1.3 Continuous Functions and Riemann Sums

Transcription

1 mth riem sums, prt 0 Cotiuous Fuctios d Riem Sums I Exmple we sw tht lim Lower() = lim Upper() for the fuctio!! f (x) = + x o [0, ] This is o ccidet It is exmple of the followig theorem THEOREM Let f be (o-egtive) cotiuous fuctio o the closed itervl [, b] The limits s! of the upper d lower Riem sums for regulr prtitios both exist d re equl Tht is, lim Lower() = lim Upper(),!! i other words, lim f (m! k )Dx = lim! f (M k )Dx The proof of this theorem is hrd It is covered i our Mth course But you hve see t lest oe exmple where this theorem holds d we will see other exmples lter The fuctio eed ot be o-egtive, but the swer will represet re oly whe the fuctio is o-egtive There is importt cosequece to Theorem Becuse f (m k ) d f (M k ) re the miimum d mximum vlues of f o the kth subitervl, if c k is y umber i the kth itervl the f (m k ) pple f (c k ) pple f (M k ) This is why we eed f to be cotiuous Cotiuous fuctios lwys hve both mx d mi o y closed itervl So I other words, Tkig limits f (m k )Dx pple Lower() pple lim Lower() pple lim!! f (c k )Dx pple f (M k )Dx f (c k )Dx pple Upper() f (c k )Dx pple lim! Upper() (6) But Theorem sys the first d lst limits re the sme, so by the squeeze theorem for limits, ll three limits i (6) must be equl Tht is, THEOREM Let f be (o-egtive) cotiuous fuctio o the closed itervl [, b] The o mtter how we select the smple poits c k i ech subitervl, lim Lower() = lim!! f (c k )Dx = lim! Upper() Trsltio of Theorem If f is (o-egtive) cotiuous fuctio o the closed itervl [, b], the we c use y coveiet poit c k i the kth subitervl to evlute f we do ot eed to choose m k or M k where the mi or mx occurs Usully the most coveiet poit to choose is x k = + kdx which is the righthd edpoit of the itervl becuse this usully produces simple summtio formul The other cosequece of this result is tht we c ow defie the re uder cotiuous curve Here we do eed the fuctio to be o-egtive

2 mth riem sums, prt DEFINITION Let f be o-egtive, cotiuous fuctio o the closed itervl [, b] The re bouded bove by the grph of f, below by the x-xis, o the left by the lie x =, d o the right by x = b is give by Are = lim! f (c k )Dx, where Dx = b d c k is y poit i the kth subitervl of the regulr prtitio of [, b] ito subitervls EXAMPLE Use Defiitio to determie the re uder the curve y = f (x) = x + x o the itervl [, ] Solutio Notice i Figure 8 tht becuse the f (x) is both icresig d decresig o the itervl, if we computed Upper(), the mx vlues of f would sometimes occur t the right-hd edpoits d sometimes t the left This mkes it hrd to compute Upper() (or Lower()) However, by Theorem, we c use y coveiet set of evlutio poits for our Riem sum As oted erlier, right-hd edpoits x k re coveiet becuse the geerl formul is firly simple I this cse d so So f (x k )= f Dx = b + k = = = x k = + kdx = + k = = k + k + k + k k The geerl form of the right-hd Riem sum is: + + k + + 8k Figure 8: The grph of the prbol f (x) = x + x o [, ] I our cse, usig our work bove Right() = f (x k )Dx = Right() = k = 8 i f (x k )Dx k 8 k = 8 ( + ) = + = By Defiitio we hve Are = lim Right() = lim!! ( + )( + ) = 6

3 mth riem sums, prt Cool! YOU TRY IT (Nottio Prctice) Fill i the followig tble for the Riem sums usig regulr prtitios d right-hd edpoits Do ot try to evlute the sums f (x) [, b] Dx x k = + kdx f (x k ) Right() = x [0, ] f (x k )Dx (x ) [, ] si(x) [0, p] For f (x) =(x ) o [, ], use lgebr to simplify Right() = f (x k )Dx The clculte the re uder f (x) =(x ) o [, ] by evlutig lim Right()! swer to you try it Are = 8 Left-hd Riem Sums We hve bee workig with right-hd Riem sums But we could use left-hd edpoit sums isted The the kth subitervl is [x k, x k ], so its left-hd edpoit is x k = +(i )Dx The form of geerl left-hd Riem sum is Left() = f (x k )Dx Becuse the expressio for the left-hd edpoit x k = +(i )Dx is more wkwrd to substitute ito fuctio, we will geerlly use right-hd edpoit sums However, otice tht by djustig the strtig d edig idices of the sum, we c mke left-hd sums s simple s right: Left() = f (x k )Dx = f (x k )Dx (7) k=0 YOU TRY IT Let f (x) =(x ) Determie the expressio for Left() by djustig the idices usig (7) bove Simplify the expressio The clculte the re uder f (x) = (x ) by evlutig lim Left() d verify tht you get the sme re s for Right() i! You Try It swer to you try it Left() = k=0 f (xk)dx = k=0 k = 5 ( )()( ) 6 = So the re is lim Left() =8! YOU TRY IT 5 This problem sks you to exted wht we hve bee doig bove Fuctios with egtive vlues fuctio f (x) eeds to be o-egtive There s o reso why i Riem sum f (c k )Dx the () Usig the two grphs of f below, drw Lower() (the lower Riem sum) d Upper() (upper sum) d evlute ech Note: some heights d res will be egtive!

4 mth riem sums, prt Lower() Upper() 0 0 (b) Usig the grph estimte Lower() d Upper() No summtio formulæ re eeded (c) Wht do these sums represet geometriclly? (d) The fuctio f (x) is stright lie i this problem Figure out the equtio of f (x) (e) Why does the sum Lower() use right edpoits? (f ) Set up d simplify Lower() thikig of it s Right() (g) Evlute lim! Lower() How is your swer relted to the two trigles i the origil grph? swer to you try it 5 () For Upper(), otice tht the bses of the rectgles re lwys o the x xis, which my ot be the bottom of the grid!!! Upper() 0 (b) U() = ( 05) = 70 (c) The rectgles below the xis produce egtive re, so the result is et re, tht is re bove the x-xis mius the re below it (d) f (x) = x + (e) f (x) is decresig so the lowest poit i ech itervl is t the right ed (f ) Dx = x k = i Simplifyig we get Lower() = (g) lim Lower() =! i + = YOU TRY IT 6 (Puttig it ll together) Suppose tht y = x + o the itervl [, ] () Wht is the formul for the right-hd Riem sum Right()? (b) Fid lim! Right() Right() = (b) lim! pple 6i swer to you try it 6 () Right() = YOU TRY IT 7 Fid the geerl formul for regulr right-hd Riem sums Right() for the followig fuctios d itervls Mke sure to simplify f (x k ) The use the summtio formulæ to simplify Right() s much s possible () f (x) =x 5 o [0, ] (b) f (x) =x x o [, ] (c) Evlute Right(00) for both Riem sums bove Determie lim! Right() for ech

5 mth riem sums, prt (c) For () Right(00) = = 66 For (b) Right(00) =5 + pple 9k + i = = 6805 Tke the limits: () lim Right() =6; (b) lim Right() =5!! = (b) # 5 i " swer to you try it 7 () webwork: Click to try Problems 8 through 9 Use guest logi, if ot i my course The Defiite Itegrl Everythig hs worked out icely, especilly for right-hd Riem sums usig regulr prtitios But we strted with very geerl Riem sums of the form f (c k )Dx k The ext defiitio reverts bck to this geerl setup DEFINITION Let f be fuctio defied t ech poit i the closed itervl [, b] Let {x 0, x,,x } be prtitio of [, b] with = x 0 < x < x < < x < x = b Let Dx k = x k x k Let c k be y smple poit i the itervl [x k, x k ] If lim ll Dx k f (c k )Dx k! 0 exists, the we sy tht f is itegrble o [, b] If the limit exists, it is deoted by Z b lim ll Dx k f (c k )Dx k = f (x) dx,! 0 where d b re the lower d upper limits of itegrtio We sy tht the limit of the Riem sums, if it exists, is the defiite itegrl of f from to b This defiitio should remid you of idefiite itegrls There is coectio tht we will see bit lter Now i light of Theorem bout upper d lower sums of cotiuous fuctios, the followig result is ot too surprisig THEOREM (Cotiuity implies Itegrbility) If f is cotiuous o [, b], the f is itegrble The proof of this result is hrd Tke Mth! But Theorem provides the ituitio Theorem sys tht for cotiuous fuctios lim! Upper() = lim! Lower() the upper d lower sums sme vlue Now these re sums usig regulr prtitios d s!, we hve tht Dx! 0 However, Defiitio pplies to y d ll prtitios So the ew theorem, Theorem, is much more geerl result th Theorem d there is some work to do i provig it Tke-home Messge Theorem sys tht if f is cotiuous, the limit for y sequece of Riem sums exists d we get the sme umber s log s the widths Dx k of ll subitervls go to 0 So to ctully compute R b f (x) dx we might s well choose the most coveiet prtitios d smple poits Typiclly these re right-hd Riem sums where the prtitio is regulr, so Dx = b d c k = x k Z x EXAMPLE Determie dx Figure 9: The grph of f (x) = x o [, ]

6 mth riem sums, prt 5 Solutio f (x) = x is cotiuous (it is polyomil, i fct lier), but it is ot o-egtive By Theorem we kow tht f is itegrble To fid the vlue of the itegrl, we use coveiet Riem sum, Right() For subitervls, Dx = b x k = + kdx = = ( ) f (x k )= x k = + k + 6k Look crefully t the drwig i Figure 0 The Riem sum rectgles AL- WAYS START o the x-xis d go up or dow to the grph of f They do ot strt t the bottom of the picture Notice tht oe of the rectgles hppes to hve height of 0 Right() = f (x k )Dx = = 6 = 6 f = 6 = 6 ( )+8 + 6k 6 + k + 6 i ( + ) Figure 0: Right() for f (x) = x o [, ] = = + 9 So Z x dx = lim Right() = lim + 9!! = Notice tht the swer to Exmple is egtive so it cot represet ordiry re which must be o-egtive The itegrl represets the et re: the re bove the x-xis mius the re below the xis I this cse, the res bove d below the x-xis re two trigles d we see tht the differece i their res is re bove the xis re below the xis = ()() ()() = which checks with the itegrl Now we did ot eed Riem sums to compute et res of trigles But s soo s we hve curved regios, itegrls (usig Riem sums) re the oly method we hve of computig such et res Filly, whe f is o-egtive, the itegrl is equl to the re i the trditiol sese I other words, we hve solved the re problem DEFINITION (Are s Itegrl) If f is cotiuous d o-egtive o the closed itervl [, b], the the re, the the re bouded bove by f (x), below by the x-xis, d

7 mth riem sums, prt 6 Are(uder f )= R b f (x) dx by the verticl lies x = d x = b is Z b Are(uder f )= f (x) dx YOU TRY IT 8 Show tht the re uder the prbol y = f (x) = [, ] is usig Z x dx = lim Right()! x o the itervl webwork: Click to try Problems 0 through 7 Use guest logi, if ot i my course b Figure : The defiite itegrl solves the re problem