5x c and 10 1 15 15xdx 225 c tave 10 0 0 by the

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Unformatted text preview: 1 (4 + 16)(4) = 40 2 2 y 16 7 4 1 2 5 x 4 (b) f (x∗ )∆x = (4 + 7 + 10 + 13)(1) = 34 k (left) x∗ = 1, 2, 3, 4; k k =1 4 f (x∗ )∆x = (7 + 10 + 13 + 16)(1) = 46; the average is 1 (34 + 46) = 40 k 2 (right) x∗ = 2, 3, 4, 5; k k =1 (c) 9. The right endpoint approximation exceeds the true area by four triangles; the true area exceeds the left endpoint approximation by four different, but congruent, triangles. 0.718771403, 0.668771403, 0.692835360 10. 0.761923639, 0.584145862, 0.663501867 11. 0.919403170, 1.07648280, 1.001028825 12. 4.884074732, 5.684074732, 5.347070728 13. 0.351220577, 0.420535296, 0.386502483 14. 1.63379940, 1.805627583, 1.717566087 Exercise Set 7.5 15. (a) (b) (c) 17. (a) 224 √ n 1/x 1/x2 sin x x ln x ex 25 0.693097198 0.666154270 1.000164512 5.336963538 0.386327689 1.718167282 50 0.693134682 0.666538346 1.000041125 5.334644416 0.386302694 1.718253191 100 0.693144056 0.666634573 1.000010281 5.333803776 0.3862964444 1.718274669 A= 1 (3)(3) = 9/2 2 (b) 1 −A = − (1)(1 + 2) = −3/2 2 y y -2 -1 x A A x 3 (c) 1 −A1 + A2 = − + 8 = 15/2 2 (d) −A1 + A2 = 0 y y A2 -5 A2 5 -1 4 A1 18. (a) A= x 1 (1)(2) = 1 2 (b) A= 1 (2)(3/2 + 1/2) = 2 2 y y 1 1 A A x 2 (c) x A1 -1 1 −A = − (1/2)(1) = −1/4 2 (d) x 1 A1 − A2 = 1 − 1/4 = 3/4 y y 1 1 A1 A2 A 2 x x 2 225 19. Chapter 7 (a) A = 2(5) = 10 (b) 0; A1 = A2 by symmetry y y 2 A1 1 c A 6 x A2 x 1 (c) 2 3 5 4 1 1 (5)(5/2) + (1)(1/2) 2 2 = 13/2 A1 + A2 = (d) 1 [π (1)2 ] = π/2 2 y 1 y 5 A -1 3 2 -1 (a) x 2 A = (6)(5) = 30 (b) y −A1 + A2 = 0 because A1 = A2 by symmetry y 6 A A2 $ x -10 (c) 1 1 (2)(2) + (1)(1) = 5/2 2 2 (d) x 4 A1 -5 A1 + A2 = x A2 A1 20. 1 1 π (2)2 = π 4 y y 2 2 A A1 A2 2 x 3 2 21. (a) 22. 0.8 (a) (b) 1 1 1 2xdx = x2 f (x)dx = 0 0 1 (b) −2.6 −1 1 1 f (x)dx = =1 0 −1 2xdx = x2 −1 = 12 − (−1)2 = 0 (c) −1.8 x (d) −0.3 Exercise Set 7.5 226 10 (c) 2dx = 2x 1 1 5 1 1/2 2 2 f (x)dx + 2 −1 4 24. 3 −1 4 f (x)dx − 1 5 f (x)dx = 1 −2 3 −2 1 1 −1 3 dx − 5 −1 0 (a) −3 2 (b) 1 f (x)dx = 1 − (−2) = 3 1 −2 3 f (x)dx + f (x)dx = −(2 − 6) = 4 1 1 − x2 dx = 1/2 + 2(π/4) = (1 + π )/2 0 3 4 1 f (x)dx = − xdx + 2 0 28. 1/2 0 f (x)dx = − (a) (b) = 12 − (1/2)2 + 2 · 5 − 2 · 1 = 3/4 + 8 = 35/4 g (x)dx = 3(2) − 10 = −4 f (x)dx − 3 27. 1 + 2x g (x)dx = 5 + 2(−3) = −1 0 26. 2dx = x2 1 5 25. 5 1 5 2xdx + 1/2 = 18 1 f (x)dx = (d) 23. 10 10 f (x)dx = −2 0 2dx + 9 − x2 dx = 2 · 3 + (π (3)2 )/4 = 6 + 9π/4 −3 dx − 3 2 −2 xdx = 4 · 4 − 5(−1/2 + (3 · 3)/2) = −4 |x|dx = 4 · 1 − 3(2)(2 · 2)/2 = −8 √ x > 0, 1 − x < 0 on [2, 3] so the integral is negative x2 > 0, 3 − cos x > 0 for all x so the integral is positive (a) x4 > 0, (b) 30. (a) (b) 29. x − 9 < 0, |x| + 1 > 0 on [−2, 2] so the integral is negative 10 31. √ 3 − x > 0 on [−3, −1] so the integral is positive 3 25 − (x − 5)2 dx = π (5)2 /2 = 25π/2 0 3 9 − (x − 3)2 dx = π (3)2 /4 = 9π/4 32. 0 3 33. (a) −3 1 4x(1 − 3x)dx ex dx (b) 0 2 34. π /2 x3 dx (a) 1 0 1 2 (3x + 1)dx = 5/2 35. sin2 xdx (b) 0 36. −2 4 − x2 dx = π (2)2 /2 = 2π n 37. (a) max ∆xk →0 n 2x∗ ∆xk ; a = 1, b = 2 k lim k =1 (b) lim x∗ k ∆xk ; a = 0, b = 1 +1 max ∆xk →0 x∗ k =1 k 227 Chapter 7 n 38. (a) ln x∗ ∆xk , a = 1, b = 2 k lim max ∆xk →0 k =1 n (b) (1 + cos x∗ ) ∆xk , a = −π/2, b = π/2 k lim max ∆xk →0 k =1 (a) 1 1 1 1 2 k+1 and x∗ = 1 + , so x∗ = x∗ + = 1 + , x∗ +1 = x∗ + = 1+ for x∗ +1 = x∗ + 1 2 1 k k k n n n nk n n k = 2, 3, · · · , n − 1 (b) f (x ∗ ) = 1 + k (c) 39. 1 n (d) 40. 1+ k =1 lim (a) k−1 , n 1 3 − 2 2n n k= k =1 = k =1 3 which is the area of the trapezoid with base 1 and sides 1 and 2 2 n n f (x∗ )∆x = k k =1 k =1 k =1 1 b−a = , x∗ = n nk n 2 1 k f (x∗ )∆x = = k n n k =1 k , n 1 n3 n k2 = k =1 1 k−1 b−a = , x∗ = , n nk n n 2 1 1 k−1 =3 f (x∗ )∆x = k n n n k =1 f (x∗ )∆x = k lim n→+∞ 1 n(n + 1)(2n + 1) ; lim n→+∞ n3 6 n f (x∗ )∆x = k k =1 2 1 = 6 3 k =1 n 1 (n − 1)n(2n − 1) ; n3 6 (k − 1)2 = k =1 1 2 = 6 3 3 3k b−a = , x∗ = , n nk n n 2 12 1 3k 3 = f (x∗ )∆x = 4− k 4n n n (right) ∆x = n k =1 k =1 n f (x∗ )∆x k lim n→+∞ (b) 1 1 1 (n − 1)n 3 1 = n+ 2 =− ; n n n 2 2 2n k−1 n 3 = 2 n (a) 1+ (left) ∆x = n 42. 3 1 1 n(n + 1) 1 n+ 2 =+ n n 2 2 2n (right) ∆x = n (b) 1 n2 1 3 + 2 2n lim n→+∞ f (x∗ ) = 1 + k n→+∞ 41. n k 1 and ∆x = n n k =1 n 1− k =1 27 4n3 n k 2 = 12 − k =1 27 n(n + 1)(2n + 1) ; 4n3 6 39 27(2) = = 12 − 4(6) 4 b−a 3 3(k − 1) = , x∗ = , k n n n n 2 1 3(k − 1) f (x∗ )∆x = 4− k 4 n (left) ∆x = n k =1 k =1 n f (x∗ )∆x = 12 − k lim n→+∞ k =1 39 27(2) = 4(6) 4 3 12 = n n n 1− k =1 27 4n3 n (k −1)2 = 12− k =1 27 (n − 1)n(2n − 1) ; 4n3 6 Exercise Set 7.5 43. (a) 228 b−a 4k 4 = , x∗ = 2 + , n nk n n 3 4 192 n(n + 1) 384 n(n + 1)(2n + 1) 256 4k f (x∗ )∆x = = 32+ 2 +3 +4 2+ k n n n 2 n 6 n (right) ∆x = n k =1 k =1 n f (x∗ )∆x = 32 + k lim n→+∞ (b) k =1 k =1 k =1 n n→+∞ f (x∗ )∆x = 32 + k...
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This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.

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