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# Calculus II MTH 252

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This 26 page Class Notes was uploaded by August Feeney on Monday October 19, 2015. The Class Notes belongs to MTH 252 at Portland Community College taught by Staff in Fall. Since its upload, it has received 40 views. For similar materials see /class/224648/mth-252-portland-community-college in Math at Portland Community College.

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Date Created: 10/19/15

12 3 Dduhle lhtagrals Over Gemlal Regan Double Integral ofvaerD If joey ls well behaved uver areetahgular repuh R then the vulume uhderl the surfaeers MM Cuhslderahuuhdedrepuhn ehelusedlhR Let N 7 my xf mum2 quoty39 n lfxylslanutnutD The duuhle lhtegral uffuver D15 M M Hm ygtdA eh can he lhterpreted as the vulume unda39 the surface Ever the buunded regluh D 18 May 2009 12 3 Dduhle lhtagrals Over General Regan Typel Type HReglorls Type l e aplahe repuhnthat cznbe expressed as DwlaxbgxSySgrK lh Then Hmym Jnndya x e m Type ll 7 aplane repuhnthat eahhe expressed as DTW rsydehlUKxShN d rm Then mm41 Jammy X T W 7 run 4 12 3 Dduhle lhtegrals ever Gemlal Regan Double Integral ofvaer D Example 17 A Type I Reglon Example l Evaluate xt zyldal vvherenlstheregluhhuuhded hy szandylx3 e S nlll nn Evaluatlng ths lhtegral Wth D as a Type l repuh we have s 71er V I foe ydydx 5200 l m yI Lr 1 and the area quhe huuhded repuh reeall 39 MTlLzszl s u x VT KZydydx v2 74 x V gwy dz gx 2 lt59 rza Kidoguchi Kenneth 12 3 Double Integrals Over General Regions Double Integral ofvaerD Example I 7 A Type I Region x71 yl x2 V ley yzl dx x771 y2x2 fx1x21x22 7 x2x22x221x x771 Find the Volume under the bounded region xl V 734 7x3 22 xllix x771 s 4 3 2 1 73x 7x 2x x x 5 4 3 2 1 2 15 mm mm m n K m 12 3 Double Integrals Over General Regions Double Integral ofvaerD Example 2 7 A Type II Region Example 2 Find the Volume of the solid that lies under z x2 y2 and above the region D in the xyplane bounded by x 2 y and x y 2 Solution Evaluating this integral With D as a Type II region We have y17 H120 V Ii Ifxydxdy WW H 49 Find the area of the bounded region recall MTH7252 3 My 2y FA Kai 2 2 v V J J x y dx dy m ya xy2 1 3quot 3 Fr E h lt gt7 V x yzx y dy 039 2 y W 0 3 xy2 7 u l u 2 x 12 3 Double Integrals Over General Regions Double Integral ofvaerD Example 2 7 A Type II Region y4 32 3 3 y 52 y y d y 3 y 24 2 y Find the Volume under the bounded region 70 4 2 52 2 72 4 V 15y 7y 96y I E T 35 18 May 2009 mm mm a W K m Kidoguchi Kenneth 12 3 Daub 1mgst Over Gemini Re ens Double Integrals and Order oflntegrauon Example 3 and the vulume quhe suhd that hes underz smo uver LhebuundedregmnD zy my 1 xgygl Evaluating Lhs as 2 Tue integral we have V I m1 dydx Evaluating Lhs as aTyge u mtegml we have a 2ng 18 May 2009 12 3 Daub 1mgst Over Gemini Regans ropemes ofDouble Integrals Assummg me fulluwmg mtegals exist Hwy s0 MA 11 N ygtdA Ilsa M I m ydA i I m M UK1y gozy Wm D men I m M 211 ea ydA 1m D v L72 Where D and D1 an nut uva lap than Hm ygtdA 1 m ygtdA I m M memeseme 12 3 Daub 1mgst Over Gemini Regans Doublelntegral ofvaer D D U L72 Example 4 Evaluate xydA where st me regmn buunded byy x71 may Zx 5 7 S nlntjnn Evaluating ths integral vmh D as aType x regun we have V Hm ygtdA 11 m ygtdA Hm ygtdA s e e IL F x W m ygtdydx I IN My D mdogucm Kenneth Venuhvmueu Fuhcnuh Dmyehhanuh aha megvanuh D fferenuauon and ntegrauon of Vectonva ued Funcuons A graph uf M2 vecmrrva ued mhmmh 71 lt2cos tcost2 cos tsmlgt s 3 shuwn m ngurel Fmdvhe venues uf 390 and r z m z 111 1 and 17 Skevch each unhe derwmwe vecmrs wwh Mew mm m We Currespundmg puhv Em 72 Thvee pvupemes u 7 0quot when 755 J mm Jan vA M was K wk a mull Jay4 pquot Jrr an M41 l cx1 u 4quotquot 1 c39 J Hgme 1 7z Mama VectorValued Function Differentiation and Integration Figure 2 shows the function 171sin1 2051 along with the velocity vector 071 71 and acceleration vector 171 I7 1 A at three different values of 1 Establish that the path described by 171 is truly circular 7quot 7 fIu 5131 4 HI L39H Q r E D Figure 2 What appears to be the relationship between 71 and 31 at every value of 1 Confirm this relationship gJ PM Ht V46 I rU39l Law N a 39 392 3 on b sii v maul Mum cu Lkisr 44 yhie MIC 1 0 ampED 47 WWI liLtllsl V19 Ya 549 75 W imam Page 2 of 8 VECLDFVahJEd Functer Dwfferenuatmn and ntegratmn gure a shows the JYvCHOYv 7rsmr cosr a ong th N12 VE OCW VECVOV and OCCE EV GHOH VZCYOV a YHV EZ dxffzrzm va uzs of t Esmbhsh that vhz path descmbzd by 7t 5 My cwaar FM W 1quot Ln Rain 2 is lt x LN w 4 Nu af u What 5 d2fmh2y not rmz about vhz r zm onsh p bszzzh v t and xrge I M V rm F Le lt ll uJC b JESi L 7 t Con rm HMS ngure 3 i Z lt 2L0 LP J IC n a 5u39 4quot It oJlC D 2 Arum n a JF 0101 2 J Show mun LI rrnamj sr qr mm m rjo 43 4N q 981 Pageama VectorValued Function Differentiation and Integration low 71 3m 2 71 Figure 4 shows the function 7f f along with the velocity vector and acceleration vector at three different values of f Upon what curve does this motion take place 1 IFue lol39 1A L3ei L l N 3 l1 Yiieiiivsf gv 44 Hi 5 lt U w 7 9m 3 J Vt quot a my 3 gt 0711 747 1 33x S e 393 What appears to be the relationship between v t and a t at every value of t Confirm this relationship 3 WW n Jug zw o W p m l quot muq Al ltGenuj J 3kHJH3gt J i l 019 7lt L HYmiUT J 19H Ibgt EMw l l 3404 geyifJ O JooH39 PageAufE VectorValued Function Differentiation and Integration Show that the speed functions for both ltsint cost and 23 32 a 3tt1 1 3r1m 71 t are constant and that the speed function for 4 f 2 cos cm 53 939 I uszulf 34 731 ltsint2 cos 1 2 is not constant j H lcns Jab 549 3 UCL quot U39LC B Cmujj Rwy37 W Z A l 31 4 3 H quot3 Z Uudf u i tum jig0 ye TPJUlt I 3 lt3 oj 041 quot2 SiL 1397 I 1 1 Wont w ma 7 K 439 Chaimfquot Page 5 of8 my 095367 x 1 A smite x M 39Eckll C X39yl39t7 ltquotzyaj LVectorValued Function Differentiation and Integration Show that if is constant then 3t 39tovt 1 fuel Lug ML 3 5 cu cum v4 ltgt Em 3 0 ms t A quot quot 4 O 71 e lt gt nggoa 5le a Eta sali sens moi1 3 Q gua s m 0 V5 32m If is constant then at any value of I what are the possible geometric relationships between 3t and 39t Emma ELMLE U 05 3195 75 Page 6 of8 VeemnVaIued FuneIIpn DnIeIenIIaIIpn and InIegIaIIpn Net Isglacement vs distance travelled In MTH 252 we IdIked dbqu Ianpn IndI mkes pIdee dung d IIne In IndI eunIexI d ppsIIIpn func un Is dsedIar funmmn If 51 Is d sedIdn ppsnIpn mneIIpn Inen vzs39t Is Ind func un s yeIpeIIy mneIIpn sInee e IanIpn Is IInedn Ine yeIpeny ydIue edn desenIbe Ine dIneeIIpn uf IanIpn WIIn IIs sIgn une dIneeIIpn currespundmg In ppsIIIye yeIpeIIy dndIne pInen In nengIye yeIpeIIy In InIs epnIexI we de ne Ine neI dIspIdeeInenI pyen Ine IIIne InIenydI abIp be 5b75a Mb 509 Is 5b75a IndIedIes Ine dIneeIIpn uf Ine neI mpyemem fur exdmpIe d ppsIIIye SIgn generaHy medns IndI Ine neI muvemem was eIInen nIgancInd ur upward Frum Me Tum Change Theurem a weak smvemem uf The Fundamenm Theurem quaIcqus we I I knuw Ind 55 7 5a 5mm 1 mph In uInen wards Imegrmmg Ine yeIueIIy mneIIun pyen abgwes usIne neI dIspIdeenenI pyen db We can enedIe d epnInIyed sIIIIdIIpn In came up WIIn Ine deIudI dIsIdnee IndyeIIed men 115 If an uf Ine IanIpn Is In Ine ppsnIye dIneeIIpn Inen Ine dIsIdnee IndyeIed Is equdI In Ine neI dIspIdeeInenI We edn farce dII pnne IanIpn InIp Ine ppsIIIye dIneeIIpn by furcmg Ine yeIpeIIy In decIys be ppsnIye Ie by mkmg Ine dbspIuIe ydIue pnne yeIpeIIy Thus Ine dIsIdnee IndyeIed Is e gwen by A vt dt sInee vt Is Me speed func un we epneIude Ind InIegndIIng Ine speed func un pyen db nesIms In IneIpIdI dIsIdneeIndyeIed durmg Ine IIIne InIenydI db Net dIspIacemem IIyer 02 0 1 2 3 4 5 Net dIspIaceIneIII wer I Z 9 9 a g I 3 g 3 Q FIguIe 5 51 4 In FIgure 5 Ine neI dIspIaeeInenI and mm dIsIdnee IndyeIed pyen 03 are nespeeIIyer IIDuztldt x Is tdt3andlls tdt5 wzsm 5 mIn mm mm mm ma Page 7 m9 VectorValued Function Differentiation and Integration n working with a vector valued function that describes motion we get similar results when W e integrating the velocity and sp vector r ther than a eed functions The only difference is that the net displacement is a scalar geometrically we can think of the net displacement as an arrow that points from the point of origin to the point of termination The function Ft sin4tcos3t is shown in figures 5 and e If we think of that function as describing a microbe moving along the curve the microbejourneys between the points labeled A and B over each of the time intervals 5 17 and 112 Let39s verify on our calculators that 43924 4 o o the net displacement vector over each time interval is indeed given by I VtdtI F tdt and let39s illustrate the net displacement as a vector subtraction on Figure 5 www I it u2 4 cosH L sin 7 5 18309563137 min Mn mm REE I film it o 49 7 u 72 u ca 4 c sin T Sunuamouum Lemmat m m nutu m E III 4quot Inquot Fl39cv bdbh 5 i a V s rimml r i r r 3i 0 J H W rntzkm VLL r quot quotq quotJ l C 3911391 3904 Ly true fuiiivi ill Jud 1I qJ F cwul t Figure 5 Ftsin4tcos3t Let39s illustrate symbolically why there is nothing even remotely surprising about the fact that o I Vtdz gives us the net displacement vector over ab L5 2le g F Ls 42 FL Ill an w lclwHuL aquot iILtl Half LLW39FU Page 8 cf 9 VectorValued Function Differentiation and Integration While our microbe travels from point A to point B over each of 517 and 7 49 the 4 24 4 24 path the little bugger takes during the 2 time intervals is very different indeed This latter fact is reflected in the distance traveled over the two time intervals Just as we did in MTH 252 we can calculate the distance traveled by integrating the speed function The difference this time around is subtle the speed function is the magnitude of the velocity function as opposed to the absaufe vaueof the velocity function Let39s use our calculator to calculate the distance traveled over the 2 time intervals and let39s draw the 2 paths followed by the microbe onto Figure 6 mailm 17 2 normcmvcosUti A T 4355927057 mu mtm INC mo as n Z4 7quot new cosm c gt T 5 34555425897 man no nimi L 4 pcthis Figure 6 71 sin 41 cos 31 UL is i rovellcd 0quotquot qr 5 vhf3 4 1li J3 I5 a Iood 44vni41 Lz UH pvthis gin439 s h Hu le 01 Tm4 whrgll ihuJuteecc is emi 3339tSv quot Page 9 cf 9 5 7Addit1ional Integration Techniques Integration Using Trigonometric Identities Example 1 Evaluate F Icosz 2012 Note that c0 v2 c0s2 1 F g cosat 1dt Hoosatm H4 sin22 t 2 2 t l e Tslntcost 3 c 20 April 2009 5 7Addit1ional Integration Techniques Integration Using Trigonometric Identities 271 Example 2 Evaluate F2 Isin2tdt Note that sin2 12 1 e cos2 2n F2 e 1 e cos22dt n 2n 2n 7 Id 7 Icos22dt n n H212 rash22 7 5 7Addit1ional Integration Techniques Integration Using Trigonometric Identities Example 2 Evaluate F th 9sec 9019 F Itm 9sec 9019 lt note that see29tan29 1 Itau 91 tau 9sec 9019 c letu tau9 du sec 9019 u 1u du u3 at c tan3 9 tan5 9 c Kidoguchi Kenneth 5 7Additional Integration Techniques Integration Using Trigonometric Identities Prove that a circle ofradius 7 has areaA 771 y lt De nition from MTH112 c De nition from MTH7112 dx ersinede c MTH7251 M Jr i u er I lr cos29d9 nz u e e r2 Isin29d9 m2 u u er Il de r2 cos2ede m m 2 HenceA ml QED 20 April 2009 Example 3 Evaluate F3 5 5 7Additional Integration Techniques Integration Using Partial Fraction Decomposition PFD dx x r 1gtc 2 L i x71x2 x2 x71 o 7 Axrl Br2 x71x2 xrlx2 0x6ABxrAZB 0x6A BlazeH23 AB0 and 287A6 and 32A o 7 2 2 xrlx2 x71 x2 Ifthe Notes Ifthe denominator has more than two linear factors at Ifthe denominat 5 7Additional Integration Techniques Integration Using Partial Fraction Decomposition PFD 5 x J ii x x71x2 x71 x2 2lngtcrl e lngtc 2 c lnllzillc 6 ram x73 m denominator has repeated linear factors 7 i i 22xel 2 22 x71 or is irreducible over reals x 7 who x 4 Kidoguchi Kenneth 5 7Additional Integration Techniques Partial Fraction Decomposition 7 The Heaviside Method Consider fr 7 lt33 Choose x 0 lt33 Choose x 1 20 April 2009 5 7Additional Integration Techniques Integration Using Long Division 2 Example 4 Evaluate FA dx 16 1 x i 1 Using long division x 1 x2 X2 X 7 ix 1 1 1 FAIxrl dx xzrxlnx1c 1 5 7Addidonal Integration Techniques Integration By Completing the Square 54 ml 2 Example 5 Evaluate F X 216 2 n Letgxx22x2 x22x1712 121 F I70 5 D x121 du Iu21 l 3 I arctanu arctanq e arctanl J54 1 uxldudx Kidoguchi Kenneth Mr Simonds MTH 252 Improper Integrals Key Concept Improper Integrals Homework Section 510 1 31 odd Definition If the fumequot y fx is commons on aoo men Ifxdxtlimmjfxdx If the function yfx is continuous on7ooathen Ifxdxtlirnmjifxdx In both cases if the limit exists we say that the improper integral converges If the limit doesn39t exist then the equality is invalid and we say that the improper integral diverges m dt Example 1 Determine the convergence status of I I e 1 Figure 1 y e 1 Page 1 of5 Mr Simonds MTH 252 Improper Integrals Defi ition If The funcTion y is conTinuous on7oooo Then m 11 I im x xil i fWW li ia XW provided ThuT boTh limiTs exisT If The Two limiTs exisT we say ThaT The improper inTegraI converges If eiTher limiT doesn39T exisT Then The equaIiTy is invalid and we say ThaT The improper inTegruI diverges dx ex 9quot Example 2 Determine the convergence status of I m m Figure 2 y 7re equot Page 2 of 5 Mr Simonds MTH 252 Improper Integrals n 10 no Example 3 Geometrically establish J 101 xdx and J xdx inition fxooThen If The funcTion y is conTinuous on ab and lim jabfxdx21ini Jffxdx If The funcTion y is conTinuous on ab and link 00 Then 17 2 In fxdx21nlija fxdx In boTh cases if The limiT exisTs we say ThaT The improper inTegraI converges If The Q doesn39T exisT Then The equaIiTy is invalid and we say ThaT The improper inTegraI diverges Page 3 of 5 Mr Simonds MTH 252 Improper Integrals d8 BMW 39 12 Example 4 Determine the convergence status of IO Figure 3 z swam Page 4 of 5 Mr Simonds MTH 252 Improper Integrals Definition If the function y fx is continuous on 10 Ucb and him 00 andor 5 I b fx 00 then J t1 nfj fxdx 1131 provnded that both limits exist lim xgtc If the two limits exist we say that the improper integral converges If either limit doesn39t exist then the equality is invalid and we say that the improper integral diverges rrZ COS d 25in071 0 Example 5 Determine the convergence status of J COS Figure 4 y 25ina 71 Page 5 of5 Key Concepts Summation notation Area and Riemann Sums Summation notation mgle 1 Find each of the following without the use of your calculator 5 Z 2 k 1 Icl LI 31 KJ KT FU i g1 l EM H b 3539 92 3 3 2211722171 LJL 1 3 1 2 2 atzttt 204141 If 1013 I quot 1m 11 2 3 39u 439 7 3 473339C1trr 3 3 4 3 310 15 1quot05 Page 1 of6 Mr Simonds MTH 252 r Summation notationarea If it were say 1986 we d be learning some sum formulas For example me2 Exa Verify by hand our historical formula when n 5 in 1434 hf KZI 15 nH l 51 L 3 AS If V Examp Verify by hand andby calculator our hisi ricul formula when n 100 I u I ol I H 1quot 1 z I D39 pa iIirT 39 i 41 quot quot I 71 3 I 7141 I I W luv r iIK 393 K 5 50 Mr Simonds MTH 252 Summation notationarea Area Examgle 4 Suppose ThaT we wunTed To know The area beTween The curve y 97 x2 and The x uxis over The inTervul 733 One way we could esTimaTe The area is To superimpose 4 recTungles of equal widTh onTo The region wiTh one side of each recTungle along The x axis and iTs parallel side drawn Through The poinT on The curve ThaT occurs uT The midpoinT of The given inTervul LeT39s do uT mgit Width of intervals y 7 x Lg339J Is Ax 7 w 39 Intervals J 3I IYj Q Coll5 c1536 4 Ctr33 9 x x 39225 775 2 39175 84375 84375 I939x 1xczV25 775 39Il5 CLSIPS H4W a 2 39375 84375 543 598625 12 65625 120 399 x x 225 775 r 5 mews 545 aigts mzc59us25 1225525 p z 75 a gt 1 is 25 Table1fx 97xZ mm W x We fx widm Ax flxXMx 212 3637 r 59m 2 i7 7437 r nrL1r 3 17 g q37r i1L LLY 4 M 342w If 47m Riemann Sum C 749 M3 37quot lzzi Page 3 of6 Mr Simonds MTH 252 Summation notationarea Generic formulae for various points along the subintervals Ax lt gtlt gt4 gt l I I I I I x I I I I I 39x x0 a 61 x2 x3 x b 1 11 39Ax 7quot 0 I AX lt IA X a is The lefT rnosT poinT of The region under consideraTion b is The righT rnosT poinT of The region under consideraTion b 7 a Ax IS The equal WIdTh of each sub InTerval Subinterval endpoints x0a0Ax x1a1Ax x2a2Ax x3a3Ax xkakAx Formula for the righthand endpoint of the kth interval XKak K a AxK Formula for the lefthand endpoint of the kth interval Xif X aAgtltltquotl Formula for the midpoint of the kth interval 07 7V I 0 70 Page 4 of 6 Mr Simoi ids MTH 252 r Summation notationarea Definition A Riemann Sum for the continuous function y fr over the interval ab is Z fxkAxkl where the interval ab has been broken into n intervals x is ki some point on the Jr interval and Ax is the width ofthe Jr interval If fx 2 0 over the entire interval ab then a Riemann Sum estimates the area between the curve y fr and the xaxis over ab Furthermore if each subinterval has equal width Ax then this area is exactly equal to lim 2 fxAx New Example 5 2 Find the lefthand endpoint kiemunn Sum formula for the function y fi 2 war the interval 723 Then use the sum to estimate the area shaded in Figure 2 using 10 20 and 100 subintervals Finally find the exact area L a 5 Ax T n n 7quotAxen EELS 391 i K39 2 V 15 7r Tabie2fee2 iiirun Ik y72 577 5quot quot i 5 igllebe E ll3 n ca39ax so s39r 7 m n A s I 4 f Mr rlt 439 g n if 1111 If Mr Simonds MTH 252 Summation notationarea Examgle 6 Z Find The midpoinT Riemann Sum formula for The funcTion yfx2 over The inTer39vaI 23 Then use The sum To esTimaTe The area shaded in Figure 3 using 10 20 and 100 subinTer39vals Finally find The exacT area 5 4 L 39 4x h ox o x K at 0x a X quot q f K i 39 1 H u iiii iiiiiiiiiiiiiiiiiiiii 1 4 I i iiii 5 6 u 1134 3 74h n XL n as i 391 UM min1 0 JLk Ahllt rlrr39l I Lr I I can j A n9 Fl 2 ff 2 H9 Page 6 of 6

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