Problem Set 9 Due 2:00PM Sep. 14 1. Evaluate the following integrals R √2 , Answer: 1/2 • 1 xdx 3 R√ 7 • 0 x2 , Answer: 7/3 R 1/2 2 , Answer: 1/24 • 0 t dt R 2a , Answer: 3a2 /2 • a xdx √ R 3b • 0 x2 dx , Answer: b/3 R1 , Answer: −14 • 3 7dx R2 , Answer: 10 • 0 5xdx R2 , Answer: −2 • 0 (2t − 3)dt R1 z • 2 1 + 2 dz , Answer: −7/4 R2 2 • 1 3u du , Answer: 7 R2 , Answer: 0 • 0 (3x2 + x − 5)dx 2. Find the indefinite integral of each of the following functions. • 4x6 − x3 ⇒ 74 x7 − 41 x4 + c • e(ex2+6x) (x + 1) • 12x2 − 6x1/2 + 3x−1/2 − x−1 3 1 ⇒ 4x3 − 4x 2 + 6x 2 − lnx + c • (x2 + 2x + 4)1/2 (x + 1) 3 ⇒ 13 (x2 + 2x + 4) 2 + c • 6e7x ⇒ 67 e7x + c • 3. Use integration by parts to integrate • R xlnxdx ⇒ 21 x2 lnx − 41 x2 1 3x1/2 +x−1/2 x3/2 +x1/2 • R x2 e2x dx ⇒ e2x ( 12 x2 − 12 x + 41 ) 4. Find the indicated integral • R 1 dx x7 • ⇒ − 6x16 + c R (3x2 − x3 − • R 1 x−a dx ⇒ log|x − a| + c • R √ 5x + 2)dx ⇒ √ 2 5 32 3 x (7x3/5 − + 2x + c √ x7 )dx ⇒ 5. Integration by parts • Find R3 2 ln xdx Z 3 Z 3 ln xdx = ln x1dx Z = x ln x|32 − 2 2 = (x ln x − 3 1dx 2 x)|32 = 3 ln 3 − 2 ln 2 − 1 • Find R x6 ln xdx Z • Find R x6 ln xdx = Z (ln x)x6 dx Z 1 1 71 = ln x( x7 ) − x dx 7 7 x Z 1 7 1 = x ln x − x6 dx 7 7 1 7 1 = x ln x − x7 + C 7 49 x2 e3x dx. 2 35 58 8 x 9 − 92 x 2 + c Let u = x2 and dv = e3x dx.R Thus du = 2xdx and v = 31 e3x . We then integrate by parts and obtain 13 x2 e3x − 32 xe3x dx. Now integrate by parts again with u = x and dv = e3x dx. Thus: Z Z 1 2 3x 2 2 3x xe3x dx x e − x e dx = 3 3 Z 1 2 3x 2 1 3x 1 3x = x e − {x e − e dx} Now have e and x seperate! 3 3 3 3 1 2 3x 2 3x 2 = x e − xe + e3x + C 3 9 27 • Show that • Find R R xn ex dx = xn ex − n R xn−1 ex dx ln x dx. x5 Let u = ln x, dv = = x−5 dx. This means that du = x1 dx and v = Z Z ln x −1 −1 1 dx = ln x( 4 ) − ( 4 ) dx 5 x 4x 4x x −4 1x −1 +C = ln x( 4 ) + 4x 4 −4 −1 x−4 ln x( 4 ) − +C 4x 16 1 dx x5 3 −1 . 4x4 Thus
© Copyright 2024 Paperzz