Business Calc Homework w answers_Part_59

E when 5 use the regions symmetry 1 so the solution is

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Unformatted text preview: grate in two parts: 0 [(2x 3 x2 [( x 2 3x) [ 2, 2] by [ 2, 2] ( 0.9286, 0.8008), (0, 0), and (0.9286, 0.8008) 0 (1 2 cos x) dx 0 1 x 2 1 sin 2x 4 /3 0 1 sec2 t 2 /3 (sec2 t 8 sin2 t) dt 0 /3 tan t 4t 2 sin 2t 0 3 4 3 4 3 3 0 x2 2 8x) dx 3x)] dx 5x)] dx ( 2x 3 8x) dx 0 14 x 2 2 0 [0 4 sin2 t dt (2x 3 (2x 3 2 2. Use symmetry: 2 0 ( x2 5x) 2 2 Section 7.2 Exercises 1. 128 15 22 15 7. Integrate with respect to y: 8 1 2 x 1 3 2x 4) dx 0 1 2 0 6. Use the region’s symmetry: x2 1 2 32 3 (0, 0) is a solution. Now divide by x. x2 4x 2) dx 0 x 3. 1 1 12 0 0 1, so that if they are 1 is zero when e x 1 12y 3) 1 14 y 4 0 1 always greater than or equal to x x 4. 5 1. From the graphs, it appears that e is ever equal, this is when e x 13 y 3 y 3) dy 0 x x dx [e dx ( y2 4x 2 (8 0 2 16)] 14 x 2 [( 8 4x 2 16) 2 0 0] 16 5 6 Section 7.2 10. Integrate in three parts: x 1 [( x 2) 2x 2 x2 4: x 2 2x 2) (x 2 4)] dx 13. Solve 7 1, so the curves intersect at 1. 2 (4 x )] dx 1 2 [(7 1 1 2 x 2) [(4 295 (x ( 3x 2 3) dx 1 2)] dx 1 1 3 x 2) dx (1 1 3 [(...
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This document was uploaded on 10/31/2011 for the course MAC 2311 at University of Florida.

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