Math3B Exam #02 Solution Fall 2017

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1 . Integrate. a) 8 MathB Eam # Solution Fall 7 e d b) ln e e d

2 . Integrate. 6 d

3 . Integrate. sin cos d

4 4. Use Simpsons Rule with n 6 to approimate sin d. Then use integration to get the eact value. 6 6

5 5. Integrate a) cos 5 sin d b) tan 5 sec d

6 6. Integrate using trigonometric substitution. a) 4 4 d b) 5 d

7 7. Solve the initial value problem using the integrating factor method. dy sec y sec tan y 4 d

8 8. Integrate. ln a) d b)

17 MathB Practice Eam #. Integrate. a) e d b) tan d

18 . Integrate. a) d (Hint: Let u ) b) 4sin cos d

19 . Use Simpsons Rule with n 6 to approimate get the eact value. cos d. Then use integration to

20 4. Integrate. a) 4 6 tan d b) sin cos d

21 5. Integrate using trigonometric substitution. a) 4 9 d b) 4 7 d (First complete the square.)

22 6. Show that the solution to the initial value problem is a hyperbolic function. Use the integrating factor method. dy d y e y

23 7. Integrate. ln a) d b) e d

31 MathB Eam # Review Problems Solutions provided by Romeo Mercado and Matthew Staley of the Learning Assistance Program at Saddleback College. March 9,. Integrate. sin a) d b) π 7 sin e e 4 ( e ) d + C c) 4 + ln ( ) ln tan d + C d) e) + 4 d e d f) tan d ln C e e + C π π ln + 6. Integrate. π e π cos ln d e b) a) ( ). Integrate. e d e + ln a) π 4 tan d b) 5 6 tan tan tan d + tan + C 5 c) + 6 d + 4 ln C e d) ln d e Integrate. a) cos(ln ) d cos(ln ) + sin(ln ) + C b) π 4 tan d ln 5 c) e d (Let u ) d) 4 e e + e + C 4 e ln d e) e π 4 6 tan d π 5 4

32 5. Integrate a) tan sec d b) sin cos d c) cot csc d 9 7 tan tan + + C cos cos + C cot cot + C 7 5 d) + 4 d e) 9 d f) ( + ) 4 9 / d C 4 9 sin + C + 4 g) d + 4 ln + ln ( + 4) tan + C 6. Integrate. (Let u tan ) a) d + sin + C tan + h) + ( )( ) d b) d sin 4cos + i) d ln + ln C ( + ) ln 5 tan tan + ln + ln + C + C 7. Use Simpsons Rule with n 6, to approimate π sin d. Then use integration to get the eact value. Compare the two results. π 8. Integrate. a) d + 9. Solve the initial value problem. dy b) e d c) ln d π π y+ cos y y y a) y e e ; y ( ) b) y ; y ( ) d c) y ; y ( ) y ln + y 4e y + sin y + π

33 Math B Eam # Review Problem Solutions Mr. Perez, Spring Solutions by Romeo Mercado and Matthew Staley, LAP, Saddleback College March 9,

34 I highly recommend that the student first attempt these problems on their own before looking at the solutions.. (a) / sin () π/6 d u du u π/6 ( (π ) ) ( ) π π u sin () u π 6 du d u. (b) e d e 4 u du sin (u) + C sin (e ) + C u e du e d du e d. (c) [ 4 + (ln ) ] d u ln u ln Note: match constant with u-sub u du d 4 + (u) du 4 + 4u du 4( + u ) du + u du tan (u) + C tan ( ) ln + C

35 . (d) + 4 d sec θ dθ 4 tan θ tan θ tan θ d sec θ dθ + 4 θ sec θ dθ 4(tan θ + ) }{{} 4 sec θ sec θ dθ sec θ sec θ dθ sec θ (sec θ + tan θ) (sec θ + tan θ) dθ sec θ + sec θ tan θ sec θ + tan θ dθ u sec θ + tan θ, du sec θ tan θ + sec θ dθ u du ln u + C ln sec θ + tan θ + C ln C ln + C

36 . (e) e d Let w dw d dw d w dw d we w dw u w dv e w dw du dw v e w [ we w ] e w dw [we w e w ] + C [ e e ] + C e [ ] + C. (f) tan d Let w, w, dw d dw d w dw d w(tan w)w dw w tan w dw Integration by parts u tan w dv w dw du v + w w

37 w tan w w + w dw Let z + w w, z, 4 z w dz w dw dz w dw Note that w w w in the integral above so that the z-substitution gives: w tan w w tan w 4 z z 4 z dz ( ( ) ) tan tan () ( z ln z dz ) 4 ( π π ) ( ) 4 ln(4) [ ln()] 4 π π 6 + ln(4) ln() (Note that ln(4) ln( ) ln()) π π 6 + ln() ln() π π 6 + ln() π π 6 + ln() 4

38 . (a) e π/ cos(ln ) d Let w ln, w, π dw d dw d e w dw d (w ln e w ) π/ e w cos w dw ( ) u cos w du sin w dw dv e w dw v e w e w cos w e w cos w π/ π/ + π/ π/ e w ( sin w) dw e w sin w dw u sin w du cos w dw dv e w dw v e w e w cos w π/ + e w sin w π/ π/ e w cos w dw }{{} Same as ( ) above, add to both sides π/ e w cos w dw e w cos w π/ + e w sin w π/ [ e π/ cos π ] [ e cos + e π/ sin π ] e sin [ ] + [e π/ ] e π/ π/ e cos w dw ( e π/ ) 5

39 . (b) e d e u dv e d du ln d v e e d + (ln ) e d e e (ln ) d e (ln ) e d } {{ } Add to both sides [ ] ( + ln ) e d e e e e e d e + ln. (a) tan d u tan dv d du + d v tan + d tan + d tan ( tan ) [ tan tan ] ([ tan ] [ tan ]) [ π ] ( π ) π 8 + π 8 π 8 π 4 6

40 . (b) tan 6 d ( Idea is to break off tan as sec ) tan 4 tan d tan 4 (sec ) d tan 4 sec d tan 4 d tan 4 sec d tan tan d tan 4 sec d tan (sec ) d tan 4 sec d (tan sec tan ) d tan 4 sec d tan sec d + tan d tan 4 sec d tan sec d + (sec ) d }{{} utan dusec d u 4 du u du + sec d d u5 5 u + tan + C tan5 5 tan + tan + C 7

41 . (c) + 6 d ( + + ) d Complete the square for + ( + + ) d ( + ) 6 d Now for a Trig Sub: I want ( + ) 6 sec θ ( + ) 6 sec θ ( + ) sec θ sec θ d 4 sec θ tan θ dθ ( + ) θ 4 ( + ) sec θ tan θ dθ 6 sec θ 6 sec θ tan θ dθ 6(sec θ ) 4 sec θ tan θ dθ 6 tan θ 4 6 sec θ tan θ dθ 8

42 4 cos θ cos θ sin θ dθ sin θ dθ csc θ dθ ln csc θ cot θ + C Use the triangle found above ln ( + ) C ln C 9

43 . (d) e ln d u ln dv d du d v / / ln e e e / ln e / ln / d e / d ( ) / e ([ ] [ ]) e/ ln e ()/ ln 4 ( e / /) 9 ( ) e/ 4 9 e/ e/ 4 9 e/ e/ [ e / + ]

44 4. (a) cos(ln ) d Let w ln ( Similar to a above) dw d dw d e w dw d (w ln e w ) e w cos w dw ( ) u cos w du sin w dw dv e w dw v e w e w cos w e w cos w + e w ( sin w) dw e w sin w dw u sin w du cos w dw dv e w dw v e w e w cos w + e w sin w e w cos w dw }{{} Same as ( ) above, add to both sides e w cos w dw e w cos w + e w sin w + C e ln cos(ln ) + e ln sin(ln ) + C cos(ln ) + sin(ln ) + C cos(ln ) d cos(ln ) + sin(ln ) + C

45 4. (b) tan d u tan dv d du + d v tan tan tan + d u +, u, du du d u d ln u ( tan tan ) (ln ln ) π 4 ln π 4 ln / π 4 ln

46 4. (c) 5 e d 4 e d u u 4 du d du d u e u du ( u e u e u + e u) + C ( 4 e e + e ) + C Derivative Integral u u + + e u e u e u e u 4. (d) e 4 e ln d e 4 e ln d u ln d e, e 4 u, 4 du d 4 u du ( u ) 4 4 ( 4 ) (6 4) 4 4

47 4. (e) π/4 tan 6 d (See b Above) ( tan 5 5 tan ) π/4 + tan [ tan 5 π 4 5 tan π 4 + tan π 4 π ] [ tan tan ] + tan [ 5 ] + π [ + ] π 4 5 π 4 5. (a) tan 6 sec 4 d tan 6 (sec )(sec ) d tan 6 ( + tan )(sec ) d (tan 6 + tan 8 ) sec d u tan, du sec d u 6 + u 8 du u7 7 + u9 9 + C tan7 + tan9 + C 7 9 4

48 5. (b) sin cos 4 d sin (sin ) cos 4 d sin ( cos )(cos 4 ) d sin (cos 4 cos 6 ) d u cos, du sin d u 4 u 6 du u 6 u 4 du u7 7 u5 5 + C cos7 cos5 + C (c) cot 4 csc 4 d cot 4 (csc )(csc ) d cot 4 ( + cot )(csc ) d (cot 4 + cot 6 ) csc d u cot, du csc d ( ) u u 4 + u 6 5 du 5 + u7 + C cot5 cot7 + C

49 5. (d) + 4 d 4 tan θ tan θ d sec θ dθ tan θ + 4 θ sec θ dθ 4 tan θ 4 tan θ + 4 sec θ dθ 4 tan θ 4(tan θ + ) sec θ dθ 4 tan θ( sec θ) sec θ dθ 4 tan θ 4 cos θ cos θ sin θ dθ cos θ 4 sin θ dθ u sin θ, du cos θ dθ 4 u du ( ) + C ( ) + C 4 u 4 sin θ ( ) 4 csc θ + C C + C 4 4 6

50 5. (e) 9 d 9 sin θ sin θ d cos θ dθ sin θ θ sin θ 9 sin θ ( cos θ) dθ 9( sin θ) 9 sin θ (cos θ) dθ 9 cos θ sin θ (cosθ) dθ cos θ sin θ (cos θ) dθ cos θ sin θ dθ sin θ sin θ dθ sin θ sin θ sin θ dθ csc θ dθ cot θ θ + C 9 sin + C 7

51 5. (f) Z / d (4 + 9)/ / Z u d (4 + 9)/ (u 9) 4 du 8 d du d 8, 8 Z 6 (u 4 9) u/ 9 Z Z 6 9 u 9, 6 du u 9 du u/ u/ 6 u / 9u / du 9 u/ 9u / / / 8 u+ u 6 6! 9! (6 + 6) [ + ] 6 8

52 5. (g) d + 4 ( + 4) d A + B + C + 4 d Partial Fraction Decomposition: Find A, B, C : + 4 ( + 4) A + B + C A( + 4) + (B + C) (A + B) + (C) + 4A [Association] A + B C 4 4A + B A B d d + Will do each integral separately: + 4 d + 4 d d ln + 4 d u du ln u ln + 4 ln( + 4) u + 4 du d du d 9

53 + 4 d want 4 tan θ tan θ tan θ tan θ d sec θ dθ sec θ dθ 4 tan θ + 4 sec θ dθ 4(tan θ + ) sec θ sec θ dθ dθ θ tan Putting it all back together gives: d d + 4 d ln + ln( + 4) tan + C

54 5. (h) ( )( + ) d A + B + + C ( + ) d ( )( + ) A + B + + C ( + ) A( + ) + B( )( + ) + C( ) A( ) + B( 6) + C( ) (A + B) + (4A B + C) + (4A 6B + C) Association gives us the following system of equations: A + B () 4A B + C () 4A 6B C () Eliminate C by () + (): A B + C 4A 6B C 6A 9B + C (4) Add 9 () + (4): 9A + 9B 9 6A 9B 5A 9 A 9 5 B 6 5

55 To find C, use equation (): 4A B + C ( ) C C 5 + C C 4 5 A + B + + C ( + ) d 9 5( ) + 6 5( + ) 4 5( + ) d d d 4 5 ( + ) d }{{ } u+, dud ln + 5 ln u du 6 ln + 5 ln + 4 ( ) + C 5 u 9 5 ln ln ( + ) + C

56 5. (i) + d Use Long Division + + d + d + d }{{} Partial Fraction Decomposition + ( ) A + B + A( ) + B + (A + B) A A + B + B B A d + + d ln + ln + C

57 6. (a) + sin d Let u tan tan u tan u du d + u Note : sin() sin cos + u u sin sin cos + sin cos d ( ) +u du ( ) ( u + +u +u ) ( ) + u + u du +u ( ) + u +u +u +u du ( ) ( + u ) + u + u + u du u + u + du du w u + dw du (u + ) ( w dw ) + C w u + + C tan + + C 4

58 6. (b) sin 4 cos d Let u tan tan u tan u du d + u + u u Note : sin() sin cos sin sin cos ( ) ( ) u + u + u u + u cos() cos sin cos() cos ( ) sin + u ( ) u u u + u ( ) ( ) ( u +u 4 u ) + u +u 6u 4+4u +u ( ) + u du ( ) + u 6u 4 + 4u + u (u u ) du du du (u )(u + ) du }{{} Partial Fraction Decomposition A u + B u + du 5

59 A(u + ) + B(u ) u(a + B) + (A B) A + B A B A B ( B) B 4B B 5B 5 B A 5 A u + B u + du 5 u du 5 }{{} wu+, dwdu u + du } {{ } zu+, dzdu 5 w dw 5 z dz 5 ln w 5 z + C 5 ln w z 5 ln u u + + C w u, z u + 5 ln tan tan + + C u tan + C 6

60 Z π sin d 7. (a) a bπ b a π π 4 n 6 6 y sin sin π 6 π 6 π 6 π π 6 π 4π 6 π π sin π π π 5π 6 6π 6 sin π6 π sin 5π 6 π 5π 6 π 6 π sin π sin π 5π 6 π 6 π 6 π π 5π π sin π π π Simpon s Rule: Z π b a [y + 4y + y + 4y + y4 + 4y5 + y6 ] n " # π π π π π 5π " # " # π 5 π π π π π 5π 4π π sin d 7

61 7. (b) π sin d u du d dv sin d v cos d cos π π ( cos ) d cos π + sin π [π cos π ( cos )] + [sin π sin ] π( ) + π Comparing the two results we see that : π

62 8. (a) + + d + + d + + d lim a a b d + lim + b + d lim a tan a + lim b tan b lim a (tan tan a) + lim b (tan b tan ) lim a ( tan a) + lim b (tan b ) ( π ) ( π ) + π + π π 8. (b) e d lim a a e d u du d dv e d v e ( lim e a a ( ) lim e e a a a ) e d lim a ((e e ) (ae a e a )) lim ( a aea + e a ) lim a aea } {{ } + lim a ea }{{} 9

63 8. (c) ln d lim a + a ln d u ln dv dv du d v ( lim ln a + ( lim ln a + ) a a d a a ) d ( lim ln a + a ) a lim (( ln a ln a) ( a)) a + lim a ln a + lim a a } + a {{}}{{ + } ln a lim a + a }{{}, LH lim a + a a lim a +( a) +

64 9. (a) y e y e y ; y() Use Separation of variables y e y + e y dy d ey ( + ) dy ( + ) d ey e y dy ( + ) d e y + + C e y + C Plug in the initial value to find the constant C: e () + C + + C C Now solve for y: e y + ) ln e y ln ( + ) y ln e ln ( + ) y ln ( + ) y ln ( +

65 9. (b) dy y ; y() Use Separation of variables d + y dy d + y ( + y) dy d + y dy d ln + y + C + + C Plug in the initial value to find the constant C: ln + () + C Solve for y: ln 4 C ln + y + ln 4 e ln +y e +ln 4 + y e e ln 4 y 4e

66 9. (c) y y + cos y ; y() π Use Separation of variables dy d y + cos y (y + cos y) dy d (y + cos y) dy d y + sin y + C y + sin y + C Plug in the initial value to find the constant C: (π) + sin π + C π + C π C y + sin y + π We cannot solve eplicitly for y here so we are done.

MATH MIDTERM 4 - SOME REVIEW PROBLEMS WITH SOLUTIONS Calculus, Fall 2017 Professor: Jared Speck. Problem 1. Approximate the integral

MATH MIDTERM 4 - SOME REVIEW PROBLEMS WITH SOLUTIONS Calculus, Fall 2017 Professor: Jared Speck. Problem 1. Approximate the integral MATH 8. - MIDTERM 4 - SOME REVIEW PROBLEMS WITH SOLUTIONS 8. Calculus, Fall 7 Professor: Jared Speck Problem. Approimate the integral 4 d using first Simpson s rule with two equal intervals and then the