8thsol - Calculus 119 7th Week1 1. 3 3 0 Area = 0 2y y 2 (y...

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Calculus 119 7th Week 1 1. Area = Z 3 0 2 y - y 2 - ( y 2 - 4 y ) dy = Z 3 0 6 y - 2 y 2 dy = [3 y 2 - 2 y 3 3 ] ± ± ± ± ± 3 0 = 27 - 18 = 9 2. (a) x = 2 y 2 and x + y = 1 Let’s determine the intersection points ( x = 2 y 2 x + y = 1 2 y 2 + y - 1 = 0 y = - 1 ,y = 1 / 2 (2 , - 1 2 , 1 2 ) Area = Z 1 2 - 1 1 - y - (2 y 2 ) dy = " y - y 2 2 - 2 y 3 3 ± ± ± ± 1 2 - 1 = ( 1 2 - 1 8 - 1 12 ) - ( - 1 - 1 2 + 2 3 ) = 9 8 (b) x = 1 - y 2 and x = y 2 - 1 1 Please email all corrections and suggestions to these solutions to htor@metu.edu.tr. All solutions are available on the web at the url http://www.metu.edu.tr/ htor. 1
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Let’s determine the intersection points ( x = 1 - y 2 x = y 2 - 1 1 - y 2 = y 2 - 1 y 2 = 1 y = 1 (0 , - , 1) Area = Z 1 - 1 1 - y 2 - ( y 2 - 1) dy = Z 1 - 1 2 - 2 y 2 dy = " 2 y - 2 y 3 3 ± ± ± ± 1 - 1 = 2[1 - 1 3 - ( - 1 + 1 3 )] = 8 3 3. First we should find the equation of the lines passes through the given three points. i)(0,5) and (5,1)
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This note was uploaded on 03/31/2010 for the course MATHEMATIC 119 taught by Professor Muhiddinuğuz during the Fall '08 term at Middle East Technical University.

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8thsol - Calculus 119 7th Week1 1. 3 3 0 Area = 0 2y y 2 (y...

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