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SampleMT1-1solutions

# SampleMT1-1solutions - Solutions to Practice Midterm#1 x1/3...

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Solutions to Practice Midterm #1 1. (a) x 1 / 3 ln x dx = ln xd 3 4 x 4 / 3 = 3 4 x 4 / 3 ln x - 3 4 x 4 / 3 d ln x = 3 4 x 4 / 3 ln x - 3 4 x 1 / 3 dx = 3 4 x 4 / 3 ln x - 9 16 x 4 / 3 + C ; (b) u + 3 ( u - 1)( u - 3) du = 3 u - 3 - 2 u - 1 du = 3 ln | u - 3 | - 2 ln | u - 1 | + c ; (c) dx 1 + sin x = (1 - sin x ) (1 - sin x )(1 + sin x ) dx = (1 - sin x ) (1 - sin 2 x ) dx = (1 - sin x ) cos 2 x dx = dx cos 2 x - sin xdx cos 2 x = tan x - sec x + C . (d) Make the substitution u = (1 + x ) 1 / 2 . Then x = ( u 2 - 1) 2 , dx = 4 u ( u 2 - 1); (1 + x ) 1 / 2 dx = 4 u 2 ( u 2 - 1) du = 4 5 u 5 - 4 3 u 3 + C = 4 5 (1 + x ) 5 / 2 - 4 3 (1 + x ) 3 / 2 + C. 2. (a) 3 1 dx x - 1 = lim t 1 + 3 t dx x - 1 = lim t 1 + (ln 2 - ln( t - 1)) = . Integral diverges. (b) First make the substitution u = e x . Then dx 1 + e x = du u ( u + 1) = du u - du u + 1 = ln | u | - ln | u + 1 | + C = ln u u + 1 + C = ln e x 1 + e x + C. 0 dx 1 + e x = lim t →∞ t 0 dx 1 + e x = lim t →∞ ln e x 1 + e t - ln e 0 1 + e 0 = ln lim t →∞ e t 1 + e t - ln 1 2 = ln 1 + ln 2 = ln 2 . (c) We use the formula | S n | ( b - a ) 5 K 180 n 4 if | f (4) ( x ) | K for a x b . First let us estimate K . (sin( x 2 )) (4) = (16 x 4 - 12) sin( x 2 ) - 48 x 2 cos( x 2 ) .

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One can choose K = 12 + 48 = 60 since | (16 x 4 - 12) sin( x 2 ) | 12 and | 48 x 2 cos( x 2 ) | 48. Then | S n | 60 180 n 4 . Apply the condition
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