# 5 ln 2 4 4x 4 4x 4 4x 4x x e2 x 4e2 e2 x x1 e2

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Unformatted text preview: x) 0 0 5 47. distance = t2 e−t dt; u = t2 , dv = e−t dt, du = 2tdt, v = −e−t , 0 5 5 distance = −t2 e−t te−t dt; u = 2t, dv = e−t dt, du = 2dt, v = −e−t , +2 0 0 distance = −25e−5 − 2te−t −5 = −25e −5 − 10e 5 5 +2 0 e−t dt = −25e−5 − 10e−5 − 2e−t 0 −5 − 2e −5 + 2 = −37e 48. u = 2t, dv = sin(kωt)dt, du = 2dt, v = − π /ω −π/ω t sin(kωt) dt = − 0 2π (−1) kω 2 k+1 = + 0 +2 1 cos(kωt); the integrand is an even function of t so kω π /ω t sin(kωt) dt = 2 5 2 t cos(kωt) kω π /ω 2 k2 ω2 sin(kωt) π /ω +2 0 0 1 cos(kωt) dt kω 2π (−1) kω 2 k+1 = 0 π /ω 49. (a) 1 2 sin3 x dx = − sin2 x cos x + 3 3 1 2 sin x dx = − sin2 x cos x − cos x + C 3 3 (b) 1 3 sin4 x dx = − sin3 x cos x + 4 4 sin2 x dx, π /4 sin4 x dx = 0 − 1 1 sin2 x dx = − sin x cos x + x + C1 so 2 2 3 1 3 sin3 x cos x − sin x cos x + x 4 8 8 π /4 0 √ √ √ √ 1 3 = − (1/ 2)3 (1/ 2) − (1/ 2)(1/ 2) + 3π/32 = 3π/32 − 1/4 4 8 50. (a) cos5 x dx = = 1 4 cos4 x sin x + 5 5 cos3 x dx = 1 41 2 cos4 x sin x + cos2 x sin x + sin x + C 5 53 3 4 8 1 cos4 x sin x + cos2 x sin x + sin x + C 5 15 15 301 Chapter 9 (b) cos6 x dx = 1 5 cos5 x sin x + 6 6 cos4 x dx = 51 3 1 cos5 x sin x + cos3 x sin x + 6 64 4 = 1 5 51 1 cos5 x sin x + cos3 x sin x + cos x sin x + x + C, 6 24 82 2 cos2 x dx 5 5 5 1 cos5 x sin x + cos3 x sin x + cos x sin x + x 6 24 16 16 π /2 = 5π/32 0 51. u = sinn−1 x, dv = sin x dx, du = (n − 1) sinn−2 x cos x dx, v = − cos x; sinn x dx = − sinn−1 x cos x + (n − 1) sinn−2 x cos2 x dx = − sinn−1 x cos x + (n − 1) = − sinn−1 x cos x + (n − 1) n sinn−2 x(1 − sin2 x)dx sinn−2 x dx − (n − 1) sinn x dx = − sinn−1 x cos x + (n − 1) sinn x dx = − 1 n−1 sinn−1 x cos x + n n sinn x dx, sinn−2 x dx, sinn−2 x dx 52. (a) u = secn−2 x, dv = sec2 x dx, du = (n − 2) secn−2 x tan x dx, v = tan x; secn x dx = secn−2 x tan x − (n − 2) secn−2 x tan2 x dx = secn−2 x tan x − (n − 2) = secn−2 x tan x − (n − 2) (n − 1) secn−2 x(sec2 x − 1)dx secn x dx + (n − 2) secn x dx = secn−2 x tan x + (n − 2) secn x dx = (b) 1 n−2 secn−2 x tan x + n−1 n−1 tann x dx = tann−2 x(sec2 x − 1) dx = = 1 tann−1 x − n−1 tan4 x dx = 1 tan3 x − 3 (b) sec4 x dx = 1 2 sec2 x tan x + 3 3 secn−2 x dx, secn−2 x dx tann−1 x sec2 x dx − tann−2 x dx tann−2 x dx (c) u = xn , dv = ex dx, du = nxn−1 dx, v = ex ; 53. (a) secn−2 x dx, tan2 x dx = xn ex dx = xn ex − n 1 tan3 x − tan x + 3 sec2 x dx = dx = xn−1 ex dx 1 tan3 x − tan x + x + C 3 1 2 sec2 x tan x + tan x + C 3 3 Exercise Set 9.2 302 x3 ex dx = x3 ex − 3 (c) x2 ex dx = x3 ex − 3 x2 ex − 2 = x3 ex − 3x2 ex + 6 xex − xex dx ex dx = x3 ex − 3x2 ex + 6xex − 6ex + C 54. (a) u = 3x, x2 e3x dx = = 1 27 u2 eu du = 1 u2 eu − 2 27 ueu du = 1 2u 2 ue − ueu − 27 27 eu du 2 2 1 2 2 1 2u u e − ueu + eu + C = x2 e3x − xe3x + e3x + C 27 27 27 3 9 27 √ (b) u = − x, 1 √ x xe− −1 dx = 2 0 u3 eu du, 0 u3 eu du = u3 eu − 3 u2 eu du = u3 eu − 3 u2 eu − 2 = u3 eu − 3u2 eu + 6 ueu − −1 2 eu du = u3 eu − 3u2 eu + 6ueu − 6eu + C, −1 u e du = 2(u − 3u + 6u − 6)e 3u 3 ueu du u 2 0 = 12 − 32e−1 0 55. u = x, dv = f (x)dx, du = dx, v = f (x); 1 1 1 x f (x)dx = xf (x) −1 −1 − f (x)dx −1 1 = f (1) + f (−1) − f (x) −1 = f (1) + f (−1) − f (1) + f (−1) 56. (a) u = f (x), dv = dx, du = f (x), v = x; b b b − f (x) dx = xf (x) a b xf (x) dx = bf (b) − af (a) − a a xf (x) dx a (b) Substitute y = f (x), dy = f (x) dx, x = a when y = f (a), x = b when y = f (b), b f (b) xf (x) dx = a f (b) x dy = f (a) f −1 (y ) dy f (a) y (c) From a = f −1 (α) and b = f −1 (β ) we get b bf (b) − af (a) = βf −1 (β ) − αf −1 (α); then β α f −1 (x) dx = β f (b) f −1 (y ) dy = α A1 f −1 (y ) dy , α b f −1 (x) dx = bf (b) − af (a) − f (x) dx a = βf −1 (β ) − αf −1 (α) − A2 x a = f –1(a) which, by part (b), yields β a f (a) f −1 (β ) f (x) dx f −1 (α) b = f –1(b) 303 Chapter 9 β Note from the ﬁgure that A1 = f −1 f −1 (β ) (x) dx, A2 = α f (x) dx, and f −1 (α) A1 + A2 = βf −1 (β ) − αf −1 (α), a ”picture proof”. 57. (a) Use Exercise 56(c); 1/2 1 x dx = sin−1 2 −1 sin 0 1 −0·sin−1 0− 2 sin−1 (1/2) sin x dx = sin−1 (0) 1 sin−1 2 1 − 2 π /6 sin x dx 0 (b) Use Exercise 56(b); e2 ln e2 ln x dx = e ln e − e ln e − 2 2 e 58. (a) ln e xex dx = x(ex + C1 ) − (b) u(v + C1 ) − 2 f −1 (y ) dy = 2e2 − e − 2 ey dy = 2e2 − e − 1 ex dx 1 (ex + C1 )dx = xex + C1 x − ex − C1 x + C = xex − ex + C (v + C1 )du = uv + C1 u − v du − C1 u = uv − v du EXERCISE SET 9.3 1. u = cos x, − 3. u = sin ax, 1 a 1 u5 du = − cos6 x + C 6 u du = 2. u = sin 3x, 1 3 u4 du = 1 sin5 3x + C 15 1 sin2 ax + C , a = 0 2a 4. cos2 3x dx = 1 2 (1 + cos 6x)dx = 1 1 x+ sin 6x + C 2 12 5. sin2 5θ dθ = 1 2...
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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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