Pre-Calc Homework Solutions 255

Pre-Calc Homework Solutions 255 - Section 6.3 23. Let dv v...

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Unformatted text preview: Section 6.3 23. Let dv v e4xdx 1 4x e 4 255 19. y x 2e4x dx x2 2x dx 1 4 u du x dx x cos x x cos x dv v sin x dx cos x Let u du y x sin x dx cos x dx sin x C (x 2) e4x 1 2 4x x e 4 1 4x e (2x dx) 4 1 xe4xdx 2 (a) 0 x sin x dx 0 x sin x dx x cos x ( 1) 0 sin x 0 Let u du y y y 20. y x dx dv v 1 1 (x) e4x 2 4 1 4x 1 4x xe e 8 32 1 4x e C 32 e4x dx 1 4x e 4 1 4x e dx 4 0(1) 0 1 2 4x x e 4 1 2 4x x e 4 x2 x 4 8 2 2 C (b) x sin x dx x cos x 2 (1) 3 0 sin x 2 x sin x dx ( 1) 0 x ln x dx ln x dv x dx (c) 0 2 Let u 2 2 2 y y y 21. y 1 1 3 du dx v x x 3 1 1 3 1 (ln x) x 3 x dx 3 3 x 1 3 1 2 x ln x x dx 3 3 1 3 1 3 x ln x x C 3 9 x sin x dx 0 x sin x dx 3 4 x dv v e e 1)e x x x sin x dx 24. We begin by evaluating (x 2 Let u du x2 (2x 1)e (x 2 Let u du (x 2 x (x 2 (x 2 (x 2 2x 2 dx 1)e x x 3x x dx. x 1 1) dx x dx x sec sec 1 1 d dv 2 Let u du d 1 2 2 (x 2 x dx 1)e x 1 1 du v x 1 (2x dv v e e 1)e x x dx dx Note that we are told in the expression for du. y y (sec 2 1, so no absolute value is needed x dx x x x 1 1 2 ) 2 sec 1 2 2 1 4 1 2 2 2 d 2 1 2 d 1 1)e 1)e 4)e (2x (2x C 1)e 1)e x x 2e 2e x x dx C 1 Let w 2 1, dw 1 1 1 2 d y y y 22. y Let y y 2 2 sec sec sec 2 2 2 1 w 1/2 dw 4 1 1/2 w C 2 1 2 1 C 2 [ 3, 3] by [ 3, 3] sec tan u du d sec sec d dv sec v sec d tan C tan d The graph shows that the two curves intersect at x where k 1.050. The area we seek is k 0 k, (x 2 x (x 2 1)e x k 0 x 2 dx x k sec ln sec 3x 4) 4)e 0 1 3 k x 3 0 Note: In the last step, we used the result of Exercise 29 in Section 6.2. ( 2.888 0.726 (0.386 0) ...
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