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

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

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.

Ask a homework question - tutors are online