Business Calc Homework w answers_Part_49

Business Calc Homework w answers_Part_49 - 241 Section 6.2...

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

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 241 Section 6.2 66. Use differential equation graphing mode. For reference, the equations of the solution curves are as Section 4.2, y1 and y2 must differ by a constant. We find follows. that constant by evaluating the two functions at x 0. 1 (0, 1): y x 1 2 (0, 2): y (0, 2 . By Corollary 3 to the Mean Value Theorem of 3 y2 3. y1 2x 1 1 x1 1): y (0, 0): y 0 [ 3, 3] by [ 2, 8] 4. [ 2.35, 2.35] by [ 1.55, 1.55] [ 10, 10] by [ 30, 30] The concavity of each solution curve indicates the sign of y. 67. (a) (b) d (ln x dx d [ln ( x) dx for x 1 for x x C) 0 Exploration 2 Two Routes to the Integral 1d ( x) x dx C] 5. The derivative with respect to x of each function graphed in part (4) is equal to 1 x 2. 1 ( 1) x 1 x 0 1 3x 2 1. C ln x C, which is a solution to the differential equation, as we showed in part (a). For x 0, ln x C ln ( x) C, which is a 3x 2 2 1. 0, we have y ln ( x) 1 dx x3 0, we have y 0 5 x 1 dx (x 1 part (b). Thus, dy dx s Section 6.2 Integration by Substitution 1 1 15 (0) 5 1)1/2 dx 1 3. dy dx 2 (x 3 dy dx 3x 5. dy dx 4(x 3 2x 2 3)3(3x 2 4x) (pp. 315–323) 6. 1 x 2 2x dx u du 2 sin (4x 5) cos (4x 5) 4 5) cos (4x 5) 7. 2 3/2 u C 3 2 (1 x 2)3/2 3 2. Their derivatives are equal: dy dx 8 sin (4x Exploration 1 Supporting Indefinite Integrals Graphically 1. dy1 dy2 dx dx 1 x 2 2x. dy dx 1 cos x 8. C 0 42 3 23 (x 1)3/2 so 3 42 . 3 32 5 3x 4. 1 for all x except 0. x 1)3/2 2 3/2 2 3/2 (4) (0) 3 3 2 16 (8) 3 3 C1, which is a solution to the differential equation, as we showed 15 (2) 5 0 5 C2, which is a ln x 23 (x 3 2 solution to the differential equation, as we showed in part (a). For x 2 3/2 u 3 u du 1 dx 15 x 5 x 4 dx 2. (d) For x x 3 2 Quick Review 6.2 1 for all x except 0. x d ln x dx 2 1 solution to the differential equation, as we showed in part (b). Thus, 2 3/2 u 3 u du 0 2. 3x 0, ln x 1 dx 1 1 (c) For x 2 x3 dy dx 1 sin x sin x cos x tan x cot x 1)3/2 5 1 242 9. Section 6.2 dy dx 1 (sec x tan x sec x tan x sec x tan x sec2 x sec x tan x sec x(tan x sec x) sec x tan x 4. u sec2 x) du dy dx dx d Check: [(7x dx 1 ( csc x cot x csc x cot x csc x cot x csc2 x csc x cot x csc x(cot x csc x) csc x cot x csc2 x) 3 du dx d Check: dx 1 sin u du 3 1 cos u C 3 1 cos 3x C 3 1 cos 3x 3 C 1 ( sin 3x)(3) 3 6. u du 1 1 r 3 9 1 3 3u 1 cos u du 4 1 sin u C 4 1 sin (2x 2) C 4 1 d1 Check: sin (2x 2) C cos (2x 2)(4x) 4 dx 4 Check: x cos (2x2) du d (6 dx 1 C r3 r3 C C) 21 r3 1 du 2 dx 1 du 2 Check: 2x d1 sec 2x dx 2 1 cos t 2 1 t sin dt 2 2 t 2 du sin dt 2 t2 t 1 cos sin dt 2 2 du dx sec 2x tan 2x dx 1 6 9r 2 7. u 1 sec u tan u du 2 1 sec u C 2 1 sec 2x C 2 1 C sec 2x tan 2x 2 2 sec 2x tan x2 u 1/2 61 2x 1 9 du 3(2)u1/2 x cos (2x 2) dx 3. u 1 3 x2 3 r3 1 du r 2 dr 3 9r2 dr 1 du 4 x dx 1 1 31 3r 2 dr du 4x dx 2)3 3u 9 sin 3x 2x 2 2. u 28(7x C 9u 2 9 du 3 9 u2 1 1 du 3 u2 1 1 tan 1 u C 3 1 x tan 1 C 3 3 d1 x Check: tan 1 C dx 3 3 3 dx 2)3(7) 4(7x 2)4 (7x dx dx 3x sin 3x dx C] C 1 dx 3 du Section 6.2 Exercises 1 du 3 2)4 u4 x 3 5. u x2 du 1 28u3 du 7 2)3 dx 28(7x csc x 1. u 2 7 dx 1 du 7 sec x 10. 7x 2 u2 du 23 u 3 2 1 3 t3 d2 1 cos dx 3 2 t2 t1 2 1 cos sin 2 22 2 t t 1 cos sin 2 2 Check: C C cos t3 2 C r3 ( 3r 2) Section 6.2 y4 8. u 4y 2 14. Let u du (4y 3 8y) dy du 3 tan x du 1 sec2 x dx 2y) dy 4( y /4 1 du 4 (y3 8( y 4 2y) dy 4y 2 /4 1)2(y 3 2y) dy 8 1 4 d24 (y dx 3 4y 2 2( y 4 4y 2 1)2(4y 3 8( y 4 4y 2 1)2( y 3 1)3 13 u 3 C 4y 2 1)3 C 1 3 8y) 15. Let u du 9. Let u 1 du x dx 1 1 x du 2 sec (x 2) dx C 2) 2 3/2 u C 3 2 3/2 (cot x) 3 C 17. Let u tan x du 2 tan x sec x dx 1/2 u du 2 3/2 u C 3 2 (tan x)3/2 3 du C 2 tan 2 2 d sec u tan u du sec u ln x 1 du dx x 6 dx e x ln x ln 6 1 du u ln 6 ln u ln (ln 6) 1 2 C tan 2 u7 du 18 u 8 1 x tan8 4 2 2 C sec 13. Let u u6 du x 2 1 2x du sec dx 2 2 x 7x tan sec2 dx 2 2 d sec 1 dx x 17 u C 7 17 (ln x) 7 18. Let u 12. Let u ln x ln6 x dx x sec x dx 2 u1/2 du sec u du tan (x C C cot x 2 tan u du 1 cos u du 3 1 sin u C 3 1 sin (3z 4) 3 cot x csc2 x dx 2 2 3 csc2 x dx du dx 11. Let u 4) dz C 16. Let u x 1 3 4 C 1 1 ( 1)3 3 dz cos (3z u 10. Let u 3z du u2 x)2 (1 1 3 dz 1 du 3 dx 1 1 (1) 3 C 2y) u2 du 1 u2 du 23 u 3 24 (y 3 Check: 1 tan2 x sec2 x dx C C C 243 244 Section 6.2 s4/3 19. Let u du 3 du 4 3 /4 cot x dx /4 /4 Let u s1/3 ds dx sin2 3x Let u 23. 4 1/3 s ds 3 s1/3 cos (s4/3 20. 3 /4 8 3 cos u du 4 3 sin u C 4 3 sin(s4/3 8) 4 8) ds du cos x dx sin x sin x cos x dx 3 /4 C /4 x 3 /4 cos x dx sin x csc2 3x dx 1 /4 u x x ln u 3 /4 x 3x du /4 3 /4 du 1 du 3 ln sin x 3 dx /4 dx 24. Let u du du cos (2t sin (2t 1 du sin (2t 2 sin (2t 1) dt cos2 (2t 1) 1) 2 2 1) dt 1 u 2 du 2 x2 25. Let u ln 2 1) 1) cos t dt 1 C C du 2x dx 1 du 2 x dx x dx 2 1 1 2 10 2 1 du u 1 ln u 2 6u 2 2 1.504 C 1x 6u 0 1 du u9 ln 9 sin t 6 cos t dt (2 sin t)2 2 2 ln u 3 du 9 2 1)(2) dt 1 2 cos (2t 1 sec (2t 2 2 ln 2 dx x 0 11 u 2 22. Let u x 2 dx 7 21. Let u 2 ln 1 csc2 u du 3 1 cot u C 3 1 cot (3x) C 3 csc2 3x dx du 1 6 sin t 10 2 1 (ln 10 2 C C ln 2) 1 ln 5 2 0.805 Section 6.2 5 26. 0 40 25 5 40 dx x 2 25 30. csc x dx x2 5 0 25 2 25 5 40 25 1 dx dx x2 5 Let u 5 du csc x cot x 0 cot x csc2 x dx 1 du u csc x dx ln u dx 5 40 dx x 2 25 cot x dx cot x 1 1 dx 5 du csc x du 0 csc x csc x csc2 x csc x cot x dx csc x cot x x 5 Let u csc x 8 (5) 5 1 0 1 u2 1 C ln csc x du 1 31. Let u 8 arctan u y cot x C 1 0 du 8(arctan 1) dy 3 8 27. dx cot 3x 4 4 y 2 0 1 4 2 3/2 u 3 1 2 3/2 2 3/2 (4) (1) 3 3 2 2 14 (8) 3 3 3 sin 3x dx cos 3x Let u cos 3x du 3 sin 3x dx 1 du 3 dx cot 3x sin 3x dx 1 3 1 3 1 3 32. Let u 1 du u ln u ln cos 3x du 5x r dr 1 1 ln sec 3x 3 0 1 u1/2 du 21 0 1 2 3/2 u 23 1 1 1 (0) (1) 3 3 r 2 dr r1 C.) 0 8 5 dx 1 du 5 dx dx 5x 8 29. sec x dx Let u du 2r dr 1 du 2 C r2 1 du C (An equivalent expression is 28. Let u u1/2 du 1 dy 33. Let u sec2 x dx 0 0 tan x sec2 x dx u du /4 1 12 u 2 sec x tan x dx sec x tan x sec2 x sec x tan x dx sec x tan x sec x sec x 1 du u 0 1 1 (0) 2 1 ( 1)2 2 1 2 tan x sec x tan x sec x dx tan x du 1 u 1/2 du 5 1 2u1/2 C 5 2 5x 8 C 5 sec2 x dx ln u 34. Let u C ln sec x tan x C 1 3 du 1 du 21 1 (4 4 r2 2r dr r dr 5r dr r 2)2 5 2 5 u 5 2 du 0 245 ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online