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# 88 Review Sheet for MA 26100 at Purdue

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This 6 page Class Notes was uploaded by an elite notetaker on Friday February 6, 2015. The Class Notes belongs to a course at Purdue University taught by a professor in Fall. Since its upload, it has received 19 views.

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Date Created: 02/06/15

Calculus Cheat Sheet Limits D ei39initions Precise De nition We say limfx L if Ha for every 5 gt 0 there is a 6 gt 0such that whenever 0 lt lxia lt 6 then fx7L lt 5 Working De nition We say limfx L Ha ifwe can make fx as close toL as we want by takingx suf ciently close to a on either side of a without letting x a Right hand limit lim fx L This has Ha the same definition as the limit except it requires x gta Lei t hand limit lim7fx L This has the Ha same de nition as the limit except it requires x lt a Limit at In nity We say lim fx L ifwe Hm can make fx as close to L as we want by takingx large enough and positive There is a similar de nition for lim fx L Hem except we require x large and negative Infinite Limit We say limfxuo ifwe Ha can make fx arbitrarily large and positive by takingx suf ciently close to a on either side of a without letting x a There is a similar de nition for limfx 7w xgta except we make fx arbitrarily large and negative Relationship between the limit and oneesided limits lgfxL gt lim fx limfxL limfx lim7fxL gt linfxL ler39n fx xlrnfx gt lgf t Does Not Exist Properties Assume limfx and limgx both exist andc is any number then Ha Ha 1 Elma clefx 2 limfxgxlinfxlgrngx Ha 3 gelxgltxgtlgegfltxgt yew 6 4 5 provided limg x 0 Ha f0 liii flx l ml l 122g x limfxwlimfxw Ha Ha lelWleiem Basic Limit Evaluations at r so Note sgna1ifagt0 and sgi1a7llfalt0 l lime uo amp lim e 0 Hm Him 2 l39 l amp l39 l 7 Kg nx co nx co 3 If r gt Othenlimi 0 Hme 4 If r gt 0 and x39is real for negativex then lim 4 0 Hem x Vlsltl lttgtl1urlalmathlamar edu fur a cumplete set at Calculus mates 5 neven lim x uo Hem 6 nodd limx uo amp lim x 7uo Hm H m 7 neven lim ax bxcsgna co Him 8 nodd limax bxcsgnauo Hm 9 nodd lim ax cxd7sgnauo Hem ZEIEIE Paul Damms Calculus Cheat Sheet Evaluation Techniques Continuous Functions L Hospital s Rule Iffxis continuous atamenliinfx fa If limfx 9 or m fx iiuo the Hagm 0 Hagm iw Continuous Functions and Composition x f fx is continuous at b and b then g a 5 a quotumber39 or 700 1 1 b Polynomials at Infinity lir39gfgx f 321 px and qx are polynomials To compute Factor and Cancel X x2 4x712 x7 2x 6 lim p factor largest power ofx out ofboth limzi 111117 Htm q x H1 x 72x H1 xx72 Hrs 8 px and qx and then compute limit liTZl 7 4 2 H x 2 3x274 x 3 Rationalize N umer atorDenominator 37 57 37 3J mi HMng Hm x2e X 781 Eggt2 781 PM Piecewise Function 2 5 If 2 x l x lt7 ligl 2 97x lg 1 where gx173t ifx 272 x 81X M x 9 Mg Compute two one sided limits H1 7 limgxlimx259 186 108 Combine Rational Expressions Hull g 7 Hull 173x 7 7 One sided limits are di erent so limgx x2 1 1 1 1 xcxh limi if lim7 7 hank xh x PHD xxh doesn t exist Ifthe two one sided limits had 7 1 L 7 1 7 1 been equal then would have existed 35 441 iaxxhii7 andhadthesame value Some Continuous Functions Partial list of continuous functions and the values of x for which they are continuous 1 POIYmmialS for 311 7 cosx and sinx for all x 2 Rational function except forx s that give division by zem tanx and sec x provided 3 3101 oddforallx wuui gygy p 4 Vnevmforallx20 2 2 2 2 5 ex for anx 9 cot x and cscx provided 6 lnx forxgt0 X quot39v27r lr07r27r Intermediate Value Theorem Suppose that fx is continuous on 0 b and let Mbe any number between fa and fb Then there exists a numberc such that a lt c ltb and fc M Vlsltl tttgtl1urlalmathlamar edu fur a cumplete set at Calculus nutes ZEIEIE Paul Damms Calculus Cheat Sheet Derivatives De nition and Notation If y fx then the derivative is defined to bef x limw ham h If y fx then all ofthe following are Ify f all of the following are equivalent equivalent notations for the derivative notations for derivative evaluated at x a df dy d df dy f xy iiifx Dfx a i i D a dxdxd fyadxwdxw f Interpretation of the Derivative If y fx then 2 f a is the instantaneous rate of 1 mf a is the slope ofthe tangent change of fx at x a line to y fx at x aand the 3 If fx is the position ofan object at equation ofthe tangent line at x a is time x then f a is the velocity of giVe by yfafyax a39 the object at xa Basic Properties and Formulas 1f fx and gx are differentiable functions the derivative exists 5 and n are any real numbers 1 Cf39cf x 5 c0 2 fig fyxgyx 6 x quotK39LPower Rule 3 fg39 f39gfg39 iProduct Rule d r 7 fgf398xg39x 439 f 2 T Quo em Rule This is the Chain Rule Common Derivatives d d d Ex1 Ecscxicscxcotx Eaxaxlna 7sinxcosx cotxicsc x exex 7 cost 7sinx i sinquotx 1 i In x 1 xgt0 dx 2 dx 17x 1 tanxsec2x ghee0 1 2 ln x i x 0 17x secxsecxtanx gaming 1 2 003101 La x gt0 vlsltmg tl1urlalmatl1lamar edu fur a cumplete set of Calculus nutes ZEIEIE Paul Damms Calculus Cheat Sheet Chain Rule Variants The chain rule applied to some specific functions 1 glimilvvwm 5 ilwslm f xsinfx 2 lemlwvem 6 ltanlflx f39lxsec2fx 3 lnfx 7 A0 flxsecfdtanm 4 girdmmomma 8 il an39va h Higher Order Derivatives The Second Derivative is denoted as The n3911 Derivative is denoted as 2 w f39x fa x 2 and is defined as f x and is defined as f39x f xy ie the derivative of the fl xf quotxy 711 me derivative of rst derivativevflxl the quot 1 derivative fl x Implicit Differentiation Find y39 if ezwy x3y2 siny11x Remembery y x here so productsquotients ofx and y will use the productquotient rule and derivatives ofy will use the chain rule The trick is to differentiate as normal and every time you di erentiate a y you tack on a y39 from the chain rule After differentiating solve for y39 ezx39gy 279y393x2y22x3yy39cosyy3911 R W 2 2 1172e 39 73xy 2e2x39gy 79y39e2x39gy My2 2x3yy cos yy 11 gt y i 7 2x3y79e2x399y 7cosy 2x3y 7 9e 39 7cos yy39 11 a 2e2x39gy a My2 IncreasingDecreasing 7 Concave UpConcave Down Critical Points x c is a critical point of fx provided either Concave UpConcave Down 1 fic0 or 2 fic doesr exist 1 If f39x gt0 for allx in an intervalIthen fx is concave up on the interval 1 IncreasingDecreasing 2 If f39x lt 0 for all x in an intervalIthen 1 If f x gt 0 for allx In an intervalIthen f is concave down on the interval fx is increasing on the interval 2 If f39x lt 0 for allx in an intervalIthen Innew39on Points x c is a in ection point of fx if the fx is decreasing on the interval 1 concavity changes at x c 3 If f39x0 for allx in an intervalIthen fx is constant on the interval 1 Vlsltmgtl1turlalmathlamar edu fur a cumplete set of Calculus nutes ZEIEIE Paul Damms mmmg awmwmmmmmumm PM W 4 W 139 W W m x us mun alvme M x k x nilSH uuSx gmwmw up Mm m as mm H mman 7m mmwmmmwumpuzw MWWPX39S ME MWJ An1 am a y gym me pzlwz Wanqu symmme Mum arminle mpglymppm 424m Hum v Hum aw z km W mm mmmdw x ndvmxmwmuwbs nusnu WWW mpxmw mmmmq w mm zgwu a p mm A unpasvlw mmumn WWW mimic W39 m39 tn mm M m m mmmawwu nus mpgqmsma Iqu x whimmqu m X mm W 1 mm 11mm ltng n lmmap Warm mmbiwjmmdw S qmm mu m mjwmzm was MW pm mum mp mm mm mamam mummvmmmm Wm WM HL nsmznzuX L n m Tmuxmmsnm mmmd an as wmvm mmwmuwam pawk H uxthxmnsuxm quot5 mm E quotmm mam A Amp may WW mumme mu m x M Mum 5 mm mm 5 X msmwmv39Imwpumuumim 4 X m WynWm mqu Nisummmmm mm umm mm mm nwmmmmmma mme mmwm MMMMMW WW mummy WM m wmmm mwm m WmmmjwmmsmmeA mum s mum mum amuuvp quotMp a Wuamumdm mm mm mm M 5mm Smrpxuunmba lmvp mm 5mm mem mum EmuVII m mug slwmlamxulwmmwn mm x PWmd u w a wwme uYWA WW mxapsn 39u m u I mm nvmum maz mm a Hm WW WWW WWWnhwmwmwwu w muquot Mmmmmmnp ummmwmm z azwmmmm mmummwuw mmmm mme w WWW WINquotum munm cmva m up mm elvmquotlnw z uaxnmwm mmmmmw w W WWNWW 0 quotm1lxnmadpwgtmm gWWWMN Z mmmm maxunnamqu mm11mmmxmm mum 1 WWWWWNMM mm x Fwadpmm m I m WNW I MN m m mm mm rpm iXsnsmwwqm msvm mmsmwzmmn a 1 mm mm mum z Amusgmjgamw mali xwmmmultxlW1ua wwpnqwmmmmmu u mmugtYlY3 mw39 z mmMum quot m39 quotgtquotquotquotquot quot Ul wadhmnwniwax mmMW m mmwmmmw a 1 Lamb Fwadhmmi 11 1 mm WW m mumu m meu39xaj1lill 3n Wmmuw Z ammmij wwmwwm a W wwwvwmu pmmwmmmumm YJawmvuqasqlm 1x uuthwnmtn Wawm quotum 1mm WWW 1 Calculus Cheat Sheet Integrals D el39initions De nite Integral Suppose fx is continuous on ab Divide ab into n subintervals of width Ax and choose x from each interval Then Ifxdx limifxAx m 11 AntinDerivative An antiderivative of f x is a function Fx such that F x fx Inde nite Integral Ifx dx F x 5 where Fx is an antiderivative of fx Fundamental Theorem 01 Calculus Part I If fx is continuous on ab then gx is also continuous on ab d and g39xjftdt t Part II fx is continuous onab Fx is an antiderivative offxie Fx Ifxdx then fxdx FbFa Variants of Part1 ilxftdtu xfux ilLMWV39xflvx I2ftdtu39xfuxrv39xfvx Properties Ifxgxdxjfxdxijgxdx Ibf xgxdxjfxdxijgxdx Iffx2gx onagxgbthen Ifxdx2Ig Iffx20 on agxgb then Ibfxdx20 cfxdxcjfxdxc is a constant Icfxdx cjfxdxc is a constant ljfltxdxlfltrgtw lfflxwlslf fle 001x IfmgfxSMon agxgb then mb7aSIfxdx Mb7a Common Integrals Ikdxkxc Icosudusinuc Ix dx x cn rl tux J W 1n junta lnlwwbhc J lnuduulnu7uc Isinudu 7cosuc 45 Iseczudutmuc Itanudulnlseculc Isecuduln secutanu c runwsmn waw Isecutanudusecuc I 1 dusin c Icscucotudu7cscuc Ie due c J csczudurcotuc Vlsltl tttgtl1urlalmathlamar edu for a complete set of Calculus notes ZEIEIE Paul Damms Calculus Cheat Sheet Standard Integration Techniques Note that at many schools all but the Substitution Rule tend to be taught in a Calculus 11 class It Substitution The substitution u gxwill convert Ifgxg39xdx ISfu du using du g xdx For inde nite integrals drop the limits of integration Ex I125 c0sjdx I125x2c0sx3dxI18c0sudx 9 2 drx xzdx 2 dwsdu sinu f sin87sin1 x1gt 131 x2 gt 238 27 27 Integration by Parts judvuvijvdu and I udvuv fl vdu Chooseu anddv from a a integral and compute du by di erentiating u and compute v using v Idv Ex I de Ex Elnde ux dv gt dudx v J xe39x dx wte39x J e39x dx ixe39xie39XJrc ulnx dvdx gt du dx vx lenxdxxlnx 7 dxxlnxix 51n5731n372 Products and some Quotients ol Trig Functions For Isin x cosquot x dx we have the following ForItan x secquot x dx we have the following 1 n odd Strip 1 sine out and convert rest to 1 n odd Strip 1 tangent and 1 secant out and cosines using sin2 x licos2 x then use CO Ve the F651 10 secants Him the substitution u cosx tan2 x sec2 x 71 then use the substitution 2 m odd Strip 1 cosine out and convert rest u secx to Sines using 9052 x 175in2 x 7 men use 2 m even Strip 2 secants out and convert rest the substitution u sinx to tangents using sec2 x 1 tan2 x then 3 n and In both odd Use either 1 or 2 use the substitution u tanx 4 n and In both even Use double angle 3 n odd and m even Use either 1 or 2 andor half angle formulas to reduce the 4 n even and m odd Each integral will be integral into a form that can be integrated dealt with differently Wig Formulm sin 2x 2sin x cosx cos2 x 1cos 20 sin2 x 1 7cos 20 5 EX Itanzxsecsxdx Ex Isln xdx cos3x 3 5 2 A 5 4 smzxzsinx Itan xsec xdx rItan xsec xtanxsecxdx J sthde J stn Xzsmxdx I 3 dx cos x cos x cos x J IncoszfstnxdC u cost cos x 7J 1ltu2 d gj 172n13u4 d n u sec2x2n cosx 7cosz xc J sec2 x71 sechtanxsecxdx Ru 71 Adu u sec x lrsec7 x7 secsxc Vlsltmg tt1turlalmathlamar edu for a complete set of Calculus notes ZEIEIE Paul Damms 1h mm mm mm mm fallmvmgmaluse m gm 9mm Mmmmmmgummmm a him a x45 a hgw as Flrsm i m i su irl anyway m 7 ma Magi x sm v wsm J uuRn Azmio Jorvx 49 MW 2km Us 1331 mg gm gnbukm X39s am numb RH sumth mmfmh mammm 1m sum3 mgdm39 mwm madw mm hubrs Ewe mam mum madw mm F39snmnmw maan mummmm Va a mp ammmmw 5a mm th puma In mam mam I551 mm f m mnde f gm rmammm mua am am 1 me dumpmmm vi mmumsm mgmmyminmmdumammPI D Faradatum mvh dammnmxwe m 1529th dumpmmm ltme m nmmm 75 amp 7739 an summwm mdzallaull kum mm 4nxuc4xw c smummem mmm Wm 4 7 c442 bArC n mnuymmmtmmnxezmbmd A L ms mum mammwmmhmmm mmsmmmsaMm wwwmy mm4xuncm mwwmmnmmm numh x lwmg zn MwmgmsA o mamWWW mmmmummu mum Em m m rpmmm a m lum lm mmmmmwwmmmm xmw mmmwwmm mmmmm Pf mlnwamcm m gemnangth m mum ch m km 2 y m w a PM 2 Afr mHmw mm mm mum mammmmm mm mummy m n 5m 5m vi W W gm mummm mm mm 5 AIxxigxa AL39Imqw AL x1xdxguy0w Vuhmuuixamhmm mwammfamlhsn Mgaqumw an m mmmmm m ch ma vi mun m m emh mm mm Anrrnmm Amwwm Lm wwfng lbmxmwzlyufhmnpmk Lm mememox Hm Munx m mwu Hm mwu Vm WWW xv4xmh vlt yv4ymw va4yrdampu xv4xm quotlt hm yum hm y hm yum hm y mm arm mummy m H x mxnnus Mox mam WHO mm mm m m wnmuwmmxwmmm Envfm umxduxnxrwa u yaSEusawnha memzdnvfm umx agt nrd Mmmmmm nguypxwpxmAImmms mmmmmmmnummw gum Caicuius Cheat Sheet Work Ifa force ofFt moves an object Average Function Value The average value 7 1 17 inagx gbthe work done is WIFxdx Inx 0 93b sfwg ia xw Arc Length Surface Area Note that this is often a Calc 11 topic The three basic formulas are 27 27 27 L I d SA I 2er d rotate about xaxis SA I 27M air rotate about yaxis a a a where d is dependent upon the form of the function being worked with as follows dr 12dx ifyfx agxgb dr 2Zdt ifxftygt astgb 2o 1j7 zay ifxfy agygb drirzzd9 ifrf9 agegb With surface area you may have to substitute in for the x ory depending on your choice ofdr to match the differential in the air with parametric and polar you will always need to substitute Improper Integral An improper integral is an integral with one or more in nite limits andor discontinuous integrands Integral is called convergent if the limit exists and has a finite value and divergent if the limit doesn t exist or has infinite value This is typically a Calc 11 topic Infinite Limit m 7 o o 1 I Monetx2e 2 Ifltxgtdx2stfltxgtdx 3 Ifxdx Ifxdx Icmfxdx provided BOTH integrals are convergent Discontinuous Integrand 1 Discont ataJ bfxdxlimIbfxdx 2 Discont atbIbfxdxlignIrfxdx 3 Discontinuity at a lt c ltb provided both are convergent Comparison Test or Improper Integrals If fx2gx 20 on 1270 then 1 IfL conv thenL gxa x conv 2 IfL gxdx divg thenL divg Useful fact If a gt 0 then Lazide converges if p gt1 and diverges for p 1 x Approximating Definite Integrals 27 For given integral x dx and a It must be even for Simpson s Rule define Ax 5 and 21 a divide ab into n subintervals xmx xhxz xwghxn with x a and x b then Midpoint Rule Ifxdx mm fx fx fx 6 is midpoint am Trapezoid Rule jabMr s fxn2fx2fx22fxHfxw Simpson s Rule Ifxdx m fxn4fx 2fx2 2fxw24fxwfxw visit mg mammal math amar edu fur a commete set of Caiemus notes ZEIEIE Pam Damms

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