ 150 = 150, −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 Exercise Set 6.7 259 52. 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 times must be 4 s: h 1080 + √ h 4 = 4, h + 270 √ h = 4320, h + 270 √ h − 4320 = 0, and by the quadratic formula √ h = −270± √ (270)2 + 4(4320) 2 , reject the negative value to get √ h ≈ 15.15, h ≈ 229.5 ft. 53. 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 = √ 30/4.9 and v = −(4.9/3) √ 30/4.9 ≈ −4.04, so the speed of the module upon landing is 4.04 m/s 54. s(t) = − 12gt2 + v0t; s = 1000 when v = 0, so 0 = v = −gt+ v0, t = v0/g, 1000 = s(v0/g) = − 12g(v0/g)2 + v0(v0/g) = 12v20/g, so v20 = 2000g, v0 = √ 2000g. The initial velocity on the Earth would have to be √ 6 times faster than that on the Moon. 55. fave = 1 3− 1 ∫ 3 1 3x dx = 3 4 x2 ]3 1 = 6 56. fave = 1 2− (−1) ∫ 2 −1 x2 dx = 1 9 x3 ]2 −1 = 1 57. fave = 1 π − 0 ∫ π 0 sinx dx = − 1 π cosx ]π 0 = 2/π 58. fave = 1 π − 0 ∫ π 0 cosx dx = 1 π sinx ]π 0 = 0 59. fave = 1 e− 1 ∫ e 1 1 x dx = 1 e− 1(ln e− ln 1) = 1 e− 1 60. fave = 1 ln 5− (−1) ∫ ln 5 −1 exdx = 1 ln 5 + 1 (5− e−1) = 5− e −1 1 + ln 5 61. (a) fave = 1 2− 0 ∫ 2 0 x2dx = 4/3 (b) (x∗)2 = 4/3, x∗ = ±2/ √ 3, but only 2/ √ 3 is in [0, 2] (c) 2 4 x y 2 3 62. (a) fave = 1 4− 0 ∫ 4 0 2x dx = 4 (b) 2x∗ = 4, x∗ = 2 262 Chapter 6 15. u = ex + 4, du = exdx, u = e− ln 3 + 4 = 1 3 + 4 = 13 3 when x = − ln 3, u = eln 3 + 4 = 3 + 4 = 7 when x = ln 3, ∫ 7 13/3 1 u du = lnu ]7 13/3 = ln(7)− ln(13/3) = ln(21/13) 16. u = 3− 4ex, du = −4exdx, u = −1 when x = 0, u = −17 when x = ln 5 −1 4 ∫ −17 −1 u du = −1 8 u2 ]−17 −1 = −36 17. u = √ x, 2 ∫ √3 1 1 u2 + 1 du = 2 tan−1 u ]√3 1 = 2(tan−1 √ 3− tan−1 1) = 2(π/3− π/4) = π/6 18. u = e−x, − ∫ √3/2 1/2 1√ 1− u2 du = − sin −1 u ]√3/2 1/2 = − sin−1 √ 3 2 + sin−1 1 2 = −π 3 + π 6 = −π 6 19. 1 3 ∫ 5 0 √ 25− u2 du = 1 3 [ 1 4 π(5)2 ] = 25 12 π 20. 1 2 ∫ 4 0 √ 16− u2 du = 1 2 [ 1 4 π(4)2 ] = 2π 21. −1 2 ∫ 0 1 √ 1− u2 du = 1 2 ∫ 1 0 √ 1− u2 du = 1 2 · 1 4 [π(1)2] = π/8 22. ∫ 6 −6 √ 36− u2du = π(6)2/2 = 18π 23. ∫ 1 0 sinπxdx = − 1 π cosπx ]1 0 = − 1 π (−1− 1) = 2/π 24. A = ∫ π/8 0 3 cos 2x dx = 3 2 sin 2x ]π/8 0 = 3 √ 2/4 25. ∫ 7 3 (x+ 5)−2dx = −(x+ 5)−1 ]7 3 = − 1 12 + 1 8 = 1 24 26. A = ∫ 1 0 dx (3x+ 1)2 = − 1 3(3x+ 1) ]1 0 = 1 4 27. A = ∫ 1/6 0 1√ 1− 9x2 dx = 1 3 ∫ 1/2 0 1√ 1− u2 du = 1 3 sin−1 u ]1/2 0 = π/18 28. x = sin y, A = ∫ π/2 0 sin y dy = − cos y ]π/2 0 = 1 29. 1 2− 0 ∫ 2 0 x (5x2 + 1)2 dx = −1 2 1 10 1 5x2 + 1 ∣∣∣∣∣ 2 0 = 1 21 30. fave = 1 1/4− (−1/4) ∫ 1/4 −1/4 sec2 πxdx = 2 π tanπx ]1/4 −1/4 = 4 π Exercise Set 6.8 263 31. fave = 1 4 ∫ 4 0 e−2x dx = −1 8 e−2x ]4 0 = 1 8 (1− e−8) 32. fave = 2 ln 3 ∫ 1 1/ √ 3 du 1 + u2 = 2 ln 3 tan−1 u ]1 1/ √ 3 = 2 ln 3 (π 4 − π 6 ) = π 6 ln 3 33. 2 3 (3x+ 1)1/2 ]1 0 = 2/3 34. 2 15 (5x− 1)3/2 ]2 1 = 38/15 35. 2 3 (x3 + 9)1/2 ]1 −1 = 2 3 ( √ 10− 2 √ 2) 36. 1 10 (t3 + 1)20 ]0 −1 = 1/10 37. u = x2 + 4x+ 7, 1 2 ∫ 28 12 u−1/2du = u1/2 ]28 12 = √ 28− √ 12 = 2( √ 7− √ 3) 38. ∫ 2 1 1 (x− 3)2 dx = − 1 x− 3 ]2 1 = 1/2 39. 1 2 sin2 x ]π/4 −3π/4 = 0 40. 2 3 (tanx)3/2 ]π/4 0 = 2/3 41. 5 2 sin(x2) ]√π 0 = 0 42. u = √ x, 2 ∫ 2π π sinu du = −2 cosu ]2π π = −4 43. u = 3θ, 1 3 ∫ π/3 π/4 sec2 u du = 1 3 tanu ]π/3 π/4 = ( √ 3− 1)/3 44. u = sin 3θ, 1 3 ∫ −1 0 u2du = 1 9 u3 ]−1 0 = −1/9 45. u = 4− 3y, y = 1 3 (4− u), dy = −1 3 du − 1 27 ∫ 1 4 16− 8u+ u2 u1/2 du= 1 27 ∫ 4 1 (16u−1/2 − 8u1/2 + u3/2)du = 1 27 [ 32u1/2 − 16 3 u3/2 + 2 5 u5/2 ]4 1 = 106/405 46. u = 5 + x, ∫ 9 4 u− 5√ u du = ∫ 9 4 (u1/2 − 5u−1/2)du = 2 3 u3/2 − 10u1/2 ]9 4 = 8/3 47. ln(x+ e) ]e 0 = ln(2e)− ln e = ln 2 48. −1 2 e−x 2 ]√2 1 = (e−1 − e−2)/2 49. u = √ 3x2, 1 2 √ 3 ∫ √3 0 1√ 4− u2 du = 1 2 √ 3 sin−1 u 2 ]√3 0 = 1 2 √ 3 (π 3 ) = π 6 √ 3 264 Chapter 6 50. u = √ x, 2 ∫ √2 1 1√ 4− u2 du = 2 sin −1 u 2 ]√2 1 = 2(π/4− π/6) = π/6 51. u = 3x, 1 3 ∫ 2√3 0 1 4 + u2 du = 1 6 tan−1 u 2 ]2√3 0 = 1 6 π 3 = π 18 52. u = x2, 1 2 ∫ 3 1 1 3 + u2 du = 1 2 √ 3 tan−1 u√ 3 ]3 1 = 1 2 √ 3 (π/3− π/6) = π 12 √ 3 53. (b) ∫ π/6 0 sin4 x(1− sin2 x) cosx dx = ( 1 5 sin5 x− 1 7 sin7 x ) ∣∣∣∣∣ π/6 0 = 1 160 − 1 896 = 23 4480 54. (b) ∫ π/4 −π/4 tan2 x(sec2 x− 1) dx= 1 3 tan3 x ∣∣∣∣∣ π/4 −π/4 − ∫ π/4 −π/4 (sec2 x− 1) dx = 2 3 + (− tanx+ x) ∣∣∣∣∣ π/4 −π/4 = 2 3 − 2 + π 2 = −4 3 + π 2 55. (a) u = 3x+ 1, 1 3 ∫ 4 1 f(u)du = 5/3 (b) u = 3x, 1 3 ∫ 9 0 f(u)du = 5/3 (c) u = x2, 1/2 ∫ 0 4 f(u)du = −1/2 ∫ 4 0 f(u)du = −1/2 56. u = 1− x, ∫ 1 0 xm(1− x)ndx = − ∫ 0 1 (1− u)mundu = ∫ 1 0 un(1− u)mdu = ∫ 1 0 xn(1− x)mdx 57. sinx = cos(π/2− x),∫ π/2 0 sinn x dx= ∫ π/2 0 cosn(π/2− x)dx = − ∫ 0 π/2 cosn u du (u = π/2− x) = ∫ π/2 0 cosn u du = ∫ π/2 0 cosn x dx (by replacing u by x) 58. u = 1− x, − ∫ 0 1 (1− u)undu = ∫ 1 0 (1− u)undu = ∫ 1 0 (un − un+1)du = 1 n+ 1 − 1 n+ 2 = 1 (n+ 1)(n+ 2) 59. y(t) = (802.137) ∫ e1.528tdt = 524.959e1.528t + C; y(0) = 750 = 524.959 + C, C = 225.041, y(t) = 524.959e1.528t + 225.041, y(12) = 48, 233, 500, 000 60. Vave = 275000 10− 0 ∫ 10 0 e−0.17tdt = −161764.7059e−0.17t ]10 0 = $132, 212.96 61. s(t) = ∫ (25 + 10e−0.05t)dt = 25t− 200e−0.05t + C (a) s(10)− s(0) = 250− 200(e−0.5 − 1) = 450− 200/√e ≈ 328.69 ft (b) yes; without it the distance would have been 250 ft Exercise Set 6.9 267 8. (a) f(ln 3) = e−2 ln 3 = eln(1/9) = 1/9 (b) f(ln 2) = eln 2 + 3e− ln 2 = 2 + 3eln(1/2) = 2 + 3/2 = 7/2 9. (a) 3π = eπ ln 3 (b) 2 √ 2 = e √ 2 ln 2 10. (a) π−x = e−x lnπ (b) x2x = e2x ln x 11. (a) lim x→+∞ [( 1 + 1 x )x]2 = [ lim x→+∞ ( 1 + 1 x )x]2 = e2 (b) y = 2x, lim y→0 (1 + y)2/y = lim y→0 [ (1 + y)1/y ]2 = e2 12. (a) y = 3x, lim y→+∞ ( 1 + 1 y )y/3 = lim y→+∞ [( 1 + 1 y )y]1/3 = [ lim y→+∞ ( 1 + 1 y )y]1/3 = e1/3 (b) lim x→0 (1 + x)1/3x = lim x→0 [ (1 + x)1/x ]1/3 = e1/3 13. g′(x) = x2 − x 14. g′(x) = 1− cosx 15. (a) 1 x3 (3x2) = 3 x (b) eln x 1 x = 1 16. (a) 2x √ x2 + 1 (b) − ( 1 x2 ) sin ( 1 x ) 17. F ′(x) = cosx x2 + 3 , F ′′(x) = −(x2 + 3) sin x− 2x cosx (x2 + 3)2 (a) 0 (b) 1/3 (c) 0 18. F ′(x) = √ 3x2 + 1, F ′′(x) = 3x√ 3x2 + 1 (a) 0 (b) √ 13 (c) 6/ √ 13 19. (a) d dx ∫ x2 1 t √ 1 + tdt = x2 √ 1 + x2(2x) = 2x3 √ 1 + x2 (b) ∫ x2 1 t √ 1 + tdt = −2 3 (x2 + 1)3/2 + 2 5 (x2 + 1)5/2 − 4 √ 2 15 20. (a) d dx ∫ a x f(t)dt = − d dx ∫ x a f(t)dt = −f(x) (b) d dx ∫ a g(x) f(t)dt = − d dx ∫ g(x) a f(t)dt = −f(g(x))g′(x) 21. (a) − sinx2 (b) − tan 2 x 1 + tan2 x sec2 x = − tan2 x 22. (a) −(x2 + 1)40 (b) − cos3 ( 1 x )( − 1 x2 ) = cos3(1/x) x2 268 Chapter 6 23. −3 3x− 1 9x2 + 1 + 2x x2 − 1 x4 + 1 24. If f is continuous on an open interval I and g(x), h(x), and a are in I then∫ g(x) h(x) f(t)dt = ∫ a h(x) f(t)dt+ ∫ g(x) a f(t)dt = − ∫ h(x) a f(t)dt+ ∫ g(x) a f(t)dt so d dx ∫ g(x) h(x) f(t)dt = −f(h(x))h′(x) + f(g(x))g′(x) 25. (a) sin2(x3)(3x2)− sin2(x2)(2x) = 3x2 sin2(x3)− 2x sin2(x2) (b) 1 1 + x (1)− 1 1− x (−1) = 2 1− x2 26. F ′(x) = 1 3x (3)− 1 x (1) = 0 so F (x) is constant on (0,+∞). F (1) = ln 3 so F (x) = ln 3 for all x > 0. 27. from geometry, ∫ 3 0 f(t)dt = 0, ∫ 5 3 f(t)dt = 6, ∫ 7 5 f(t)dt = 0; and ∫ 10 7 f(t)dt = ∫ 10 7 (4t− 37)/3dt = −3 (a) F (0) = 0, F (3) = 0, F (5) = 6, F (7) = 6, F (10) = 3 (b) F is increasing where F ′ = f is positive, so on [3/2, 6] and [37/4, 10], decreasing on [0, 3/2] and [6, 37/4] (c) critical points when F ′(x) = f(x) = 0, so x = 3/2, 6, 37/4; maximum 15/2 at x = 6, minimum −9/4 at x = 3/2 (d) F(x) x -2 2 4 6 2 4 6 8 10 28. fave = 1 10− 0 ∫ 10 0 f(t)dt = 1 10 F (10) = 0.3 29. x < 0 : F (x) = ∫ x −1 (−t)dt = −1 2 t2 ]x −1 = 1 2 (1− x2), x ≥ 0 : F (x) = ∫ 0 −1 (−t)dt+ ∫ x 0 t dt = 1 2 + 1 2 x2; F (x) = { (1− x2)/2, x < 0 (1 + x2)/2, x ≥ 0 30. 0 ≤ x ≤ 2 : F (x) = ∫ x 0 t dt = 1 2 x2, x > 2 : F (x) = ∫ 2 0 t dt+ ∫ x 2 2 dt = 2 + 2(x− 2) = 2x− 2; F (x) = { x2/2, 0 ≤ x ≤ 2 2x− 2, x > 2 Exercise Set 6.9 269 31. y(x) = 2 + ∫ x 1 t1/3dt = 2 + 3 4 t4/3 ]x 1 = 5 4 + 3 4 x4/3 32. y(x) = ∫ x 1 (t1/2 + t−1/2)dt = 2 3 x3/2 − 2 3 + 2x1/2 − 2 = 2 3 x3/2 + 2x1/2 − 8 3 33. y(x) = 1 + ∫ x π/4 (sec2 t− sin t)dt = tanx+ cosx− √ 2/2 34. y(x) = ∫ x 0 tet 2 dt = 1 2 e−x 2 − 1 2 35. P (x) = P0 + ∫ x 0 r(t)dt individuals 36. s(T ) = s1 + ∫ T 1 v(t)dt 37. II has a minimum at x = 12, and I has a zero there, so I could be the derivative of II; on the other hand I has a minimum near x = 1/3, but II is not zero there, so II could not be the derivative of I, so I is the graph of f(x) and II is the graph of ∫ x 0 f(t) dt. 38. (b) lim k→0 1 k (xk − 1) = d dt xt ] t=0 = lnx 39. (a) where f(t) = 0; by the First Derivative Test, at t = 3 (b) where f(t) = 0; by the First Derivative Test, at t = 1, 5 (c) at t = 0, 1 or 5; from the graph it is evident that it is at t = 5 (d) at t = 0, 3 or 5; from the graph it is evident that it is at t = 3 (e) F is concave up when F ′′ = f ′ is positive, i.e. where f is increasing, so on (0, 1/2) and (2, 4); it is concave down on (1/2, 2) and (4, 5) (f) F(x) x -1 -0.5 0.5 1 1 2 3 5 40. (a) x -1 1 -4 -2 2 4 erf(x) (c) erf ′(x) > 0 for all x, so there are no relative extrema (e) erf ′′(x) = −4xe−x2/√π changes sign only at x = 0 so that is the only point of inflection (g) lim x→+∞ erf(x) = +1, lim x→−∞ erf(x) = −1 272 Chapter 6 14. (a) 1 n n∑ k=1 √ k n = n∑ k=1 f(x∗k)∆x where f(x) = √ x, x∗k = k/n, and ∆x = 1/n for 0 ≤ x ≤ 1. Thus lim n→+∞ 1 n n∑ k=1 √ k n = ∫ 1 0 x1/2dx = 2 3 (b) 1 n n∑ k=1 ( k n )4 = n∑ k=1 f(x∗k)∆x where f(x) = x 4, x∗k = k/n, and ∆x = 1/n for 0 ≤ x ≤ 1. Thus lim n→+∞ 1 n n∑ k=1 ( k n )4 = ∫ 1 0 x4dx = 1 5 (c) n∑ k=1 ek/n n = n∑ k=1 f(x∗k)∆x where f(x) = e x, x∗k = k/n, and ∆x = 1/n for 0 ≤ x ≤ 1. Thus lim n→+∞ n∑ k=1 ek/n n = lim n→+∞ n∑ k=1 f(x∗k)∆x = ∫ 1 0 exdx = e− 1. 15. Since f(x) = 1 x is positive and increasing on the interval [1, 2], the left endpoint approximation overestimates the integral of 1 x and the right endpoint approximation underestimates it. (a) For n = 5 this becomes 0.2 [ 1 1.2 + 1 1.4 + 1 1.6 + 1 1.8 + 1 2.0 ] < ∫ 2 1 1 x dx < 0.2 [ 1 1.0 + 1 1.2 + 1 1.4 + 1 1.6 + 1 1.8 ] (b) For general n the left endpoint approximation to ∫ 2 1 1 x dx = ln 2 is 1 n n∑ k=1 1 1 + (k − 1)/n = n∑ k=1 1 n+ k − 1 = n−1∑ k=0 1 n+ k and the right endpoint approximation is n∑ k=1 1 n+ k . This yields n∑ k=1 1 n+ k < ∫ 2 1 1 x dx < n−1∑ k=0 1 n+ k which is the desired inequality. (c) By telescoping, the difference is 1 n − 1 2n = 1 2n so 1 2n ≤ 0.1, n ≥ 5 (d) n ≥ 1, 000 16. The direction field is clearly an even function, which means that the solution is even, its derivative is odd. Since sinx is periodic and the direction field is not, that eliminates all but x, the solution of which is the family y = x2/2 + C. 17. (a) 1 · 2 + 2 · 3 + · · ·+ n(n+ 1)= n∑ k=1 k(k + 1) = n∑ k=1 k2 + n∑ k=1 k = 1 6 n(n+ 1)(2n+ 1) + 1 2 n(n+ 1) = 1 3 n(n+ 1)(n+ 2) (b) n−1∑ k=1 ( 9 n − k n2 ) = 9 n n−1∑ k=1 1− 1 n2 n−1∑ k=1 k = 9 n (n− 1)− 1 n2 · 1 2 (n− 1)(n) = 17 2 ( n− 1 n ) ; lim n→+∞ 17 2 ( n− 1 n ) = 17 2 (c) 3∑ i=1   2∑ j=1 i+ 2∑ j=1 j   = 3∑ i=1 [ 2i+ 1 2 (2)(3) ] = 2 3∑ i=1 i+ 3∑ i=1 3 = 2 · 1 2 (3)(4) + (3)(3) = 21 Chapter 6 Supplementary Exercises 273 18. (a) 14∑ k=0 (k + 4)(k + 1) (b) 19∑ k=5 (k − 1)(k − 4) 19. For 1 ≤ k ≤ n the k-th L-shaped strip consists of the corner square, a strip above and a strip to the right for a combined area of 1+(k−1)+(k−1) = 2k−1, so the total area is n∑ k=1 (2k−1) = n2. 20. 1 + 3 + 5 + · · ·+ (2n− 1) = n∑ k=1 (2k − 1) = 2 n∑ k=1 k − n∑ k=1 1 = 2 · 1 2 n(n+ 1)− n = n2 21. (35 − 34) + (36 − 35) + · · ·+ (317 − 316) = 317 − 34 22. ( 1− 1 2 ) + ( 1 2 − 1 3 ) + · · ·+ ( 1 50 − 1 51 ) = 50 51 23. ( 1 22 − 1 12 ) + ( 1 32 − 1 22 ) + · · ·+ ( 1 202 − 1 192 ) = 1 202 − 1 = −399 400 24. (22 − 2) + (23 − 22) + · · ·+ (2101 − 2100) = 2101 − 2 25. (a) n∑ k=1 1 (2k − 1)(2k + 1) = 1 2 n∑ k=1 ( 1 2k − 1 − 1 2k + 1 ) = 1 2 [( 1− 1 3 ) + ( 1 3 − 1 5 ) + ( 1 5 − 1 7 ) + · · ·+ ( 1 2n− 1 − 1 2n+ 1 )] = 1 2 [ 1− 1 2n+ 1 ] = n 2n+ 1 (b) lim n→+∞ n 2n+ 1 = 1 2 26. (a) n∑ k=1 1 k(k + 1) = n∑ k=1 ( 1 k − 1 k + 1 ) = ( 1− 1 2 ) + ( 1 2 − 1 3 ) + ( 1 3 − 1 4 ) + · · ·+ ( 1 n − 1 n+ 1 ) = 1− 1 n+ 1 = n n+ 1 (b) lim n→+∞ n n+ 1 = 1 27. n∑ i=1 (xi − x̄) = n∑ i=1 xi − n∑ i=1 x̄ = n∑ i=1 xi − nx̄ but x̄ = 1 n n∑ i=1 xi thus n∑ i=1 xi = nx̄ so n∑ i=1 (xi − x̄) = nx̄− nx̄ = 0 28. S − rS= n∑ k=0 ark − n∑ k=0 ark+1 = (a+ ar + ar2 + · · ·+ arn)− (ar + ar2 + ar3 + · · ·+ arn+1) = a− arn+1 = a(1− rn+1) so (1− r)S = a(1− rn+1), hence S = a(1− rn+1)/(1− r) 274 Chapter 6 29. (a) 19∑ k=0 3k+1 = 19∑ k=0 3(3k) = 3(1− 320) 1− 3 = 3 2 (320 − 1) (b) 25∑ k=0 2k+5 = 25∑ k=0 252k = 25(1− 226) 1− 2 = 2 31 − 25 (c) 100∑ k=0 (−1) (−1 2 )k = (−1)(1− (−1/2)101) 1− (−1/2) = − 2 3 (1 + 1/2101) 30. (a) 1.999023438, 1.999999046, 2.000000000; 2 (b) 2.831059456, 2.990486364, 2.999998301; 3 31. (a) If u = secx, du = secx tanxdx, ∫ sec2 x tanxdx = ∫ udu = u2/2 + C1 = (sec2 x)/2 + C1; if u = tanx, du = sec2 xdx, ∫ sec2 x tanxdx = ∫ udu = u2/2 + C2 = (tan2 x)/2 + C2. (b) They are equal only if sec2 x and tan2 x differ by a constant, which is true. 32. 1 2 sec2 x ]π/4 0 = 1 2 (2− 1) = 1/2 and 12 tan2 x ]π/4 0 = 1 2 (1− 0) = 1/2 33. ∫ √ 1 + x−2/3dx = ∫ x−1/3 √ x2/3 + 1dx; u = x2/3 + 1, du = 2 3 x−1/3dx 3 2 ∫ u1/2du = u3/2 + C = (x2/3 + 1)3/2 + C 34. (a) ∫ b a n∑ k=1 fk(x)dx = n∑ k=1 ∫ b a fk(x)dx (b) yes; substitute ckfk(x) for fk(x) in part (a), and then use ∫ b a ckfk(x)dx = ck ∫ b a fk(x)dx from Theorem 6.5.4 35. left endpoints: x∗k = 1, 2, 3, 4; 4∑ k=1 f(x∗k)∆x = (2 + 3 + 2 + 1)(1) = 8 right endpoints: x∗k = 2, 3, 4, 5; 4∑ k=1 f(x∗k)∆x = (3 + 2 + 1 + 2)(1) = 8 36. (a) x∗k = 0, 1, 2, 3, 4 4∑ k=1 f(x∗k)∆x = ( e0 + e1 + e2 + e3 + e4 ) (1) = (1− e5)/(1− e) = 85.791 (b) x∗k = 1, 2, 3, 4, 5 4∑ k=1 f(x∗k)∆x = ( e1 + e2 + e3 + e4 + e5 ) (1) = e(1− e5)/(1− e) = 233.204 (c) x∗k = 1/2, 3/2, 5/2, 7/2, 9/2 4∑ k=1 f(x∗k)∆x = ( e1/2 + e3/2 + e5/2 + e7/2 + e9/2 ) (1) = e1/2(1− e5)/(1− e) = 141.446 37. fave = 1 e− 1 ∫ e 1 1 x dx = 1 e− 1 lnx ]e 1 = 1 e− 1 ; 1 x∗ = 1 e− 1 , x ∗ = e− 1