Business Calc Homework w answers_Part_50

Business Calc Homework w answers_Part_50 - 246 Section 6.2...

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Unformatted text preview: 246 Section 6.2 35. Let u 39. 3 1/2 d 2 du 2 du 3 1 1/2 dy ( y 5)(x 2) dx dy (x 2)dx y5 Integrate both sides. d 10 3/2 2 (1 0 3/2 1 ) dy 2 (10) 3 d y 2 2 u du 2 let u y 1 20 1 32 5 du 1 20 3 2) dx On the left, 1 20 1 u 3 (x 5 1 2 dy 1 du u 10 3 12 x 2 12 x 2 12 x 2 ln u 2x C 2x C 3 sin x 2x C 5 5 e (1/2)x y du 4 ln y y 36. Let u 5 e Ce(1/2)x 3 cos x dx 1 du 3 cos x dx cos x dx 1 3 1/2 u du 0 2 2x eC or C We now let C 4 2 2x C e C, depending on whether (5t 4 du 1 t5 4 2t 2) dt 2t (5t 4 u1/2 du 2) dt 2 2x C e (1/2)x 2 2x value C 3 2 3/2 u 3 2 3 27 2 3 40. dy dx dy cos 2 sin 2 d 5. y x dx y cos2 y Integrate both sides. dy x dx y cos2 sin 2 d 3 2 2x y cos2 x 2 sin 2 d /6 0 gives a solution to the original differential Ce (1/2)x y cos 2 1 du 2 5 equation), we may write the solution as 0 2 3/2 (3) 3 0 C e (1/2)x Since C represents an arbitrary constant (note that even the 0 du 5 3 0 38. Let u 5) is positive or negative. Then y t5 37. Let u (y y 3 sin x 4 y On the left, let u 1 2 1 2 1/2 u 3 du 1 11 42 1 (3) 4 1 u2 2 2 1 3 4 y 1 1/2 y dy 2 du 1/2 2 du 1/2 y dy 1 12 x C 2 12 2 sec2 u du x C 2 12 2 tan u x C 2 12 2 tan y x C 2 12 tan y x C 4 2 du cos2 u (Note: technically, C is now C y y tan tan 2 1x 4 1x C 2 2 4 C C . But C’s are generic.) 2 247 Section 6.2 41. dy dx dy dx dy ey (cos x)(e y esin x) dy dx dy cos x esin x dx Integrate both sides. (cos x)e y sin x 44. dy On the right, let u ln x dx x 4 y cos x e sin x dx du y ln x x ln x 4 dx x y Integrate both sides. dy ey 4 On the right, let u sin x du 2y1/2 4 u du 2y1/2 cos x dx ln x 1 dx x 4 u2 1 2 e y e u du e y e u e y e sin x C y (ln x) C e sin x C y [(ln x)2 C]2 y e 1/2 C y 1/2 C]2 1 (1 2 1 C 0 e sin x) x y e xe y e C [(ln e)2 y (e) ) e ln (C dy 42. dx dy dx dy ey C.) 2 sin x ln (C y 2(ln x) 2y (Note: technically C is now C C 2 C) y (ln x)4 Note: Absolute value signs are not needed because the original problem involved ln x, so we know that x 0. e x dx 45. (a) Let u x 1 Integrate both sides. dy ey y e e y e dx x ln (e C) dy 43. 2xy 2 dx dy 2x dx y2 dy 2x dx y2 y 1 y x2 x y (1) C C 3 y dy1 x x dx 1 and antiderivatives of dy2 x dx 1. x 1, so both are 1. x 0 1 2 y1 0.25 4 0 1.219 2.797 y2 4.667 y2 y1 1 2 x (c) Using NINT to find the values of y1 and y2, we have: C 1 C C C 1 1 C (b) By Part 1 of the Fundamental Theorem of Calculus, 1 2 u1/2 du 1 dx 2 3/2 u C 3 2 (x 1)3/2 3 d2 Alternatively, (x 1)3/2 dx 3 C x dx x x e dy y du e x dx 3 4.667 3.448 4.667 1.869 4.667 2 3 C 4 (d) C y1 y2 x x x 1 dx 0 x x 1 dx x 1 dx 0 3 x x 0 x 1 dx 3 3 1 dx 3 4 4.667 6.787 0 2.120 4.667 4.667 248 46. (a) Section 6.2 d [F(x) dx x4 47. Let u C] should equal f (x). 1 (b) The slope field should help you visualize the solution curve y F(x). 10 x 3 dx (a) x4 0 9 x (c) The graphs of y1 F(x) and y2 4x 3 dx. 9, du f (t) dt should 0 9 1 2 1 2 1 1/2 1 1/2 u du u 4 2 1 10 9 2 3 10 0.081 2 10 9 differ only by a vertical shift C. x3 (b) (d) A table of values for y1 y2 should show that y1 y2 C for any value of x in the appropriate domain. x 4 9 1 1/2 u 2 (e) The graph of f should be the same as the graph of NDER of F(x). x 0 x3 x 4 9 dx u1/2 x2 1 48. Let u C 9 1 2 x4 9 10 1 2 1 2 2x dx. x4 1 2 1, du 1 1/2 u du 2 dx 1 C 1 2 1 x2 (f) First, we need to find F(x). Let u x2 1 1/2 u du 4 dx 10 3 2 C 1 0 1 cos 3x, du (1 9 0.081 3 sin 3x dx. cos 3x) sin 3x dx /3 x2 Therefore, we may let F(x) (a) 1. 2 /6 a) d ( dx x2 1 1 C) 12 (2) 6 (2x) 2 x2 x x2 1 1 u du 3 1 f (x) 1 (b) (1 12 (1) 6 12 u 6 cos 3x)2 /3 (1 cos 3x) sin 3x dx /6 c) 1 (1 6 12 (2) 6 49. We show that f (x) f (x) [ 4, 4] by [ 3, 3] f (x) d) x 1 2 y1 1.000 1.414 2.236 3.162 4.123 y2 y1 0 0.000 0.414 1.236 2.162 3.123 1 1 1 y2 3 1 4 1 f (3) e) [ 4, 4] by [ 3, 3] tan x and f (3) cos 3 5. cos x cos 3 d ln 5 cos x dx d (ln cos 3 ln cos x 5) dx d ln cos x dx 1 ( sin x) tan x cos x cos 3 ln 5 (ln 1) 5 5 cos 3 ln 1 1 2 C 1 (1 6 [ 4, 4] by [ 3, 3] 2 1 u du 3 cos 3x) sin 3x dx b) 12 u 6 C /3 cos 3x)2 /6 12 (1) 6 5, where 1 2 Section 6.3 50. (a) u 2 csc2 2 d cot 2 , du 1 u du 2 1 u2 C 2 2 u2 C 4 1 cot2 2 C 4 csc2 2 cot 2 d F1( ) csc 2 , du 2 csc 2 cot 2 d 1 u du 2 1 u2 C 2 2 u2 C 4 1 csc2 2 C 4 F2 ( ) d (sin2 x C) 2 sin x cos x dx d ( cos2 x C) ( 2 cos x)( sin x) dx 1 1 d cos 2x C sin 2x (2) 2 dx 2 s Section 6.3 Integration by Parts (pp. 323–329) Exploration 1 Integrals 1. u Evaluating and Checking dx and dv x ln x ⇒ du 2 csc 2 cot 2 (d) F1( ) F2( ) b 51. (a) u sin x, du 1 (csc2 2 cot2 2 ) 4 1 1 cos2 2 1 sin2 2 sin2 2 4 4 sin2 2 2. cos x dx cos x, du 2 sin x cos x dx dx x ln x 1 4 d (x ln x dx x) x ln x 3. The slope field of 2 2u du u C 2 sin x C dy dx ( 2u) du C cos2 x C C x 1 x 1 ln x ln x shows the direction of the curve as it is graphed from left to right across the window. sin x dx u2 x. Thus, v du x ln x b dx ⇒ v u dv uv 1 csc2 2 4 2 sin x cos x dx (b) u ln x dx b 1 cot2 2 4 2 sin x cos x 2 sin x cos x 1 cot 2 ( 2 csc2 2 ) csc2 2 cot 2 2 1 csc 2 ( 2 csc 2 cot 2 ) 2 (c) F1 ( ) sin 2x dx sin 2x 1 csc2 2 4 F2( ) 2 dx 1 sin u du 2 1 cos u C 2 1 cos 2x C 2 (d) csc2 2 cot 2 du 2x, du 2 sin x cos x dx 1 cot2 2 4 (b) u (c) u 249 [0, 6] by [ 2, 5] 250 Section 6.3 4. The graph of y2 x ln x x appears to be a vertical shift of the graph of y1 1 ln t dt (down 1 unit). Thus, y2 appears to be an antiderivative of ln x which supports x ln x x dy dx x sin x dy 9. x (x sin x)dx Integrate both sides. C as the set of all antiderivatives of dy ln x. (x 12 x 2 y y (0) sin x) dx cos x 10. [0, 6] by [ 2, 5] dy dx (x 3)(cos 2x)(2) 2x 3 cos 2x 2. dy dx 1 1 (2x)2 2 1 4x 2 dy dx 5. y tan tan y 1)(2e 2x) 2 x du dv 1 3) 2. Let u du cos x dx sin x sin x C C) (cos x)( 1) cos x x2 1 2x dx 1 cos e 2x dx Integrate both sides. C sin x 2 x sin x dx 1 2[ x cos x (x 2 cos 0 (2x)( sin x) sin x] 2)sin x d Check: [2x cos x dx cos x e 2x x 2 sin x 2x cos x 1 ( 1) cos x dx v x 2 sin x (x 2 C C 2)sin x (2 cos x)(1) (sin x)(2x) 2 x 2 cos x e 2x dx dv Using the result from Exercise 1, 0 1 2x e 2 x cos x x 2 cos x dx 1 y cos x 2 3x 1 dy v dx sin x dx ( x)( sin x) (x sin x dx dy 1x e sin x 2 Section 6.3 Exercises x cos x cos 1 (x 1) x1 cos y 1 dy dx 1x e sin x 2 1 2 1x e cos x 2 cos x) e x (sin x d Check: ( x cos x dx 0 8. sin x) x sin x 1 7. cos x) x sin x dx 1 tan y 3 y cos y x 3 1) 3x x 6. cos x 1 1 3 d1x e (sin x dx 2 1x e (cos x 2 1x e cos x 2 1. Let u ln (3x 3x 1 3e 2x 2e 2x ln (3x 3x 1 dy 3. dx 4. (sin 2x)(3x 2) 3x 2 sin 2x 3 (e 2 x) 2 e x sin x Quick Review 6.3 1. 12 x 2 C C y 1 C (x 2 C] 2)(cos x) ...
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