Pre-Calc Homework Solutions 247

# Pre-Calc Homework Solutions 247 - Section 6.2 dy dx dy dx...

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Unformatted text preview: Section 6.2 dy dx dy dx dy ey dy ey 247 41. (cos x)e y sin x 44. (cos x)(e y esin x) cos x esin x dx dy dx dy y dy y 4 y ln x x ln x 4 dx x ln x dx x Integrate both sides. 4 Integrate both sides. cos x e sin x dx sin x On the right, let u du 2y1/2 2y1/2 2y 1/2 ln x On the right, let u du e e e e y y y y 1 dx x cos x dx e u du e e e u sin x sin x 4 u du 4 u2 2(ln x) (ln x) 2 2 1 2 C C C C]2 C]2 C) 2 C C C C.) sin x y y ) 1/2 [(ln x)2 (Note: technically C is now C y y dy 42. dx dy dx dy e y dy e y y(e) [(ln e)2 (1 1 1 0 ln (C ln (C e x y y e e sin x) C e xe e x dx y (ln x)4 Note: Absolute value signs are not needed because the original problem involved ln x, so we know that x 0. 45. (a) Let u x dx x 1 dx u1/2 du 1 Integrate both sides. e x dx e dx x x x du e dy e y y y e C C) ln (e 2 3/2 u C 3 2 (x 1)3/2 3 d 2 Alternatively, (x 1)3/2 dx 3 C C x 1. dy 43. 2xy 2 dx dy 2x dx y2 dy 2x dx y2 (b) By Part 1 of the Fundamental Theorem of Calculus, dy1 dx x 1 and dy2 dx x 1. 1, so both are y y 1 x2 1 x 2 antiderivatives of x C (c) Using NINT to find the values of y1 and y2, we have: C 1 y(1) 1 C y C 3 1 C 0.25 x y1 y2 y1 y2 4 2 3 0 0 4.667 4.667 1 1.219 3.448 4.667 2 2.797 1.869 4.667 3 4 4.667 6.787 0 2.120 4 1 x 2 4.667 4.667 3 C (d) C y1 x 0 x y2 x x x 0 3 1 dx 3 3 x x x 1 dx 1 dx 1 dx 1 dx x 0 ...
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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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