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10252009 Fx1 Math 104 7 Immner 21 Sequmees 121 Sequences A seguence is an ordered list of numbers In this class we will deal A sequence can be finite or infinite with infinite sequences infinitely many countably many numbers in the list numbers in the list Note the sequence doesn39t have to Notation start at n 1 apaza3awanp an or an1 l l 1 1st 2 d hm quotle51 a formula for the n H term term term term term 1 2 3 4 5 n 4 9 16 25 36quotquot n12 5 111 112 113 114 n 21 Math 104 7 Immner g 12 Surnames n input positive integers n 12 output terms of the sequence Li 1231 33 41 Silva 4 9 16 25 36 These isolated points make up the graph of the sequence It seems as though the te1ms of the sequence are approaching 0 as n gt co 025 a n12 new 112 2n1 0 hrgwi 0 1 21 lim now lim 0 p and q polynomials am When degnum lt degdenom FE Math 104 7 Rimmer 2 In general if the terms of the Q 1215mm sequence are approaching L as n a 00 then an L When this limit exists and is finite we say the sequence is convergent When the lim an does not exist or is infinite the sequence is called divergent 117139 COS 0 1010 101 2 2 cos is divergent since the lim cos does not exist 2 n lijZw limMw p andqpolynomials n2 Hwn2 W Iinl 2 when degnum gt degden0m 2 n is divergent n2 So basically finding the limit of a sequence 51 mmmima a IZJSequenees boils down to being able to find limits at infinity Tools Section 22 Limit Laws Section 44 Limits at Infinity Section 78 Indeterminate forms and L39Hopitals Rule Thoerems l Squeeze Theorem 4 lime L aWSCWSb forallngtN and DRfiamfy amfiLl and gt limo L f is contin atL mmethme amhgbmL W m m 5 Every bounded and increasing 2 hm sequence and every bounded and A decreasing sequence is convergent an 0gtlirnam0 convergent if 71 lt r 1 3 The sequence 1 1s d1vergent for all other values of r 0 if71ltrlt1 hmr new 1 rfrl 10252009 Math 104 e Kimmer IZJSequenbes Determine whether the sequence converges or diverges If it converges find the limit 2 3 571 V a m m n 72 hm m m m 71 Hm bmx bm4x 7 rblxb0 when degrmm deg denom 35n2 5 lim an lim 2 an is convergent with lim an 5 mam Determine whether the sequence converges or diverges If it converges find the limit n W convergent if71ltril a 3 The sequence 1 IS n divergent for 311 other values ofr 7r 1 W7 0 if71ltrlt1 quot3933 7 1 ifr1 1 1 1 lima hm rizi aw aw 7 7 3 an is convergent with lim an O aw Determine whether the sequence converges or diverges If it converges find the limit 41 lima L n1 a and limfaL n 9 f is contini at L 1 n1 1 am amxm alxaoiam lima hm 7 quot1 i7 1127 MM 1 MM 9n1 imagn H 9 3 bmx bmx bxbo m 1 when degrmm degden0m an is convergent with lim an 7 mam 3 1quot n 123 4 5 a n 1 2 5 10 17 26 711 712 713 114 715 altemating sign term 2 liin a 0gtjma0 a 1 hm a hm n O EPO pandqpolynomials n21 new newn21 whendegnumltdegd2nom an is convergent with lim an O aw 10252009 10252009 Determine whether the sequence converges or diverges W39hlmekimmer a m If it converges find the limit 16 IZJSequenbes n1 n 1 1 quot1 n quot mquot an I113 an I113 7 1 Indetermrnate form 71 1 1 Lety1im jmyh1hm 1j j111yhmlnnrm new n aw n max n h1ylimnln7n1j quot00 0 Wm H Indeterminate form n1 111 7 0 lnylim 0 Indeterminate form that n we can use L39Hopitals Rule on Wm 8quot n lny lrm l n quot1 11 n n n l L LH 1 n2 I n1 quot2 I n1 n 11m 1 1 hm 11m F n2 lny 1 139 zany 144 e my 7 1s convergent n n1 Math 10A 7 Rimmer 121 Seqmmes In general 11m 15j ek aw n n1 L gtlim j 1 Ingeneral n n1 H n1 8 kn 1 k 1 1 11m17 e lim17 e 1 H91 11 new n Putting these together kn lim1 emquot I L m Determine whether the sequence converges or diverges If it converges find the limit 1quotsinn2 11 Math 104 7 Rimmer 1 21 121 Seqmmes 1 sinn2 m Consider hm an 2 11m quotam quotam nl sincen is posi t ilve n n 2 0 s1nn 1 1 2 7lt7lt7 hm00 and hm 0 03 s1nn lt1 n n n Hm Hm sin n2 so by the Squeeze theorem hmi 0 Ram n hm an 03hman 0 nam new 10252009 Math 104 7 Rimmer 1 we a on B Section 81 Integration By Parts Goal To be able to integrate Chapter 8 Techniques of Integration Xi39 lig m m Integration using can be thought of as the in reverse Integration by parts can be thought of as the in reverse negoo gmgame ml r 42 31 ln39eya on By Pam Ifxg xdx fxg xIf xg xdx Mfx Vgx duf xdx dvg xdx Big Picture We are trading in Goal To get a for integral than the original one 1 Choose u to be a function that becomes simpler when 2 Make sure dv can be readily Math 10A 7 Rimmer wide 31 ln39eya on By Pam traded in 2 5 for J x e xdx th 15 th1s Wrong Way u dv 5 Ixze dx du v traded in this 39 Correct Way A for hlE d u V 2 5 IX 639 dx IBP again du V 7 dv u 2 5 2 5 5 Ixe dxxex Jxexdx du v 5 5 Ixze dxxze 5 5 Ixze dxxze 25 1 2515 2 5 2 5x Uxe dX gxe gxe e C 3 MatthArRimmer 6 31 megxmonsymu m Works when you have one of the following two situations 1 J polynomial exponential dx 2 Jpolynomialtrigdx J xzeSde Step 1 Differentiate the polynomial down to 0 Step 2 Integrate the trig or exponential the same amount of times Int Step 3 Multiply along diagonals going down to the right applying an alternating sign starting with 25x 1 2515 5 L5 Uxe dX gxe zsxe 1256 C 3 Math 10A Rimmer wide 31 ln39eya on By Farm 1 Ixcos7rx dx ixcos7rxdx O m 7 Rimmer 2 31 ln39eya on By Farm 9 In J iy dy 1 Shortcut doesn39t work here 4 J Wrong Way f y712 f 39 u dv Correct Way dv J yrlZdy du v In 2 12 de 2y 111 yIdey y 3 Math 10A Rimmer wide 31 ln39eya on By Pam J3 Jarctany dx 1 Shortcut doesn39t work here 1 x Only Way f mmn u dv f du v Jarctandxxarctan dx J arctamdxxarctanlnx2 1 farctan dxxarctanh1xz 0133 I6t sintdt u Sint dv 6 dt du costdt v 6 u cost dv 6 dt du Sintdt v 6 Math 104 Rimmer 81 Integration By Parts g gp x Shortcut doesn t work here gt J6 sintdt 2 6 sint J t costdt gt J6 sint dt 6 sint J6 cos tdt a J 6 cos t J6t sintdt gt J6 sintdt 2 6t sint 6t COSl J6t sintdt J6t sintdt J6t sint dt 2 6 sint dt 2 6 sint 6t cost I6t sin t dt 6l sin t 6quot cos Find the volume of the solid of revolution formed by rotating the region bounded by the curves yCOS y0 03x31 aboutthey I I U2 U n2 DA DE DE 1 Try disk method Use Shells radius height Problem Math 104 Rimmer 81 Integration By Parts W axis 7 V J27239radius height dx Ll o 7 Rimmer 2 31 ln39eya on By Farm 1 V Z JXCOS O 2 Math 104 Rimmer 62 Volumes a wgp 62 Volumes Goal To find the Method the solid into many find the volume of the pieces and to find the total volume The pieces are treated are 0 The base of each cylinder is called a 62 Volumes Math104 Rimmer wag gs 62 Volumes The volume of each cylinder is found by taking the area of the cross section A xf and multiplying by the height Ax x Thummn Hamel Human The volume of the solid can be approximated by the sum of all cylinders V a Taking the limit as the number of cylinders goes to infinity gives the exact volume of the solid V gtV Math 104 Rimmer 62 Volumes A solid has a circular base of radius 4 If every plane cross section perpendicular the xaxis is a square then find the volume of the solid 39 quotIEquot1 21 Cg 41 39r39 quot39 r 1 Math 104 Rimmer 432 62 Volumes 39W When you revolve a plane region about an axis the crosssections are circular and the solid generated is called a If there is between the axis of rotation and the region then the method used is called the a b nnnnnnnnnnnnnnnnnnn an If there is between the axis of rotation and the region then the method used is called the W Math 104 Rimmer 62 Volumes Disk Method with axis of rotation not necessarily the x axis Cross sections are circular Ax 19 Volume 2 IAx dx a b nnnnnnnnnnnnnnnnnn an Math 104 Rimmer 62 Volumes Calculate the volume of the solid generated by rotating the region between the curves y 4 x2 and y 0 about the xaxis Math 104 Rimmer W 62 Volumes Disk Method with axis of rotation not necessarily the yaxis Cross sections are circular Ay 9 Volume 2 IAydy nnnnnnnnnnnnnnnn an Math 104 Rimmer 62 Volumes Calculate the volume of the solid generated by rotating the region between the curvesy 12 andx 2 2 andy 4 about the line x 4 x w Math 104 Rimmer saggy 62 Volumes Washer Method with axis of rotation not necessarily the xaxis Draw a radius from the axis of rotation to the outer curve and call this Draw a radius from the axis of rotation to the inner curve and call this I b 2 2 V01umejAxdx7rJ dx 0 a outer radius as inner radius as a function of x a function of x Thomson may Enucalmn E Math 104 Rimmer 55 wigs 62 Volumes 49 Calculate the volume of the solid generated by rotating the region between the curves y 4 x2 and y 0 about the line y 2 lg Math 104 Rimmer 62 Volumes Washer Method with axis of rotation not necessarily the y axis Draw a radius from the axis of rotation to the outer curve and call this Draw a radius from the axis of rotation to the inner curve and call this Volumej Aydy7j 2 2 dy a outer radius as inner radius as a function of y a function of y y Rldr i lty e Thomson Higher Education w Math 104 Rimmer ragga 62 Volumes Calculate the volume of the solid generated by rotating the region between x the curves y E and y J about the yaxis 11112009 12ml Math 1m 7 Rimmer ii 2 WW W A power series is a series of the form n zcnx n0 where a b For each fixed x the series above is a series of constants that we can test for convergence or divergence A power series may converge for some values of x and diverge for other values of x Math 1m 7 Rimmer WA 123 Power Series The sum of the series is a function whose is the set of allx for which the series converges f x is reminiscent of a but it has infinitely many terms If all cquot 39s1 we have 00 fx 1xx2 x Zx n0 This is the with The power series will converge for and diverge for all other x 11112009 PM Math 1m 7 Rimmer 123 Power Series In general a series of the form is called a power series or a power series about a We use the to find for what values ofx the series converges solve for ixia to get ixia lt R 5 7R lt x 7 a lt R 5 a 7R lt x lt a R This is called the Plug in the endpoints to check for convergence 100 or divergence at the endpoints Find the radius of convergence and the interval of convergence swig m iatge39ms quot 2 n m 1 n x 2 2n nl x x Radius of convergence Interval of convergence 7 Find the radius of convergence and the interval of convergence m 3quotx4quot Z n1 n x x PM Math 10A 7 Kimmer 123 Power Series Find the radius of convergence and the interval of convergence n 2 4x 1 2 n1 n Check endpoints x x w Math 10A 7 Kimmer Wye 123 Power Series 11112009 PM Math 1m 7 Rimmer 123 Power Series Sometimes the Root Test can be used just as the Ratio Test When an can be written as 17 then the Root Test should be used 3quot x 5quot 2 H n1 an1 an lim 03 ngtm Math 1m 7 R39mmer Wye 123 Power Series 11112009 Find the radius of convergence m 1gtquotnzgt2 quot1 2n PM Math 1m 7 Rimmer 123 Power Series Radius of convergence if 11112009 Math 104006 Chapter 82 Trigonometric Integrals Outline For Today Integration of sinmxcos x Integration of secmxtan x Integration of sinmxsinnx Integration of sinmxcosnx Integration of cosmxcosnx Integrate sin2k1xcosmx Use sin2x 1c052x to get 112ka cosmocmx JCOSmX1 COS2Xk sinxdx Integrating sin2k1xcosmx Continued Then use u cosx du sinx dx To get JCOSmX1 cos2 xquot sinxdx J um1 L12quot du Lets Do An Example Lets find JSiII3x COS2xdx Jeos2xsin3xdx jcos2x1 cos2xsinxdx so if we let u cosx du sinxdx Jcoszx1 cos2xsinxdx J u2l u2du u33u55C cosx3 3 cosx5 5 C Integrate sinmxc052k1x Use c052x 1sin2x to get smmxcos2k1xdx j rsinmx1 sin2xk cosxdx Integrating sinmxc052k1x Continued Then use u sinx du cosx dx To get I sinmx1 sin2 xk cosxdx J um 1 L12 quot du Now You Try One What is sinmcyzx A cosx cos3 x cos5 x C D sinx 3in3 x sin5 x C B sinx sin3 x sin5 x C sinx 5in3 x sin5 x C C cosx cos3 x cos5 x C F c0sx 3033 x g cos5 x C Now You Try One What is sin5xazx A 00306 g 0033 x 0055 x D sinx 3in3 x sin5 x C B sinx sin3 x sin5 x C E sinx 3in3 x sin5 x C cosx cos3 x 0035 x C F C0Sx c033 x g cos5 x C Integrate sin2mxc052 x Use the identities c052x 121cos2x sin2x 121cos2x sinxcosx 112sin2x Another Example What is Ism2x C052xdx A 2x sin2x C D 2x cos2x C 4x sin4x C x sin2x C 1 r 1 C 4x cos4x C F g 2x cos2x C Another Example What is Ism2x 0052xdx A 2x sin2x C D 2x cos2x C B xsin2xC 1 v 1 C 4x cos4x C F g 2x cos2x C Integrate sec2k2xtanmx Use sec2x 1tan2x to get I sec2k2xtanmxdx tanmx1 tan2xk sec2xdx Integrating sec2k2xtanmx Continued Then use u tanx du sec2 x dx To get Jtanmx1 tan2xk sec2xdx j um1 u2kdu Integrate secm1xtan2k1x Use ta n2x sec2x1 to get Jsecm1x tan2k1xdx Jsecm xsec2 x 1k sec39x tanxdx Integrating secm1xtan2k1x Continued Then use u secx du secxtanx dx To get I seem xsec2 x 1k secx tanxdx J 2 ltle 1k du Another Example What is j tan2xsec4xdx A gsec3xsecsxC D tan3xsec5xC B 1 2 1 2 C E lsec3x ltan5xC Esec xZsec x 3 5 C gtan3x tan5xC F gtan3xtan5xC Another Example What is j tan2xsec4xdx A g3 03x5605x C D tan3x secsxC 1 3 1 5 B sveczxseczxC E 3360 x tan xC 1 3 1 5 C 3ltan3x tan5xC F gtanxgtan xC The Other Cases Are Less Clear To find Jtanxdx Use u cosx To get I tanxdx du sinxdx J smx dx cosx idu u In u C In l C u In secx C Integral of secx This requires a trick isecwx isecltxgt 22211 3333 J sec2 x secx tanx secx tanx dx Integral of secx Continued We then use u secx tanx du sec2xtanxdx to get Jsec2xsecxtanx dx du secxtanx 17 I C n u In secx tanx C Integral of sec3x First we integrate by parts with u secx dvsec2xdx dusecxtanx vtanx I sec3 xdx secx tanx I tan2 x secxdx secx tanx j sec2 x 1 secrxdx secxtanx e sec3vxdx j secxdx Integral of sec3x Continued So 2 I see3 xdx secx tanx J secxdx secx tanx In secx tanx C and we therefore get J sec3 xdx secx tanx In secx tanx C Other Trig Identities sinA 00B sinA B sinA B sinA sinB COSA B 00A B 00A 00B 005A B 00A B Other Trig Identities Continued We can use these identities to get I sinmx cosnxdx j sinmx nx sinmx nxdx j sinmx sinnxdx j cosmx nx cosmx nxdx I cosmx cosnxdx Icosmx nx cosmx Yet Another Example What is sin2xcosxazx A c0slt3xgt cosltxgtc D C083xcosxC B s139n3x sinx C E cosBx cosx C C COS3xcosxC cos3x sinxC Yet Another Example What is sin2xcosxazx A 0033x COSXC cosBx cosxC B s139n3x sinx C E cosBx cosx C cos3xi cosx C cos3x sinx C Math 104 Rimmer WEN 88 Improper Integrals Hmr r e Infinite Upper Limit Ifxdx iigjfxdx X b b I 2x 2x o 1 2x 1 I6 dlelmje dx 11m 6 11m 2x b gtltgtltgt b gtoo 2 bgt o 26 1 1 1 1 1 11m 2b b gt 26 2 2 1 1 2 7 Slnceggg 0 e 3 Math 104 Rimmer 88 Improper Integrals Infinite Upper Limit x b x 6 o 6 u1e 7 Z I dx 11m dx 11ex b gt 11ex 12211nlt1exgt1f 1im1nlt1 eb 1n 1 e DIVERGENT b gtoo xzmggimrgti Inf1n1te Lower L1m1t b b 2 i if xdx giggly xdx x ex 1 1 1 1ex I 366de lim Ixexdx 6111xex exa 0 ex lim e e aea ea 2 lim e 1 a 000 indeterminate lim 1 3 L39Hospital a X e 00 L39H lim 1 xzmtgsmrgstt Infinite Upper and Lower Limit IfxdxI fxdxIfxdx c any real number Ifxdxalimojfxdxmlifxdx oo 2 0 2 b 2 ux3 x x x 2 1 2 I 6 dx 2 11m 6 dx11m 6 dx du3xdx sduxdx 1x HWa1x 1H 01x OO lim arctan aa oo 3 0 3 b x 11marctanx a baoo 0 a3 1im arctan b3 baoo 3 lim arctan ale 00 6 lt gt lt ole Math 104 Rimmer W 3 quot 88lmproperlntegrals an infinite Dibscontinuity at Lower Limit Ifxdx 1iInJ fxdx f aainfinite discontinuity 9 9 9 dx dx 9 12 I 11mj 11m x dx 11m2xm 0 1 x 1amp0 I Ix za0 I zaO Z f0ainfinite 9 discontinuity X 12 zao I lim 6 2J2 E za0 w Math 104 Rimmer 88 Improper Integrals bInfinite Discontinuity at Uppeij Limit Ifxdxli11jfxdx f bainfinite discontinuity 8 I J dx 1 J dx 1m 0 38x 1amp8 0 38 35 f 8ainfinite discontinuity ta I1 738t23823 8132 since 2 23 0 n 1 2 3 A 5 E 7 E g Math 104 Rimmer 88 Improper Integrals Infinite Discontinuity inside the interval JfxdxJfxdxJfxdx limJfxdxlinJfxdx f cainfinite discontinuity altcltb 3 0 3 t 3 dx dx dx dx dx 4 J 4J 4 21111 411m 4 2 x 2 x 0 x taO 2 x taO t x f 0 gtinfinite t 3 discontinuity 1 1 2lt0lt3 2 11m 3 11111 3 t gt0 3x 2 t gt0 3x t 5 Both are DIVERGENT actually only need one of them to be divergent for the entire integral to be divergent 393 Math 104 Rimmer ism 88 Improperlnte rals Doubly Improper 9 00 C 00 C b Jfxdx JfxdxJf xdx lim JfxdxlimJfxdx a gt0 b gt 0 0 C a C f 0 gtinfinite discontinuity 0 1X 1 1 oo 1 b 6 dx 6 x e x 1 x 1 x 2 2J 2 dxJ 2 dxzhm 2 dx11m dx 0 x x x 6H0 x Hm x 1 a 1 f0 gt1nfm1te 1 b discontinuity u 2771 e 1x e 1x duxl de jeudueuc a gt0 a b m 1 1 b 11m 11m a e a 7 00 e 1 l 61 galaX 11 11 21 0 limb lima 617 eagtOJr Comparison Theorem Suppose that f x and g x are continuous functions with f x 2 g x 2 0 for x 2 a a If I f x dx is convergent then Ig x dx is convergent a b If Ig xdx is divergent then I f xdx is divergent See problems 4954 in section 88 x 49 dx x3 1 00 J arctan X 1 Math 104 Rimmer 88 Improper Integrals v 0 26 2 e x 1 2 50 j dx 53 1 sec xdx 1 x 0 Mg x1 5 sin2 x 51 dx 54 1 dx 00 Math104 Rimmer x 88lmproperlntegrals 49 I 3 dx 0 x 1 x x lt 1 3 S 3forx21 3 2 f0rxgt1 x 1 x LL L f i 0 Thus 341 is convergent by the Comparison Theorem 1 x 1 00 x x x 0 x3 1 dx E x3 1 dx 4 x3 1dx Ix3x1dx is convergent constant convergent 0 ml Nhlhlo Rimme M l amp 50 Illa r quotMar aslmproperlmeguals 1 X x 2 2e 2 2 627f0rx21 2 f0rx21 x x L 35 f g I I Math 104 Rimmer I u I w 129 Functions as Power Series The very first function we have seen represented as a power series is the geometric series with a 1 and r x 1 co Zx lxx2x3w xltl 1 x n20 We can find the power series representation of other functions by algebraically manipulating them to to be some multiple of this series 1 1 n The interval of convergence remains unchanged 2 Z Z x lxl lt 1 since this is still at e of eometric series 1 x l x 0 yp g l n Z l x lxl lt1 1 x n20 l xx2 x3 129 chgonslgznlirower Series S y A an S 8 55 RV Thumwn vahar mam m Math 35 129 F Power Series Represent the function as a power series and determine the interval of convergence 43362 brils 3L4lt11rgt12 11fj L 4 104 Rimmer unctions as so the interval of convergence is lxl lt 2 Ina Math my mm 1 qzmmnmmm Math 104 Rimmer a 129 Functions as Power Series f x CnOc a 2 CO Clc aczc a2 C3c a3 If the polx er series representation off x has a radius of convergence R gt 0 we can obtain a power series representation for f 39x by term by term differentiation f x 2 CO Clx aczx a2 C3c a3 mcl2c2ltx a3cgxa2 fX X ayl iCnX an gncn xan 1 withthe sameradius quot0 of convergence R starts at n 1 we can obtain a power series representation for I f x dx by term by term integration If dxCCoxc1 xa c2 xa 3xa 2 3 4 JECnCX ayl dx JCan ayl dx withthesameradius n0 n0 n0 n 1 of convergence R C is a constant of integration MW Math 104 lemer unctlons as Power Serles 35 9F Represent the function as a power series and determine the radius of convergence gx ix withR1 quot0 gm lt1 xgt 2 gm 1lt1 xgt 2lt 1gt 8 06 1 2 22m WithR1 1 X n1 3 co 00 x 2 x3gx x3ann 1 anmz 1x n1 n1 3 co quot Math 104 Rimmer 129 Functions as Power Series Represent the function as a power series and determine the radius of convergence f x arctan x 1 n n 1x2 1 x2 w1thR1 1 x2x x6 1 co 2n1 J 2ix n withR1 1x n20 2n1 co 2n1 arctanxCZ 1n x arctanOC00gtC0 n20 2n1 00 n x2n1 arctanx 1 2n1 withR1 x3 x5 x7 x 357 Math 104 Rimmer 35 129 Functions as Power Series Represent the function as a power series and determine the radius of convergence fx1n1 x inx withR1 1x n0 1 co n1 I dxC l x n20n1 co xn1 1 1 C 1n1 0C0gtC0 n x gn1 co n1 1n1 x Z withR1 Fonxz x3 x4 x 2n1 Do n X t 1 thR1 arcanx 2n1 WI 3 5 7 x x x arctanxx 3 5 7 1 1 1 arctan11 3 5 7 1n1 x x witthl 1n1 x x 32 x3x7j 1n1 44m 1nlt 4124 1n1 1n2 34124 11122122 3123 4124 Algebraically manipulate 1 x2 the same way we manignlated inf 1 with R 1 x n21 Represent 4 3 2 as a power series and determine the radius of convergence x 1 1 1 1 1 0 3 H 2 n x 4 3x2 41 Tx 2 16 1 Tx2 1Tx2 I 4 n n E 1 gt i x lt1 Ziin n 1i 00 n3 1x 1 4 lt 4 l 16 M 4 42 7H 4H lxl lt i 3 co n3n 1xn 1 3 3 I 4n 12 A1 gebraically manipulate 1n 1 x the same way we manipulated X 00 n1 x 1 1 39tthl n x I n1 W1 Represent 1n 3 2x as a power series and determine the radius of convergence arm 6 x 2x 1n32x 1n312T 1n31 1n1 3 n1 lt1gtxlt1 2x 3 1n 3 ln1 since 1nab1nalnb 00 2x quot1 3 3 1n3 Z xlt a R 3 PO n1 1 1n11n2 co 1 2n1 n1 1n32x1n32 x R23 n20 3quot1 n 1 2 lIig LN U H W J quoti 1 quotquot011 m 7 v H Surface Area of a Cone If the length ofthe slant line is L and the base radius is r we get a section of a circle of some angle 9 The are of such a section is AV2L29 Surface Area a Band We want to find the surface area of a band with band Large Radius r2 Small Radius r1 Length ofthe Fu Cone L1L Length of Cone Removed L1 Length of Band L Surface Area a Band Continued The Surface Area is then Surface Area a Band Continued And by similar triangles Oi Llr2 L11quot1 Lr1 Surface Area a Band Continued So A 7rr1L r211 And if We let P 2 A27rLr Approximation to Surface Area 1 Are a 4955392018 Let P x I I Then the surface area of g the Iine connectinil 391 rotated about i band with radius r12yiyti i Approximation to Surface Area Con nued So the surface area of the the line connecting Pi1 and Pi rotated about the xaxis is SAW 2 yiyi113131 NowRBl 7r 1f x2Ains and when Axis small we have yifximfxi and 39 LV 7 r R v H g H H f k r Try An Example Find the surface area obtained by rotating the curve y x2 from x 1 to 2 about the yaxis A 17 5 D 39 2 j 5 of the above Gabriel s Horn Lets find the surface area obtained by rotating the curve y lx from 1 to so about the xaxis 11 Mathlmelhmner 121 Sequences 2Sequmees A is an ordered list of numbers A sequence can be or 39 In this class we will deal q with s equences Note the sequence doesn39t have to start at n 1 61 or an 1 Notation apaza3awanp l l a formula for the n H term I L Mathlmemlmner 51 1215am input 2 n output H 23 33 41 Silva 4 9 16 25 36 These isolated points make up the graph of the sequence 025 a It seems as though the temis of the sequence are approaching as n a co lim Hquot n12 lim 0 p and q polynomials nae q F35 Math 104 7 Rimmer In general if the terms of the any 111Sequemes sequence are approaching L as n a 00 then Egan L When this limit exists and is finite we say the sequence is When the lim an does not exist or is infinite the sequence is called H 117139 COS 2 2 cosi is Since the 11m cosi i 2 W 2 2 n hm quot2 w limM w p and q polynomials n2 Hwn2 HWIiquot 2 when n 2 ls i n 2 So basically finding the limit of a sequence boils down to being able to find limits at infinity Tools Section 22 Limit Laws Section 44 Limits at Infinity Section 78 Indeterminate forms and L39Hopitals Rule Thoerems l Squeeze Theorem 4 lime L and Eligfwm f igga a Sc 17 forallngtN Mtgthelimitinsh le and 3 hmc L f is contin atL ljmamlimbmL W MW MM 5 Every bounded and increasing sequence and every bounded and 2 lima 0 gt lime 0 A decreasing sequence is convergent convergent if 71 lt r 1 3 The sequence 1 1s divergent for all other values of r 0 if 71lt r lt1 1 if r 1 MW Mr 1 Math 104 7 Rimmer IZJSequenbes Determine whether the sequence converges or diverges If it converges find the limit 35n2 nn2 Determine whether the sequence converges or diverges Mathlmrkimmer 339 WWW IZJSequenbes If it converges find the limit n1 9n1 1n71 an 2 n Determine whether the sequence converges or diverges I W39hlmekinmer M 121 Seqnenbes If it converges find the limit n n1 I l 1m Math 1m 7 R39mmer v 1215qu Math 104 7 R39mmer 3 121 Seq lame Determine whether the sequence converges or diverges W39hlovkinmer my IZJSequenbes If it converges find the limit 1quotsinn2 n m m minim quotAggy 83 Mg Substitut m Folmm mm mm l 2 Xasin6 555 a X xazix2 ncos x 39 J37 Z I 2 2 ngsecg OS9S3urIIS9S7 x I 7 xi 7 u a an H I It W XIHtang 33953 Cl X 1 112 x nsec Mmm smocxx u xanzsA 111 aazsase unz7 er3639 m1 NblhanrRimmer dx x 281116 A E3Yri suhslilulinn 13 4 c2 dx200s d 4 x22005 9 x24sinlg 444mm 41rsm29 4w52920059 Limit Sw tcch umg Us mg Tr amg ca 7r xx gtJ 25in0 musk 307 M i M 4 I 29939016 x1gt125inl9 sin0 DEBE 4SlnA9 W 1 1 from the ubxt 1 4sin am 4 w l 1 0019E 1 chs02 dH Z00tg l 14 62 l 5 4 J5 1 l cost l1 4 1 4 sing 4 am 4 lJ csczt9 d6 4 C N m quot m Ii i a i I dx m 1m J xii x sec mgtang dxgsec tan d MWWgmg IM Isec6d6 ln sec tan6 c w 5 x 5x 5609 i ck 1V4x x2 2 2 ck 4 x 2Y ux 2 du dx 0 4 0 cm 4 142 i 4 142 Ziloes lees6 zid a 0 E mp lms scum 4x x2 x2 4x Math 104 Rimmer 83 Trig Substitution air 4x x2 4 x 22 2 x2 4xii x 22i x2gtu0 x1gtu 1 u dbl 2sin6 4 2 220086 20056516 x4 4sin262 41 sin26 4coszl92cosl9 u0gt02gng sm6O gt60 1 u 1gt 12in6 sin6 262 2 2 6 Find the volume of the solid generated by revolving the region bounded by Math 1 439Rimme39 83 Trig Substitution the curves y 24 y0x0andx2aboutthex axis 39 x 4 no gap bw axis of rotation and the region gt Disk Method Radius rx 2 4 X Volume 2 Ej r x zdx u 2 2 4 2 2 d 16zi dx 2 0 x24 V01ume7ZJ 2 dx 167 JI xZZtang 0 x 4 0x 4 dx 2sec2 6d6 116 Mam 4 2 x244tan26441tan2 94se02 9 COS x2 42 16sec4 639 0 SCCZH 0 x2322tan0 x0302tan0 27Z39Il1cos 26d6 7r1cos 26d6 02 O tant91 tanl90 7T 1 75 75 f l gtl9 gtl90 7r8sin26 ZSIHEDO 714 2 4 Math 104006 Chapter 91 Arc Length Outline For Today 39 Arc Length Approximate Arc Length Suppose we have a curve given by y x and we want to approximate its length on ab We do this by breaking up ab into n intervals approximating the cune by a line on the interval and then adding up the lengths Length of 3 Segment Let Pi xi yi The length of a segment is xi xi12 yi yi l2 13131 Joe xi12ltfltxgt fxi12 tsz Ay2 Length of a Segment Continued By the mean value theorem there is an xquot i such that Ay fxifxi1 f x Ax Hence IlaPHI Mummy 1f39xi2Ax Arc Length So the arc length is n 39 2 ggnmzl l 31g 1fxzAx i1 j 1f39x2dx Try An Example Find the arc length from 00 to a point 2 27 of y 2x 132 I O A ggJE gm D g l7 B aim e470 E J1 5 7 C 44 0 1 J10 F None ofthe Above Try An Example Find the arc length from 00 to a point 2 27 of y 2x 132 B gJ g E LiJE C JZTO17 J1O F None ofthe Above 35 Arc Length for x fy If x gy CSXSd and g y is continuous then the arc length is given by d L N1g39y 2dy General Arc Length If y fx is a curve we call the function which calculates the arc length of a curve the arc length function 11f39t2dt sx H So 3 1f39x2 2 1 General Arc Length Continued We can rewrite this as ds 1dex Or as W aboth General Arc Length Continued We then get the arc length can be expressed as Example 2 trick Lets nd the arc length from the point 11 to the point tft where fxx2 18lnx and x 2 1 Example 2 Continued trick d 1 Then L 2x so dx x 8x L iJl2xx2dx t J1 4x2 22x 6x2dx J1 4x2 6362513 J4x2 6x2dx ll Example 2 Continued trick
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