ass1 answrs - MAT1332, Winter 2011, Assignment 1 Solutions...

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MAT1332, Winter 2011, Assignment 1 Solutions Total=7 points. 1. (1 point) (a) Use the substitution u = 3 - 14 t . Then du dt = - 14, so dt = du - 14 . Thus Z 1 3 - 14 t dt = - 1 14 Z 1 u du = - 1 14 ln | u | + C = - 1 14 ln | 3 - 14 t | + C (Don’t forget to resubstitute - and don’t forget the absolute value signs in the logarithm or the + C !) (b) Z 3 - 3 ( y 7 - 2 y 9 ) dy = ± y 8 8 - y 10 5 ² 3 - 3 = ± 3 8 8 - 3 10 10 ² - ± ( - 3) 8 8 - ( - 3) 10 10 ² = 0 . 2. (a) Z π - π [ x 2 - 30 cos( x )] dx = ± x 3 3 - 30 sin x ² π - π = ± π 3 3 - 0 ² - ± ( - π ) 3 3 - 0 ² = 2 π 3 3 . (b) (1 point) First approach First find the indefinite integral by using the substitution u = 3 π ( x - 5). Then du dx = 3 π , so dx = du 3 π . Hence 3 Z sin(3 π ( x - 5) dx = 3 Z sin u du 3 π = - 1 π cos u + C = - 1 π cos(3 π ( x - 5)) + C . Then evaluate 3 Z 5 2 sin(3 π ( x - 5) dx = - 1 π cos(3 π ( x - 5)) ³ ³ ³ 5 2 = - 1 π [cos(0) - cos( - 9 π )] = - 1 π [1 + 1] = - 2 π . Second approach Transform the limits of integration first. When x = 2, u = 3 π (2 - 5) = - 9 π . When x = 5, u = 3 π (5 - 5) = 0. Then the integral after substitution becomes 3 Z 5
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ass1 answrs - MAT1332, Winter 2011, Assignment 1 Solutions...

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