Business Calc Homework w answers_Part_49

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

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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 ...
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