Calculus: Early Transcendentals, by Anton, 7th Edition,ch07

Calculus - Early Transcendentals

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278 CHAPTER 7 Applications of the Definite Integral in Geometry, Science, and Engineering EXERCISE SET 7.1 1. A = 2 1 ( x 2 + 1 x ) dx = ( x 3 / 3 + x x 2 / 2) 2 1 = 9 / 2 2. A = 4 0 ( x + x/ 4) dx = (2 x 3 / 2 / 3 + x 2 / 8) 4 0 = 22 / 3 3. A = 2 1 ( y 1 /y 2 ) dy = ( y 2 / 2 + 1 /y ) 2 1 = 1 4. A = 2 0 (2 y 2 + y ) dy = (2 y y 3 / 3 + y 2 / 2) 2 0 = 10 / 3 5. (a) A = 4 0 (4 x x 2 ) dx = 32 / 3 (b) A = 16 0 ( y y/ 4) dy = 32 / 3 5 1 (4, 16) y = 4 x y = x 2 x y (4, 4) (1, -2) x y y 2 = 4 x y = 2 x – 4 6. Eliminate x to get y 2 = 4( y + 4) / 2 , y 2 2 y 8 = 0, ( y 4)( y + 2) = 0; y = 2 , 4 with corresponding values of x = 1 , 4. (a) A = 1 0 [2 x ( 2 x )] dx + 4 1 [2 x (2 x 4)] dx = 1 0 4 xdx + 4 1 (2 x 2 x + 4) dx = 8 / 3 + 19 / 3 = 9 (b) A = 4 2 [( y/ 2 + 2) y 2 / 4] dy = 9
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Exercise Set 7.1 279 7. A = 1 1 / 4 ( x x 2 ) dx = 49 / 192 1 4 (1, 1) x y y = x 2 y = x 8. A = 2 0 [0 ( x 3 4 x )] dx = 2 0 (4 x x 3 ) dx = 4 2 x y y = 2 x 3 – 4 x 9. A = π/ 2 π/ 4 (0 cos 2 x ) dx = π/ 2 π/ 4 cos 2 x dx = 1 / 2 3 6 -1 1 x y y = cos 2 x 10. Equate sec 2 x and 2 to get sec 2 x = 2, 1 2 x y y = sec 2 x ( # , 2 ) ( 3 , 2 ) sec x = ± 2, x = ± π/ 4 A = π/ 4 π/ 4 (2 sec 2 x ) dx = π 2 11. A = 3 π/ 4 π/ 4 sin y dy = 2 3 9 x y x = sin y (2, 4) (–1, 1) x y y = x 2 x = y – 2 12. A = 2 1 [( x + 2) x 2 ] dx = 9 / 2
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280 Chapter 7 13. A = ln 2 0 ( e 2 x e x ) dx = 1 2 e 2 x e x ln 2 0 = 1 / 2 2 4 x y ln 2 y = e 2 x y = e x 14. A = e 1 dy y = ln y e 1 = 1 1/ e 1 1 e x y 15. A = 1 1 2 1 + x 2 − | x | dx = 2 1 0 2 1 + x 2 x dx = 4 tan 1 x x 2 1 0 = π 1 -1 1 1 2 x y 16. 1 1 x 2 = 2 , x = ± 3 2 , so A = 3 / 2 3 / 2 2 1 1 x 2 dx = 2 sin 1 x 3 / 2 3 / 2 = 2 3 2 3 π 0.5 1 1.5 2 x y 2 3 - 2 3 1 - x 2 1 y = y = 2 (–5, 8) (5, 6) y = 3 – x y = 1 + x y = – x + 7 1 5 x y 17. y = 2 + | x 1 | = 3 x, x 1 1 + x, x 1 , A = 1 5 1 5 x + 7 (3 x ) dx + 5 1 1 5 x + 7 (1 + x ) dx = 1 5 4 5 x + 4 dx + 5 1 6 6 5 x dx = 72 / 5 + 48 / 5 = 24
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Exercise Set 7.1 281 18. A = 2 / 5 0 (4 x x ) dx + 1 2 / 5 ( x + 2 x ) dx = 2 / 5 0 3 x dx + 1 2 / 5 (2 2 x ) dx = 3 / 5 (1, 1) 2 5 8 5 ( , ) x y y = 4 x y = x y = - x + 2 19. A = 1 0 ( x 3 4 x 2 + 3 x ) dx + 3 1 [ ( x 3 4 x 2 + 3 x )] dx = 5 / 12 + 32 / 12 = 37 / 12 4 -8 -1 4 9 -2 -1 3 20. Equate y = x 3 2 x 2 and y = 2 x 2 3 x to get x 3 4 x 2 + 3 x = 0, x ( x 1)( x 3) = 0; x = 0 , 1 , 3 with corresponding values of y = 0 , 1 . 9. A = 1 0 [( x 3 2 x 2 ) (2 x 2 3 x )] dx + 3 1 [(2 x 3 3 x ) ( x 3 2 x 2 )] dx = 1 0 ( x 3 4 x 2 + 3 x ) dx + 3 1 ( x 3 + 4 x 2 3 x ) dx = 5 12 + 8 3 = 37 12 21. From the symmetry of the region A = 2 5 π/ 4 π/ 4 (sin x cos x ) dx = 4 2 1 -1 0 o 22. The region is symmetric about the origin so A = 2 2 0 | x 3 4 x | dx = 8 3.1 -3.1 -3 3
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282 Chapter 7 23. A = 0 1 ( y 3 y ) dy + 1 0 ( y 3 y ) dy = 1 / 2 1 -1 -1 1 24. A = 1 0 y 3 4 y 2 + 3 y ( y 2 y ) dy + 4 1 y 2 y ( y 3 4 y 2 + 3 y ) dy = 7 / 12 + 45 / 4 = 71 / 6 4.1 0 -2.2 12.1 25. The curves meet when x = ln 2, so A = ln 2 0 (2 x xe x 2 ) dx = x 2 1 2 e x 2 ln 2 0 = ln 2 1 2 0.5 1 0.5 1 1.5 2 2.5 x y 26. The curves meet for x = e 2 2 / 3 , e 2 2 / 3 thus A = e 2 2 / 3 e 2 2 / 3 3 x 1 x 1 (ln x ) 2 dx = ( 3 ln x sin 1 (ln x ) ) e 2 2 / 3 e 2 2 / 3 = 4 2 2 sin 1 2 2 3 1 2 3 5 10 15 20 x y 27. The area is given by k 0 (1 / 1 x 2 x ) dx = sin 1 k k 2 / 2 = 1; solve for k to get k = 0 . 997301.
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