# 1011T3S - MAT137Y 2010-2011 Winter Session Solutions to...

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Unformatted text preview: MAT137Y, 2010-2011 Winter Session, Solutions to Term Test 3 1. Evaluate the following integrals. (a) (7%) Z e 1 ln( x ) x 3 dx Here we use integration by parts: u = ln( x ) dv = 1 x 3 dx du = 1 x dx v =- 1 2 x 2 . Z e 1 ln( x ) x 3 dx =- 1 2 x 2 ln( x ) e 1 + 1 2 Z 3 1 1 x 3 dx =- 1 2 x 2 ln( x )- 1 4 x 2 e 1 =- 1 2 e 2 ln( e )- 1 4 e 2-- 1 2 ln(1)- 1 4 =- 3 4 e 2 + 1 4 = e 2- 3 4 e 2 . (b) (9%) Z 1 (1- x 2 ) 5 / 2 dx Let x = sin( θ ), dx = cos( θ ) dθ so that the integral becomes Z 1 (1- x 2 ) 5 / 2 = Z 1 (cos 2 ( θ )) 5 / 2 cos( θ ) dθ = Z 1 cos 4 ( θ ) dθ = Z sec 4 ( θ ) dθ = Z sec 2 ( θ ) ( tan 2 ( θ ) + 1 ) dθ Let u = tan( θ ), du = sec 2 ( θ ) dθ = Z ( u 2 + 1) du = 1 3 u 3 + u + C = 1 3 tan 3 ( θ ) + tan( θ ) + C = x 3 3(1- x 2 ) 3 / 2 + x √ 1- x 2 + C. (c) (8%) Z x 2 ( x- 1) 2 ( x + 1) dx Here we use partial fractions: x 2 ( x- 1) 2 ( x + 1) = A x- 1 + B ( x- 1) 2 + C x + 1 which simplifies to x 2 = A ( x- 1)( x + 1) + B ( x + 1) + C ( x- 1) 2 . When x = 1 this equation reads 1 = 2 B = ⇒ B = 1 2 . When x =- 1 the equation reads 1 = 4 C = ⇒ C = 1 4 . When x = 0 we see that 0 =- A + B + C = ⇒ A = B + C = 3 4 . Thus x 2 ( x- 1) 2 ( x + 1) = 3 4 x- 1 + 1 2 ( x- 1) 2 + 1 4 x + 1 which means that Z x 2 ( x- 1) 2 ( x + 1) dx = 3 4 Z 1 x- 1 dx + 1 2 Z 1 ( x- 1) 2 dx + 1 4 Z 1 x + 1 dx = 3 4 ln | x- 1 | - 1 2(...
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1011T3S - MAT137Y 2010-2011 Winter Session Solutions to...

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