Business Calc Homework w answers_Part_70

Business Calc Homework w answers_Part_70 - 346 Section 8.4...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 346 Section 8.4 5. x 2 6 x2 x2 5x x 2 8 x2 x2 5x 6 (x 5x 6 x 5x 2 A( x (A 3)(x x A 2 B(x B 5 and 2A 3B 17, B x2 x 2 x2 0 and 2A 5x 6. y 2 y3 y2 6 4 y3 y3 x 0y 2 0y 2 9. y 2 (1 x2 1 1 1 x)(1 1 B y B( y 1 3) B)y (A 3B) 3B 0. 3 ,B 4 Solving the system simultaneously yields A y dy 2y 3 y 3/4 dy 3 3 ln y 4 1 1/4 dy y1 1 ln y 1 4 3 x) B(1 B)x y2 (A A y 4 A( y 1) 1 . 4 C By (A B) y(y 4 y y x) y y y2 10. x 0 and A B 1. B)y A 1 . 2 1/2 1/2 dx dx 1x 1x 1 1 ln 1 x ln 1 2 2 1 1x ln C 2 1x 1) B y 1 Equating coefficients of like terms gives Solving the system simultaneously yields 1 A ,B 2 dx 1 x2 1) 1 and A Equating coefficients of like terms gives A 3 B x A(1 (A y A( y x) A 1) B 3 B y2 x2 3)(y Equating coefficients of like terms gives The rational function cannot be decomposed any further. 7. 1 1 . 2 C y 4y 1 y2 4 y (y A 2 y 1 0 1 0y 4y 4y 3 y 2y y2 A 1 4 2y 2 (A x 1 ,B 2 1/2 dx 2 x 1 1 ln x ln x 2 2 1 x ln C 2 x2 12 3 1 1/2 dx x 2 17 1 2A 2x 12. 8 B)x dx 3B) Solving the system simultaneously yields A Bx Solving the system simultaneously yields A 3) ( 2A B Equating coefficients of like terms gives A 2 2) Equating coefficients of like terms gives B 3 B x (A 6 2) A 2) A( x 2 B)x 2 A x 1 2) 5x x2 x(x 2x x2 5x 6 2x 1 5x 1 5x x2 8. x 2 1 8 6 2 0x 5x 5x A B 1 and A 4. Solving the system simultaneously yields A x C y y2 4 dy y 4 dy y 4 ln y y 3 dy 1 3 ln y 1 C 4, B 3. Section 8.4 11. t 3 t2 1 t2 t3 1 t(t 2 2t A t 2t A( t 2)(t A( t 2 t (A t 2) t(t B 1) 2 t 1 5x 2 5x 2 x 1 B(t)(t 1) C(t)(t B( t 2 t) C(t 2 2t) C)t 2 (A B 2C)t 2A x 5x 2 x 2 5x 5 x2 x 1 5 1 5 5 5 5 5x 5x 5x 2) 2) B 15. x 2 C t 1) 2)(t 5x 2 dx x x1 5 dx 2 x 5 5x B C 0, A B 2C 0, and 2A 1 1 1 ,B ,C . 2 6 3 1/2 dt dt t t 2 2t 1 ln t 2 t 3 12. 2t 3 2t(t 2 8t t3 2t 3 8t t A t 3 4) 1/6 dt t2 1 ln t 6 t t 2A(t 2)(t 2A(t 2 4) (2A x2 1/3 dt t1 1 ln t 1 3 2 x C 2)(t 2) 2B 2C 2B(t 2 2) 0, 4B 3 ,B 8 3 dt 8t t 2t 3 13. s 2 4 s 3s 3 s3 2C(t 2 2t) (4B 4C)t 2t) 8A 4C 1, 8A s2 s2 5 1 ,C . 16 16 3/8 5/16 dt dt t t2 3 5 ln t ln t 2 8 16 s 4 s3 (x 12 2 1 2 (x x x x2 4 ds 1/16 dt t2 1 ln t 16 2 1)(x (s 2 13 s 3 2 x 2 ln (s ds (A 4) 2s s2 1 1 s 2 1) ds s ln (s 2 1/2 3/2 2x 1 tan 1 3 5 1) tan 2x 1 1 3 2x 1) (x 3 C 1) 1)2 2 x 2x x C (x 1 1)2 1)(x 1) B)x 2 B 1 B(x 1)2 1)(x A 2x A( x A( x 4 ( 3/2)2 x 1 tan 3 1)(x 2 (x 2 1/2)2 3 1 1) x2 s2 2 dx 3/4 dx C 16. (x s ds 1 1 2 1) 5 ln (x 2 2 5x 1) C(x 1) 2 1) C(x 1) A B B(x (2A C)x C Equating coefficients of like terms gives s2 1 4 3 2 1 s + 0s + 0s 2s + 0 s 4 + s3 s 2+ s + 0 s 2 2s + 0 s2 + s 1 2s 1 s 4 2s s2 s2 1 s 4 2s ds s2 1 3 4 1/2 1/2)2 5x 2 dx x1 4s s2 4 12 s 2 14. s 2 x 1/2 dx 1/2)2 3/4 The second integral was evaluated by using Formula 16 from the Brief Table of Integrals. 3 s 4s 4s 4s 4s 3 4 dx 1 ln x 2 2) 12 2 x x1 dx 1/2)2 3/4 x2 s3 1 3 4 1 ln (x 2 2 2C(t)(t Solving the system simultaneously yields A x 1 4 x 1 ln (x 2 2 2B(t)(t dx 1 (x 2) 2 2C)t 2 2B x2 1 1 x2 Equating coefficients of like terms gives 2A x C 2 x (x 2t(t B 1 x2 dx 1 To evaluate the second integral, complete the square in the denominator. Solving the system simultaneously yields A x x 5 1. 1 x2 Equating coefficients of like terms gives A A B 1, 2A C 0, and A B C 0. Solving the system simultaneously yields 1 ,B 4 A 1 2s s2 2s ds s2 1 1) 347 s2 1 ds s tan 1 2 1 C 1 . 2 x 2 dx 1)(x 2 2x (x 1 x 1 s 3 ,C 4 1/4 dx 1 1 ln x 4 1 1) x 3/4 dx 1 3 ln x 4 (x 1 1/2 dx 1)2 1 2(x 1) C C 348 Section 8.4 17. (x 2 1)2 1)2(x (x 1 1)2 A (x 2 1)2 21. x 3 B C 1)2 1)(x 1)2 B(x 1)2 1)(x 2 x2 2x 2 A 1)2 x A(x A(x 3 1 x2 C)x x2 x 2B B C 2x 1)2 x3 1)2(x x2 1) C 2D)x 2x C A 0, 2B A B C C 2D D) x (A B D 18. x 2 5x x x2 x 6 4 5x 6 4 1 ,D 4 C 1, B 2 6)(x A x B C D 1 A( x 1) 2 1 dx B)x 1 (x A 4x 4 x3 x2 1 4x r 2 2 dr 2r A 1, r2 4r 5 3 dr 4r 5 1) C x C x 1 1) B)x 2 A 3, B x2 B (Bx (A C 2, C 4x x3 C)(x B 1) C)x 4, and A 4 (A C) C 4. 1 1. 3 dx x 1 dx 3 ln x 3x 2 2x 12 (x 2 4)2 3x 2 4. 2x Ax x2 x2 1 2x 1 dx x1 ln (x 2 x 1) B)(x 2 4) (Cx D) Bx 2 (4A C)x 4B C D 4)2 (Ax 12 Cx (x 2 Ax 3 23. B 4 D Equating coefficients of like terms gives 5/7 dx x1 5 ln x 1 7 6 A C 0, B 2r 1 2 dr (r 1)2 1 2 tan 1 1)2 (r 1 (r A 0, B r2 4r (r 3 dr 2)2 4 1 1 2)2 (r 3 tan 1 (r 2 (x 2 1 1) 3, 4A C 2, and 4B D 12 Solving the system simultaneously yields C 3, C 2x 12 4)2 dx 2, D 0. 3 x2 4 dx 3 x tan 1 2 2 2x dx (x 2 4)2 1 C x2 4 The first integral was evaluated by using Formula 16 from the Brief Table of Integrals. 20. Complete the square in the denominator. r2 x 1) x2 A(x 2 4 B A 2/7 dx x6 r2 2 2. Equating coefficients of like terms gives 3x 2 C) Solving the system simultaneously yields 19. Complete the square in the denominator. 2r x 1 6B) 2 ln x 7 r2 (A C ln (x 2 Bx x Solving the system simultaneously yields 2 5 A ,B . 7 7 x4 dx x 2 5x 6 C)x 1) 2x 1 dx x2 x 1 1 A x 6) 6B B 1 dx 1 1)(x 2 1 . 4 (A (A C)(x 2, and A x 2 1 B(x (Bx 1. 22. x 3 1) x 1 1) C ln x B 6 x 2, C 2x x3 1/4 1/4 1/4 dx dx dx (x 1)2 x1 (x 1)2 1 1 1 ln x 1 C 4 4(x 1) 4(x 1) (x 1 and 1, A A D) Equating coefficients of like terms gives B B x B C x B)x 2 (A (A A A( x 2 2 0, 0, and A x2 1 Solving the system simultaneously yields Solving the system simultaneously yields 1 1 A ,B ,C 4 4 dx (x 2 1)2 1/4 dx x1 1 ln x 1 4 x 2x A 1) 2 Equating coefficients of like terms gives A 1 1) Equating coefficients of like terms gives 1) D(x 2 1) (A (A (x C(x B(x 2 1) 3 1 x Bx (A x C(x 3 (A (x D x D(x 1 (x 1 1 2) C Section 8.4 24. x 3 2x 2 2 (x 2 1)2 x3 2x 2 Ax (x 2 Cx (x 2 B 1) (Ax B)(x 2 Ax 3 2 Bx 2 D 1)2 1) (Cx (A 1 dy y2 y 1 dy y(y 1) 1 y(y 1) 27. C)x D) (B D) Equating coefficients of like terms gives A 1, B 2, A C 0, B D e x dx e x dx A y 1 ex B y 1 A( y 1) (A 2 C B(y) B)y A Solving the system simultaneously yields Equating coefficients of like terms gives A A 1, B 2, C 1, D 0. x 3 2x 2 2 dx (x 2 1)2 x x2 1 ln (x 2 2 25. x (x 2 1)2 2 x2 1) 1 y(y dx dx 1 2 tan (x 2 1 1 1 2(x 2 1) x 1 1 d 0 0 1 1 2 0 1 1 1 y ex 1 1, B 1. dy 1 C2 C 2. C or C ln y 1 1 ln 2 dy ex 1 sin 1 ln 2. d 1)2 1 dy sin d ( y 1)2 1 cos C y1 Substitute x 0 1 0 2 ,y 0. C or C 1 The solution to the initial value problem is 1 y 2 1 d 2 d 0 2 0 2 tan 0 2 1 1 1 1 tan 2 1 1 cos 1 1 y d cos 1 1 2 x2 dx 3x 2 3x 2 (x 29. dy 0 1 cos 1 y 1 2 1 1 ln y 1 0, y 0 (y ln 2 2 2 ln 2 1 1 +1 1 1 2 2 ln y Substitute x C d 1 0 1 2 ln y 1 ln 1 dx 28. d 2 1 dy y dy 1) ln y 1 0 1 The solution to the initial value problem is 1 2 1)2 1 1 A ln y x 1 +1 1 1 1 0 and 1 dx 1 B Solving the system simultaneously yields A 2 dx 1 x x2 26. 349 x 2 2)(x x2 1 3x 1 A( x 1) B(x 1 (A B)x A 1) A 2 B x 2 x 1 2) 2B Equating coefficients of like terms gives A B 0, A 2B 1 Solving the system simultaneously yields A dy y dx 3x x2 ln x 2 Substitute x 0 0 dx 2 x 2 ln x 1 C 3, y ln 2 dx x 1 0. C or C ln 2 The solution to the initial value problem is y ln x 2 ln x 1 ln 2 1, B 1. 350 Section 8.4 ds 2 dt t 2 2t 1 ds 2s 1 2t t(t 2) 2t A t t 1 A( t 2) Bt Let u 1 30. (A B)t 2A with x 2s 2 ds 2s t2 1 t2 32. (a) Complete the square in the denominator. x2 C1 B 0 and 2A 2t 1 ,B 2 1/2 dt 2 t ln s 1 ln t ln t Substitute t 1, s 0 ln 3 1 ln s 2 C or C s ln t 1 ln 2 1 t ln t t 6t 2 ln 3 2 ln 6 . x with x (x 2 1 (x 2 4x (x 2)2 2 so du B 0.5 5 (b) x2 x d1 ln x dx 2a a a 1 1 2a x a 1 2a (x a 1 x2 C C a a (x 1 a2 3 3 1 5 ln x xa a)(x a) a2 x2 dx x2 3x 0.5 x x) Bx B)x 0 and 3A 3A 1 2.5 dx 3x x 2 3 0.5 3 1 x x 9 2.5 dx x 0.5 dx 3 x 2.5 C 1d ln x 2a dx 2.5 dx 2 Solving the system simultaneously yields A 9 du 9 u2 1 u ln 6 u 1 x ln 6 x xa a)(x a) 1 . 3 x 2)2 B 3 A(3 2.5 3. 1 ,B 3 1 Equating coefficients of like terms gives A dx, and then use Formula 18 x2 x) (A 4) C 1 1/a 2a 3 1 (x/a)2 (a 2 3x A x x2 C 1 a2 9 x(3 4x) u and a dx 4x 2 1 3x 9 x x2 x 2)2 33. Volume 1) a2 1 2a 4 a 2 x 2 2a 2 1 2a 2 (a 2 x 2)2 3x 5 1 x tan 1 2a 3 a x 2) 0.5 9 Let u (1 2.5 ln 6 2 x2 4x dx and then use Formula 17 du u 2)2 1 u tan 1 u C 2 2(1 u 2) x1 1 tan 1 (x 2 2(x 2 2x 2) 2)2 1 a2 2a 2 (a 2 31. (a) Complete the square in the denominator. 5 1)2 1. a2 x2 2a 2(a 2 x 2)2 C 6t ln 1 1 1 (a 2 x 2) x(2x)] (a 2 x 2)2 2a 2 The solution to the initial value problem is ln s u and a x d dx 2a 2(a 2 1. ln 2 (x 1 so du 1 . 2 (b) 1 1 ln t ln t 2 C2 2 2 1 1 1 ln t ln t 2 C3 2 2 1 ln s 2 dx 2x (x 2 x 1 1/2 dt t dt 1)2 1) 2 Solving the system simultaneously yields A t2 2 B Equating coefficients of like terms gives A 2x 2x 1 1 (x 2 (x 1 ln s 2 ln x 2.5 ln 3 x 0.5 3 (ln 2.5 ln 0.5 3 ln 25 6 ln 5 0.5 ln 0.5 ln 2.5) ...
View Full Document

Ask a homework question - tutors are online