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Calculus II

by: Miss Emerald Langosh

Calculus II MATH 2242

Miss Emerald Langosh
GPA 3.94

Yi Lin

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Yi Lin
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This 15 page Class Notes was uploaded by Miss Emerald Langosh on Monday October 12, 2015. The Class Notes belongs to MATH 2242 at Georgia Southern University taught by Yi Lin in Fall. Since its upload, it has received 14 views. For similar materials see /class/222046/math-2242-georgia-southern-university in Mathematics (M) at Georgia Southern University.

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Date Created: 10/12/15
Math 2242 Review 3 Name Eagle ID last four digits only Instructor Yi Lin Spring 2009 Show your work to receive credit7 and your nal answer in every computation 1 a 5pts Replace the following polar equation by equivalent Cartesian equations 7 cos07 2sin0 Solution Recall that we have x rcost97 y rsint97 r2 x2 y2 Multiply the both sides of the equation 7 cos0 7 2sin0 by r we get r2 rcos0 7 2rsin0 It follows x2 y2 z 7 2y b 4pts Write down two different polar coordinates for each of the following points given by Cartesian coordinates 722 01 3 7 Solution Polar coordinates for 722 2V2 7227 Polar coor dinates for 01 17g7 7 7737 c 6pts Find the length of the curve given by polar equation Solution Note that 20 The length of the curve 2 dTZ 2 7 L 0 17quot d6 d6 2 V64402d6 0 2 x62624d6 0 2 6x624d6 0 1 Let y 02 4 Then dy 29dt97 0d6 Edy7 and b 8 1 gdy 4 8 1 2lt2gt4 716 78 3 wlw 2 3 gtlt 5 15pts Evaluate the limit for the following sequences a 5pm u an 5 Solution Note that an 757171 WWW 5 Thus lim an5 lim W5 Here we used the important limit limyH00 5 1 0 5m Solution Note that 5 5 1 1177 1177 n na n n nn n By LHopitahs rule7 we have x 7 x 135 A 53 71 m F 1 5 5 7 m lt1 95 9 53 71 g 5 Ht 5 ma 1 i lt x 75 Thus hm In an 75 It follows that hm an 6 5 c 5pm 3 1 an m 71 Solution By LHopitahs rule7 1 3 hm an hm naoo na 71 1 31n n2 7 hm 61mm hm mace n hm 7 naoon 3 2 gtlt 5 10pts Explain that the following two series are convergent7 and nd the sum of them a co Zlttan71 n 7 tan 1n 1 n1 Solution Let 5 be the n th partial sum of the series Then 5 tanil 7tan 1 2 tan 1 2 7 tan 1 3 tan 1 3 7 tan 1 4 tan l n 7 tan 1n 1 1 7 tan 1 2 tan 1 2 7 tan 1 3 tan 1 3 7tan 1 n tan l n 7 tan 1n 1 tan 117 tan 1n 1 tan Therefore lirnyH00 5n lirnnnootan 1 7tan 1n 1 de nition we conclude that the series converges to 7 0 Solution This is a geometric series with the initial term 0 1 and ratio 72 i 72 i Since lt 17 1t converges to i 1 we a n1 Solution The n th term of the series an Note that lanl 71 n 1 l nl n i 17 7 We have that lirnyH00 lanl 1 Thus the 1 n 1 n 1 n 7 sequence can does not converge to 0 Hence the series has to be divergent Solution The n th term of the series an However7 the integration 00 oo ZcosQn 2 nan 1 2 COS n Note that n21 cos2n 1 lanll27l 727 71 1 71 1 1 1 dz lirn ids 1 ham 1 21 i b blir 10tan 1 x1 lirn tan 1 b tan 1 1 baoo 7T 4 77139 2 if 4 By integral test7 we know that the series 00 Z n1 n21 converges Then by comparison test7 we conclude that the series COHVSI gSS 00 Z n1 cos 2n n2 1 oo 1 EM 1 1 ln 71 Solution Note that the integration 1 a 1 1lnx dx 1 1 lirn dx baoo 1 x 1 lns 1nb 1 lirn dy Sety1lnz baoo 1 y blirn ln y nb blirn ln1 ln b ln1 OO diverges By the integral test7 the series 1 Z n1lnn n1 diverges s1nn1 Solution The n th term of the series an sin lt 1 First note that n 1 1 1 sin lim 78 1 Hoe 7 n1 Here we used the important limit sinx 1 lim 410 z 1 1 1 1 00 i 1 00 1 So by limit comparison test the series Zn sin n 1 and Zn converges or diverges at the same time However7 the integration 1 b 1 7 dx lim 7 dx 1 1 baoo 1 1 blim ln OO 1 diverges By integral test7 we conclude that the series 221 di 71 verges Therefore the series 221 sin lt 1 diverges n e 00 3 7 71 Z 713 n Solution Note that 3 i n 3 n 1 1 my Z Z n1 n1 n1 The geometric series 1 37 M8 ll n 1 converges since its ratio is g which is less than 1 The series 1 2 n1 converges because its n th term 1 1 7 lt 7 ngnil 7 3n71 1 and because the geometric series 221 w converges Comparison Test 71 57 n1 Solution Recall that for any a gt 0 we have an lirn 7 0 Waco ml 71 1 i 1 Thus the n th term 57 ST of the above series diverges to 00 as n a 00 H We conclude that the series diverges i2 n1 2n V L Solution The n th term of the series an Thus 71 TLn i n Vain W quot2W2 By LHopital7s rule7 1 320 m E EEOW mace 2 It follows from the root test that the series converges 5 5X7 35 Determine the radius and the interval of convergence for the following power series a 00 ZQ a 1 m0 Solution The coef cient of the n th term of the power series is an 1 We 1 an 39 power series is 1 The power series is convergent for all z with lx a 1 lt 17 ie7 0 lt z lt 2 At the endpoint z 07 the series 24 m0 have lirnnnool It follows that the radius of convergence of the diverges since its n th terrn 71 71 does not converge to zero At the endpoint z 27 the series 00 1 m1 diverges for the same reason Thus the interval of the convergence is 0 2 m2 m 1 Solution The coef cient of the n th term of the power series is an We 71 have lirn lawn W 1H1 7 mace an mace n1 mace n1 a Thus the radius of convergence is R 1lirnyHoo am convergent for all z with lx a 2 lt 17 ie7 1 lt z lt 3 At the endpoint z 1 the series 1 The series is converges by the Alternating Series Test At the endpoint z 37 the series f L m1 W diverges by the integral test So the interval of convergence is 13 0 Sn 2 2 nln 71 Solution The coef cient of the n th term of the power series is 1 an7 71an71 We have an 7 71an71 an n 1 ln2n 1 7 n ln2 n Tn1 ln2n139 Using L7Hopital7s rule7 it is easy to see that hm i hm Ni H00 ln2n 1 mace 21nn 1 1 31 1quot n1 71 1 lirn mace n 1 1 1 a 34 n 1 Thus am1 71 ln2 n lirnl llirn lirn2 mace an mace n 1 mace 1n n 1 a So the radius of convergence R 1lirn7H00 l 1 As a result7 the series an converges for all z with lt 17 ie7 71 lt z lt 1 At the endpoint z a1 the series converges by the Alternating Series Test At the endpoint z 17 the series 1 2271an71 converges because the integration 1 b 1 izds lirn izds 2 zln z baoo 2 zln z 1 11117 1 7 1m 7 y 7400 1112 y lnb lirn ham z 1112 1 1 1 biog m2 lnb L T ln2 converges The integral test d 00 Z x2n 12z 1n n1 Solution The coef cient of the n th term of the power series is an 1271 1 We have x2n3 an1 an W 7 2n1 7 2 1271139 an1 an As a result7 11 lirn l Hoe an1 Hence the radius of convergence R 1 lirnyH00 l l 1 The power series converges for all z with l2x ll lt 17 ie7 71 lt 7r lt 0 At the endpoint z 71 the series 00 Z x2n 191 n1 diverges since its n th term does not converge to zero At the endpoint z 0 the series ixanLl n1 Math 2242 Review 2 Name Eagle ID last four digits only Instructor Yi Lin Spring 2009 Show your work to receive credlit7 and your nal answer in every computation 1 14 gtlt 7 98 points Evaluate the following integrals 1 sin 1z dx rs sin 2x dx 62m cos 3x dx COS3 z sinz dx 7r4 0034 dz 0 sin 4x cos 2x dx 0 sinz 3x dx 3 d 2z76 24 z 71 72d9 lt10 3 21 2 WM 11 327z9 x39 Ch 1 dx WW 2 lt13 14 3 7dz 29


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