05PFsolution

05PFsolution - y = p √ x-1 L= R 16 1 q 1 p √ x-1 2 dx =...

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SOLUTIONS to PRACTICE FINAL 1. y 0 = 5 cos(25 x 2 ) + sin x cos(cos 2 x ) 2. a. R e s cos( e s ) ds = (let u = e s ) = sin( e s ) + C b. R 1 0 ( e - e x ) xdx = R 1 0 ex - xe x = ex 2 / 2 - xe x + e x | 1 0 = e/ 2 - 1 c. R ln 2 x x dx = (let u = ln x ) = 1 / 3 ln 3 x + C d. R 2 0 t 1 + t 2 dt = (let u = 1 + t 2 )= 1 / 3(1 + t 2 ) 3 / 2 | 2 0 = (5 5 - 1) / 3 e. R sin(3 x ) e 2 x dx = 1 / 13 e 2 x ( - 3 cos(3 x ) + 2 sin(3 x ) + C f. R 2 1 x 2 - 1 x dx = (let x = sec θ ) = R tan 2 θdθ = R sec 2 - 1 = tan θ - θ = x 2 - 1 - sec - 1 x | 2 1 = 3 - π/ 3 g. R sin 2 x cos 3 xdx =(let u = sin x )= 1 / 3 sin 3 x - 1 / 5 sin 5 x + C h. R 1 x 2 +4 x +8 dx = R 1 ( x +2) 2 +4 (let x + 2 = 2 tan θ or ( u = ( x + 2) / 2) = 1 / 2 tan - 1 ( x +2 2 ) + C i. R 1 x 2 +5 x - 6 = 1 / 7 R 1 x +6 - 1 x - 1 = 1 / 7(ln | x + 6 | - ln | x - 1 | ) + C j. R 1 x 2 - 6 x +10 dx = R 1 ( x - 3) 2 +1 = tan - 1 ( x - 3) + C k. R 1 0 x ln xdx = CAREFUL!!! lim t 0 + Z 1 t x ln x = lim t 0 + 1 / 2 x 2 (ln x - 1 / 2) | 1 t = lim t 0 + - 1 / 4 - 1 / 2 t 2 (ln t - 1 / 2) = - 1 / 4 For lim t 0 + - 1 / 2 t 2 ln t write lim t 0 + - 1 / 2 ln t t - 2 and use L’Hospital’s Rule = lim t 0 + - 1 / 2 t - 1 - 2 t - 3 = 0 l. R 1 x 2 +1 3 x 7 +3 x 3 +7 x 7 dx R 1 x 2 3 x 7 +3 x 7 +7 x 7 = R 1 1 3 11 1 x 1 / 3 . Diverges. 3. W = R 50 0 1000 - 2 xdx = 47500 ft · lbs 1
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Unformatted text preview: y = p √ x-1. L= R 16 1 q 1 + ( p √ x-1) 2 dx = 4 / 5 x 5 / 4 | 16 1 = 124 / 5 b. SA = R 16 1 2 πx q 1 + ( p √ x-1) 2 dx = 8 π/ 9 x 9 / 4 | 16 1 = 4088 π/ 9 5. A = R 1-2 (4-x 2 )-( x + 2) dx = 9 / 2 x = 2 / 9 R 1-2 x [(4-x 2 )-( x + 2)] dx =-1 / 2 y = 2 / 9 R 1-2 1 / 2[(4-x 2 ) 2-( x + 2) 2 ] dx = 12 / 5 6. π R 1 4 2-(4-e x/ 3 ) 2 = 24 e x-3 / 2 e 2 x/ 3 | 1 = 24 e 1 / 3-3 / 2 e 2 / 3-45 / 2...
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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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05PFsolution - y = p √ x-1 L= R 16 1 q 1 p √ x-1 2 dx =...

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