Pre-Calc Homework Solutions 296

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

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Unformatted text preview: 296 16. Section 7.2 18. Solve y 2 y 2: y2 y 2 (y 2. 1 3 y 3 8 3 1 2 2 1 2)(y 1) 0, so the curves intersect at y 2 1 and y 1 2 y 2 [ 2, 12] by [0, 3.5] (y 1 2 y 2) dy 2y 4 4 1 2 y 4 y 4 The curves intersect at three points: x 1, x 4 and x 9. Because of the absolute value sign, break the integral up at x 0 also: 0 1 2 9 2 2 1 3 x 5 9 6 x 4 x dx 0 x 5 6 x dx 19. Solve for x: x Now solve y2 4 y2 4 1 and x y 4 4. 5 0, x 4 6 5 dx 1 2 x 2 1 9 4 1 (y 4: y2 4 1 2 x 2 y2 6x 5 2 3/2 x 3 y 20 5)(y 4) 0. 5. 6x 5 0 2 ( x)3/2 3 1 2 x 2 The curves intersect at y 0 5 4 4 and y dy 2 3/2 x 3 6x 5 4 y 4 5 4 4 y2 4 y2 8 25 8 y 4 y2 4 1 5 dy 5 0 11 10 2 3 189 10 1 6 5 3 16 3 32 5 32 5 2 3 16 3 0 y3 12 125 12 5y 4 18 13 30 16 15 25 16 3 2 20 243 8 30 3 8 1 20. Solve for x: x y2 3 y 2 and x 3 2y 2. Now solve 1. 17. 2y 2: y 2 1, so the curves intersect at y Use the region's symmetry: [ 5, 5] by [ 1, 14] 1 2 0 (3 2y 2 y 2) dy 1 2 6 0 1 0 (3 (1 3y 2) dy y 2) dy 1 3 y 3 1 3 1 0 The curves intersect at x 0 and x 4. Because of the absolute value sign, break the integral up at x 2 also (where x 2 4 turns the corner). Use the graph's symmetry: 2 6y 2 0 x2 2 2 4 2 (4 4 x 2) dx x 2 2 4 2 2 x2 2 4 (x 2 4) dx 6 1 y 2 and x 2 3y 2: y 2 0 4 2 0 3x dx 2 2 2 8 dx 4 2 x3 2 2 0 2 2 x3 6 32 3 8x 2 21. Solve for x: x 4 3 64 3 1 3 2 3y 2. 1, so the curves intersect Now solve at y 1 y2 2[4] 32 16 21 1. Use the region's symmetry: (2 1 0 2 0 3y 2 (1 y 2) dy 4y 1 2 0 (2 1 2y 2) dy 4 1 0 4 y 2) dy 1 3 y 3 1 3 0 8 3 ...
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This note was uploaded on 10/05/2011 for the course MAC 1147 taught by Professor German during the Spring '08 term at University of Florida.

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