MATH2007* Partial Answers to Review Exercises Fall 2004

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1 MATH27* Partial Aswers to Review Eercises Fall 24 Evaluate each of the followig itegrals:. Let u cos. The du si ad Hece si ( cos 2 )(si ) (u 2 ) du. si u 2 cos 7 u 7 du Please fiish this. 2. We use itegratio by parts: Please fiish this. ( d ( ) l + 52 ( ) ( l ) l ). We use trigoometric substitutio 2 si t. The 2 cos t dt ad 2 4 si 2 I t 2 cos t dt cos t 4 si 2 t dt 2 ( cos 2t) dt 2t si 2t + C Notice that t si (/2) ad si 2t 2 si t cos t 2 (2 si t) 4 (2 si t) So the fial aswer is I 2 si C. 4. Let u +. The du 2 ad + u 2 du 2 u / du Please fiish this.

2 5. Use the partial fractios decompositio Please fiish this ( )( + 5) ( 8 ) Use the trigoometric substitutio (/2) ta t. Please cotiue ad fiish this. 7. Use the partial fractios decompositio of the form Please cotiue ad fiish this. 4 + (4 2 + ) A + B + C Use the trigoometric substitutio cos t. Please cotiue ad fiish this. 9. Note that 4 ( ) ( )( ). Use the partial fractios decompositio of the form A + B + C + D Please cotiue ad fiish this. Determie which of the followig improper itegrals coverge:.. Diverget. ( + ) / Coverget.. e l. Diverget. 4. e 2. Coverget. 5.. Diverget. 6. π/2 sec. 2 Diverget.

3 Sketch the curves described by parametric or polar equatios ad write dow the itegrals represetig their legths (do ot evaluate these itegrals):. We have /dt e t, dy/dt 2e 2t. So the arc legth is (/dt)2 + (dy/dt) 2 dt e 2t + 4e 4t dt. 2. /dt 6 cos t, dy/dt 6 si t. So the arc legth is π/2 6 2 cos 2 t si 2 t dt 4. We have r cos θ cos θ cos θ ad y r si θ cos θ si θ. So /dθ si θ cos θ cos θ si θ, dy/dθ si θ si θ + cos θ cos θ So the required arc legth is L 2π 2π The required area is 2 ( si θ cos θ cos θ si θ)2 + ( si θ si θ + cos θ cos θ) 2 dt 2π 9 si 2 θ + si 2 θ dθ + 8 si 2 θ dθ. C r 2 dθ 2 π/6 π/6 cos 2 θ dθ. Please compute this itegral. Solve each of the followig differetial equatios:. Separate variables ad itegrate: e y dy si. Please fiish it. 2. Usig the itegratiig factor e 2, we have d 2y) (e ( + )e 2 +2 or e 2 y ( + )e Let u ( + ) The last itegral becomes e u du 2 2 eu + C. Please fiish this.

4 . This is a homogeeous equatio. As usual, let u y/. The y u, which gives dy/ u + (du/). The origial equatio becomes Separatig variables, we have u + du + u u + u 2 u u du u 2 or du u2. u. Please fiish. 4. Aswer: y C e ( 2+7) + C 2 e ( 2 7). Notice that r 2 ± 7 are roots of r 2 + 4r. 5. Aswer: y + C e + C 2 e. 6. Write M e si y + 2 ad N e cos y + 2y. We ca check M/ y N/ e cos y. Hece this equatio is eact. We look for a fuctio u u(, y) satisfyig u M e si y + 2, u y N e cos y + 2y. From the first equatio we have u (e si y + 2) e si y h(y). Substitutuig this i the secod equatio, we obtai h (y) 2y, from which we get h y 2 + C. So the fial aswer (i the form u C) is e si y y 2 C. 7. The geeral solutio to the homogeeous equatio y + y 2y is y H C e 2 + C 2 e. The fial aswer is of the form y y H + y S. To fid a special solutio y S, try y S A cos(2) + B si(2). Please complete these steps. 8. The fial aswer is y y H + y S, where y H is the same as the oe i Questio 7 above. To fid a particulra solutio y S, try y S Ae. Please complete this. 9. The fial aswer is y y H + y S. where y H is the same as the oe i Questio 7 above. To fid a particulra solutio y S, try y S Ae. Please complete this. Determie whether the give sequece is coverget or diverget:. { ( ) e 2 }. The sequece diverges to ifiity. 4

5 2. {( ) } 2. The sequece diverges to ifiity i view of as.. { 2 }. The sequece coverges to zero. Determie whether the give series is coditioally coverget, absolutely coverget, or diverget:. e e ( ) e <. So the series coverges <. So the series coverges ( 4) <. So the series coverges. 4. The series coverges. Actually, 2+! 9! e9. 5. This series is absolutely coverget. 6. Notice that ( ) as. Sice /2 <, the root test tells us the covergece of the give series. 7. Note that cos π is othig but ( ). So cos π + 2 ( ) + 2 which is coverget i view of the alteratig series test. This series is ot absolutely coverget. Ideed Note that 2 ( ) as. Hece the series is coverget. 5

6 9. Notice that, for large, si π π ( because ) si lim. So si π, comparable to the diverget series, diverges as well.. Use the itegral test ad compute 2 l. The series diverges.. The series is coditioally coverget. 2. Sice / l as, the series diverges. I each part, fid all for which the give series is coverget:. ( )! Aswer: (, ) ( 5) Aswer: (4, 6) 5 5 Aswer: [ 5, 5]. (4 + ) 2 Aswer: [ /2, ]. 5. ( 4) Aswer: [, 5). Fid the Taylor series of f() at a ad its iterval of covergece: Let s recall a list of well-kow series epasios that oe should lear by heart: l 6 e!

7 si ( ) 2+ (2 + )! cos If possible, we make use of ay oe of these idetities. ( ) 2 (2)!.. I the followig maipulatio, keep i mid that we are tryig to use the secod idetity above l l( ( )) l ( ( /)) l + l( ( /)) l ( /) (Note that / (( )/) ( ).) 2. With a, we have l ( ) ( ). e 2+ e 2(+)+ e e 2(+) e 2 ( + )! e 2! ( + ).. Agai we try to ue the secod idetity 4. At π/, we have l(2 + 2 ) l 2( ( 2 /2)) l 2 l ( /2) l 2 ( /2) l 2 ( ) 2. ( π ( si si + ( ) 2 (2)! π )) si π ( cos π ) + cos π ( si π ) ( π ) ( ) ( π ) 2+. (2 + )! 5. This is easy: e! +! ( )!. 6. We use the double agle formula si cos 2 si(2) 2 2 ( ) 2 2+ (2 + )! ( ) (2) 2+ (2 + )! 2+ ( ) 2 2 (2 + )! 2+. 7

8 Fid the biomial series of the give fuctio We use the biomial epasio formula ( + ) α. ( ) ( + ( )) So we have ( ) ( ) ( ) with ( )( )( 2) ( + )! 4 5 ( + 2) ( ) 2 ( ) α to each part: ( + )( + 2) ( ) 2 ( + ( )) ( ) ( + )( + 2) ( ) ( ) 2 ( + )( + 2) ( + ( )) /2 ( ) /2 ( ) ( /2 ) ( ). For, /2(/2 )(/2 2) (/2 + ) ( )! ( ) ( ) (2 )!!! 2.! 8

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