X 2 1 1 n 2 1 n 2 1 1 n 2 321 31 212

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Unformatted text preview: = [cos(π/8) + cos(3π/8) − cos(3π/8) − cos(π/8)] (π/4) = 0. 4 f (x∗ )∆x = k (c) Right endpoints: k=1 √ 2/2 + 0 − √ 2/2 − 1 (π/4) = −π/4. 31. (a) 0.718771403, 0.705803382, 0.698172179. (b) 0.692835360, 0.693069098, 0.693134682. (c) 0.668771403, 0.680803382, 0.688172179. 33. (a) 4.884074734, 5.115572731, 5.248762738. (b) 5.34707029, 5.338362719, 5.334644416. (c) 5.684074734, 5.515572731, 5.408762738. 35. ∆x = 3∗ 3 1 1 , xk = 1 + k ; f (x∗ )∆x = x∗ ∆x = k k n n 2 2 n f (x∗ )∆x = k k=1 A = lim n→+∞ 3 2 n k=1 3 3 1+ 2 2 1 + n 1+ n k=1 1 n 1+ 3 k n 3 31 3 = + 2k , n 2n n 3 3 31 3 3n+1 k= 1 + 2 · n(n + 1) = 1+ , n2 2 n2 2 2n = 3 2 1+ 3 2 = 15 . 4 136 Chapter 5 37. ∆x = 3∗ 3 , x = 0 + k ; f (x∗ )∆x = k nk n n n f (x∗ )∆x = k k=1 9−9 k=1 27 − A = lim n→+∞ 39. ∆x = 27 n3 k2 n2 9−9 3 27 = n n 3 , n n 1− k=1 n k=1 k2 n2 1 3 k 2 = 27 − 27 k2 n2 = 27 − f (x∗ )∆x k k=1 32 = n n k=1 6 1+ n n k=1 12 k+ 2 n n k2 , k=1 = 18. 4 4 4∗ , x = 2 + k ; f (x∗ )∆x = (x∗ )3 ∆x = 2 + k k k nk n n n 27 n3 n 8 k+ 3 n 3 3 = 32 8 6 12 1 + k + 2 k2 + 3 k3 , n n n n n 2 k=1 4 32 2 = 1+ k n n n k3 = k=1 32 61 12 1 81 = n + · n(n + 1) + 2 · n(n + 1)(2n + 1) + 3 · n2 (n + 1)2 = n n2 n6 n4 = 32 1 + 3 (n + 1)(2n + 1) (n + 1)2 n+1 +2 +2 , n n2 n2 A = lim 32 1 + 3 1 + n→+∞ 41. ∆x = +2 1+ 1 n 2+ 1 n +2 1+ 1 n 2 = 32[1 + 3(1) + 2(1)(2) + 2(1)2 ] = 320. 3∗ 13 9 3 1 1 33 , x = 1 + (k − 1) ; f (x∗ )∆x = x∗ ∆x = = + (k − 1) 2 , 1 + (k − 1) k nk n 2k 2 nn 2n n n f (x∗ )∆x = k k=1 1 2 39 + 24 A = lim n→+∞ 43. ∆x = 1 n n k=1 3 9 + n n2 1− 1 n n 1 3 9n−1 91 , 3 + 2 · (n − 1)n = + 2 n2 2 4n (k − 1) = k=1 = 15 39 += . 24 4 3∗ 3 (k − 1)2 3 , xk = 0 + (k − 1) ; f (x∗ )∆x = 9 − 9 , k n n n2 n n n f (x∗ )∆x k = k=1 k=1 (k − 1)2 3 27 9−9 = n2 n n A = lim = 27 − 27 n→+∞ 1 3 n k=1 (k − 1)2 1− n2 27 = 27 − 3 n n 54 k+ 3 n n 2 k=1 k=1 k− 27 , n2 + 0 + 0 = 18. 4(n − 1) 4n 2 6 10 4n − 6 4n − 2 48 , ,..., , = 4, and midpoints , , , . . . , , . Approximate the area with nn n n nnn n n 4k − 2 4 16 n(n + 1) 2 =22 − n → 16 (exact) as n → +∞. n n n 2 45. Endpoints 0, n the sum k=1 47. ∆x = 1∗ 2k − 1 (2k − 1)2 1 k2 k 1 , xk = ; f (x∗ )∆x = = 3 − 3 + 3, k 2 n 2n (2n) n n n 4n n f (x∗ )∆x = k Using Theorem 5.4.4, A = lim n→+∞ 49. ∆x = k=1 2∗ 2k , x = −1 + ; f (x∗ )∆x = k nk n 4 n(n + 1) 2 = −2 + 2 + , A = lim 2 n→+∞ n 2 n −1 + f (x∗ )∆x = k k=1 1 n3 n k=1 k2 − 1 n3 n k+ k=1 1 4n3 n 1. k=1 1 1 +0+0= . 3 3 2k n 2 2 k = − + 4 2, n n n n f (x∗ )∆x = 0. k k=1 n n k=1 f (x∗ )∆x = −2 + k 4 n2 n k=1 k = −2 + Exercise Set 5.4 137 The area below the x-axis cancels the area above the x-axis. 51. ∆x = 2k 2∗ ,x = ; f ( x∗ ) = k nk n n f (x∗ )∆x k 2, A = lim n→+∞ k=1 2k n 2 −1 f (x∗ )∆x = k k=1 8 n3 n k=1 k2 − n 2 n 1= k=1 1 1+ n b∗ b b4 , xk = k ; f (x∗ )∆x = (x∗ )3 ∆x = 4 k 3 , k k n n n 2 n f (x∗ )∆x = k k=1 b−a ∗ b−a , xk = a + k ; f (x∗ )∆x = (x∗ )3 ∆x = k k n n 2 2 3 3a(b − a) 2 (b − a) 3 b − a 3 3a (b − a) k+ a+ k+ k, n n n2 n3 k=1 b4 n4 n k3 = k=1 = b4 /4. (b) First Method (tedious): ∆x = n 8 n(n + 1)(2n + 1) − n3 6 16 2 = −2= . 6 3 53. (a) With x∗ as the right endpoint, ∆x = k b4 b4 (n + 1)2 , A = lim n→+∞ 4 4 n2 n 2 8k 2 2 = 3− , n n n a+ b−a k n 3 b−a = n 3 n+1 1 (n + 1)(2n + 1) 1 (n + 1)2 f (x∗ )∆x = (b − a) a3 + a2 (b − a) + a(b − a)2 + (b − a)3 , k 2 n 2 n2 4 n2 n A = lim n→+∞ k=1 1 1 3 f (x∗ )∆x = (b − a) a3 + a2 (b − a) + a(b − a)2 + (b − a)3 = (b4 − a4 ). k 2 4 4 Alternative method: Apply part (a) of the Exercise to the interval [0, a] and observe that the area under the curve 1 1 and above that interval is given by a4 . Apply part (a) again, this time to the interval [0, b] and obtain b4 . Now 4 4 1 subtract to obtain the correct area and the formula A = (b4 − a4 ). 4 m 55. If n = 2m then 2m + 2(m − 1) + · · · + 2 · 2 + 2 = 2 k=1 k = 2· m(m + 1) n2 + 2 n = m(m + 1) = ; if n = 2m + 1 2 4 m+1 then (2m + 1) + (2m − 1) + · · · + 5 + 3 + 1 = n2 + 2 n + 1 (m + 1) = . 4 k=1 m+1 (2k − 1) = 2 k=1 m+1 k− k=1 1 = 2· (m + 1)(m + 2) − (m + 1) = 2 2 57. (35 − 34 ) + (36 − 35 ) + · · · + (317 − 316 ) = 317 − 34 . 1 1 −2 2 2 1 59. n 61. (a) k=1 = 1 2 (b) + 1 1 −2 2 3 2 + ··· + 1 1 = (2k − 1)(2k + 1) 2 1− 1 3 + 11 − 35 + n k=1 1 1 −2 2 20 19 = 1 399 −1=− . 2 20 400 1 1 − 2k − 1 2k + 1 11 − 57 + ··· + = 1 1 − 2n − 1 2n + 1 = 1 1 n 1− = . 2 2n + 1 2n + 1 n 1 =. n→+∞ 2n + 1 2 lim n n n n 1 63. (xi − x) = ¯ xi − x= ¯ xi − nx, but x = ¯ ¯ n i=1 i=1 i=1 i=1 65. Both are valid. n n xi , thus i=1 n xi = nx, so ¯ i=1 i=1 (xi − x) = nx − nx = 0. ¯ ¯ ¯ EXERCISE SET 5.5 1. (a) (4/3)(1) + (5(4/3)(1) + (5/2)(1) + 6 (b) 1. (a) /2)(1) + (4)(2) = 71/ (4)(2) = 71/6(b)2 2 (b) 2 1. (a) (4/3)(1) + (5/2)(1) + (4)(2) = 71/6 1. (a) (4/3)(1) + (5/2)(1) + (4)(2) = 71/6 (b) 2 √ √ √ √ π/ √ √ √ 2. (a) (√ 2/2)(π /2) (√(−2)(ππ2) ++−1)(3/2) ++√ 2/2)(2) + (= 2/√...
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