MAT1332W10A1S - MAT1332B, 1332C, Winter 2010, Assignment #1...

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Unformatted text preview: MAT1332B, 1332C, Winter 2010, Assignment #1 Solutions 1. We use the substitution u = 1 + 4 t . Then du dt = 4, so dt = du 4 . Thus Z 1 1 + 4 t dt = 1 4 Z 1 u du = 1 4 ln | u | + C = 1 4 ln | 1 + 4 t | + C. (Dont forget to resubstitute.) 2. Z 2- 2 ( y 3 + 4 y 5 ) dy = 1 4 y 4 + 4 6 y 6 | 2- 2 = 16 4 + 4 64 6 - 16 4 + 4 64 6 = 0 . 3. Z / 2- / 2 [ x 2- 20sin( x )] dx = 1 3 x 3 + 20cos( x ) | / 2- / 2 = " 3 24 + 0 #- "- 3 24 + 0 # = 3 12 4. First approach: First find indefinite integral by substitution u = 2 ( x- 2). Then du dx = 2 so dx = du 2 . Hence Z cos(2 ( x- 2)) dx = Z cos( u ) du 2 = 1 2 sin( u ) + C = 1 2 sin(2 ( x- 2)) + C. Then evaluate Z 5 2 cos(2 ( x- 2)) dx = 1 2 sin(2 ( x- 2)) fl fl fl fl 5 2 = 1 2 (sin(6 )- sin(0)) = 0 . Second approach: Transform the limits of integration first. When x = 2, u = 2 (2- 2) = 0. When x = 5, u = 2 (5- 2) = 6 . Then the integral after substitution becomes Z 5 2 cos(2...
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MAT1332W10A1S - MAT1332B, 1332C, Winter 2010, Assignment #1...

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