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Pre-Calc Homework Solutions 296

# Pre-Calc Homework Solutions 296 - 296 16 Section 7.2 18...

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16. [ 2 2, 12] by [0, 3.5] The curves intersect at three points: x 5 2 1, x 5 4 and x 5 9. Because of the absolute value sign, break the integral up at x 5 0 also: E 0 2 1 1 } x 1 5 6 } 2 ˇ 2 w x w 2 dx 1 E 4 0 1 } x 1 5 6 } 2 ˇ x w 2 dx 1 E 9 4 1 ˇ x w 2 } x 1 5 6 } 2 dx 5 3 1 } 2 3 } ( 2 x ) 3/2 4 1 3 2 } 2 3 } x 3/2 4 1 3 } 2 3 } x 3/2 2 4 5 3 0 2 1 2 } 1 1 1 0 } 1 } 2 3 } 24 1 31 } 3 5 2 } 2 } 1 3 6 } 2 2 0 4 1 31 18 2 } 1 1 8 0 9 } 2 2 1 } 1 3 6 } 2 } 3 5 2 } 24 5 } 1 3 3 0 } 1 } 1 1 6 5 } 1 } 1 6 } 5 } 5 3 } 5 1 } 2 3 } 17. [ 2 5, 5] by [ 2 1, 14] The curves intersect at x 5 0 and x 5 6 4. Because of the absolute value sign, break the integral up at x 5 6 2 also (where ) x 2 2 4 ) turns the corner). Use the graph’s symmetry: 2 E 2 0 31 } x 2 2 } 1 4 2 2 (4 2 x 2 ) 4 dx 1 2 E 4 2 31 } x 2 2 } 1 4 2 2 ( x 2 2 4) 4 dx 5 2 E 2 0 } 3 2 x 2 } dx 1 2 E 4 2 1 2 } x 2 2 } 1 8 2 dx 5 2 3 } x 2 3 } 4 1 2 3 2 } x 6 3 } 1 8 x 4 5 2[4] 1 2 31 2 } 3 3 2 } 1 32 2 2 1 2 } 4 3 } 1 16 24 5 } 6 3 4 } 5 21 } 1 3 } 18. Solve y 2 5 y 1 2: y 2 2 y 2 2 5 ( y 2 2)( y 1 1) 5 0, so the curves intersect at y 5 2 1 and y 5 2. E 2 2 1 ( y 1 2 2 y 2 ) dy 5 3 } 1 2 } y 2 1 2 y 2 } 1 3 } y 3 4 5 1 2 1 4 2 } 8 3 } 2 2 1 } 1 2 } 2 2 1 } 1 3 } 2 5 } 9 2 } 5 4 } 1 2 } 19. Solve for x : x 5 } y 4 2 } 2 1 and x 5 } 4 y } 1 4. Now solve } y 4 2 } 2 1 5 } 4 y } 1 4: } y 4 2 } 2 } 4 y } 2 5 5 0, y 2 2 y 2 20 5 ( y 2 5)( y 1 4) 5 0. The curves intersect at y 5 2 4 and y 5 5. E 5 2 4 31 } 4 y } 1 4 2 2 1 } y 4 2 } 2 1 24 dy 5 E 5 2 4 1 2 } y 4 2 } 1 } 4 y } 1 5 2 dy 5 3 2 } 1 y 2 3 } 1 } y 8 2 } 1 5 y 4 5 1 2 } 1 1 2 2 5 } 1 } 2 8 5 } 1 25 2 2 1 } 1 3 6 } 1 2 2 20 2 5 } 24 8 3 } 5 30 } 3 8 } 20. Solve for x: x 5 y 2 and x 5 3 2 2 y 2 . Now solve y 2 5 3 2 2 y 2 : y 2 5 1, so the curves intersect at y 5 6 1. Use the region’s symmetry: 2 E 1 0 (3 2 2 y
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