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Unformatted text preview: Solution set for Math 155, of Assignment 3 Prob. 1 Given y = x 2 , y = 2 x , y = First, we find the point of intersection by equating the two equation . x 2 = 2 x x 2 + x 2 = ( x 1 )( x + 2 ) = x = 1 , x =  2 Area under the curve in the 1st quadrant is calculated as Area = Z 1 ( 2 x ) x 2 dx = [2 x x 2 2 x 3 3 ] 1 = [2 1 2 1 3 ] = 7 6 Prob. 2 Given y = x 2 , y = ( x 2 ) 2 , y = and < x < 2 . First we compute the intersection point of equation y = x 2 andy = ( x 2 ) 2 x 2 = ( x 2 ) 2 x 2 = x 2 4 x + 4 4 x = 4 x = 1 Thus we can split the integral into two parts and calculate the area . Area = Z 1 x 2 + Z 2 1 ( x 2 ) 2 dx = [ x 3 3 ] 1 + [ ( x 2 ) 3 3 ] 2 1 = 1 3 + 1 3 = 2 3 1 Prob. 3 Given y = x , yx = 1 , y = 1 2 Since, we have to integrate in terms of y, we first calculate y in terms of x. x = y , x = 1 y , y = 1 2 We calculate the point of intersection of the curves x = y , x = 1 y y = 1 y y 2 = 1 y = ± 1 Since we need to calculate the area in 1st quadrant only, we have y...
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 Spring '09
 AlanM
 Math, Prob., ar ea

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