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 = ² 1 5 y 5 + 5 4 y 4 ³ | 2 2 = ² 32 5 + 80 4 ³ ² 32 5 + 80 4 ³ = 64 5 =12 . 8 2. ± π/ 2 2 [ x 2 20 sin( x )] dx = ² 1 3 x 3 + 20 cos( x ) ³ | 2 2 = ´ π 3 24 +0 µ ´ π 3 24 µ = π 3 12 3. First approach: First fnd indefnite integral by substitution u =2 π ( x 2). Then du dx 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: Trans±orm the limits o± integration frst. When x , u = 2 π (2 2 )=0 . Wh en x =5 , u π (5 2 )=6 π . Then the integral a±ter 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 = xv ± = cos(2 x ) u ± =1 v = sin(2 x ) 2 . Thus ± 2 0 x cos(2 x ) dx = x 2 sin(2 x ) 2 0 ± 2 0 sin(2 x ) 2 dx = · x 2 sin(2 x cos(2 x ) 4 ¸ 2 0 = π 4 sin(
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This note was uploaded on 02/23/2012 for the course MAT MAT1332 taught by Professor Arianemasuda during the Spring '09 term at University of Ottawa.

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

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