05PM2solutions

# 05PM2solutions - 2 π Z π 2 x cos 2 x = π Z π 2 x cos(2...

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SOLUTIONS to MIDTERM II Practice Sheet 1. a) Z e 1 ln( x 4 ) dx = Z e 1 4 ln( x ) dx f = 4 ln ( x ) g 0 = dx f 0 = 4 /xdx g = x = 4 x ln( x ) | e 1 - 4 Z e 1 dx = [4 x ln( x ) - 4 x ] | e 1 = 4 b) Z x 5 e x 2 f = x 4 g 0 = xe x 2 dx f 0 = 4 x 3 dx g = 1 / 2 e x 2 = 1 / 2 x 4 e x 2 - 2 Z x 3 e x 2 = e x 2 (1 / 2 x 4 - x 2 + 1) + C c) Z π/ 2 0 cos 3 t sin 4 tdt = Z π/ 2 0 (1 - sin 2 t ) sin 4 t cos tdt u = sin t du = cos tdt Z 1 0 (1 - u 2 ) u 4 du = [1 / 5 u 5 - 1 / 7 u 7 ] | 1 0 = 2 / 35 d) I = Z ( e - 2 x sin 2 x ) dx f = sin(2 x ) g 0 = e - 2 x dx f 0 = 2 cos(2 x ) dx g = - 1 / 2 e - 2 x I = - 1 / 2 e - 2 x sin(2 x ) + Z e - 2 x cos(2 x ) dx 1

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2 f = cos(2 x ) g 0 = e - 2 x dx f 0 = - 2 sin(2 x ) dx g = - 1 / 2 e - 2 x I = - 1 / 2 e - 2 x sin(2 x ) - 1 / 2 e - 2 x cos(2 x ) - Z e - 2 x sin(2 x ) dx = - 1 / 2 e - 2 x sin(2 x ) - 1 / 2 e - 2 x cos(2 x ) - I I = - 1 / 4 e - 2 x (sin(2 x ) + cos(2 x )) + C e) Z x 2 + 8 x - 3 x 3 + 3 x 2 dx = Z - 2 x + 3 + 3 x - 1 x 2 = - 2 ln | x + 3 | + 3 ln | x | + 1 /x + C f) Z 2 / 2 0 x 2 1 - x 2 dx x = sin θ dx = cos θdθ Z sin 2 θ cos θdθ p 1 - sin 2 θ = 1 / 2 Z 1 - cos(2 θ ) = 1 / 2( θ - sin θ cos θ ) = 1 / 2(sin - 1 x - x p 1 - x 2 ) | 2 / 2 0 = ( π - 2) / 8 g) Z 0 e - 4 x dx = lim t →∞ Z t 0 e - 4 x = lim t →∞ - 1 / 4 e - 4 x | t 0 = lim t →∞ - 1 / 4( e - 4 t - 1) = 1 / 4
3 h) Z 0 -∞ e x e x + 1 dx = lim t →-∞ Z 0 t e x x x + 1 dx u = e x + 1 du = e x dx Z 1 /udu = ln( u ) lim t →-∞ ln( e x + 1) | 0 t = lim t →-∞ ln(2) - ln( e t + 1) = ln (2) 2) By trapezoidal rule: 1 / 2 * 1 / 4(sin 2 (0) + 2 sin 2 ( . 25) + 2 sin 2 ( . 5) + 2 sin 2 ( . 75) + sin 2 (1)) . 2774 By midpoint rule: 1 / 4(sin 2 ( . 125)+sin 2 ( . 375)+sin 2 ( . 625)+sin 2 ( . 875) . 2703 3) about y=axis:

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Unformatted text preview: 2 π Z π/ 2 x cos 2 ( x ) = π Z π/ 2 x cos(2 x ) + x (after integration by parts) = π/ 2[ x sin(2 x ) + 1 / 2 cos(2 x ) + x 2 | π/ 2 = π/ 2( π 2 / 4-1) about x-axis: π Z π/ 4 cos 4 ( x ) = π Z π/ 2 ( cos (2 x ) + 1) 2 / 4 = π/ 4 Z π/ 2 (cos(2 x )+1) / 2+2 cos (2 x )+1 = π/ 8(sin(2 x ) / 2+ x +2 sin(2 x )+2 x ) | π/ 2 = 3 π 2 / 16 4) Yes. (2 + cos ( u )) /u 2 ≤ 3 /u 2 , and R ∞ 1 3 /u 2 du converges. 4 5) We know F = kx , so R 1 kx = 12, so k = 24. Thus we get W = R 3 / 4 24 x = 27 / 4...
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## This note was uploaded on 12/27/2011 for the course MATH 5B taught by Professor Rickrugangye during the Fall '07 term at UCSB.

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05PM2solutions - 2 π Z π 2 x cos 2 x = π Z π 2 x cos(2...

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