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CALCULUS 2 PACK - MAT1332 Winter 2008 Assignment#1...

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MAT1332, Winter 2008, Assignment #1 Solutions 1. 2 2 ( y 4 + 5 y 3 ) dy = 1 5 y 5 + 5 4 y 4 | 2 2 = 32 5 + 80 4 32 5 + 80 4 = 64 5 = 12 . 8 2. π/ 2 π/ 2 [ x 2 20 sin( x )] dx = 1 3 x 3 + 20 cos( x ) | π/ 2 π/ 2 = π 3 24 + 0 π 3 24 + 0 = π 3 12 3. First approach: First find indefinite integral by substitution u = 2 π ( x 2). Then du dx = 2 π so dx = du 2 π . Hence cos(2 π ( x 2)) dx = cos( u ) du 2 π = 1 2 π sin( u ) + C = 1 2 π sin(2 π ( x 2)) + C. Then evaluate 5 2 cos(2 π ( x 2)) dx = 1 2 π sin(2 π ( x 2)) 5 2 = 1 2 π (sin(6 π ) sin(0)) = 0 . Second approach: Transform the limits of integration first. When x = 2, u = 2 π (2 2) = 0. When x = 5, u = 2 π (5 2) = 6 π . Then the integral after substitution becomes 5 2 cos(2 π ( x 2)) dx = 6 π 0 cos( u ) du 2 π = 1 2 π sin( u ) 6 π 0 = 1 2 π sin(6 π ) 1 2 π sin(0) = 0 . 4. Using integration by parts, we have u = x v = cos(2 x ) u = 1 v = sin(2 x ) 2 . Thus π/ 2 0 x cos(2 x ) dx = x 2 sin(2 x ) π/ 2 0 π/ 2 0 sin(2 x ) 2 dx = x 2 sin(2 x ) + cos(2 x ) 4 π/ 2 0 = π 4 sin( π ) + cos( π ) 4 [0 + cos(0) 4 ] = 0 1 4 0 1 4 = 1 2 .
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