Business Calc Homework w answers_Part_51

Business Calc Homework w answers_Part_51 - Section 6.3 3....

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Unformatted text preview: Section 6.3 3. Let u ln y dv 1 dy y du 12 y 2 6. Let u 12 y 2 v 12 y ln y 2 12 y ln y 2 12 y ln y 2 y ln y dy y dy 1 sin dv 1 du 1 dy y sin d d v 2 1 1 sin 1 1 2 1 Let w 1 y dy 2 12 y C 4 2d 1 1 1 d 1 1 sin 2 sin sin sin 12 d12 Check: y ln y y C 4 dy 2 12 1 1 y (ln y)(y) y 2 y 2 d 2 1 dw 1 u 1 dy y tan 1 dy v y y y 1 y2 du 2y dy 1 y Check: 5. Let u du dy y2 tan 1 y x dv dx v x sec2 x dx 1 y 1 y 1 y 1 2 1 2 1 2 1 ln (1 y 2) 2 1 1 (2y) 2 1 y2 d y tan 1 y dy y 1 y2 1 y tan y tan y t2 7. Let u y tan y tan [ 1, 1] by [ 0.5, 2] dy y2 1 dw dv y2 1 y dy Let w y 1 du tan 1 tan 1 dw w ln (1 y 2) Let u C tan C x tan x t cos t (2 cos t dt sin t 2 t cos t dt 2t sin t 2 sin t dt 2t sin t 2 cos t t 2) cos t 2t sin t C dv v C csc2 t dt cot t tan x dx x tan x v 2 tan x 2 (cos t)(t) dt dv t 2 cos t sec2 x dx t 2 cos t dt du y cos t t t 2 cos t 1 sin t dt v t 2 sin t dt ln w C dv 2t dt sin x dx cos x x tan x ln cos x C [ 3, 3] by [0, 8] 8. Let u du t dt t csc2 t dt [ 1.5, 1.5] by [ 1, 4] t cot t cot t dt t cot t cos t dt sin t t cot t [0, 3] by [ 4, 4] 1 1 2 y ln y 4. Let 251 ln sin t C dw w 1/2 C 1 2 w C 252 Section 6.3 9. Let u ln x 1 dx x du x 3 dx dv 14 x 4 v 14 x ln x 4 14 x ln x 4 14 x ln x 4 x 3 ln x dx f (x) and its derivatives f (x) and its derivatives 1 41 x dx 4 x 13 x dx 4 14 x C 16 3x2 (–) 6x x and g (x) (+) 6 (–) e. x4 x 3e e–x (+) 4x 3 (–) (+) e–x 24x (–) –e–x (+) e–x 24 x 4e x x xe (x 4 4x e 4x 3 x2 11. Let u du (x 2 3 x 2 12x e 12x 2 24)e dv 24e C 5x)e x x C du e y sin y dy u 2x du (x 2 (x 2 5 5x)e x (x 2 5x)e x (x 2 7x e x(2x x e 5)e x 5)e x (2x C e du Let u e y cos y 1y e (sin y 2 y e C e du 2e x dx 2e x v e y cos y sin y e y sin y sin ye y dy e y sin y cos y) C dv y e y e cos y dy y 2e y dy v sin y e y e cos y dy cos y dy cos y dy e ysin y e ycos y dy 5) dx (2x 7)ex e dx v 2 dx 5x)e x x dv cos y dy 5) dx 14. Let u Let e dy e y sin y dy e x(2x cos y e y dy dv y 2 e y sin y dy ex v (x 2 x x e x dx 5) dx 5x)e x dx 24xe 24x 5x (2x x cos y e y cos y ey C sin y dy v e y sin y dy Let u 3 2x e 8 dv e y dy du dx 4 ey 13. Let u –e–x 0 dx 1 3 2x 3 2 2x 3 xe xe xe 2x 2 4 4 3 2 x 3x 3x 3 e 2x C 2 4 4 8 –e–x 12x 2 2x . 1 –2x e 2 1 –2x e 4 1 –2x e 8 1 –2x e 16 0 g(x) and its integrals 2x e–2x (+) x e g(x) and its integrals x3 4 10. Use tabular integration with f (x) x 3 and g(x) 12. Use tabular integration with f (x) sin y y dv y y e e sin y dy v dy cos y sin y y dy sin y 1y e (sin y 2 e ycos y e y e cos y cos y) C y cos y dy Section 6.3 x 2 and g(x) 15. Use tabular integration with f (x) f (x) and its derivatives sin 2x. g(x) and its integrals x2 (+) 2x (–) 2 (+) sin 2x 1 cos 2x 2 1 sin 2x 4 1 cos 2x 8 0 12 x cos 2x 2 x 2 sin 2x dx 1 x sin 2x 2 2x 2 cos 2x 4 1 /2 1 x 2 sin 2x dx 0 1 cos 2x 4 x sin 2x 2 2x 2 cos 2x 4 C C x /2 sin 2x 2 0 2 1 2 2 ( 1) 4 2 8 1 2 0 0.734 Check: NINT x 2 sin 2x, x, 0, 0.734 2 x 3 and g(x) 16. Use tabular integration with f (x) f (x) and its derivatives 1 (1) 4 0 cos 2x. g(x) and its integrals x3 (+) 3x2 (–) 6x (+) 6 (–) cos 2x 1 sin 2x 2 1 cos 2x 4 1 sin 2x 8 1 cos 2x 16 0 x 3 cos 2x dx 13 x sin 2x 2 x3 2 /2 3x sin 2x 4 x3 2 x 3cos 2x dx 0 0 3 4 32 x cos 2x 4 3x 2 4 3 ( 1) 8 32 16 Check: NINT x 3cos 2x, x, 0, 3 cos 2x 8 3x 2 4 3x sin 2x 4 32 16 3 x sin 2x 4 0 1.101 2 1.101 3 cos 2x 8 C 3 /2 cos 2x 8 0 3 (1) 8 253 254 Section 6.3 e 2x 17. Let u dv 2e 2x dx du e 2x cos 3x dx Let 1 sin 3x 3 1 1 (e 2x) sin 3x sin 3x (2e2x dx) 3 3 1 2x 2 2x e sin 3x e sin 3x dx 3 3 v e2x u du cos 3x dx dv 2e 2x dx sin 3x dx 1 cos 3x 3 2 2x (e ) 3 v 1 2x 1 1 e sin 3x cos 3x cos 3x (2e 2x dx) 3 3 3 1 2x 4 2x e (3 sin 3x 2cos 3x) e cos 3x dx 9 9 13 2x 1 2x e cos 3x dx e (3 sin 3x 2 cos 3x) 9 9 1 2x e 2x cos 3x dx e (3 sin 3x 2 cos 3x) 13 3 1 2x 3 e 2x cos 3x dx e (3 sin 3x 2 cos 3x) 13 2 2 e 2x cos 3x dx 16 [e (3 sin 9 13 16 [e (2 cos 9 13 2 cos 9) e 4 3 sin 9) e 4 (3 sin ( 6) (2 cos 6 2 cos ( 6)] 3 sin 6)] 18.186 Check: NINT(e 2x cos 3x, x, 18. Let u e du 2x e 2x 2, 3) 18.186 dv 2x 2e sin 2x dx (e v dx 2x sin 2x dx 1 cos 2x 2 1 cos 2x 2 ) 1 cos 2x ( 2e 2x dx) 2 1 2x e cos 2x 2 Let u e du 2x dv 2x 2e 2x e cos 2x dx cos 2x dx 1 sin 2x 2 v e sin 2x dx 2x 2e 2x 2 e 1 2x e cos 2x 2 sin 2x) 1 2x e (cos 2x 2 sin 2x dx e sin 2x dx 2x (e 1 2x e (cos 2x 2 2x e 1 2 (cos 2x e sin 2x) sin 2x) 4 (cos 2x sin 2x) 4 e sin 2x dx C 4 3 sin 4) e6 [cos ( 6) 4 (cos 4 sin 4) e6 (cos 6 4 4 e 2 (cos 4 4 125.028 2x 2x C 2x e 3 Check: NINT(e 1 sin 2x ( 2e 2x dx) 2 ) sin 2x 2x 4 sin 2x dx 2x sin 2x, x, 3, 2) 125.028 sin ( 6)] sin 6) 255 Section 6.3 23. Let x 2e4x dx 19. y x2 Let u du y 1 4 1 2 4x xe 4 y y dx y v 1 1 (x) e4x 2 4 1 4x 1 4x xe e 8 32 1 4x e C 32 Let u 2 x cos x du 2 v 1 x dx 2 1 sec 2 2 Let w 2 y ) sec 2 2 y 22. y Let y y 2 sec x2 (x 2 2 ln sec x 2x v x 1)e 2 dx 1)e x e v 2 x 1)e x (2x 1)e x (x 2 1 dx dx (x 2 d dx x e x 1)e x (2x 1)e x (x 2 1 x dx x e (2x dv x x e dx. dx 1 1)e x 1)e dv x x x 1 x 1)e du 1 3x 4)e x x 2e 2e x dx C C [ 3, 3] by [ 3, 3] dv sec v sec tan d The graph shows that the two curves intersect at x where k 1.050. The area we seek is k (x 2 x 1)e k x x 2 dx 0 d tan 4 1) dx (2x Let u d sec x (x 2 12 2 x sin x dx 0 Let u C Note: In the last step, we used the result of Exercise 29 in Section 6.2. 2 x sin x dx 24. We begin by evaluating (x 2 1 w 1/2 dw 4 1 1/2 w C 2 1 2 1C 2 sec tan u du d sec 0 3 2d 1 sec ( 1) 0 d 12 2 2d 1 sec 2 0 x sin x dx (x 2 12 2 1 4 1 2 y 0 2 2 (c) 1, so no absolute value is needed 1, dw 1 0 0(1) 2 dv 1 (sec 0 x sin x dx sin x 2 (1) in the expression for du. y x sin x dx du Note that we are told y sin x 2 (b) C d 1 du x sin x dx 0 ( 1) 1 4x e 4 1 4x e dx 4 dv 1 sec C x cos x 1 13 du dx v x x 3 1 13 1 (ln x) x 3 x dx 3 3 x 13 12 x ln x x dx 3 3 13 13 x ln x x C 3 9 sec sin x 3 ln x 21. y cos x cos x dx e4x dx x ln x dx Let u y x sin x dx 0 2 20. y y (a) dv 1 2 4x xe 4 1 2 4x xe 4 x2 x 4 8 y x cos x x cos x 1 xe4xdx 2 x du v dx sin x dx 1 4x e (2x dx) 4 (x 2) e4x Let u dv x sin x dx 1 4x e 4 v x du e4xdx dv 2x dx u 0 (x 2 ( 2.888 0.726 3x 4)e 4) xk 0 (0.386 13k x 30 0) k, ...
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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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