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SEÇÃO 15.2 INTEGRAIS ITERADAS 3 1. 20 x 2y3 dy = x 2 14 y 4 2 0 = 4x 2 , 1 0 x 2y3 dx = y3 13 x 3 1 0 = 1 3 y 3 2. 20 2xy − 3x 2 dy = xy 2 − 3x 2y 20 = 4x − 6x 2 , 1 0 2xy − 3x 2 dx = yx 2 − x 3 10 = y − 1 3. 20 xe x + y dy = xe x [ey ]20 = x e x +2 − ex = xe x e2 − 1 1 0 xe x + y dx = ey 10 xe x dx = ey [xe x − ex ]10 = e y 4. 2 0 x y2 + 1 dy = x tg− 1 y 20 = x tg − 1 2, 1 0 x y2 + 1 dx = 1 y2 + 1 x 2 2 1 0 = 1 2 (y2 + 1) 5. 40 2 0 x y dx dy = 4 0 y 1 2 x 2 2 0 dy = 4 0 2 y dy = 43 y 3 / 2 4 0 = 323 6. 20 3 0 e x − y dy dx = 20 −e x − y 3 0 dx = 20 e x 1 − e− 3 dx = e2 − e− 1 − 1+ e− 3 7. 1 − 1 1 0 x 3 y2 + 3xy 2 dy dx = 1− 1 1 4 x 3 y4 + xy 3 y=1y=0 dx = 1 − 1 1 4 x 3 + x dx = 116 x 4 + 12 x 2 1 − 1 = 0 Solução Alternativa: Aplicando o Teorema de Fubini, a integral torna-se igual a 1 0 1 − 1 x 3 y2 + 3xy 2 dx dy = 10 1 4 y 2 x 4 + 32 y 2 x 2 x =1x =− 1 dy = 1 0 0 dy = 0 8. 10 2 1 x 4 − y2 dx dy = 21 1 0 x 4 − y2 dy dx = 21 x 4y − 13 y 3 y=1 y=0 dx = 2 1 x 4 − 13 dx = 15 x 5 − 13 x 2 1 = 88 15 9. pi/40 3 0 sen x dy dx = 3 pi/4 0 sen x dx = 3 [− cos x ] pi/4 0 = 3 1 − 12 10. pi/20 pi/2 0 sen x cos y dy dx = pi/20 sen x d x pi/ 2 0 cos y dy (como no Exemplo 5) = [− cos x ]pi/20 [sen y] pi/ 2 0 = − (0 − 1) (1 − 0) = 1 11. 30 1 0 x + y dx dy = 3 0 2 3 (x + y) 3 / 2 x =1 x =0 dy = 23 3 0 (1 + y) 3 / 2 − y3 / 2 dy = 23 2 5 (1 + y) 5 / 2 − 25 y 5 / 2 3 0 = 415 32 − 3 5 / 2 − 1 = 415 31 − 9 3 12. pi/ 2 0 pi/2 0 sen (x + y) dy dx = pi/ 20 [− cos (x + y)] y=pi/ 2 y=0 dx = pi/ 20 cos x − cos x + pi 2 dx = sen x − sen x + pi2 pi/2 0 = (1 − 0) − (0 − 1) = 2 13. 2 1 3 0 2y 2 − 3xy 3 dy dx = 21 2 3 y 3 − 34 xy 4 y=3 y=0 dx = 21 18 − 243 4 x dx = 18x − 243 8 x 2 2 1 = − 585 8 14. 32 0 − 1 xy 2 + yx − 1 dy dx = 32 1 3 xy 3 + 12 y 2x− 1 y=0y=− 1 dx = 3 2 1 3 x − 1 2 x − 1 dx = 16 x 2 − 12 ln x 3 2 = 5 6 + ln 2 3 15. pi/60 4 1 x sen y dx dy = pi/6 0 sen y dy 4 1 x dx = 1 − 32 15 2 = 15 (2− 3) 4 16. 1 0 1 1+ y dy 2 − 1 (1 + x ) dx = [ln (1 + y)]10 x + 1 2 x 2 2 − 1 = (ln 2) 2+ 2+ 1 − 12 = 9 2 ln 2 17. R xye y dA = 20 1 0 xye y dy dx = 20 x d x 1 0 ye y dy = 12 x 2 2 0 [e y (y − 1)]10 (por partes) = 12 (4 − 0) 0+ e 0 = 2 18. 1 0 1 0 xe xy dy dx = 10 [e xy ]y=1y=0 dx = 1 0 (e x − 1) dx = [ex − x ]10 = e − 2 19. 1 0 2 1 1 x + y dx dy = 10 [ln ( x + y)] x =2 x =1 dy = 10 [ln ( 2+ y) − ln (1 + y)] dy = [(2 + y) ln ( 2 + y) − (2 + y)] − [(1+ y) ln ( 1+ y) − (1 + y)] 1 0 por partes, separadamente, para cada termo ou pela Tabela de Integrais = (3 ln 3) − 3 − (2 ln 2) + 2− [(2 ln 2 − 2) − (0 − 1)] = 3 ln 3 − 4 ln 2 = ln 2716 20. V = R (2x + 5y + 1) dA = 4 1 0 − 1 (2x + 5y + 1) dx dy = 41 x 2 + 5xy + x x =0x =− 1 dy = 4 1 5y dy = 5 2 y 2 4 1 = 752 15.2 SOLUÇÕES Revisão técnica: Ricardo Miranda Martins – IMECC – Unicamp 4 SEÇÃO 15.2 INTEGRAIS ITERADAS 21. V = R x 2 + y2 dA = 3− 3 2 − 2 x 2 + y2 dx dy = 3− 3 1 3 x 3 + y2x x =2x =− 2 dy = 3 − 3 16 3 + 4y 2 dy = 163 y + 4 3 y 3 3 − 3 = 2 (16 + 36) = 104 22. V = 31 1 − 1 y 2 − x 2 dx dy = 2 31 1 0 y 2 − x 2 dx dy = 2 31 y 2x − 13 x 3 x =1 x =0 dy = 2 3 1 y 2 − 13 dy = 23 y 3 − y 31 = 16 23. A (R ) = pi2 · 1 = pi 2 , assim f med= 1 A (R ) R f (x, y) dA = 1 pi/2 pi/ 2 0 1 0 x sen xy dy dx = 2pi pi/2 0 [− cos xy ] y=1 y=0 dx = 2pi pi/2 0 (1 − cos x ) dx = 2pi [x − sen x ] pi/2 0 = 1 − 2 pi
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