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Unformatted text preview: A = 4/5 26 (x) = 3 (x 1), g(x) = x 1 3 (x 1) = x 1 (x 1) = (x 1) 3 Intersection points x = 0, 1, 2 A = 1 [(x 1) ( x  1) 1/3 ]dx + 1 2 [(x 1) 1/3 ( x  1)]dx A = [(x 1) 2 /2 3/4( x  1) 4/3 ] 1 + [3/4(x 1) 4/3 ( x  1) 2 /2] 1 2 A = 1/2  +  A = 0 28 (y) = y(2 y) , g(y) = y y(2 y) = y y(3 y) = 0 y = 0,3 A = 3 [(y) g(y)]dy A = 3 [y(2 y) (y)]dy A = 3 (3y y 2 )dy A = ((3/2)y 2 (1/3)y 3 ) 3 A = 27/2 9 A = 9/2 30 (y) = y/(16 y 2 ), g(y) = 0, y = 3 A = 3 (y/(16 y 2 ) dy 16 y 2 = t therefore, 2ydy = dt when y = 0, t = 16 Then y = 3, t = 7 A = 1/2 16 7 t1/2 dt A = 1/2 [2t 1/2 ] 16 7 A =  [t] 16 7 A = 7 +4 A = 4  7...
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This note was uploaded on 09/29/2010 for the course MAT MAT 231 taught by Professor Hannah during the Spring '10 term at Thomas Edison State.
 Spring '10
 Hannah
 Calculus

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