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

# Calculus - Early Transcendentals

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278 CHAPTER 7 Applications of the DeFnite Integral in Geometry, Science, and Engineering EXERCISE SET 7.1 1. A = Z 2 1 ( x 2 +1 x ) dx =( x 3 / 3+ x x 2 / 2) ¸ 2 1 =9 / 2 2. A = Z 4 0 ( x + x/ 4) dx = (2 x 3 / 2 / x 2 / 8) ¸ 4 0 = 22 / 3 3. A = Z 2 1 ( y 1 /y 2 ) dy y 2 / 2+1 /y ) ¸ 2 1 =1 4. A = Z 2 0 (2 y 2 + y ) dy = (2 y y 3 / y 2 / 2) ¸ 2 0 = 10 / 3 5. (a) A = Z 4 0 (4 x x 2 ) dx = 32 / 3 (b) A = Z 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 , 4. (a) A = Z 1 0 [2 x ( 2 x )] dx + Z 4 1 [2 x (2 x 4)] dx = Z 1 0 4 xdx + Z 4 1 (2 x 2 x dx =8 / 3+19 / 3=9 (b) A = Z 4 2 [( 2+2) y 2 / 4] dy

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Exercise Set 7.1 279 7. A = Z 1 1 / 4 ( x x 2 ) dx = 49 / 192 1 4 (1, 1) x y y = x 2 y = x 8. A = Z 2 0 [0 ( x 3 4 x )] dx = Z 2 0 (4 x x 3 ) dx =4 2 x y y = 2 x 3 – 4 x 9. A = Z π/ 2 4 (0 cos 2 x ) dx = Z 2 4 cos 2 xdx =1 / 2 36 -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 = ± 4 A = Z 4 4 (2 sec 2 x ) dx = π 2 11. A = Z 3 4 4 sin ydy = 2 3 9 x y x = sin y (2, 4) (–1, 1) x y y = x 2 x = y – 2 12. A = Z 2 1 [( x +2) x 2 ] dx =9 / 2
280 Chapter 7 13. A = Z ln 2 0 ( e 2 x e x ) dx = µ 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 = Z e 1 dy y = ln y i e 1 1/ e 1 1 e x y 15. A = Z 1 1 µ 2 1+ x 2 −| x | dx =2 Z 1 0 µ 2 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 ,x = ± 3 2 ,so A = Z 3 / 2 3 / 2 µ 2 1 1 x 2 dx sin 1 x ¸ 3 / 2 3 / 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 x, x 1 , A = Z 1 5 ²µ 1 5 x +7 (3 x ) ¸ dx + Z 5 1 ²µ 1 5 x (1 + x ) ¸ dx = Z 1 5 µ 4 5 x +4 dx + Z 5 1 µ 6 6 5 x dx = 72 / 5+48 / 5=24

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Exercise Set 7.1 281 18. A = Z 2 / 5 0 (4 x x ) dx + Z 1 2 / 5 ( x +2 x ) dx = Z 2 / 5 0 3 xdx + Z 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 = Z 1 0 ( x 3 4 x 2 +3 x ) dx + Z 3 1 [ ( x 3 4 x 2 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 x = 0, x ( x 1)( x 3) = 0; x =0 , 1 , 3 with corresponding values of y , 1 . 9. A = Z 1 0 [( x 3 2 x 2 ) (2 x 2 3 x )] dx + Z 3 1 [(2 x 3 3 x ) ( x 3 2 x 2 )] dx = Z 1 0 ( x 3 4 x 2 x ) dx + Z 3 1 ( x 3 +4 x 2 3 x ) dx = 5 12 + 8 3 = 37 12 21. From the symmetry of the region A Z 5 π/ 4 4 (sin x cos x ) dx =4 2 1 -1 0 o 22. The region is symmetric about the origin so A Z 2 0 | x 3 4 x | dx =8 3.1 -3.1 -3 3
282 Chapter 7 23. A = Z 0 1 ( y 3 y ) dy + Z 1 0 ( y 3 y ) dy =1 / 2 1 -1 -1 1 24. A = Z 1 0 £ y 3