Business Calc Homework w answers_Part_57

Business Calc Homework w answers_Part_57 - Chapter 6 Review...

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Unformatted text preview: Chapter 6 Review 10. Let u 3x du 1 dx x 1 du u dx x ln x dx ln u dx 3x ln x du 3 dx 1 du 3 3 15. Let u 4 1 u 1/3 du 3 1 3 2/3 u C 32 1 (3x 4)2/3 C 2 4 dt du 1 du 2 t2 3/2 12. Let u 1 2 sec 2 1 du u 1 ln u 2 ln t 2 5 ln (t 2 5) f (x) and its derivatives tan (+) cos x 3x2 (–) sin x 6x (+) –cos x 6 1 2 (–) –sin x d 1 d sec u tan u du 0 C 1 C x 3 sin x 1 dy y 18. Let u du sin u du 1 dw w ln w C ln cos u C ln cos (ln y) 14. Let u du C x e e x dx e x sec (e x) dx 6x sin x ln x dv 1 dx x v x 4 ln x dx cos u dw 3x 2 cos x 6 cos x x 4 dx 15 x 5 tan u du sin u du cos u Let w cos x x 3 cos x dx ln y tan (ln y) dy y cos x. g(x) and its integrals x3 1 x 3 and g (x) C sec du C 17. Use tabular integration with f (x) C C sec u 13. Let u C t 1 2 1 2 1 2 1 1/2 2t t dt du dt 5 2t dt t dt t2 5 C dt t 3/2 tt t 11. Let u C ln ln x 16. 281 sec u du ln sec u tan u C ln sec (e x) tan (e x) C 15 x ln x 5 15 x ln x 5 15 x ln x 5 151 x dx 5 x 14 x dx 5 15 x C 25 C 282 Chapter 6 Review e 3x 19. Let u e sin x dx e 3x sin x dx v 3e 3x dx du 3x dv cos x 12 x 12 x dx x 2 1 x dx x 1 x2 2 dx x2 13 x 2x x 1 C 3 1 21C1 3 dy dx 22. x dy 3 cos x e 3x dx cos x dy Integrate by parts again. 3e 3x Let u dv 9e 3x dx du v y cos x dx y sin x y (1) e 3x 10 e 3x e e sin x dx sin x dx 3x x2 20. Let u du cos x cos x dv 2x dx dx 1 2 3x xe 3 dv Let u du e sin x 3e 3x sin x 9e 3x 3x sin x dx 4 3 1 1 3 C C x3 3 y 2x 1 x Graphical support: dx 2 e 3xx dx 3 e 3x dx [ 2, 2] by [ 10, 10] 23. 1 3x e 3 1 2 3x 2 1 1 xe xe 3x e 3x dx 3 3 3 3 1 2 3x 2 2 xe xe 3x e 3x dx 3 9 9 1 2 3x 2 2 xe xe 3x e 3x C 3 9 27 x2 2x 2 e 3x C 3 9 27 dx C C 1 3x e 3 v x 3x 3e 3x 1 [ e 3x cos x 3e 3x sin x] 10 3 sin x cos x 3x e C 10 10 e 3x sin x dx x 2e 3x dy dt 1 t 4 1 dy v t dy y dt 4 1 t ln t 4 dt 4 y ( 3) C ln (1) C ln (t 2 2 y C 4) 2 Graphical Support: x2 1x 2 x2 dy 1x dx 2 2 x dy 1x dx 2 12 13 yx x x C 2 6 dy 21. dx y (0) y C x3 6 x 24. dy d csc 2 cot 2 dy 1 x2 2 [ 4.5, 5] by [ 2, 5] csc 2 cot 2 d 1 dy csc 2 cot 2 d Graphical support: 1 csc 2 C 2 1 C1 2 y y C [ 4, 4] by [ 3, 3] y 4 3 2 1 csc 2 2 3 2 [0, 1.57] by [ 5, 3] 1 3 283 Chapter 6 Review d(y ) dx 25. d (y ) d (y ) x2 y y (1) x 2 1 C 26. d(r ) dt cos t d (r ) 1 x2 1 2x dx x2 1 2x dx x2 2x cos t dt d (r ) cos t dt r C r (0) 1 sin t C r C sin t d(r ) y (1) 2 3 C C y x (x 2 dy y 1 1 1 x r 1) dx 13 x ln x 3 1 01 3 r (0) x3 3 t C C C 0 r cos t r ln x 1) dt 1 2 dr (cos t sin t 2 t t2 2 2) dt C 2t 1 We first show the graph of y r r C sin t 1 t2 2 Graphical support: 2 . 3 x t 2t r (0) 2 3 x ln x 1 C 0 2 3 x3 3 cos t C We first show the graph of y x 1 ( sin t x Graphical support: Let f (x) 1 1 x2 y C f (x) 0, along with the slope field for y x2 x f (x) 1 1, 2x with the slope field for y sin t r 1 along cos t. 1 . x2 [ 6, 4] by [ 3, 3] [ 0.5, 4.21] by [ 9, 21] Next, we show the graph of y We now show the graph of y for y f (x) x 2 x f (x) along with the slope field 1 with the slope field for y r cos t r sin t t 2 along 1. 1. [ 6, 4] by [ 3, 3] [ 0.5, 4.21] by [ 9, 21] Finally we show the graph of y r along with the slope field for y r [ 6, 4] by [ 8, 2] sin t cos t t2 2 2t t 2. 1 284 27. Chapter 6 Review dy y2 dx dy dx y2 dy dx y2 ln y 2 2 y Since 1 C C x x2 2 C g (x)] dx x C. 2 f (x) dx 2(1 2 g (x) dx x) (x x C 2 4e x C C is an arbitrary constant, we may write the 31. [2f (x) 2 x) 1 4 y f (x) dx (1 indefinite integral as 2 C x dx x 2 x2 2 x Ce x f (x)] dx 2 Ce x y y (0) 30. [x 2 32. [g (x) 4] dx x g (x) dx Graphical support: C 4 dx (x 2) 4x 2 Since 2 2) 3x C C C is an arbitrary constant, we may write the indefinite integral as 3x C. [ 5, 5] by [ 5, 20] 28. dy dx dy y1 dy y1 33. We seek the graph of a function whose derivative is (2x 1)(y 1) Graph (b) is increasing on [ (2x 1) dx (2x ln y 1 x2 y 1 Ce x 2 sin x is positive, x and oscillates slightly outside of this interval. This is the 1) dx x , ], where sin x . x correct choice, and this can be verified by graphing C NINT x sin x , x, 0, x . x 34. We seek the graph of a function whose derivative is e x . Since e x 0 for all x, the desired graph is increasing for all x. Thus, the only possibility is graph (d), and we may verify that this is correct by graphing NINT(e x , x, 0, x). 2 2 y Ce x x 1 1 2 1 2 y ( 1) C C 2 y 2e x 35. (iv) The given graph looks like the graph of y 2 x satisfies 1 36. Yes, y Graphical support: 37. (a) dy dx 2 dv [ 3, 3] by [ 10, 40] 29. 2t 1 Since 1 4 x) x C x C. 0 4 C C is an arbitrary constant, we may write the indefinite integral as 6t) dt 3t 2 C Initial condition: v f (x) dx (1 6t (2 2t 4 when t C v f (x) dx 1. x is a solution. dv dt v 2x and y (1) C 3t 2 4 1 1 v (t) dt (b) 0 3t 2 (2t 0 t2 t3 4) dt 1 4t 0 6 0 6 The particle moves 6 m. 0 x 2, which Chapter 6 Review 38. 285 40. Set y1 (2 y)(2x 3) and use IMPEULT with intial values x 3 and y 1 and step size 0.1 for 20 points. x y 3 2.9 x 0.6680 2.8 [ 10, 10] by [ 10, 10] 39. Set y1 y cos x and use EULERT with initial values x 0 and y 0 and step size 0.1 for 20 points. 1 0.2599 2.7 0.2294 y 2.6 0.8011 0 0 2.5 1.4509 0.1 0.1000 2.4 2.1687 0.2 0.2095 2.3 2.9374 0.3 0.3285 2.2 3.7333 0.4 0.4568 2.1 4.5268 0.5 0.5946 2.0 5.2840 0.6 0.7418 1.9 5.9686 0.7 0.8986 1.8 6.5456 0.8 1.0649 1.7 6.9831 0.9 1.2411 1.6 7.2562 1.0 1.4273 1.5 7.3488 1.1 1.6241 1.4 7.2553 1.2 1.8319 1.3 6.9813 1.3 2.0513 1.2 6.5430 1.4 2.2832 1.1 5.9655 1.5 2.5285 1.0 5.2805 1.6 2.7884 1.7 3.0643 1.8 3.3579 1.9 3.6709 2.0 4.0057 x x 41. To estimate y (3), set y1 initial values x 0 and y points. This gives y (3) 1 and step size 0.05 for 60 0.9063. 42. To estimate y (4), set y1 with initial values x x2 2y x 1 and y 60 points. This gives y (4) 43. Set y1 e (x x 0 and y 2y and use IMPEULT with 1 y 2) 1 and use EULERT 1 and step size 0.05 for 4.4974. and use EULERG with initial values 2 and step sizes 0.1 and 0.1. (a) [ 0.2, 4.5] by [ 2.5, 0.5] ...
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