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Unformatted text preview: SOLUTIONS FOR HOMEWORK PROBLEMS 16.9 MATH 132  WI00 4 Let’s forget about the graphs ... first thing, where do these functions intersect? x = x ( x 2) 2 → x x ( x 2) 2 = 0 → x [1 ( x 2) 2 ] = 0 → x [(1 ( x 2))(1+( x 2))] = 0 (difference of squares, where 1 = 1 2 right?). x (3 x )( x 1) = 0 → x = 0 , x = 1 , x = 3. So we have 3 intersection points, which forces us to compute the integral on two intervals separately, and ”add” the results in the very end (in the sense that we ignore the sign of whatever we get there, and add only the positive values together). R 1 [ x x ( x 2) 2 ] dx = R 1 [ x x ( x 2 4 x + 4)] dx = R 1 ( x x 3 + 4 x 2 4 x ) dx = R 1 ( x 3 + 4 x 2 3 x ) dx = ( x 4 4 + 4 x 3 3 3 x 2 2 )  1 = ( 1 4 + 4 1 3 3 1 2 ) 0 = something R 3 1 [ x x ( x 2) 2 ] dx = ... = ( x 4 4 + 4 x 3 3 3 x 2 2 )  3 1 = something Now ... if you want to use the graph, it will tell you that the result you get for the first integral is negative, and for the second it’s positive ... since x...
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This note was uploaded on 07/26/2011 for the course MATH 132 taught by Professor Staff during the Spring '08 term at Ohio State.
 Spring '08
 Staff
 Math, Calculus

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