Pre-Calc Homework Solutions 248

Pre-Calc Homework Solutions 248 - 248 Section 6.2 d [F(x)...

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Unformatted text preview: 248 Section 6.2 d [F(x) dx 46. (a) C] should equal f (x). 47. Let u 1 x4 x 3 dx x4 9, du 10 4x 3 dx. 1 1/2 1 1/2 u du u 4 2 1 10 9 2 3 10 0.081 2 10 9 (b) The slope field should help you visualize the solution curve y F(x). x (a) 0 9 9 (c) The graphs of y1 F(x) and y2 f (t) dt should 0 1 2 1 2 differ only by a vertical shift C. (d) A table of values for y1 y2 should show that y1 y2 C for any value of x in the appropriate domain. (e) The graph of f should be the same as the graph of NDER of F(x). 1 (b) x3 x 4 9 dx 1 1/2 u du 4 1 1/2 u 2 1 2 C 9 1 x4 x4 10 10 C (f) First, we need to find F(x). Let u x x2 1 x2 x3 x 4 1, du 2x dx. 0 9 dx 1 2 1 2 1 2 9 0 dx 1 1/2 u du 2 1 2 3 2 9 0.081 u1/2 x2 1 C x2 1 2 x2 x x2 1 1 48. Let u /3 1 (1 /6 cos 3x, du 3 sin 3x dx. 2 Therefore, we may let F(x) a) d ( dx 1. (a) cos 3x) sin 3x dx 1 1 u du 3 1 2 u 6 2 1 x2 1 C) (2x) f (x) (b) (1 cos 3x) sin 3x dx 1 2 (2) 6 1 u du 3 1 2 u 6 1 (1 6 /3 1 2 (1) 6 1 2 b) C cos 3x)2 1 (1 6 C /3 /6 [ 4, 4] by [ 3, 3] (1 /6 cos 3x) sin 3x dx cos 3x)2 1 2 (1) 6 1 2 c) 1 2 (2) 6 49. We show that f (x) [ 4, 4] by [ 3, 3] tan x and f (3) 5, where f (x) f (x) 1 1.414 0.414 1 2 2.236 1.236 1 3 4 ln d) x y1 y2 y1 y2 0 1.000 0.000 1 3.162 4.123 2.162 3.123 1 1 f (3) e) 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 [ 4, 4] by [ 3, 3] ...
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This note was uploaded on 10/05/2011 for the course MAC 1147 taught by Professor German during the Spring '08 term at University of Florida.

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