Ch 5 solutions

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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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/√...
View Full Document

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern