Math 12 Final Exam, May 11, 2011 ANSWER KEY. 2sinh(2x) = lim. 1 x. lim e. x ln. = e. (x+1)(1) x(1) (x+1) 2. (2secθ) 5 2sec2 θ dθ.

Size: px

Start display at page:

Download "Math 12 Final Exam, May 11, 2011 ANSWER KEY. 2sinh(2x) = lim. 1 x. lim e. x ln. = e. (x+1)(1) x(1) (x+1) 2. (2secθ) 5 2sec2 θ dθ."

Nancy Butler
5 years ago
Views:

1 Math Fial Exam, May, ANSWER KEY. [5 Poits] Evaluate each of the followig its. Please justify your aswers. Be clear if the it equals a value, + or, or Does Not Exist. coshx) a) L H x x+l x) sihx) x x L H x) x coshx) b) e e ) x x x ) x ) x+ el x+ e ) ) x x+ x l x+ x xx+) x L H e ) l x+)) x) x+) x ) x e xx+) e ) x x ) x+ e e ) x+ ) x x+ e e x xl x x+) ) x x+. [ Poits] Evaluate each of the followig itegrals. a) dx sec θ dθ sec θ dθ x +) 5 ta θ +) 5 sec θ) 5 sec θ) 5 sec θ dθ secθ) 5 sec θ dθ sec θ sec 5 θ dθ sec θ dθ cos θ dθ 6 cos θcosθ dθ si θ)cosθ dθ w ) dw ) w w +C ) siθ si θ +C x x + ) ) x +C x + ) x 6 x + x +C x +)

2 x taθ Trig. Substitute dx sec θdθ Stadard w substitutio for odd trig. itegral w siθ dw cosθ dθ x + θ cos θ dθ techique: x b) x arcsix dx x arcsix x arcsix x arcsix x arcsix si θ x dθ cos θ cosθ cosθ) x x arcsix dθ x arcsix x dx arcsix si θ cosθ si θ cosθ dθ si θ cosθ dθ x arcsix cosθ)dθ [θ ]+C siθ) x arcsix θ + 8 siθ)+c x arcsix θ + x siθcosθ +C 8 arcsix arcsix+ x x +C si θdθ u arcsix dv xdx du x dx v x Trig. Substitute x siθ dx cosθdθ x θ x x +x +7x +8x+7 x +x +7x +8x+7 c) x +x dx +x+ x+)x dx +) x + x++ x+)x +) dx x++ x+ + x x + dx x++ x+ + x x + x x dx + +x+l x+ + l x + x ) arcta +C Log divisio yields:

3 x+ x +x +x+)x +x +7x +8x+7 x +x +x +x) x +x +x+7 x +x +x+) x + Partial Fractios Decompositio: x + x+)x +) A x+ + Bx+C x + Clearig the deomiator yields: x + Ax +)+Bx+C)x+) x + Ax +A+Bx +Cx+Bx+C) x + A+B)x +B +C)x+A+C so that A+B, B +C ad A+C Solve for A, B ad C. [ Poits] For each of the followig improper itegrals, determie whether it coverges or diverges. If it coverges, fid its value. a) 7 Substitute x 6x+5 u x du dx t t u +6 t dx t 7 x 6x+5 du t x 7 u x t u t t 6 t dx t 7 t u du t 6 + ) x ) +6 v + dv complete the dx square Substitute v u t arctav dv du dv du t u v u t v t arcta t t OR you could skip all the substitutio steps ad go straight to t )) x t t 7 x ) dx arcta t 7 usig the formula x +a dx x ) a arcta +C a ) ) arcta ) ) 6

4 b) 9 x s dx s 9 x x ) dx arcsi s s ) arcsi arcsi) s s. [ Poits] Fid the sum of each of the followig series which do coverge): a) ) Here we have a ice geometric series with a 6 ad r 9 As a result, the sum is give by 6 a r ) b) l7+ l7)! l7)! + l7)! l7)5 5! +... e l7) 7 c) ) + 9 )! ) 9 )! ) )! ) cos ) ) 5. [ Poits] I each case determie whether the give series is absolutely coverget, coditioally coverget, or diverges. Justify your aswers. a) ) + First, we show the absolute series is diverget. Note that + which is a diverget p-series with p. Next, Check: + + which is fiite ad o-zero. Therefore, these two series share the same behavior. Sice is diverget, the is also diverget by Limit Compariso + Test. As a result, we have o chace for Absolute Covergece. Secodly, we are left to examie the origial alteratig series with the Alteratig Series Test.

5 b + > + < b + b x + x +) < whe x >. because the related fuctio fx) x x + has egative derivative f x) Therefore, the origial series coverges by the Alteratig Series Test. Fially, we ca coclude the origial series is Coditioally Coverget b) ) ) 5 Each piece is a costat multiple of a coverget geometric series, the first oe r 5 < ad the secod oe has r <. Therefore, the orgial series is coverget, as a sum of two coverget 5 series. OR we ca boud the terms of the origial series < ) 5 5 Note ) is coverget, sice it s a costat multiple of a coverget geometric series 5 5 r <. Fially the orgial series is coverget by CT sice we bouded the terms ad the 5 larger series coverges. c) arcta+ 7 + There are a couple ways to work this series. First we ca split it up ito two coverget pieces. arcta+ arcta We ca boud the terms for the first piece arcta 7 + < ad 7 + < 7 arcta 7 + is coverget sice it s a costat multiple of a coverget p-series, p 7 >. 7 5

6 Fially, sice the terms are bouded, ad the larger series coverges, the smaller series coverges by CT. We ca boud the terms for the secod piece arcta < 7 + < 9 ad is a coverget p-series, p 9 >. Fially, sice the terms are bouded, ad the 9 larger series coverges, the smaller series 7 coverges by CT. + Fially, sice each piece is coverget, the sum of them is also coverget. *********************************************************************** OR we ca boud arcta+ 7 < If we show + 7 coverges, the the orgial series coverges by CT. + We see that p 9 >. Check: which is a coverget p-series which is fiite ad o-zero. Therefore, these two 9 series share the same behavior. Sice coverges by LCT. 9 coverges, the the series also *********************************************************************** also d) ) )!!) l Try Ratio Test: ) + +))! +) + + +)!) l+) ) )!!) l 6

7 ) + ) +)! )! +) + +)+)+))! )! +)+)+) +)+)+) +) Here we used ) + e ad l L H l+) + +!) +)!) +)+) ) + +) ) + + The series is Diverget by the Ratio Test. l l+)!) +)!) +) l l+) 7 e > l l+) l l+) 6. [ Poits] Fid the Iterval ad Radius of Covergece for the followig power series. Aalyze carefully ad with full justificatio. Use Ratio Test. ) +x )+ +)6 + ) x ) 6 + ) x ) x ) + x ) 6 + The Ratio Test gives covergece for x whe x x 6+ 6 < or x < 6. That is 6 < x < 6 < x < 9 < x < 9 Edpoits: ) ) 9 x 9 The origial series becomes ) ) ) which 6 is a costat multiple of the coverget alteratig harmoic series. We check usig AST here:. b >. b. b + < b sice + < 7

8 x The origial series becomes ) ) ) ) 6) which is a costat multiple of the diverget harmoic series p, ad therefore it s also diverget. Fially, Iterval of Covergece I, 9 ] with Radius of Covergece R. 7. [5 Poits] Write the MacLauri Series for fx) e x. Use this series to determie the fourth ad fifth derivatives of fx) e x at x. e x x! e x x )! ) x! x! + x! x6! + x8! +... I geeral, the MacLauri Series for ay f is give as f)+f )x+ f ) x + f ) x + f) ) x + f5 ) x !!! 5! Match coefficiets of like degreed terms: f 5) ) sice the is o x 5 term. f ) )!! f) )!! 8. [ Poits] Please aalyze with detail ad justify carefully. a) Fid the MacLauri series represetatio for fx) xarctax. Your aswer should be i sigma otatio. arctax OR Memorize arctax xarctax x +x dx ) x + + ) x + + x ) dx x ) ) x ) x + + ) x+ + b) Use the MacLauri series represetatio for fx) xarctax from Parta) to Estimate xarctax dx with error less tha. 8

9 Justify i words that your error is ideed less tha. xarctax dx 5 7 ) ) ) ) x + + dx ) x + +)+) ) ++...) x x5 5 + x Note this is a alteratig series. Use the Alteratig Series Estimatio Theorem. If we approximate the actual sum with oly the first term, the error from the actual sum will be at most the absolute value of the ext term, 8. Here 8 < as desired. 9. [5 Poits] Cosider the regio bouded by y cosx, y six, x ad x. Rotate the regio about the y axis. Compute the volume of the resultig solid usig the Cylidrical Shells Method. Sketch the solid, alog with oe of the approximatig cylidrical shells. See me for a sketch. V I.B.P) radius height dx xcosx six) dx I.B.P. here or ca split ito two u x dv cosx six dx du dx v six + cosx xsix + cosx) six + cosx dx ) si ) + cos ) ) ) cosx + six) si ) + cos ) ) ) cos ) + si )) )) cos + si) ) ) ) ) + ) + + ) ). [5 Poits] Cosider the Parametric Curve represeted by x e t t ad y e t/. a) Compute the arclegth of this parametric curve for t. First, dx dt et ad dy dt et/. dx ) ) dy L + dt dt dt e t ) + e t/) dt e t e t + + e t dt 9

10 e t + e t + dt e t + dt e t + t e t + ) dt e + ) e + ) e + e b) Set-up, BUT DO NOT EVALUATE, the defiite itegral represetig the surface area obtaied by rotatig this curve about the y-axis, for t. S.A. dx ) ) dy xt) + dt dt dt e t t ) e t + ) dt from parta) above. xt) e t ) + e t/) dt. [5 Poits] Compute the area bouded iside the polar curve r + cosθ ad outside the polar curve r. Sketch the Polar curves. These two polar curves itersect whe +cosθ cosθ cosθ θ or θ. Usig symmetry, we will itegrate from θ to θ ad double that area. ) Area A outer r) ier r) ) dθ + cosθ) ) dθ ) + 8cosθ + cos + cosθ) θ 9 dθ 5 + 8cosθ + dθ 5 + 8cosθ + + cosθ) dθ + 8cosθ + cosθ) dθ

11 θ + 8siθ + siθ) + 8 ) + + 8si ) + si )) + 8si + si) ) [ Poits] Fid the geeral solutio for each of the followig differetial equatios. a) dy dx lx) y Separable y dy lx dx Atidifferetiate: dy y lx dx xlx x x dx xlx dx u lx dv dx du x dx v x arcsiy xlx x + C by Itegratio by Parts Fially, y sixlx x + C) Note, y trivial solutio here. b) x dy dx y x e x Liear First Order dy Liear Form: dx x y xex Px) dx Itegratig Factor: Ix) e e x dx e l x e l x ) x Take Ix) x Multiply Diff. Eq. i its liear form by Ix): dy xdx xy ex Recogize left side as a product rule derivative: ) x y e x Atidifferetiate: x y ex + C

12 Fially, y xe x + C)

Math 142, Final Exam. 5/2/11.

Math 142, Final Exam. 5/2/11. Math 4, Fial Exam 5// No otes, calculator, or text There are poits total Partial credit may be give Write your full ame i the upper right corer of page Number the pages i the upper right corer Do problem