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# Exam 1 Study Guide 5.1-5.6 & FOT Math 1045

Marketplace > The University of Cincinnati > Math > Math 1045 > Exam 1 Study Guide 5 1 5 6 FOT
Gerry Otto
UC
Applied Calculus II
Ryan Therkelsen

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COURSE
Applied Calculus II
PROF.
Ryan Therkelsen
TYPE
Study Guide
PAGES
10
WORDS
KARMA
50 ?

## 1

1 review
"I'm really struggling in class and this study guide was freaking crucial. Really needed help, and Gerry delivered. Shoutout Gerry, I won't forget!"
Talia Connelly

## Popular in Math

This 10 page Study Guide was uploaded by Gerry Otto on Thursday September 17, 2015. The Study Guide belongs to Math 1045 at The University of Cincinnati taught by Ryan Therkelsen in Fall 2015. Since its upload, it has received 159 views. For similar materials see Applied Calculus II in Math at The University of Cincinnati.

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## Reviews for Exam 1 Study Guide 5.1-5.6 & FOT

I'm really struggling in class and this study guide was freaking crucial. Really needed help, and Gerry delivered. Shoutout Gerry, I won't forget!

-Talia Connelly

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Date Created: 09/17/15
ha g mt em ufchmgeofu uqmn es Example 2 Hemofsalles ngm permitofanewn aagamehmmTa leil ssmd gthatthemm s amemedmghm the attainm ofgmneeao ddmmg39 1 Table 52 Mainframes1an Timeweekst lol 5 1o 15 20 mwmgmpam oljoslimlm5235o Solution lithe rate of 531 is constant we have Total poles Rate of sales pot week N39ombet of weelie Howma ygamwa39eawlddmingthe rat veweek mnghhia ma eawm om toi Wineowl Ifweasmmeitut gammm d m u h eg mm e n e mt e mmw r wmof 555 gamele weal 5 weeks 2925 guinea im arm mm n fot39eaohof39 m ve mkpaiodagirwanmum mateformem heZD mkpaiod Dt39et39est imate for total solea 585 5 892 5 l5 2350 5 28510 games mmmmmt mn mmmfmmwmmmufnm t e w perm Untiereattmmte for total aaloa 5 535 5 39 2 5 HTS 395 1ot oo games molemo amoromgammmgme ED uaeekpedodisbemrean l a d l w gmdsi gle mm ofteta aal istheat emge ofttl etwommheta 39 391 r Total sales at El m 339 m 22635 gmnes Improving the Approximation The Rollo of n and lit Ingmaitotetda lohange mat mc onf It t gtkotateof mgath me mmn am m m nthea eoftheF immolpuoodmeo 39totqmaemthemtmh ofanbo enelsof teuglh LIutbetollo39etimgotmplaweoaehowdactemg 5 mm ljimptot39ex ieaomacpo iae a Example 3 If isiohoma sincelhemtofal hompuiod atooteriapappla onincmeaatzo Irate gitElth f ti 3 lllt39 millions ofopctp139loerltopt Mattemmd ee mate of thetotalcha geiothammher moo a or at ltouta or E home 5 or 1 how Solution 3 themteofchangeio f rt 3i ltllfweuse or dlspneumoniathermostattlttmmantfliii 2W4 aninglhe mm atacatu ate It toeohtathhhle 3 nuodem matefcrlhe chaogedmingthe mdhmmis 31 lltillloolltonf 4 oors 13 millionEomhioingtheomm tm ana ma thembi tmo s ginlame ln a 3 Total change a 439 46 94 944 114 l 3864 a 2511 million hoc39toria 52 The De nite Integral Taktng the Limit to Obtain the De niite tlntegral ff gamufmurmqummmmmmanm m 39T wvm4mm u 39 Ihetotalchangetlntheqnan lyhrmmt m msflheappmxima misimpmtedbyin easimg 1ewhmufnTo md letotalchangeexac watake largmrandllargenmucfn andlndkattthewlmanpmmhedby mle a d ghlmThisiscalledtakimtheifm ofmeaemsasn guesmjn nitjrand smitlmh ingf fismtimms xqgt theEmjtsu tele andmighhhan me stan wwwmde e Megrafis neonmmmlim oxfthmm Snppoaefismm mmns ra f bihedie n tei m egmloff martinimttm figf it is relhn af ele hm dor ght hammmswi nn mbdi simsofa 5 as iftr f1 n are mendpuinisof mmh sims e t fnf fr at align Left hand stun E Liza a Ar J I f f 3 div93 Righthandsm quotling ZfEIIJAIL n i1 139 Fachafhaemism eda jemm sum f iscallad re integrand and a and areaa ed m mifs anegm on The f mia mmmes m1moDd shjmampd 3 mhichstands or m in msametwy39thatz daes139he dr tntheintegmlms nmthe ctor ALNo nethmttheljmith EE symbolare mdn l furthalle hmdmamdlaudu ght handmmhammemmmf signing 5311 Whenfl39 isposi x wle mdright hmdmmreprmmdbjmemofareasufrectanrgles quot39 quotquot quotquotquot quot I g quot quot Manama 521 estimate the area under a curve using a Riemann sum Example Compute It39 I m and reprxesem this integral as an area 30 ltuttom Usi g a calmer or rampu oer we fde 1 3 f 9 39 mm 1 a Thei tegral represents the are betweenr a d r t under thecurt e ft tquot r See Figure 317quot 1 1 Figure 511 Shaded JET 393 f c f a39r a Taking the Limit to Obtain the De nite llntegral Eff ham dhanrgeuimeqnan t thenthele hmdmamme ghhhndmappim malte iemhlehmgeintheqnan inormmt m msflheappmina misimputedhyin eas ng ewhmufnTo md lemtalchangelmc ywemlke largetandllaxga39valinauc n mdhakatthe ralmwatedbythele md ghtmThistcalledtakingthe miiuf msemsasn guesmin nityandismi mliiglanf ism mums xag r gb hehm sufttele md ghhhandmse stand am eqmlllhedg niie Megrafis mmmm m tof lmmm Supposefismu mnns ragtgblihed inminmgnloff mambmi nn a f f i ii isi m n toef lelaB hzmdor ght hammmswi ui mbdivisimsufh b asquot gelsatb xmilylargelnoih mxlsifro r1 Fquot b quot quot4 i f f fr air 4141011 Left hand sum 21141 3 Eng 7 go a Aral 139 a f f r if Mlin Right hand sum llings Eachoflhesesnmsism eda iemm 5m f amusementsng anch and areas ed le mf qj39mggm on The noh mma mmo shimai ss mhichmds x m inmesamewavyttntz dues139lle dr iltheinmwm mlhe mhar ALNo mtlllattheljm sm nz symblamOandn l Mttek hmdmamdlmdn fertile rim handmmhaeasmglijmf signateaau lb Muff ispoen39timnihele andright hmdsumsamreptmmtedbjmemofm mammmmkmmmwmm 522estimate de nite integrals using areas of regions on a graph Theme xii isimF gme 513m f lij 39 i Cl Figure EH13 lEelJimate f bf 13 in c Selutinn we cane the using le wrighthand ms with n 3 an E Figuree 519am give Left hmcl sum HT 2 f2 2 394 E 212 11 5 i2 T4 32 2f 43 22f 396 2 1l325E3E 38 R1 glut hand 511111 we Badman the mgt by tel1mg average f f xii a Rs 333 55 Cl 523decide if an estimate of a de nite integral is an overestimate or an underestimate hell if it 2 l equot 7 in a it it I39ll H11quot tiiitipiiiti39le if lIhi39 MIMIiiviaiutm Hirie lri lr39la ilw iti iih 39illlt39 j EillHlll HIquot lithe Infill ltirnrllali il Hill Fr 1 lslllwl filial l l39Hllm i l full new l lhim iJ lIll ill39lltll ll39u isiigll lluilil llt Hilliill Hi1 luluquot right Riemann HIHJITH ill ill fill jail lquotljlg l a illm i ii If the graph is increasing on the interval then the left sum is an underestimate of the actual value and the rightsum is an overestimate If the curve is decreasing then the right sums are underestimates and the leftsums are underestimates 53 The De nite Integral as Area Relationship Between Definite linteg rel and Area If f I is cm inuom and posi me each term f 10 AI f Iii AI in a Ie mt right hand Riemann the area Ufa rectangle See Piglet 52T As lithe width Alquot f the rectangles Winches the rectangles t the tune of the graphmme exad y and the of their areas gets cleaner to the area under the mine shaded m 528 In o ner wards 1quot i xp l md bi L Area under graph ef f between a and E f f xi tilquot I 39 l39 Fii g u re 5 2 Nee of ireotaingl es appiroaui mating lilii e area under lilii e Ullllr u E39 531 compute the area under a curve by using a de nite integral and a calculator The area under a curve between twa paints tart be teenttl by aleing a lettinite integral between the bee eethte Te find the area under the eu rue 5r rst between a a and it b integrate 3r fix between the llirnita ef a anal b Ettarrtelle What te the area behaee the euree are a3 at and the it atria The ehatted area lie the area that we want We earl easilly wart net that the lLIWE era ea the a aaiia when a e m 2 and a 2 Te rn t the are therefare we integrate the funetian between 42 and E E I F1334 the 3 Ax a a lt a 3 2 3 3 1W3 a 115 quotlate the area ia megatiee aw int heeaaee it ie helew tlhe arr atria 1 retreat abate the teaaie ah the ether barrel give paettiae reewlte 532 nd the exact value of a de nite integral from a graph using areas of triangles rectangles and semicircles Eaarrrplle Lrttltblfltgt the titttgtetl i1 lrrterpret the integral an area arrrt Had he eaaet ealee Flatintate the integtat using a ealle111atnrr Seltttierr that the eqaatien allquot a eirele eeateeett at the ertgtrr and 39as39it h ditra t he given by t2 4F ll Seleing Fer y we hart y ll 2 32 The fuaetiee 2 l at eerreaee ette the the upper aertrteirele and the t39rreetiee y i 1 l 32 etrrreapertrta tie the learner aertttetr ele See trquot39igttre Figure ill iiiluwe that Lite gimm tiliriegrtii ii39iiilquotti ti mli it are Hf the 114mmquot Hemliiiiiiftile Hi hi litritiittrit lial in riiilitii Hing a Ill lx tiii etairiiiniwlquot H111 iim friitil Li 1 iii 12 i I f 54 Interpretations of the De nite Integral The Mutation and U nits tr the Definite Integrai Fuel as the Leibniz netatien 1quot if it far the dertrative he that the derivative is the limit of a quetiend of iiEereneee the notation for the de nite integral 1 r f fife cit 39 reminds ue that an integral is alimil efa sum Sinee titquot eanbe thought efae a small i erenne in 139 the terms eeing adtied are products efthe form times a diEerwee in 139 We have the fellawing reen it 1 Theunitnfmeaemememtferf f tquot 1quot139 isthepmtincteftheunuiteferf andtheunuil efuri V 39 Ex ampi If and f If i have the same units thenthe f Hf zil ismeaetmedinsquam em 533 Cm x cm E mama igaswe xpm mam integral mmeenIEanarea Ex ampIe51frfri isveleeitjrinmetemfeecend emit istimeineumds hentheintegtal it air hasme quotniteterefeec X3 Ij sec n1etereaewem5mm the in gfal Wm Change in peeitien 639 541 interpret the value of a de nite integral using units when appropriate in Let C ll represent the cast per the tn heat new Irene in rlellars per its where it is time meannerl in days and if E El correspnnrls to January 1 hillit f C If t it t Snlutiinn W K The units fer the integral f f If till itquot are t tlnllns Jl rings l I rlnrs l 2 sellers The represents the nest in Ilnllsrs tn heat your iltnuse fer the rst 9 clays of 2014 Iii namely the menths elf I February and We saw in Section 51 that the integral of a rate nf change gives tetal chnrge If i ll is a rate vnfelnnge nfa quantitynhen f t at Tnnl clings in nantin betseen r 5 and r to quot fquot The units centspend it If iquot I is a nfchange 1smith wits efqnsntityltine i it h lquot anrl the de nite integral have units if C ttMttit time l time CHLlMllllF 55 Total Change amp Fundamental Theorem of Calculus 551 recall the Fundamental Theorem of Calculus FTC The Fundamental Theerern n f Calculus It E wit 1 is centimetre ter n g r 5 it then s 7 f F it stiff lth i39quot in Inururds The de nite integral ntthe derivatiwe of a inction gives the total change in the 552estimate total costs from marginal costs and xed costs Marglnal Cost and Change in Total Beet Suppose C qu represents the met if producing 7 items The derivative C J i is the cost Sinee marginal crest C i I all isthe rate rif change of the east respect to byquot the Fundamean the integral 399 r t f f t qr e39q t represents the tale change in the met E C Q q E t and q E b In other wands the integral gives the amount it costs tn increase preductien hem t units te in units The east efprorlueing ll units is the xed cost C it 391 The area under the maiginal east curve between q E U and q E b is the betel increase in cost between a productien if 39U and a nuductien 0ft This is callerl the total WinElle 051 Adding to the xed cost gives the total cost to produce in units In IfCiIthII isamanginaleest inctienand C ill isthe xerleost Cost to increase preclucrlen from rt units to l Lnilrs C h 391 C int f C I q cr39q V 5 w I a i f 39 Total variable nest in produce I mint5 f C t qt rtqr 3 Trent cost of producing in units Fined crest Ternl variable feet 5 r C Ij39 flf C qudq 39 39U 56 Average Value Average Function Value The average value of a mction f over the interval 61 b is given by Hewte Visualize the Average 1 3 Graph Thede mm39 quotan efamagevahlete seetthml g Average value If b ej 22f aim L e terg einlegm ae emm er mgmphef fff iepee iiemm maemgevmnffie mheigt efammrglewheeeheseie b e amde husemisihesameasihem betwem he SeeFJigme 5quot1 3quot Areaundereuwe 39 Areeefreelmgle 9 m H J39i 4 H Figure 531 1 Area and eeeregrevelue J H H II I l i 7 I l t l I I J 561 compute the average value of a function on an interval The pepulatien ef Texas can he medeled by the lnetien P a r1 are if Mimi where F is in theusands ef peeple and t is in years since 200i Use this mctien to predict the average papalmien ef Meatler hehveen the years 202039 and 2040 Solutin We vvant the average value of f it 39i r l and r Using a calculator to evalnate the integral we get a Average inelanlatien f l39 ti it ljquot 3 4e b5 bml 3 2 li l l The average population ef Me llen between 2020 and 204039 is predicted tn he aheut 173339 thensantt penple 562 estimate the average value of a function graphically Fer the nletten i n i graphed 111 Frgnre 572 evaluate f I ti alt 39 a l J 1 le I E Figure 52 Estimate f V if I ah in Find the average value at in en the interval 1 tel 539 Cheek vein answer graphically Solution 7 3 Since ii 1 2 390 the de nite is the area ef the regien nuder se graph of f 71 between 1 l1 and 1 5 52 sheets that this region censcists ef t3 rll gitd squares and 4 half grid squares eaeh git square of area 1 fer a total area of 15 an 5 f i r 15 a In The average value of f If i an the ern II tn 5 a given byquot 1 5 l Average value W a f if r l ni39i13939r5 1quot t5 3 To check the graphit39alljr draw a horizontal lime at itquot 3 en the graph eff quoti See Figure 53 Then observe that between I 2 U and r E 5 the area MITth the graph ef in is equaltethe area atquot the rectangle with height 3 illll r39 l has wide arseh area ei slated restangle Ill 1 quot3 i i l 77 Figure 573 Aversgevallneet fee Focus on Theory 5e1 Recall the Second Fundamental Theorem of Calculus FTC 2 Second Fundamental Theorem of Calculus The second fundamental theorem of calculus holds for f a continuous function on an open interval I and a any point in f and states that if F is de ned by the integrall antiderivative F Lt ff trldn Fquot txi f tr then at each point in f where F39 x is the derivative of F I Fundamental Thenrem 0f Calculus Part 11 Suppose f is a continuous tnction on 5139 and also suppose that F is any antiderivative for f Then a Ifxjdx xj eta ha 5e2 differentiate a function that is de ned by a de nite integral The first part of the tneprern shpvvs that indenite integratinn can be reversed by differentiated It alsp prpves that fut everv tdntinudus functidn there is an anttderivative integral Let F he anv anttderivative nr indefinite tntegrat fnrf pn taint Iff39is a tnntinunns funtttdn and is defined by Fnji than then an 2 5m and therefore if a fins flirt are at H 1 a Every tnntinnpus functipn f has an anttderivativer and there is she antiderivative Ftv given by a de nite integral pf fwtth a variahle upper npundarvd varying the lower heundarvnfthe integrand witl 1pndduce ether antiderivatives

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