1 1 + x 2 dx. tan 1 (2) = ] ] x 3. Solution: Recall that the given integral is improper because. x 3. 1 x 3. dx = lim dx.

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1 . Use Simpson s rule wih n 4 o esimae an () +. Soluion: Since we are using 4 seps, 4 Thus we have [ ( ) f() + 4f + f() + 4f 3 [ ] 5 [ ]. 5. Our funcion is f() +. ( ) ] 3 + f(). Evaluae he improper inegral Soluion: Recall ha he given inegral is improper because lim Therefore, lim Now, lim ( 3) ( 3) lim 3 +( 3).

2 3. Which of he following inegrals corresponds o he lengh of he shorer arc of he ellipse 4 + y 9 (shown in he picure a righ) from he poin (4, ) o he poin (4, ). Soluion: 4 4, , 4 This par of he curve can be wrien in he form g(y). We have 4 + y y 4(9 y ) ± 9 y. Since he -values are all posiive on his par of he graph, we have ha his par of he graph is on he curve 9 y. We ge dy y 9 y ( dy ) 4y 9 y We have ha he arc lengh of curve for y ( ) dy + + 4y dy 9 y dy Alernaively Noicing he opions we are given we realize ha we use b ( ) + dy dy a 9 + 3y 9 y dy Page

3 for he arc lengh. We see from he graph ha a and b. To calculae dy we use implici differeniaion. d ( 4 + y ) d (9) dy + y y dy dy 4y dy 4y. Noe from he given equaion ha +4y y. Therefore he arc 4 lengh is given by + 6y dy + 6y 36 4y dy + 6y 4(9 y ) dy + 4y 9 y dy 9 y + 4y 9 y dy 9 + 3y 9 y dy 4. Evaluae he improper inegral e /. Soluion: Recall ha lim e / e. / Page 3

4 To calculae we use u-subsiuion. e / implies ha Now, e / e / e / e du u e u du e u e / e / + e /. Noe ha lim e / e / e. Le u hen du. This. This gives us 5. Use Euler s mehod wih sep size. o esimae y(.4) where y() is he soluion o he iniial value problem y ( + y), y(). Soluion: Here F (, y) ( + y), h. and he iniial poin is (, ). Therefore, y y + hf (, y ) +. F (, ). Now, y y + hf (, y ) +. F (., ). (.) Find he soluion of he differenial equaion: wih iniial condiion y(). Soluion: Firs we separae he variables: dy +, e y e y dy ( + ). Page 4

5 Inegraing boh sides yields To solve for y we ake he log of boh sides: e y / + + C. y ln( / + + C). Finally, we use he iniial condiion y() o solve for C. Thus ln(c) C e. y ln( / + + e ). Noe ha ln( / + + e ) ln / + + e since / + + e >. 7. Find he general soluion of he differenial equaion: ( y y + ). Soluion: This is a firs order linear differenial equaion which is already in sandard form. We find he inegraing facor: I() e R ( /) e ln ( ). Muliplying he equaion above by I() is he same as muliplying by. we ge y / y/ + giving us ha d y ) ( +. This gives ha y Muliplying across by, we ge ( + ) ln + + C. y (ln + + C). 8. Deermine if he sequence given by a n an (n) converges or diverges, and if i converges n find an (n) lim. n n Page 5

6 Soluion: As n, an (n) π/. Hence he numeraor of he sequence approaches π/ while he denominaor approaches +. This means ha a n as n. 9. Consider he following sequences: (I) { ( ) n n n } n (II) { ( ) n n n } n (III) { } ( ) n n ln(n). n Which converge, and which diverge? Soluion: (I): By applying L Hospial s Rule o he funcion f() we can see ha lim f() n. Thus lim. Bu for n, n n n n so he sequence (I) also converges o. n ( )n, n n (II): lim n n /, so as n grows large, he epression ( ) n n n oscillaes beween values close o +/ (when n is even) and values close o / (when n is odd). Thus he sequence (II) diverges. (III): As n, n ln(n) grows arbirarily large. The facor of ( ) n in sequence (III) makes he values oscillae beween posiive values of large magniude and negaive values of large magniude. Thus he sequence (III) diverges.. Find he sum of he following series: ( ) n n+ 3 n n Soluion: This is a geomeric series of he form ar n a + ar + ar + n { converges o a if r < r diverges if r. Page 6

7 (echnically we should check if a n+ /a n is a consan r in order o check his.) We can idenify a by calculaing he firs erm wih a. When n, we ge When n, we ge Now we have a a ( ) ar a ( ) r a ( 3 ) ( ) / a 3 3 ( 3 )( 3 ) 3 3. This means a 4 and r. Then r < so he series converges o 3 3 a r Find he family of orhogonal rajecories o he family of curves given by y k( 3 ) Soluion: Firs we compue y y k 3 /3. We now solve for k in y k( 3 ). Doing so we ge k y 3. So y 3 y 3 /3 y 3. If y we ake he negaive inverse, we ge ha he family of orhogonal rajecories saisfies y 3 y. We hen separae o ge y y 3 Inegraing we ge y dy 3. So y 3 + C. Page 7

8 Rearranging we finde y + 3 C. If here is an such ha y() hen y is idenically we have he family y + 3 C and y. Find he arc lengh of he curve y f() from he poin (, /3) o he poin (, e3 +e 3 ) 6 where f() e3 + e 3. 6 Soluion: This curve is a funcion of so we can use Now dy f () e3 e 3, so L arc lengh arc lengh + ( ) dy ( ) e 3 e + 3 ( e 6 + e e e 6 4 (e ) 3 + e 3 e 3 + e 3 e3 + e 3 6 e3 e 3 6 ) 3. (a) Which of he picures below show he direcion field for he differenial equaion dy (4 y)(4 + y). (b) On he direcion field you have seleced above, skech he graph of he soluion wih iniial condiion y() 3. Page 8

9 (c) For he soluion you have skeched in par (b), use he direcion field o deermine lim y()? Soluion: (a) When y > 4, dy (4 y)(4 + y) < so he slopes should be negaive for all poins above y 4. Similarly when y < 4, dy (4 y)(4 + y) < so all poins below y 4 should also be negaive. When 4 < y < 4, dy (4 y)(4 + y) > so all poins in beween should have posiive slope. This is answer (IV). (b) The poin (, 3 ) is in he middle porion y should slowly curve up o y 4. Page 9

10 (c) This means ha for his iniial condiion lim y() 4 Page

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