Itegrtio, Prt 5: The Me Vlue Theorem for Itegrls d the Fudmetl Theorem of Clculus, Prt The me vlue theorem for derivtives tells us tht if fuctio f() is sufficietly ice over the itervl [,] the t some poit its istteous rte of chge is equl to its verge rte of chge. Viewed geometriclly, this mes tht the slope of the tget lie t tht poit is equl to the slope of the sect lie coectig the edpoits. (See picture.) There is similr theorem for itegrls. Not surprisigly, it is clled the Me Vlue Theorem for Itegrls. While the Me Vlue Theorem for derivtives tells us tht the rte of chge chieves its verge vlue, the Me Vlue Theorem for Itegrls essetilly tells us tht the fuctio itself chieves its verge vlue. We stte the theorem elow d the tke look t just wht we might me y the verge vlue of fuctio. Me Vlue Theorem for Itegrls If f() is cotiuous fuctio o [,] the there is t lest oe poit * i [,] such tht f ( ) d ( ) f ( *). At first glce this my seem to hve othig to do with verge vlues, ut the theorem is essetilly syig tht f(*) is the verge vlue of the fuctio over the itervl [,]. Geometric iterprettio: We kow tht f ( ) d is the siged re etwee the grph of f() d the -is. Are is determied y se d height. The se of this regio is the itervl [,] log the -is. The height vries sice height is determied y the fuctio vlues f(). We c thik of the verge fuctio vlue s the verge height of our re. The otio of verge is essetilly oe of eveig thigs out. We kow we clculte the verge of list of umers y ddig them ll up d dividig y the umer of terms. For emple, if oe wom hd $50, oe hd $75, d oe hd $00, we c clculte the verge s follows: $50 $75 $00 $5 $75. The verge 3 3 60
mout of moey the three wome hs is $75. Note tht this is the mout tht EACH wom WOULD hve if they still hd $5 etwee them ut ll hd the sme mout. This could e ccomplished if the rich wom were to give $5 to the poor wom. Averges re the Commuism of mth. Geometriclly, we c visulize the ide of eveig thigs out. As we c see i the emple elow, the height of f() my vry gretly over the itervl [,]. This mes tht the re uder the curve is ot evely distriuted there s more re i the regios where the height is ig. Imgie tht the re is mde up of cly or some other mllele sustce. Eveig thigs out would cosist of squshig the plces where the fuctio is uduly tll. This would force some of the re to spill ito the other regios util ll regios hd the sme height. The et result would e rectgle the totl re would e the sme, ut ow it would e evely distriuted over the itervl [,]. We c thik of the height of this rectgle s the verge height over the itervl. Algeric iterprettio : The lgeric otio of me or verge is the sum divided y the umer of terms. Clcultig me is two-step process:. Add up ll the terms.. Divide y the umer of terms. Thus the verge y-vlue of fuctio o itervl should just e the sum of ll of the y-vlues divided y the umer of y-vlues. The prolem is tht there re ifiite umer of y-vlues fuctio chieves o y give itervl--oe for ech -vlue. (Note tht if fuctio hs the sme y-vlue t two or more differet -vlues we shll still cout these s distict vlues. Similrly, i clcultig verge of fiite umer of terms it is possile tht two or more of the terms will e the sme.) This fct cretes prolem i oth steps of clcultig the me: we c t dd up ifiite umer of y- vlues d we c t divide y ifiity. While we c t dd up ifiite umer of terms, we hve see tht we c tke the limit s we dd up more d more terms. This is wht we do i tkig the limit of Riem sum. Usig similr logic, we might thik of pproimtig the verge y This is i fct very simplistic iterprettio of Commuism d I could sped pges critiquig it, ut I will limit myself to poitig out oe ovious flw. Oe of the teets of Commuism is (I m prphrsig here) from ech ccordig to [her] ility, to ech ccordig to [her] eed. This does ot me tht everyoe will hve ectly the sme mout of moey sice people could coceivly hve differet eeds. However, if we dd the hypothesis tht the three wome i our emple ll hve the sme eed for moey, the this would e (etremely simplified) emple of Commuism t work. I m ideted to Dr. Al Weistei of UC Berkeley for this epltio. 6
vlue y ddig up LOTS of y-vlues d dividig y the umer of terms. The we could clculte the true verge y tkig the limit s the umer of terms wet to ifiity. To keep thigs (reltively) simple, let us suppose tht we hve chopped the itervl [,] ito equl pieces d we will choose represettive -vlue from ech resultig suitervl. We c the clculte the correspodig y-vlues, f( * k ), d tke the verge of them. This will ivolve ddig the y-vlues d the dividig y the umer of terms,. The first step, ddig up the y-vlues, is very similr to wht we do i clcultig Riem sum. The oly differece is tht ow we re ddig just the heights, ot the res, so we do ot multiply y the se efore ddig the y-vlues. We get: me y-vlue k f * ( k ) Of course if we ve oly couted y-vlues whe there re ifiitely my, we ve missed quite few o mtter how lrge is. Idelly we wt to cout ALL the y-vlues. So it seems logicl to let go to ifiity. If we follow this logic * f( k ) k me y-vlue= lim As is so ofte the cse, this limit ecomes esier to evlute if we multiply y coveiet form of ; i this cse we multiply y i the form d the rerrge the multiplictio to get: * * f ( k) f ( k) k k * f k k lim lim lim ( ) Sice is costt (it s just the reciprocl of the legth of our itervl) we c fctor it out of the limit to get lim ( * f k ) k Filly, we recogize tht the remiig sum c e iterpreted s sum of res of rectgles with se d height f( * k ). As such, it is Riem sum formed usig regulr prtitio d so its limit s goes to ifiity is equl to defiite itegrl. Thus * me y-vlue = lim f ( k) f ( ) d k limit of Riem sum Note tht this is cosistet with our geometric iterprettio: * k 6
f ( ) d totl re me y-vlue = verge height = () se The ove discussios do t prove ythig. The otio of me or verge vlue is oly defied for fiite umer of terms so it is ot possile to prove wht it ought to e whe there re ifiitely my vlues to cosider. We eed ew defiitio. We will tke () to e our defiitio of the me y-vlue of cotiuous fuctio o [,]. The preceedig discussios DO idicte tht this choice for defiitio is good oe, sice it is cosistet with our epecttios for wht verge ought to me. With this defiitio of me y-vlue we see tht the Me Vlue Theorem simply sttes tht there must e some -vlue i the itervl where this verge vlue is chieved. This we c prove, d we do so elow: Proof of the Me Vlue Theorem for Itegrls: Sice f() is cotious o [,], the Etreme Vlue Theorem tells us tht it chieves miimum vlue m d mimum vlue M o this itervl. Thus m f ( ) M for ll i [,] If we cosider the costt fuctios y = m d y = M we c pply Theorem 5.5.6 to see tht md f ( ) d Md (3) (Recll tht Theorem 4.5.6 tells us tht the fuctio with the igger height will hve the igger re. See picture.) re of little rectgle < re uder curve < re of ig rectgle 63
As the picture shows, the itegrl of costt fuctio is just the re of rectgle so we c evlute the first d lst itegrls esily. (3) is equivlet to: m(-) f ( ) d M(-) (4) Dividig ech piece y (-) we get f ( ) d m M (5) Tht is, our me y-vlue is trpped etwee m d M. Now m d M re oth y-vlues tht f() chieves o the itervl [,]. Tht is, there must e some poit m such tht f( m) m d other poit M such tht f ( M ) M, so iequlity (5) is equivlet to f ( ) d f ( m) f ( M) Further, we kow tht f() is cotious o the etire itervl [,], d thus o the (possily smller) itervl etwee d --ote tht this itervl will e either, or, m M M m m depedig o which -vlue is igger. I the picture elow I illustrte the first possiility. M Itermedite Vlue Theorem t work: it is ot possile to coect the dots with cotiuous fuctio without crossig the dotted lie. 64
Thus the two hypotheses of the Itermedite Vlue Theorem re met. Sice cotious fuctios c t get from here to there without goig through wht s i etwee we kow y IVT there is poit * etwee d such tht f ( ) d f(*)=. This is precisely wht the Me Vlue Theorem for Itegrls sttes: there is some -vlue t which the verge y-vlue is chieved. Note tht while we ssumed tht <, the theorem holds if < s well. I this cse (-) is egtive, which is pproprite sice if < we re itegrtig right to left d thus coutig the se s egtive. If = oth sides of the MVT reduce to 0, so the theorem holds i this cse s well. (However, i this cse we c t divide y - d must e sure to use the oed versio of the theorem. Note tht i this cse the theorem is orig; it tlks out the verge of sigle vlue.) Emple : ) Fid the verge vlue of the fuctio f()= o the itervl [0,4]. ) Verify tht the Me Vlue Theorem for Itegrls holds for this fuctio y fidig -vlue i [0,4] whose fuctio vlue is the vlue you fid i prt (). () Accordig to our defiitio (), the verge y-vlue is 4 4 3 4 3 3 4 0 8 0 totl re d d 0 0 3 0 4 3 3 3 se 4 0 4 4 4 4 3 () We eed to fid -vlue etwee 0 d 4 whose y-vlue is 4 3 : 4 Solve for y squrig oth sides. 3 6. Note tht this is etwee 0 d 4. 9 The Fudmetl Theorem of Clculus, prt Prt of the Fudmetl Theorem of Clculus tells us tht for cotiuous fuctio f(), f ( ) d F( ) F( ), where F() is tiderivtive of f(). Tht is, whe we itegrte derivtive, we get the origil fuctio ck, evluted t the edpoits. Omittig the fier detils, loose trsltio of the FTC prt is the itegrl of the derivtive is the origil. Tht is, the processes of itegrtig d differetitig re i some wy opposite processes tht udo ech other. We hve come cross the otio of opposite processes udoig ech other 3 3 previously, i delig with iverse fuctios. For emple, y d y = re iverse fuctios. To cue somethig is the opposite of tkig its cue root d these processes udo ech other i the sese tht if I perform oe d the the other, I will e ck to the origil umer I strted with. We epress this through the compositio rules: m M 65
f ( f ( )) d f ( f ( )) Whe we compose iverse fuctios IN EITHER ORDER they ccel ech other out, levig us with our origil iput. It is iterestig to see whether similr sttemet holds for the processes of itegrtig d differetitig. FTC, prt, roughly spekig, sys the itegrl of the derivtive is the origil. Tht is, if you differetite first d the itegrte, you ll get ck to the fuctio you strted with. Let us cosider wht hppes if we reverse the order of our opertios: wht is the derivtive of the itegrl? If itegrtio d differetitio re ideed opposite processes, we would epect to get the origil fuctio ck i some wy. Before we ivestigte, we must clrify the questio. There re two types of itegrls to cosider: defiite d idefiite. The defiite itegrl of fuctio is just umer. Tkig the derivtive of umer is ot iterestig we c view umer s costt fuctio, d the derivtive of y costt is 0. O the other hd, the idefiite itegrl is ritrry tiderivtive of our fuctio. It is certily true tht if we tke the derivtive of y tiderivtive of f() we will get f(); this is simply the defiitio of tiderivtive. So here we c see tht the derivtive of the (idefiite) itegrl is the origil. We c ctully fid stroger coectio etwee itegrtio d differetitio if we costruct specific fuctio usig itegrtio. We ll cll such fuctio re fuctio d deote it A(). Costruct it s follows:. Let f(t) e y cotiuous fuctio. We will thik of this s our height fuctio. Note tht it is fuctio of the vrile t, while the re fuctio we re ultimtely defiig will e fuctio of differet vrile,.. Sice the height fuctio is fuctio of t, we re viewig the horizotl is s t is. Imgie lyig dow is directly o top of this t is so tht poits with the sme umeric vlue overlp. Tht is, 3 is i the sme loctio whether we re thikig of this loctio s the plce where =3 or the plce where t=3. This will llow us to use the sme picture to descrie oth f(t), our height fuctio, d A(), our re fuctio. Pick strtig -vlue d cll it. 3. Defie the re fuctio to e A() = f () t dt. Tht is, A() is the re we get y itegrtig our height fuctio from to. Our iput is our edig poit, the upper limit of itegrtio. Our output is the re we get. Everythig else is fied. The strtig poit does ot chge. Neither does the height. This is ot to sy tht the height fuctio does ot vry, ut tht oce the grph of the height fuctio hs ee drw, it s fied. Thik of it this wy: you drw the grph of fuctio f(t) d pick strtig poit,. If you the show your picture to your fried, she will view oth the fuctio d the strtig poit s gives. If you sk her to pick other -vlue to e the edig poit you c the clculte the re uder the curve etwee your strtig poit () d edig poit (the poit your fried picked). The re you get will deped o which poit your fried picks. 66
Note tht we must use DIFFERENT vrile for the height d re fuctios. Tht s ecuse the two vriles ply very differet roles. is our edig poit. We itegrte s t vries from our strtig poit to our edig poit. We c t use oe vrile for oth roles. If we did this vrile would hve to simulteously e our edig poit d tke o ll vlues etwee our strtig d edig poits. This does ot mke sese. Emple: Let f(t) = t d pick =. We c the defie A()= t dt. The vlues of A() for =, =, d =0 re show elow with the correspodig res shded. 3 0 3 0 t t 8 7 0 A() t dt 0 A()= t dt A(0)= t dt 3 3 3 3 3 3 3 3 (o re) (egtive re sice we re itegrtig right to left) The secod prt of the Fudmetl Theorem of Clculus dels with the derivtive of such re fuctio. Fudmetl Theorem of Clculus, Prt If f(t) is cotiuous fuctio o d itervl cotiig the f ( t) dt f ( ) d o tht itervl. I words, the derivtive of the re fuctio = the height fuctio evluted t the edpoit Thus if f(t) is cotiuous o itervl it will hve tiderivtive o this itervl, mely, the re fuctio: A()= f () t dt where is y poit i the itervl. This is slightly more iterestig wy of syig the derivtive of the itegrl is the origil. Thus differetitio d itegrtio DO ct s opposite processes d ccel ech other out. Note tht we do t quite get ck the origil height fuctio. The height fuctio ws fuctio of t d we get f s fuctio of. Similrly i the first prt of the Fudmetl Theorem, we did t quite get the origil fuctio ck, ut the origil fuctio evluted t the edpoits. These detils rise ecuse we re delig with more complicted opertios th simple fuctios. Differetitio d 67
itegrtio re oth opertios i which we plug i FUNCTION d get out either umer or fuctio depedig o which opertio we perform. Noetheless, the similrity to iverse fuctios is oteworthy d highlights the sic priciple: itegrtio d differetitio re essetilly opposites. We will ultimtely prove prt of the Fudmetl Theorem. First, let s look t couple of emples to get more comfortle with ll of the ottio d cocepts. Emple : Let f(t) = Let = Defie the re fuctio A()= tdt ) Use geometry to fid formul for A() tht does ot ivolve the itegrl sig. ) Differetite the formul you get i prt () d verify tht the derivtive is i fct. () We kow tht A() represets the re etwee the lie t d the t-is over the itervl from to. The pictures elow show tht this c tke oe of three forms, depedig o the loctio of : > re=trpezoid positive re sice height is positive d se is positive (we re itegrtig left to right) 0 re=trpezoid (degeerte trpezoid if = 0 or ) egtive re sice height is positive ut se is egtive (we re itegrtig right to left) <0 re=two trigles trigle etwee 0 d is egtive sice height is positive ut se is egtive trigle etwee 0 d is positive sice height d se re oth egtive Sice ll the res re fmilir geometric shpes, we c use formuls from geometry to descrie the re fuctio without usig itegrl ottio. I the clcultios elow I lwys tret the se d height s if they re POSITIVE vlues d the multiply y if the re is supposed to cout s egtive. for >: re of trpezoid = (se)(verge height) 68
( ) for 0 : -re of trpezoid = -(se)(verge height) for <0: re of left trigle re of right trigle ( se)( height) ( se)( height) ( )( ) ()() Sice i ll cses we get the sme formul, we see tht A ( ). ) A () = d. It worked! d ( ) This is the origil height fuctio with t replced y Emple 3: A height fuctio f(t) d strtig poit re show o the grph elow. Defie the re fuctio A()= f () t dt. I this cse, we wo t prove tht the derivtive of the re fuctio is ectly equl to f(), ut we will show tht A () d f() must lwys hve the sme sig. [Note: if we re just delig with the sigle fuctio f, it mkes o differece whether we cll it f() or f(t) these re the sme fuctio with differet mes. The vrile is importt whe we re lso workig with A(). We c see redily where f is positive d where it is egtive d we idicte this o the umer lie elow: f is egtive f is positive f is positive f is egtive c 69
While it is certily true tht f is positive o the etire itervl (c,) so tht there is o eed to rek it ito two prts: from c to d from to, it will e more coveiet for us to do so. I clcultig the derivtive of the Are fuctio we re mesurig the rte of chge i the re etwee d. This will e positive if the et re icreses s we move left to right d egtive if the et re decreses s we move left to right. Cse : >. The grph elow show tht the re etwee d will cout s positive d the re to the right of will cout s egtive. Suppose tht is etwee d d imgie movig little it to the right s show elow. I ll of the pictures tht follow I depict h s smll positive umer so tht +h is little it to the right of. Comprig res, we see tht there is more positive re if we use poit little it to the right of. Tht is, the et re icreses s moves to the right. Hece the derivtive will e positive if is etwee d. Oce is pst, we see tht if we move just little further to the right we re ddig o more NEGATIVE re, so tht the et re is decresig d the derivtive of the re fuctio will e egtive if f is to the right of. 70
Wht hppes t? As we pss through we stop ddig o positive re d strt ddig o egtive re. This suggests tht right t the et re either icreses or decreses, so tht the derivtive of the re fuctio must e either 0 or udefied. Lookig more closely, we see tht er the height fuctio (f) is very smll. This mes tht chge i re er, whether positive or egtive, is lso very smll. I other words, while the re is chgig, it is t chgig very quickly, so the derivtive will e smll. At it chges from eig smll positive to smll egtive umer. It is resole to thik tht the derivtive t will e 0. Cse : < This cse is slightly more complicted. The picture elow shows tht the re to the left of c will cout s positive d the re etwee c d will cout s egtive. This is ecuse the re fuctio is defied s itegrl FROM TO. If is to the left of we re itegrtig right to left, so the se of our re will cout s egtive. However, i clcultig the derivtive of the re fuctio s fuctio of, we re still skig how the re chges s we move from left to right. Suppose is to the left of c d imgie movig just little to the right. As we see elow, this results i coutig LESS of the positive re, so tht the et re decreses. Thus the derivtive of the re fuctio should e egtive to the left of c. 7
If is etwee c d d we move little to the right we see tht this results i coutig LESS NEGATIVE re. This ctully mkes the et re icrese, so the derivtive of the re fuctio should e positive here. Agi, sice the height vlues er c re very smll, it seems resole tht the derivtive t c should e 0 sice s we pss through c the derivtive vlues will chge from eig smll d egtive to eig smll d positive. Cse 3: = As we pss through we stop coutig LESS NEGATIVE re d egi coutig MORE POSITIVE re. Sice oth thigs result i mkig the et re lrger, it seems resole tht the derivtive t should e positive. f is egtive f is positive f is positive f is egtive A is egtive c A is positive A is positive A is egtive Lookig t our results, we see tht A () is positive precisely whe f() is positive d egtive precisely whe f() is egtive. Further, our lysis shows tht A () should e ig whe f() is ig d smll whe it is smll. We hve geometric evidece tht there is strog coectio etwee A () d f() so it should ow seem resole tht these two fuctios re i fct equl. Proof of the Fudmetl Theorem of Clculus, prt Let f(t) e cotiuous fuctio o some itervl cotiig d defie the re fuctio: A() = f () t dt. We will try to clculte the derivtive of this fuctio. Sice we hve ot yet lered y shortcuts for clcultig the derivtive of such fuctio, we must fll ck o the limit defiitio of the derivtive, tht is, tht the derivtive is the limit of the slope of the sect lie (see picture.) 7
picture : picture : sect lie o the grph of A() o the grph of f(t) we see tht through the poits (,A()) d (+h, A(+h)) the differece etwee A() d A(+h) is just the re etwee d +h A( h) A( ) A'( ) lim lim the re etwee d +h is the sum of the res etwee d d etwee d +h See picture. h0 h0 h defiitio of A() slope of sect lie through (,A()), d (+h, A(+h)) lim h f ( t) dt f ( t) dt h h f ( t) dt f ( t) dt f ( t) dt f ( t) dt h0 h0 h h lim h Sice f is cotiuous o the itervl etwee d +h, y the Me Vlue Theorem for itegrls we kow tht f ttis its verge vlue t some poit i this itervl. Cll this poit t* d recll tht the the defiite itegrl from to +h is equl to f ( t*) ( h) f ( t*) h. Note tht this is true for ll o-zero vlues of h ut tht there my e differet t* vlue for differet vlues of h, sice t* must lwys lie etwee d +h. Pluggig this result ito our limit we see h f () t dt f ( t*) h A'( ) lim lim lim f ( t*) f ( ) h0 h0 h0 y the work ove h MVT for h eplied elow itegrls The uderlied phrse ove remids us tht t* is i fct vrile whose vlue depeds o h. Becuse t* must lie etwee d +h, s h gets closer to 0, t* gets closer to. Sice f is cotious, this mes tht f(t*) gets closer to f(), d thus our fil limit evlutes to f(). Alterte proof to the Fudmetl Theorem, prt : Suppose f d A re defied s ove, d suppose tht F(t) is tiderivtive of f(t), tht is, tht F (t)=f(t). The 73
d d A'( ) f ( t) dt F( ) F( ) F '( ) 0 F '( ) f ( ) d FTC, prt d ote tht the derivtive this is of costt costt is 0 This seems much icer proof of prt. It is ideed shorter d esier, ut it ssumes tht we re le to fid ti-derivtive for f(t). I fct, very importt prt of the Fudmetl Theorem, prt, is the sttemet tht every cotiuous fuctio hs tiderivtive, mely, the re fuctio. The first proof demostrtes this fct; the secod ssumes it is true. Homework for sectio 5: Sectio 4.6, p. 30: 47-55 odd Sectio 4., p. 70: 9-3 odd (directios precede prolem 9) Sectio 4.8, p. 335: -5 odd,, 3 For the remider of chpter 4 we ll retur to usig your regulr tet. The homework prolems for the remiig sectios re give elow. Sectios 4.3 d 4.9: 4.3, p. 85: Quick Check d, -4 odd 4.9, p. 340: Quick Check -3,, 3- (igore directios; just evlute ech itegrl), 3-33 odd Additiol prolems:. e cos( e ) d si. cos 3. 4. 0 l 4 l 6 d l e cos e d Aswers (worked solutios egi pge 76):. si( e ) c. l 3. l c d 4. 3 74
Sectio 4.7, p. 39: -7 odd, 3-37 odd 75