# Cos2 d 2 sin 2 c 2 2 sin cos c 4 ex 42 ex

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Unformatted text preview: c = 693.8597, d = 299.2239. d 650 (a) 1 + a2 b2 sinh2 bx dx (b) L = 2 0 = 1480.2798 ft -300 300 0 (d) 82◦ (c) x = 283.6249 ft b 1 + (12.54 − 0.82x)2 dx = 196.306 yd 14. y = 0 at x = b = 30.585; distance = 0 15. Let u = ax then du = a dx, √ 1 du = − a2 u2 1 1 du = a2 − u2 a √ 1 x2 −1 √ 1 du = a2 + u2 √ a dx = a2 + a2 x2 √ 1 dx = sinh−1 (u/a)+ C, 1 + x2 dx = cosh−1 (u/a) + C , u > a, 1 tanh−1 (u/a) + C, |u| < a a a+u 1 1 ln +C = dx = 1 1 − x2 2a a−u coth−1 (u/a) + C, |u| > a a (a) sinh−1 (x/2) + C (b) cosh−1 (x/3) + C (c) (d) 1 √ √ √ tanh−1 (x/ 2) + C, |x| < 2 2 1 √ coth−1 (x/√2) + C, |x| > √2 2 √ 1 dx =√ 5 16 + 5x2 √ 1 2+x = √ ln √ +C 22 2−x 1 dx = √ sinh−1 5 16/5 + x2 dx √ 5x 4 +C Chapter 8 Supplementary Exercises 290 16. (a) cosh 3x = cosh(2x + x) = cosh 2x cosh x + sinh 2x sinh x = (2 cosh2 x − 1) cosh x + (2 sinh x cosh x) sinh x = 2 cosh3 x − cosh x + 2 sinh2 x cosh x = 2 cosh3 x − cosh x + 2(cosh2 x − 1) cosh x = 4 cosh3 x − 3 cosh x x x (b) from Theorem 8.8.2 with x replaced by : cosh x = 2 cosh2 − 1, 2 2 1 2x 2x = cosh x + 1, cosh = (cosh x + 1), 2 cosh 2 2 2 x x 1 cosh = (cosh x + 1) (because cosh > 0) 2 2 2 x x (c) from Theorem 8.8.2 with x replaced by : cosh x = 2 sinh2 + 1, 2 2 1 x x x 1 2 sinh2 = cosh x − 1, sinh2 = (cosh x − 1), sinh = ± (cosh x − 1) 2 2 2 2 2 1 1 = k , k = 2, W = 2 4 17. (a) F = kx, 1/4 kx dx = 1/16 J 0 L kx dx = kL2 /2, L = 5 m (b) 25 = 0 150 (30x + 2000) dx = 15 · 1502 + 2000 · 150 = 637,500 lb·ft 18. F = 30x + 2000, W = 0 1 19. (a) F = ρx3 dx N 0 (b) By similar triangles 4 x w(x) = , w(x) = 2x, so 4 2 h(x) = 1 + x 0 ρ(1 + x)2x dx lb/ft2 . F= 1 w(x) x 2 (c) A formula for the parabola is y = 0 −10 9810|y |2 125 (y + 10) dy N. 8 (b) r = 1 when t ≈ 0.673080 s. r 20. (a) 82 x − 10, so F = 125 4 (c) dr/dt = 4.48 m/s. 2 1 t 1 291 Chapter 8 y 21. (a) (b) The maximum deﬂection occurs at x = 96 inches (the midpoint of the beam) and is about 1.42 in. x 100 200 -0.4 (c) The length of the centerline is 192 -0.8 1 + (dy/dx)2 dx = 192.026 in. 0 -1.2 -1.6 22. The x-coordinates of the points of intersection are a ≈ −0.423028 and b ≈ 1.725171; the area is b (2 sin x − x2 + 1)dx ≈ 2.542696. a 23. Let (a, k ), where π/2 < a < π , be the coordinates of the point of intersection of y = k with y = sin x. Thus k = sin a and if the shaded areas are equal, a a (k − sin x)dx = 0 (sin a − sin x) dx = a sin a + cos a − 1 = 0 0 Solve for a to get a ≈ 2.331122, so k = sin a ≈ 0.724611. k x sin x dx = 2π (sin k − k cos k ) = 8; solve for k to get 24. The volume is given by 2π 0 k = 1.736796. a+2 25. (a) √ a x dx 1 + x3 (b) Use the result in Exercise 24, Section 7.9, to obtain d da a+2 a √ x dx = 1 + x3 a+2 The maximum work is a a+2 1 + (a + √ 2)3 −√ a = 0; solve for a to get a ≈ 0.683772. 1 + a3 x dx ≈ 1.347655 J. 1 + x3 CHAPTER 9 Principles of Integral Evaluation EXERCISE SET 9.1 1. u = 3 − 2x, du = −2dx, 1 2 − 1 9 2. u = 4 + 9x, du = 9dx, 1 1 u3 du = − u4 + C = − (3 − 2x)4 + C 8 8 2 3/2 2 u +C = (4 + 9x)3/2 + C 3·9 27 u1/2 du = 1 1 tan u + C = tan(x2 ) + C 2 2 3. u = x2 , du = 2xdx, 1 2 sec2 u du = 4. u = x2 , du = 2xdx, 2 tan u du = −2 ln | cos u | + C = −2 ln | cos(x2 )| + C 5. u = 2 + cos 3x, du = −3 sin 3xdx, 6. u = 3 3x , du = dx, 2 2 2 3 1 dx, x 1 du = 2 4 + 4u 6 10. u = x2 , du = 2xdx, 1 2 √ − √ 1 7 − u5 du = − 16 1 u + C = − cos6 7x + C 42 42 √ 1 + 1 + sin2 x 1 + 1 + u2 du + C = − ln +C = − ln u sin x u u2 + 1 √ 12. u = sin x, du = cos x dx, 15. u = eu du = −eu + C = −ecot x + C 1 1 du = sin−1 u + C = sin−1 (x2 ) + C 2 2 2 1−u 11. u = cos 7x, du = −7 sin 7xdx, 14. u = tan−1 x, du = 1 1 du = tan−1 u + C = tan−1 (3x/2) + C 2 1+u 6 6 sec u tan u du = sec u + C = sec(ln x) + C 9. u = cot x, du = − csc2 xdx, 13. u = ex , du = ex dx, 1 1 du = − ln |u| + C = − ln(2 + cos 3x) + C u 3 3 1 3 sinh u du = cosh u + C = cosh ex + C 7. u = ex , du = ex dx, 8. u = ln x, du = − √ du = ln u + 4 + u2 1 dx, 1 + x2 1 dx, x − 2, du = √ 2 x−2 16. u = 3x2 + 2x, du = (6x + 2)dx, u2 + 4 + C = ln ex + eu du = eu + C = etan 2 1 2 −1 x +C √ x−2 eu du = 2eu + C = 2e cot u du = 292 e2x + 4 + C +C 1 1 ln | sin u| + C = ln sin |3x2 + 2x| + C 2 2 293 Chapter 9 17. u = √ 18. u = ln x, du = 19. u = √ √ 2 cosh u du = 2 sinh u + C = 2 sinh x + C 1 x, du = √ dx, 2x du = ln |u| + C = ln |ln x| + C u dx , x 1 x, du = √ dx, 2x 2 du =2 3u 20. u = sin θ, du = cos θdθ, 21. u = − dx = ln x + x2 − 3 x2 − 3 + C 23. u = e−x , du = −e−x dx, 24. u = ln x, du = 1 dx, x 26. u = x−1/2 , du = − 2 1 ln 16 ln 4 csch2 u du = 1 2 1 coth u + C = coth + C 2 2 x 2+u 1 2 + e−x 1 du + C = − ln = − ln +C 4 − u2 4 2−u 4 2 − e−x ex dx = 1 − e2x 1 dx, 2x3/2 1 2 − 1 du = sec u 2 √ 28. 2u = ex , 2du = ex dx, 2 − √ 27. u = x2 , du = 2xdx, 29. 4−x = e−x 1 2 cos u du = sin u + C = sin(ln x) + C 25. u = ex , du = ex dx, −...
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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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