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SalasSV_11_04_ex_ans

# 2n c n 1x dx ln 2n ln n ln 2 section 115

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Unformatted text preview: {n, n + 1, n + 2, . . . , 2n} be the regular partition of the interval [n, 2n] into n subintervals; form the Riemann sum by evaluating f (x) = 1/x at the right endpoints (= the lower sum). 2n (c) n (1/x) dx = ln (2n) − ln (n) = ln 2. SECTION 11.5 1. −1 + x + 1 x2 − 2 9. P0 (x) = 1, m 2k 14 x 24 3. − 1 x2 − 2 14 x 12 5. 1 − x + x2 − x3 + x4 − x5 P3 (x) = 1 − x + 3x2 + 5x3 k 7. x + 1 x3 + 3 n 25 x 15 P1 (x) = 1 − x, P2 (x) = 1 − x + 3x2 , n 11. k =0 ( − 1)k xk k! 23. |x| < 1. 513 13. k =0 n x where m = and n is even (2k !) 2 (79/48 ∼ 1. 646) = |c| < |x| 35. 27. 5/6 −4 sin 2c 5 x, 15 41. 15. k =0 rk x k! 17. 0.00002 29. 13/24 4 19. n = 9 (13/24 ∼ 0. 542) = 2 21. n = 6 31. 0. 17 25. 79/48 33. 39. 4e 5 x, 15 2c (5/6 ∼ 0. 833) = |c| < |x| 37. 3 sec c − 2 sec c 3 x, 3 43. |c| < |x| |c| < |x| 3c 2 − 1 3 x, 3(1 + c2 )3 (b) 2 |c| < |x| (c) 999 ( − 1)n+1 e−c n+1 x, (n + 1) ! |c| < |x| 1 xn+1 , (1 − c)n+2 45. (a) 4 47. (a) 1. 649 (b) 0. 368 ∞ 49. For...
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