# soln3 - MATH 138 Calculus 2 Solutions to Assignment 3 1(a...

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Unformatted text preview: MATH 138 Calculus 2, Solutions to Assignment 3 1: (a) Find Z √ 3 x 2 dx (4- x 2 ) 3 / 2 . Solution: Make the substitution 2 sin θ = x so 2 cos θ = √ 4- x 2 and 2 cos θ dθ = dx to get Z √ 3 x 2 dx (4- x 2 ) 3 / 2 = Z π/ 3 (2 sin θ ) 2 2 cos θ dθ (2 cos θ ) 3 = Z π/ 3 tan 2 θ dθ = Z π/ 3 sec 2 θ- 1 dθ = h tan θ- θ i π/ 3 = √ 3- π 3 . (b) Find Z 3 1 x 4 + 3 x 3 + 6 x 3 + 4 x 2 + 3 x dx . Solution: We use long division to obtain x- 1 x 3 + 4 x 2 + 3 x x 4 + 3 x 3 + 0 x 2 + 0 x + 6 x 4 + 4 x 3 + 3 x 2- x 3- 3 x 2 + 0 x + 6- x 3- 4 x 2- 3 x x 2 + 3 x + 6 This shows that x 4 + 3 x 3 + 6 x 3 + 4 x 2 + 3 x = x- 1+ x + 3 x + 6 x 3 + 4 x 2 + 3 x . Note that x 3 +4 x 2 +3 x = x ( x +1)( x +3). In order to get A x + B x + 1 + C x + 3 = x + 3 x + 6 x ( x + 1)( x + 3) for all x we need A ( x +1)( x +3)+ Bx ( x +3)+ Cx ( x +1) = x 2 +3 x +6 for all x . Put in x = 0 to get 3 A = 6 so A = 2, put in x =- 1 to get- 2 B = 4 so B =- 2, and put in x =- 3 to get 6 C = 6 so C = 1. Thus Z 3 1 x 4 + 3 x 3 + 6 x 3 + 4 x 2 + 3 x dx = Z 3 1 x- 1 + 2 x- 2 x + 1 + 1 x + 3 dx = h 1 2 x 2- x + 2 ln x- 2 ln( x + 1) + ln( x + 3) i 3 1 = ( 9 2- 3 + 2 ln 3- 2 ln 4 + ln 6 )- ( 1 2- 1- 2 ln 2 + ln 4 ) = 2 + 3 ln 3- 3 ln 2 = 2 + 3 ln 3 2 . 2: (a) Find Z 1 x 2 p 2- x 2 dx . Solution: Make the substitution √ 2 sin θ = x so that √ 2 cos θ = √ 2- x 2 and √ 2 cos θ dθ = dx to get Z 1 x 2 p 2- x 2 dx = Z π/ 4 ( √ 2 sin θ ) 2 · √ 2 cos θ · √ 2 cos θ dθ = Z π/ 4 4 sin 2 θ cos 2 θ dθ = Z π/ 4 sin 2 2 θ dθ = Z...
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## This note was uploaded on 09/18/2008 for the course MATH MAth 138 taught by Professor Marshman during the Spring '08 term at Waterloo.

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soln3 - MATH 138 Calculus 2 Solutions to Assignment 3 1(a...

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