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8. Chapter 7 Review
x y 1 implies y (1 x)2 1 2 x x. 12. [ 0.5, 2] by [ 0.5, 1]
1 2 , 3 2 by [ 3, 3] The area is
0 (1 2 x x) dx x
1 . 6 4 3/2 x 3 1 2 x 2 1 0 The area is (2 sin x
0 sin 2x) dx 2 cos x 1 cos 2x 2 4.
0 13. 9. x 2y 2 implies y x . 2 [ 5, 5] by [ 5, 5] [ 1, 19] by [ 1, 4] The curves intersect at x
2.1281 2.1281 2.1281. The area is (4 x 2 cos x) dx, 8.9023. The curves intersect at x
18 18. The area is
18 4 x 3/2 which using NINT evaluates to 14. 3
0 3 x dx 2 3x
2 3 y 3
3 3 2 18, 0 or 10. 4x x 0 2y 2 dy y2 18.
0 4 implies x 4. 1 2 y 4 1, and 4x y 16 implies
[ 4, 4] by [ 4, 4] 1 y 4 The curves intersect at x
0.8256 0.8256. The area is (3
0.8256 x sec2 x) dx, 2.1043. which using NINT evaluates to
[ 6, 6] by [ 6, 6] 15. Solve 1 The curves intersect at (3,
5 4 5 4 cos x 2 cos x for the xvalues at the two 2
3 4) and (5.25, 5). The area is ends of the region: x , i.e., dy symmetry of the area:
7 /3 5 7 or . Use the 3 3 1 y 4 4
1 2 y 4 1 2 y 4 1 y 4 1 2 y 8 38 3 1 5 dy
5 2
2 [(1
7 /3 cos x) (2 1) dx cos x)] dx 2
2 (2 cos x x 1 3 y 12 425 24 5y
4 2 2 sin x 30.375. 16.
/3 7 /3 2 243 8 2
5 /3 3 [(2
5 /3 2 3 1.370. (1 cos x)] dx 11. cos x) (1
/3 2 cos x) dx
5 /3 /3 x
[ 0.1, 1] by [ 0.1, 1]
/4 2 sin x 3
4 3 2
1 2 x 2
2 7.653 /4 The area is
0 (x sin x) dx cos x
0 2 2 32 1 0.0155. ...
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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.
 Spring '08
 GERMAN

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