Chapter 5. Integration

Transcription

1 Chpter 5 Itegrtio Itrodutio The term "itegrtio" hs severl meigs It is usully met s the reverse proess to differetitio, ie fidig ti-derivtive to futio A ti-derivtive of futio f is futio F suh tht its derivtive F' stisfies F' = f o some suitle domi I mehis the veloity of ojet i motio my e determied from its elertio y ti-differetitio I Busiess, reveue futio my e omputed from give Mrgil Reveue futio y usig ti-differetitio I these d other emples, the questio of how to fid ti-derivtive of prtiulr futio f leds to other meig of the term "itegrtio" relted to re uder grph, defiite itegrtio These two meigs uder pproprite oditio re lied y the so lled "fudmetl theorem of lulus" There re other sutle mthemtil formultios lso lled itegrtio some of whih re geerliztios of the defiite itegrtio 5 Ati-derivtive Defiitio Let I e ope itervl Suppose F :I R is differetile o I d tht F' = f The F is lled ti-derivtive, idefiite itegrl or primitive of f Remr For give f :I R, there my e my or oe suh F with F' = f If F' = f, the let G = F + C, where C is ostt The G' = F' = f Thus y two ti-derivtives differ y ostt We reple I i Defiitio y y itervl if we te the derivtives t the ed poits of I to e the pproprite oe-sided limits (See Chpter 3 for osequees) Theorem If F is prtiulr ti-derivtive of f o ope itervl I, the every ti-derivtive of f o I is give y F() + C, where C is ostt, tht is, the set of ll ti-derivtives of f o I is { F() + C: C R} Proof Suppose G is other ti-derivtive of f o I Let h = G F o I The h' = G' F'= 0 o I Therefore, y Theorem 6 Chpter 4, h is ostt futio, sy h = C The h = G F = C d so G = F + C We ormlly write f d F C to deote fidig ti-derivtive of f, d wheever futio F with F'() = f () or d F f is foud Not every futio hs ti-derivtive s the followig emple will show Emple 3 The futio f :R R defied y f, 0 does ot hve, 0 ti-derivtive This is see s follows Suppose f hs ti-derivtive F :R R with F' = f The for > 0, F'() = f () = Therefore, y Theorem, for > 0, F() = + for some ostt i R For < 0, F'() = f () = Thus, lso y Theorem, for < 0, F() = + for some ostt i R Thus, lim F d lim F Sie F is differetile, F is otiuous d is therefore otiuous t 0 0 Ng Tze Beg

2 Chpter 5 Itegrtio = 0 Thus lim F lim F F 0 Hee = =, sy Therefore, F() = But the this implies tht F is ot differetile t = 0 sie the futio is ot differetile t = 0 This otrditio shows tht f does ot hve ti-derivtive The futio f i the ove emple is ot otiuous t = 0 But ot every disotiuous futio does ot hve ti-derivtive s the followig emple shows Emple 4 The futio f :R R defied y f si ( ) os( ), 0 0, 0 hs ti-derivtive F :R R, defied y F si( ), 0 eve though f 0, 0 is ot otiuous t = 0 We te the sequee ( ), where The 0 The f si ( ) os( ) for eh i P Therefore, f ( ) f (0) Hee the sequee ( f ( )) does ot overge to f (0) Therefore, y Defiitio of Chpter 3, f is ot otiuous t = 0 Theorem 5 d C; If f hs ti-derivtive o ope itervl I, the for y ostt, f lso hs ti-derivtive o I d f d f d 3 If f d f hve ti-derivtives o I, the for y rel umers d, [ f f ]d f d f d Theorem 5 is proved y diret verifitio d is left to the reder Theorem 6 For y rtiol umer, d C, Proof For, d d Remr Usig Theorem 5 d Theorem 6, we fid ti-derivtive of y polyomil futio Emple d 3 d 5 d 3 C 5 C 3 5 3C 5C 3 5 C, where C 3C 5C d C C d C C 4 d 3 d 3 3 C 5 5 C We shll ow eted our method of fidig ti-derivtive This is pplitio of the hi rule Suppose I d J re ope itervls, g: I R d F : J R re differetile futios with g(i ) J The F d g re omposle The Chi Ng Tze Beg

3 Chpter 5 Itegrtio Rule sys (F g )'() = F'(g())g'() for every i I (Referee: Theorem 8 Chpter 4) If we write f for F', we hve (F g )'() = f (g())g'() for every i I This shows tht F g is ti-derivtive of f (g())g'() I this wy, owig the ti-derivtive of f will ele us to determie the ti-derivtive of more omple futio f (g())g'() However, to pply this method of fidig ti-derivtive, we eed to put our futio i the form of f (g())g'() for suitle differetile futio g d futio f whose ti-derivtive is ow Hee, we hve the followig theorem Theorem 8 (Chge of Vrile, Sustitutio) Suppose g: I R is differetile d f : J R is suh tht f hs ti-derivtive F : J R d g(i ) J The f g g d F g C Moreover, F e hose to e y ti-derivtive of f If we write y = g (), the the theorem is usully lso rememered s f y dy d d f y dy = F y C Remr This is theorem out ti-derivtive It gives us etesio to fidig ti-derivtives y elrgig our "looup" tle through differetitio of omposite futios Emple 9 dy Fid 3 4 d Let y = The d 3 Here we let f (y) = y d g()= 3 +4 A ti-derivtive of f is F y d g'() = 3 y 3 y d d 3 f 3 4 g d 3 f g g d 3 3 g 3 C C Fid 3 d Let g d so g'()=, g 3 d d d g g g d g g g d y 3 y dy y Theorem 8, where y = g() is to e sustituted, y y C 5 y 5 3 y 3 C 3 3 C 3 Fid si d Let g() = d f () = si() The g'() = 3 Therefore, si d si d si g g d os g C os C sie ti-derivtive of si() is os() Ng Tze Beg 3

4 Chpter 5 Itegrtio 5 Riem Itegrls Are uder grph my e omputed y Riem itegrl If the veloity of ojet is ow umerilly over time itervl from time = 0 to sy, time = T, the diste trveled i this time itervl my e omputed y determiig the re uder the veloity-time grph of the ojet usig Riem itegrtio This pproh is differet from usig ti-derivtive of the veloity futio, whih my ot e ville Wht is ow lled the Riem itegrl of futio f is mthemtilly rigorous formultio of the ituitive otio of the re uder the grph of f Are uder urve We ow give desriptio of proedure to ompute the re of regio uder urve Cosider otiuous futio f defied o [, ] whose grph is give s follows y = = 0 The re uder the urve etwee = d = d y = f () is pproimted y the retgles show elow the grph of f We hve prtitioed [, ] ito smller itervls [ i, i+ ], i =0,,, with P: = 0 < < < = We ow defie the lower sum with respet to the 'prtitio' P to e the qutity give y i m i, where m i = mi{ f (): [ i-, i ]} By the Etreme Vlue Theorem, there eists i i [ i-, i ] suh tht m i = f ( i ) Now we let the prtitio e eqully sped, sy i for ll i The the lower sum with respet to the prtitio is f i f i If we te the ommo width of the retgles to e s smll s we wish we shll evetully pproimte the re uder the urve Tht is to sy, the re uder the urve A lim f i lim 0 f i Ie Give > 0, there eists positive iteger N suh tht for y iteger N f i A f i A Note here tht the i 's re geerlly differet for differet This gives pproh of fidig re from elow Ng Tze Beg 4

5 Chpter 5 Itegrtio Similrly we defie the upper sum with respet to the 'prtitio' P to e the summtio i M i, where M i = m{ f (): [ i-, i ]} = f (d i ) for some d i i [ i-, i ] y the Etreme Vlue Theorem If we te the prtitio to e regulr prtitio s efore, suh tht i, the re uder the urve my e give y A lim f d i lim 0 f d i This is pproh to fidig re uder urve from ove Emple 0 Fid the re of the regio ouded y y = d the -is d the lie = 3 Let f () = Suppose is positive iteger Let 0 = 0 d = 3 [We re goig to prtitio [0,3] ito equl prts] Let 3 0, Sie i 0 i 3i is iresig futio o [0, ), m i mi f :, i f 3 i () Therefore, A lim f lim 3 y () i 3 lim 7 3 i lim 7 3 i lim lim 3 9 lim 9 9 y = y 0 3 Typilly Emple 0 illustrtes the method of fidig re from elow Similrly, y tig the upper sum we should get the sme vlue for the re of the regio i Emple 0 Now if isted of tig the miimum or mimum i eh suitervl [ i-, i ], we simply hoose y vlue e i i [ i-, i ], the y the otiuity of f, the sum i f e i will lie etwee the upper sum d the lower sum I this wy, s the size of the suitervls, teds to 0, the sum i f e i would ted to the sme vlue s the re uder the urve This pproh hs some dvtges Firstly we do ot eed to fid the mimum or the miimum i eh suitervl Seodly we do ot eed to ofie to otiuous futios so log s the futio is defied However we eed to impose o hvig the sum Ng Tze Beg 5

6 Chpter 5 Itegrtio i f e i to e ouded d tht for meigful defiitio of re, the sum should ehve just lie the se for the otiuous futio Thus we wt the sum i f e i lled Riem sum to pproh fiite vlue s the prtitio of the itervl [, ] gets fier d fier Riem Sum d Riem Itegrl We ow formlize the ove disussio We will eve e etedig the defiitio to disotiuous futios The dvtge of the followig defiitio med fter the Germ mthemtii George Friedrih Berhrd Riem (86-866), is tht it voids the eed to fid the mimum or miimum of the futio over eh suitervl Let f e defied o the losed itervl [, ] We shll sudivide [, ] ito suitervls y hoosig - itermedite poits etwee d Defiitio A prtitio of [, ] is fiite set of umers { 0,,, } suh tht : = 0 < < < = Let i e the legth of the i-th suitervl [ i-, i ], ie i = i i- The orm or mesh of the prtitio is m{ i : i =,, } d is deoted y Thus = m{ i i- :i =,, } Suppose ow : = 0 < < < = is prtitio Let e poit i [ i-, i ], for i =,,, The Riem sum S of f with respet to is defied y S f i i f i We deote S y R( f,, ), where =(,, ) is the hoie of i i [ i-, i ] We sy the futio f is Riem itegrle o [, ] if d oly if there eists umer L suh tht give y > 0, there eists > 0 suh tht for ll prtitio of [, ] with orm <, we hve for every Riem sum S for S L <, ie, f i i L R f,, L where is the prtitio : = 0 < < < = d =(,, ) is give y y hoie of i [ i-, i ] If f is Riem itegrle o [,], we the write limit is deoted y f lim 0, f i i Ld this Ng Tze Beg 6

7 Chpter 5 Itegrtio Defiitio If f : [, ] R is itegrle o [, ], the the defiite itegrl of f or Riem itegrl of f from to is deoted y, f d lim f i i 0 is lled the lower limit, is lled the upper limit d f () is lled the itegrd Remr Though Defiitio is remrle elegt defiitio, it is ot esy to use it to prove the properties of the Riem itegrl Thus, we shll loo t pproh y Drou usig his so lled Drou sums, whih re geerliztio of the upper d lower sums This pproh is prtiulrly useful for verifitio of the properties of the Riem itegrl We shll develop this fter the followig emple Emple 3 0 d Te y prtitio: P: 0 =0 < < < = Let i e y poit i [ i-, i ] Let f () = The Riem sum, R f, P, Therefore, Liewise, R f, P, f i i i i It follows tht f i i f i i R f, P, i i f i i i i i i i P P R f, P, P These two iequlities () d () implies tht () i i i i P R f, P, P () P P Now give y > 0, just te = The for y prtitio P with P < R f, P, P [0, ] d 0 f d Hee, y Defiitio, f is Riem itegrle o Ng Tze Beg 7

8 Chpter 5 Itegrtio Let f : [, ] R e futio We rell tht the futio f is ouded if its rge is ouded This mes there eist rel umers d suh tht f () for ll i [, ] or equivletly, there eists positive rel umer K suh tht f () K for ll i [, ] If f is ot ouded, the oe hve ritrrily lrge Riem sum or ritrrily egtively lrge Riem sum d so f would ot e Riem itegrle (Referee: my rtile, "Riem Itegrl d Bouded Futio" o My Clulus We ) Thus y uouded futio is ot Riem itegrle 53 Upper d Lower Drou Sums, Upper d Lower Itegrls Suppose ow f : [, ] R is ouded futio Let P : = 0 < < < = e prtitio for [, ] The upper Drou sum with respet to the prtitio P is defied y U f, P M i i, where M i = sup{ f () : [ i-, i ]} Note tht sie f is ouded o [, ], f is ouded o eh [ i-, i ] d so the supremum M i eists for eh i Lie wise for eh i, m i = if{ f () : [ i-, i ]} eists sie f is ouded o eh [ i-, i ] We defie the lower Drou sum with respet to the prtitio P y L f, P m i i Beuse for eh iteger i suh tht i, m i M i, L( f, P) U( f, P) Sie f is ouded, there eist rel umers m d M suh tht m f () M for ll i [, ] Hee m M i M d m m i M for i =,,, Therefore, for y prtitio P the upper Drou sum U f, P M i i m i m Hee the set of ll upper Drou sums (over ll prtitios of [, ]) is ouded elow y m( ) Liewise, the lower Drou sum L f, P m i i M i M We olude tht the set of ll lower Drou sums (over ll prtitios of [, ]) is ouded ove y M( ) We my ow me the followig defiitio followig Drou Defiitio 4 Suppose f : [, ] R is ouded futio The the upper Drou itegrl or upper itegrl is defied to e U f if U f, P : P prtitio of, The lower Drou itegrl or lower itegrl is defied to e L f sup L f, P : P prtitio of, Ng Tze Beg 8

9 Chpter 5 Itegrtio Note tht y the ompleteess property of the rel umers, the upper itegrl eists, euse the set of ll upper Drou sum is ouded elow d the lower itegrl eists euse the set of ll lower Drou sum is ouded ove We ow oserve some ovious properties of the Drou sums with respet to prtitio Suppose Q d P re prtitios of [, ] We sy Q is refiemet of P if eh prtitio poit of P is lso prtitio of Q, more preisely if P Q Lemm 5 The Refiemet Lemm Suppose f : [, ] R is ouded futio Suppose Q d P re prtitios of [, ] suh tht Q is refiemet of P The L( f, P) L( f, Q) d U( f, Q) U( f, P) Proof First we shll prove the lemm whe Q otis just oe dditiol poit y th P Let P e deoted y P : = 0 < < < = Suppose y ( j-, j ) for some j etwee d The Q is the prtitio Q : = 0 < < < j- < y < j < < = Let m j ' = if{ f () : [ j-, y]}, m j '' = if{ f () : [y, j]} The m j = if{ f () : [ j-, j ]} m j ', m j '' Therefore, j L f, P m i i m i i m j y j m j j y m i i i j j m i i m j y j m j j y i j m i i L f, Q Let M j ' = sup{ f () : [ j-, y]}, M j '' = sup{ f () : [y, j]} The M j = sup{ f () : [ j-, j ]} M j ', M j '' Therefore, U f, P j j M i i M i i M j y j M j j y i j M i i M j y j M j j y i j M i i M i i U f, Q This proves the lemm for the se whe Q hs just oe dditiol prtitio poit th P For the geerl se, if Q otis poits ot i P, the there is sequee of prtitios, P = P 0, P, P,, P = Q where Q is otied y ddig oe poit t time Tht is P i+, is otied y ddig oe poit i Q ot i P i to P i Thus y the speil se, L( f, P)=L( f, P 0 ) L( f, P ) L( f, P ) ( f, P ) =L( f, Q) d U( f, P)=U( f, P 0 ) U( f, P ) U( f, P ) U( f, P ) =U( f, Q) This ompletes the proof The et result is oservtio tht y lower Drou sum is less th or equl to y upper Drou sum Ng Tze Beg 9

10 Chpter 5 Itegrtio Theorem 6 Suppose f : [, ] R is ouded futio Suppose Q d P re prtitios of [, ] L( f, P) U( f, Q) Proof The prtitio P Q is refiemet of oth P d Q Therefore, y Lemm 5, L( f, P) L( f, P Q) d U( f, P Q) U( f, Q) Therefore, L( f, P) L( f, P Q) U( f, P Q) U( f, Q) Theorem 7 Suppose f : [, ] R is ouded futio The L f U f Proof By Theorem 6 for y prtitio P of [, ], L( f, P) is lower oud of {U( f, Q): Q prtitio of [, ] } Therefore, L f, P if U f, Q : Q prtitio of, U f Thus the upper itegrl U f is upper oud of {L( f, P): P prtitio of [, ] } Therefore, U f sup L f, P : P prtitio of, L f 54 Drou Itegrl For ouded futio we hve defied the upper d lower itegrls So isted of oe itegrl we hve two Arhimedes hd devised strtegy to ompute the re of o-polygol geometri ojet y ostrutig outer d ier polygol pproimtios of the ojet We hve defied the upper d lower itegrls, similr i philosophy to the pproh of fidig suessive outer d ier polygol pproimtios of geometri ojet Ulie the o-polygol geometri ojets, where the outer d ier polygol pproimtios will ted to the sme vlue, there is o gurtee tht the two itegrls re the sme Whe they re the sme it gives meig to geerliztio of re Defiitio 8 Suppose f : [, ] R is ouded futio We sy f is Drou itegrle if the lower d upper itegrls re the sme, ie, L f U f Remr This is Drou's versio of the itegrility of ouded futio ompletig the formultio of Riem We shll show tht this is equivlet to Riem itegrility Emple 9 The Dirihlet futio h : [0, ] R, defied y h 0, if is rtiol, if is irrtiol is ot Drou itegrle Let P : 0 = 0 < < < < = e y prtitio for the itervl [0, ] By the desity of the rtiol umers d irrtiol umers, i eh of the suitervl, Ng Tze Beg 0

11 Chpter 5 Itegrtio [ i -, i ], (i =,, ) we lwys fid rtiol umer d irrtiol umer Hee for i =,,, {h(): [ i -, i ]} = {0, } It follows tht, for eh i =,,, M i (h, P) = sup{h(): [ i -, i ]} = sup{0, }= m{0, }= d m i (h, P) = if{h(): [ i -, i ]} = if{0, }= mi{0, }= 0 Therefore, the upper Drou sum with respet to the Prtitio P is U h, P M i h, P i i d the lower Drou sum with respet to the Prtitio P is L h, P m i h, P i 0 The ove sttemet is true for y prtitio P for [0, ] Hee the lower Drou itegrl of h, L h sup L h, P : P is prtitio for 0, m 0 0 d the upper Drou itegrl of h, U h if U h, P : P is prtitio for 0, m Therefore, the lower Drou itegrl L h is ot equl to the upper Drou itegrl U h d so h is ot Drou itegrle over [0, ] y Defiitio 8 We oserve lso tht y the ove remr out desity of the rtiol d irrtiol umers, for y prtitio P with P < we hoose rtiol poits i for eh i d so the Riem sum R(h, P, ) = 0 d if we hoose irrtiol poits i for eh i, the Riem sum R(h, P, ) = Hee y Defiitio, h is ot Riem itegrle The futio f : [0, ] defied y f () = is Drou itegrle Let P : 0 = 0 < < < < = e y prtitio for the itervl [0, ] For eh i =,,, M i ( f, P) = sup{ f (): [ i -, i ]} = i d m i ( f, P) = if{ f (): [ i -, i ]} = i - Therefore, the upper Drou sum with respet to the Prtitio P is U f, P M i f, P i i i 3 i i 3 i i 3 6 i P i P 3 6 P P euse i i Therefore, sie we hoose the orm of the prtitio P ritrrily smll, U 0 f 3 Similrly the lower Drou sum with respet to the prtitio P L f, P m i f, P i i Ng Tze Beg

12 Chpter 5 Itegrtio 3 i i 3 i i 3 6 i 3 i P 3 P Therefore, L 0 f L f, P 3 P Sie P e hose to e ritrrily smll, It follows the from Theorem 7 tht L 0 f 3 U 0 f L 0 f sie L 3 f U f 55 Itegrility Criteri Drou itegrility is equivlet to Riem itegrility s we shll show i this setio It is ot lwys esy to ompute the lower d upper itegrls just to he whether they re the sme to deide o itegrility This mouts to tully omputig the itegrl It is suffiiet to he o the ehviour of the lower d upper sums to see if they re very lose to some vlue or ltertively if their differees re gettig smller d smller This is ruil to formulte itegrility riteri We strt y givig et very useful hrteriztio of the upper d lower itegrls i terms of sequees Propositio 0 Suppose f : [, ] R is ouded futio The there eists sequee of prtitios (P ) of [, ] suh tht P P +, L f, P L f d U f, P U f lim P 0 d Proof By defiitio of the lower itegrl d upper itegrl, there eist prtitios P ' d P '' of [, ] suh tht L f L f, P L f d U f U f, P U f Let P e ommo refiemet of P ' d P '' for whih P < The y the Refiemet Lemm (Lemm 5) L f L f, P L f d U f U f, P U f Similrly, there eist prtitios P ' d P '' of [, ] suh tht d L f L f, P L f U f U f, P U f Ng Tze Beg

13 Chpter 5 Itegrtio Let P e ommo refiemet of P, P ' d P '' for whih P < / The d L f L f, P L f U f U f, P U f We ow defie the sequee {P } y repetig the ove proess Suppose we hve defied prtitio P suh tht P - P, P < / By the defiitio of the lower d upper itegrls, there eist prtitios P + ' d P + '' of [, ] suh tht L f d L f, P L f U f U f, P U f Let P + e ommo refiemet of P, P + ' d P + '' for whih P + < /(+) The y the Refiemet Lemm, L f L f, P L f d U f U f, P U f d P P + I this wy we oti the sequee of prtitios {P } of [, ] suh tht P P + d lim P 0 I prtiulr, y the defiitio of overgee of sequee or y the Compriso Test (Propositio 8 Chpter ), L f, P L f d U f, P U f This ompletes the proof Theorem Suppose f : [, ] R is ouded futio The followig sttemets re equivlet () f is Drou itegrle () There is sequee (P ) of prtitios of [, ] suh tht lim U f, P L f, P 0 Furthermore, for y suh sequee, L f, P f d U f, P (3) Give > 0, there eists prtitio P for the itervl [, ] suh tht the differee U( f,p) L( f, P) < (4) f is Riem itegrle, ie, there eists umer L suh tht give y > 0, we fid > 0 suh tht for y prtitio P for [, ] with orm P <, d for y Riem sum S with respet to P, S L < f Ng Tze Beg 3

14 Chpter 5 Itegrtio Proof () () Suppose f is Drou itegrle, tht is, the lower d upper itegrls re the sme Te the sequee of prtitios (P ) of [, ] give y Propositio 0 suh tht L f, P L f d U f, P U f Sie L f U f, lim U f, P L f, P 0 () () Suppose there is sequee (P ) of prtitios of [, ] suh tht lim U f, P L f, P 0 For eh iteger, y Theorem 7, L f, P L f U f U f, P Therefore, 0 U f L f U f, P L f, P d so 0 U f L f lim U f, P L f, P 0 Hee, U f L f 0 d so L f U f d f is Drou itegrle Moreover, sie 0 U f, P U f U f, P L f, P for eh iteger, d so y the Compriso Test, U f, P U f f Similrly, sie 0 L f L f, P U f, P L f, P for eh iteger, y the Compriso Test for sequees, L f, P L f f () (3) Suppose there is sequee (P ) of prtitios of [, ] suh tht lim U f, P L f, P 0 The, give > 0, there eists positive iteger N, suh tht N U( f,p ) L( f, P ) < Let P = P N d we hve U( f,p ) L( f, P ) < (3) () Suppose for y > 0, there eists prtitio P for the itervl [, ] suh tht the differee U( f,p) L( f, P) < Thus, for y > 0, 0 U f L f U f, P L f, P Therefore, U f L f It follows the y Theorem 7 tht L f U f d so f is Drou itegrle (4) (3) Assume f is Riem itegrle Give > 0, the there eists > 0 suh tht for y prtitio P for [, ] with orm P <, d for y Riem sum S with respet to P, S L < /4 Let T ={ Riem sum S: S hs the sme prtitio P} The for y S i T, L /4< S < L + / () Ng Tze Beg 4

15 Chpter 5 Itegrtio Let P: = 0 < < < = e prtitio of [, ] with P < Let M i = sup{ f () : [ i-, i ]} for i =,, By the defiitio of supremum, for eh i, i, there eists i i [ i-, i ] suh tht f ( i ) > M i 4 Usig this iequlity, the Riem sum R f, P, C f i i M i 4 i M i i U f, P, 4 4 where, C=(,, ) Therefore, U f, P R f, P, C L L () y iequlity () Now let m i = if{ f () : [ i-, i ]} for i =,, By the defiitio of ifimum, for eh i suh tht i, there eists d i i [ i-, i ] stisfyig f (d i ) < m i 4 The Riem sum R f, P, d f d i i m i 4 i m i i L f, P 4 4 where d=(d,,d ) It follows tht L f, P R f, P, d L L (3) y iequlity () Iequlity () d (3) implies tht, U( f, P) L( f, P) < Hee () follows () (4) Suppose there is sequee (P ) of prtitios of [, ] suh tht lim U f, P L f, P 0 This mes give > 0, there eists prtitio P j : = 0 < < < L = with L > suh tht U f, P j L f, P j Deote the prtitio P j y P Let K = mi { i i- : i =,, L } Sie f is ouded, there eists M 0 suh tht f () < M First we shll speify our for the prtitio Let R: = y 0 < y < < y N = e y prtitio suh tht R, where mi K, ML The euse R < K, the umer of suitervls of R, N must e stritly igger th L, whih is the umer of suitervls of P Moreover euse K = mi { i i- : i =,, L } d for eh i =,,, N, y i y i- R < K, eh suitervl [y i-, y i ] oti t most oe poit from P Therefore, eh i for i =,,, L must elog to oe d oly oe suitervl y ji, y ji for some j i, j i N, ie, for i =,,, L, y ji i y ji Let I = {j i : i =,,L } I j i : i,, L The for y Riem sum with respet to the prtitio R, R f, R, N f i y i y, where i [y i-, y i ], =(,, ) f i y i y L f ji y ji y ji i I i I Ng Tze Beg f i y i y L f i y ji y ji 5 L f ji f i y ji y ji

16 Chpter 5 Itegrtio i I f i y i y L L f i y ji i f ji f i y ji y ji L f i i y ji Note tht the reted term is Riem sum S for the prtitio R P Thus R f, R, S L f ji f i y ji y ji U f, R P L M y ji y ji U f, P L M y ji y ji y the Refiemet Lemm L f, P L M R L f, P ML R L f ML L sie ML f R ML We hve thus proved tht R f, R, L f (4) Similrly, R f, R, N f i y i y S L f ji f i y ji y ji L f, R P L f ji f i y ji y ji L f, P L f ji f i y ji y ji U f, P L M y ji y ji U P, f L M R U f ML R U f ML ML U f ie R f, R, U f (5) Sie L f U f C (euse () ()), it follows from iequlities (4) d (5) tht R f, R, L We hve thus show tht there eists rel umer C suh tht for y prtitio R with orm R < d for y Riem sum S =R( f, R, ) with respet to R, S C Hee, f is Riem itegrle, Remr Note tht we hve give the proof of Theorem i more th oe wys Most oftely used prt of the theorem will e the equivlee of sttemets () to (3) Note tht (4) () Theorem is very useful tool for further developmet of the itegrl 3 I view of Theorem, we shll simply sy futio is itegrle wheever yoe of the equivlet oditios i Theorem is met Itegrility of Mootoe Futio Theorem Suppose f : [, ] R is mootoe futio The f is Riem itegrle Proof Sie f is mootoe o the losed d ouded itervl [, ], f is ouded Suppose tht f is iresig Let P : = 0 < < < = e prtitio for [, ] The Ng Tze Beg 6

17 Chpter 5 Itegrtio U f, P L f, P M i m i i Sie f is iresig, M i = f ( i ) d m i = f ( i- ) Therefore, U f, P L f, P P P M i m i f i f P f f If f () = f (), the give y > 0, for y prtitio P, U( f, P) L( f, P) = 0 < If f () f (), te y prtitio P suh tht P < /( f () f ()) The U f, P L f, P P f f Therefore, y Theorem, f is Riem itegrle The proof is similr whe f is deresig Itegrility of Cotiuous Futio Net we hve theorem whih ofirms tht defiite itegrls do eist t lest o otiuous futios Theorem 3 Every futio whih is otiuous o the losed itervl [, ] is (Riem) itegrle o [, ] Proof If f : [, ] R is otiuous, the it is lso uiformly otiuous (Referee: Theorem 9 Chpter 3) Therefore, give y > 0, there eists > 0 suh tht for ll, y i [, ], y < f () f (y) < Let P : = 0 < < < < = e prtitio with orm P < whih is give y () ove For i =,,, let M i = sup{ f () : [ i-, i ]} Sie f is otiuous o [ i-, i ], for eh i, y the Etreme Vlue Theorem, M i = f ( i ) for some i i [ i-, i ] Similrly for eh i =,,, let m i = if { f () : [ i-, i ]} Agi y the Etreme Vlue Theorem, for eh i =,,, there eists d i i [ i-, i ] suh tht m i = f (d i ) The upper Drou sum with respet to P is U f, P M i i d the lower Drou sum with respet to P is L f, P It follows tht the differee U f, P L f, P m i i f i f d i i f i i f d i i f i f d i i Ng Tze Beg 7

18 Chpter 5 Itegrtio y () sie i - d i P i i Therefore, U f, P L f, P i 0 Thus, y Theorem, f is Riem itegrle This ompletes the proof Remr More geerlly, suppose f : [, ] R is ouded futio, tht is, for some rel umers M d L, M < f () < L for ll i [, ] Suppose f is otiuous eept possily o suset of "mesure zero" The f is itegrle Ideed the overse is lso true: f is itegrle implies tht f is otiuous eept o suset of "mesure zero" This result is stted y Drou ut is ofte referred to s Leesgue Theorem By set of mesure zero we me tht the poits of disotiuity e elosed y t most outly ifiite set of itervls whose totl legth is ritrrily smll Ay fiite or outle set is of mesure zero The set of rtiol umers is of mesure zero (See Chpter 4 for defiitio of the Leesgue mesure of set) This gives other wy of provig Theorem sie y mootoe futio o [, ] hve t most outle umer of disotiuities Emple 4 [,] f 0, 0, 0 f is ot otiuous t 0 ut f is itegrle o Let P : = 0 < < < < = e prtitio of [,] Let i = i i- Let R f, P, f i i f i i () e Riem sum with respet to the prtitio P d =(,, ), where i is i [ i-, i ] If i 0 for i =,,, the f ( i ) 0 for ll i =,,, d so R f, P, f i i 0 If j 0 for some j, the f ( i ) 0 fo i j, j or i j, j+, whihever mes sese The Riem sum R f, P, f j j f j j or R f, P, f j j f j j Sie f () for ll i [, ], R f, P, Give y > 0, we te y prtitio P with j j P or j j P P < Therefore, f is Riem itegrle o [, ] d f 0 d R( f, P, ) 0 Ng Tze Beg 8

19 Chpter 5 Itegrtio Emple 5 d Proof Let f () =, Let P : = 0 < < < < = e prtitio of [, ] Sie for y ostt futio, mimum d miimum re the sme d so the upper Drou sum d the lower Drou sum for the prtitio P re the sme d it follows tht y ostt futio is Riem itegrle Therefore, f is Riem itegrle d the itegrl is give y the limit of sequee of lower sums of sequee of prtitios y Theorem Ay lower sum L f, P m i i i 0 e (P ) to e sequee of prtitios suh tht P 0 The L( f, P ) ( ) sie L( f, P ) is ostt sequee d so d 56 Properties of the Riem Drou itegrl Theorem 6 Suppose f : [, ] R is ouded futio If f is ouded d itegrle o the losed itervls [, ] d [, ], where < <, the f is itegrle o [, ] d f d f d f d Proof By Theorem, there eists sequee (P ) of prtitios of [, ] suh tht lim U f, P L f, P 0 d L f, P f Similrly, there eists sequee ( Q ) of prtitios of [, ] suh tht lim U f, Q L f, Q 0 d L f, Q f The (P Q ) is sequee of prtitios of [, ] Note tht U( f, P Q ) = U( f, P ) + U( f, Q ) d L( f, P Q ) = L( f, P ) + L( f, Q ) Therefore, lim U f, P Q L f, P Q lim U f, P U f, Q L f, P L f, Q lim U f, P L f, P lim U f, Q L f, Q It follows y Theorem tht f is Riem itegrle o [, ] Moreover, L f, P Q L f, P L f, Q f Hee f d f d f d f Theorem 7 Suppose f : [, ] R is ouded futio If f is itegrle o [, ], the for y i (, ), f is itegrle o [, ] d o [, ] d Ng Tze Beg 9

20 Chpter 5 Itegrtio f d f d f d Proof By Theorem, sie f is Riem itegrle, there is sequee (P ) of prtitios of [, ] suh tht lim U f, P L f, P 0 We my ssume tht the poit elogs to eh prtitio P We epli this s follows Let Q = P {}, the y the Refiemet Lemm, for eh positive iteger, U( f, Q ) L( f, Q ) U( f, P ) L( f, P ) Therefore, y the Compriso Test for sequees, lim U f, Q L f, Q 0 If P, we reple P y Q Hee we my ssume tht elogs to prtitio P for eh positive iteger Therefore, eh prtitio P is uio of prtitio P ' for [, ] d prtitio P '' for [, ] I prtiulr, U( f, P ) = U( f, P ' ) + U( f,p '' ) d L( f, P ) = L( f, P ' ) + L( f,p '' ) Thus, U( f, P ) L( f, P ) = U( f, P ' ) L( f, P ' ) + U( f,p '' ) L( f,p '' ) d so U( f, P ' ) L( f, P ' ) U( f, P ) L( f, P ) d U( f, P '' ) L( f, P '' ) U( f, P ) L( f, P ) Therefore, y the Compriso Test for sequees, lim U f, P L f, P 0 d lim U f, P L f, P 0 It follows y Theorem tht f is Riem itegrle o [, ] d o [, ] Moreover, L f, P L f, P L f, P Thus, f d f d f d f f Net we shll list the properties of the Drou sums whih we shll use lter Lemm 8 Suppose f : [, ] R d g : [, ] R re ouded futios Let P : = 0 < < < < = e prtitio of [, ] The L( f, P) + L( g, P) L( f + g, P) d U( f, P) + U( g, P) U( f + g, P) (A) Furthermore, for y rel umer, L( f, P) = L( f, P) d U( f, P) = U( f, P) if 0 U( f, P) = L( f, P) d L( f, P) = U( f, P) if (B) Proof These re oservtio regrdig properties of ifimum d supremum First we set up some ottio Suppose h : [, ] R is ouded futio d P : = 0 < < < < = is prtitio of [, ] For eh i =,,, let M i (h, P) = sup{h(): [ i -, i ]} d m i (h, P) = if{h(): [ i -, i ]} We egi y emiig the ompoets of the lower sums Ng Tze Beg 0

21 Chpter 5 Itegrtio m i ( f, P) = if{ f (): [ i -, i ]} is lower oud of { f (): [ i -, i ]}d m i (g, P) = if{ g(): [ i -, i ]} is lower oud of { g(): [ i -, i ]} Therefore, m i ( f, P) + m i (g, P) f () + g() for ll [ i -, i ] It follows tht m i ( f, P) + m i (g, P) is lower oud of { f () + g(): [ i -, i ]} Therefore, m i ( f, P) + m i (g, P) if { (f + g)() : [ i -, i ]} = m i ( f + g, P) () y the defiitio of ifimum, d so L( f, P) + L( g, P) = m i f, P i m i g, P i m i f g, P i L f g, P This proves the first iequlity of (A) We proeed with the seod iequlity of (A) similrly By defiitio M i ( f, P) is upper oud of { f (): [ i -, i ]} d M i (g, P) is upper oud of { g(): [ i -, i ]} Therefore, for ll [ i -, i ], M i ( f, P) + M i (g, P) f () + g() It follows tht M i ( f, P) + M i (g, P) is upper oud for the set { (f + g)() : [ i -, i ]} Hee, y the defiitio of supremum, M i ( f, P) + M i (g, P) sup{ (f + g)() : [ i -, i ]} = M i ( f + g, P) () The U( f, P) + U( g, P) = M i f, P i M i g, P i M i f g, P i U f g, P y () This proves the seod iequlity of (A) To prove the iequlity (B), ote tht for ouded set of rel umers, A, for y 0, if( A) = if(a), sup(a) = sup(a) Also for < 0, if( A) = sup(a), sup(a) = if(a) It follows tht for eh i =,,, m i ( f, P) = m i ( f, P) d M i ( f, P) = M i ( f, P) if 0 d m i ( f, P) = M i ( f, P) d M i ( f, P) = m i ( f, P) if < 0 Therefore, it follows from this set of equlities tht L( f, P) = L( f, P) d U( f, P) = U( f, P) if 0 ; U( f, P) = L( f, P) d L( f, P) = U( f, P) if 0 Theorem 9 Suppose f : [, ] R is ouded futio If f is Riem itegrle, the for y rel umer, f is Riem itegrle d f d f d Proof By Theorem, sie f is itegrle, there is sequee (P ) of prtitios of [, ] suh tht lim U f, P L f, P 0 d tht L f, P d f U f, P f If 0, the lim U f, P L f, P lim U f, P L f, P lim U f, P L f, P 0 Ng Tze Beg

22 Chpter 5 Itegrtio If < 0, the lim U f, P L f, P lim L f, P U f, P lim U f, P L f, P 0 Therefore, y Theorem, f is itegrle d L f, P L f, P f 0 d L f, P U f, P f if < 0 d so f f if Theorem 30 Suppose f : [, ] R d g : [, ] R re ouded futios If f d g re itegrle, the f + g is itegrle d f g f g Proof Sie f d g re itegrle, y Theorem there eist sequee (P ) of prtitios of [, ] suh tht lim U f, P L f, P 0 d sequee (Q ) of prtitios of [, ] suh tht lim U g, Q L g, Q 0 Te the sequee of prtitio (R ), where R = P Q for eh positive iteger By the Refiemet Lemm, for eh positive iteger, L( f, P ) L( f, R ) U( f, R ) U( f, P ) d L( g, Q ) L(g, R ) U( g, R ) U( g, Q ) It follows tht 0 U( f, R ) L( f, R ) U( f, P ) L( f, P ) () d 0 U( g, R ) L(g, R ) U( g, Q ) L( g, Q ) () Beuse of iequlities () d (), lim U f, P L f, P 0 d tht lim U g, Q L g, Q 0, y the Compriso Test for sequees, lim U f, R L f, R 0 d L g, R 0 Cosequetly, (see for emple Theorem ()), we hve tht L f, R, f U f, R f L g, R Now y Lemm 8, for eh positive iteger, g, U g, R (3) g L( f, R ) + L( g, R ) L( f + g, R ) U( f + g, R ) U( f, R ) + U( g, R ) (4) Therefore, 0 U( f + g, R ) - L( f + g, R ) U( f, R ) L( f, R ) + U( g, R ) L( g, R ) ---- (5) Sie lim U f, R L f, R U g, R L g, R lim U f, R, L f, R lim U g, R L g, R 0 it follows from the iequlity (5) d the Compriso Test for sequees, lim U f g, R L f g, R 0 Therefore, y Theorem (), f + g is itegrle By the iequlity (4), the limits i (3) d the Squeeze Theorem for sequees, L f g, R But f g L f g, R f g d so f g f g Remr A proof usig Theorem (4) is preseted i Ng Tze Beg's "Clulus, itrodutio", 943 Theorem 3 Suppose f : [, ] R is ouded futios If f 0 d itegrle, the f 0 Ng Tze Beg

23 Chpter 5 Itegrtio Proof Sie f is itegrle, y Theorem (), there eists sequee (P ) of prtitios of [, ] suh tht L f, P f Sie f 0, L( f, P ) 0 for eh positive iteger th or equl to 0 f eig the limit of the sequee (L( f, P ) ) is therefore greter Theorem 3 Suppose f : [, ] R d g : [, ] R re ouded futios If f d g re itegrle o [, ] d f () g() for ll i [, ], the g f Proof Sie f () g() for ll i [, ], f g 0 d so y Theorem 3 f g 0 But y Theorem 9 d Theorem 30, f g f g f g It follows tht f g 0 d so f g Remr Oe give proof usig either the upper Drou sum or the lower Drou sum d Theorem () We et stte useful oservtio Propositio 33 Suppose f : [, ] R is ouded d itegrle If g : [, ] R is futio suh tht g = f eept o fiite set of poits i [, ], the g is itegrle d g f Proof g() f () = 0 for ll ut fiite set of poits i [, ] This mes g f is lier omitio of futios of the type osidered i emple 4 Therefore, y Emple 4, Theorem 9 d 30, g - f is itegrle d g f 0 Sie g = (g f ) + f d oth (g f ) d f re itegrle, y Theorem 30, g is itegrle d g f g f f g f f 0 f This ompletes the proof Remr I ft muh more is true By Leesgue Theorem, if g = f eept o set of "mesure" 0 i [, ], d if f is itegrle, the g is itegrle d g f Theorem 34 Suppose f : [, ] R is ouded futio Suppose f is otiuous o (, ) The f is itegrle d the itegrl does ot deped o the vlues t the ed poits of the itervl Proof We shll use the equivlet oditio (3) for itegrility i Theorem Tht is, we shll show tht give y > 0, there is prtitio P for the itervl [, ] suh tht the differee U( f,p) L( f, P) < Sie f is ouded, there eists K > 0 suh tht f () K for ll i [, ] Let ( ) e deresig sequee i (, ( +)/) suh tht Let ( ) e iresig sequee i (( +)/, ) suh tht Give y > 0, there eists positive iteger N suh tht N implies tht 0 < < / (8K) d 0 < < Ng Tze Beg 3

24 Chpter 5 Itegrtio / (8K) Let m e y fied iteger Sie f is otiuous, f is otiuous o the losed d ouded suitervl [ m, m ] Therefore, y Theorem 3, the restritio of f to [ m, m ] is itegrle By Theorem (3), there eists prtitio P for the itervl [ m, m ] suh tht U f m,m, P L f m,m, P / Addig i the two ed poits d we get prtitio Q for [, ] Tht is, Q = P {, } The the upper Drou sum with respet to Q is U f, Q sup f :, m m U f m,m, P sup f : m, m K m K m U f m,m, P /4 U f m,m, P Liewise, the lower Drou sum with respet to Q is L f, Q if f :, m m L f m,m, P if f : m, m K m L f m,m, P K m /4 L f m,m, P Therefore, U f, Q L f, Q / U f m,m, P L f m,m, P / / It follows y Theorem (3) tht f is itegrle By Propositio 33 the itegrl is idepedet of the vlues of f t the ed poits of the itervl We show muh more Give y iteger i P, there eist itegers d m suh tht 0 8K d 0 m By Theorem (3), there eists prtitio P for the 8K itervl, m suh tht U f, m, P L f, m, P Without loss of geerlity we my ssume tht < + d m < m + for eh iteger i P Let Q = P {, } The, we show similrly s ove tht U f, Q L f, Q Hee lim U f, Q L f, Q 0 Therefore, y Theorem (), f U f, Q But U f, Q sup f :, U f, m, P sup f : m, m d sup f :, sup f : m, m 4 Sie lim, y the Compriso Test, 4 0 lim (sup f :, sup f : m, m ) 0 It follows tht lim U f, m, P lim U f, Q It is ler tht U f, m, P is idepedet of the ed poits of the itervl [, ] Similrly, we show tht L f, m, P f Sie L f, m, P m sequees, lim f f m f f U f, m, P, y the Squeeze Theorem for Ng Tze Beg 4

25 Chpter 5 Itegrtio Remr Note tht the oly use of the otiuity of f o (, ) i the proof ove is to dedue the itegrility of f o the suitervls [ m, m ] Emple 35 () Let f, 0, 0 By Propositio 33, f d d ssumig the Fudmetl 0 Theorem of Clulus () Let f, 0 3, f d 0 f d f d y Theorem 34 d Theorem 6 0 Clulus 0 d 3 d d ssumig the Fudmetl Theorem of (3) 03 d 0 d d 3 d 0 0d d 3 d y Propositio Riem Sums Covergee Theorem We hve see i Theorem () tht if f : [, ] R is itegrle, the the itegrl is the limit of sequee of lower Drou sums with respet to sequee of prtitios of [, ] We shll show et tht the itegrl is the limit of y sequee of Riem sums with respet to sequee of prtitios {P } of [, ] so log s P 0 Theorem 36 Suppose f : [, ] R is Riem itegrle o [, ] d (P ) is sequee of Prtitio of [, ] Let R( f, P, C ) e Riem sum with respet to the prtitio P with C eig y hoie of poits i the suitervls of P If P overges to 0 s teds to ifiity, the the sequee of Riem sums teds to f (R( f, P, C ) ) Proof Sie f is itegrle, y Theorem (4) give y > 0, we fid > 0 suh tht for y prtitio P for [, ] with orm P <, d for y Riem sum S with respet to P, S f Sie P overges to 0, we fid iteger N suh tht for y N, P < Therefore, euse eh R( f, P, C ) is Riem sum with respet to P, R f, P, C f for ll N This mes the sequee (R( f, P, C ) ) overges to s teds to ifiity Remr The usul pplitio of the ove theorem is to sequee of prtitios ( P ), where P = ( )/ s is the se whe P is regulr prtitio d the f Ng Tze Beg 5

26 Chpter 5 Itegrtio Riem sum R( f, P, C ) with respet to P is hose suh tht C the hoie of poits i the suitervls of P is hose either to e the ed poit or egiig poit of the suitervl I more detil, the pplitio is stted elow Corollry 37 Suppose we re give limit of the form lim g i If we write g i f i, where either i i or i i, the lim g i lim f i, f d if f is Riem itegrle o [, ] Therefore, if we fid ti-derivtive F of f, the lim g i F F y the Fudmetl Theorem of Clulus Emple 38 5 i lim 3 lim 0 5 i 5 d 3 [ 5 3/ ] / 5 3/ Drou Sums Covergee Theorem It is turl questio to s tht if the futio f is itegrle o [, ], the does the sme olusio s i Theorem 36 holds true for either the upper Drou sums or lower Drou sums, tht is, does the sequee of either upper Drou sum or lower Drou with respet to y sequee of prtitios overge to the itegrl of f if the orm of the prtitios ted to 0 Theorem 39 Suppose f : [, ] R is itegrle d (P ) is sequee of prtitios of [, ] If P overges to 0 s teds to ifiity, the lim U f, P L f, P 0 d osequetly L f, P f d U f, P f Proof Sie f is itegrle, y Theorem (4), give > 0, the there eists > 0 suh tht for y prtitio P for [, ] with orm P <, d for y Riem sum S with respet to P, S L < /4 Let T = { Riem sum S : S hs the sme prtitio P} The for y S i T, L /4 < S < L + / () Let P: = 0 < < < = e prtitio of [, ] with P < Let M i = sup{ f () : [ i-, i ]} for i =,, By the defiitio of supremum, for eh i, i, there eists i i [ i-, i ] suh tht f ( i ) > M i 4 The, usig this iequlity the Riem sum Ng Tze Beg 6

27 Chpter 5 Itegrtio R f, P, C f i i M i 4 i M i i U f, P, 4 4 where, C=(,, ) Therefore, U f, P R f, P, C L L () y iequlity () Now let m i = if{ f () : [ i-, i ]} for i =,, By the defiitio of ifimum, for eh i, i, there eists d i i [ i-, i ] suh tht f (d i ) < m i 4 The, usig this iequlity, the Riem sum R f, P, d f d i i m i, 4 i m i i L f, P 4 4 where d=(d,,d ) Therefore, L f, P R f, P, d L L (3) It follows from () d (3) tht if P <, the U( f, P) L( f, P) < Now sie P 0, there eists iteger N suh tht N P < Therefore, y wht we hve just proved, N U( f, P ) L( f, P ) < This mes lim U f, P L f, P 0 Remr The rgumet i the proof of Theorem 39 is i the proof of Theorem (4) (3) d is reprodued here for oveiee Me Vlue Theorem Theorem 40 Me Vlue Theorem for Itegrls Suppose f is otiuous o the losed itervl [, ] with < The there eists poit i [, ] suh tht f f Proof Sie f is otiuous o [, ], y the Etreme Vlue Theorem, there eists, d i [, ] suh tht m = f () f () f (d) = M for ll i [, ] Therefore, y Theorem 3, m f M d so f f f d Dividig the ove iequlity y ( ), we get f f f d Ng Tze Beg 7

28 Chpter 5 Itegrtio Therefore, sie f is otiuous, y the Itermedite Vlue Theorem, there eists etwee d d, ie i [, ] suh tht f (The left hd side of the equtio ove is ow s the me vlue) Hee f f Emple 4 We estimte Itegrls It sys there eists i f 3 si 8 d usig the Me Vlue Theorem for 6 6, 3 suh tht 6 si 8 Sie the futio sie is iresig o 6, 3, 3 si 8 d si si 6 si si 3 3 It follows tht si 8 d so si 8 3 d 4 6 Therefore, 3 si 8 7 d Fudmetl Theorem of Clulus The most importt theorem i Clulus is the fudmetl theorem euse it provides the li etwee differetitio d Riem itegrtio Our et result is theorem of Drou whih gives the fudmetl theorem without the ssumptio of the otiuity of the derived futio f ' of f Theorem 4 Drou Fudmetl Theorem of Clulus Suppose F : [, ] R is futio otiuous o [, ] d differetile o (, ) Suppose the derived futio F' : (, ) R is itegrle i the sese tht y etesio of F' to the ed poits d is itegrle The F F F Proof For eh iteger let the regulr prtitio P for [, ] e P : = 0 < < < =, with P = ( )/, for = 0,,, For eh =,,, sie F is otiuous o [, ] d differetile o (, ), y the Me Vlue Theorem (Theorem 5 Chpter 4), there eist i (, ) suh tht F' ( ) ( ) F( ) F( ) () Let R( F', P, C ) e the Riem sum with respet to the prtitio P, where C =(,, ) is the hoie of i i [ i-, i ] give y () The R F, P, C F F F F F 0 F F Sie P = ( )/ 0, y Theorem 36, the sequee of Riem sums (R( F', P, C ) ) teds to F But the sequee is ostt d so R( F', P, C ) F() F() Therefore, F F F The followig is the more fmilir form of the fudmetl theorem of lulus Theorem 43 (First Fudmetl Theorem of Clulus) Ng Tze Beg 8

29 Chpter 5 Itegrtio Suppose F : [, ] R is futio otiuous o [, ] d differetile o (, ) Suppose F' : (, ) R is otiuous d ouded The F F F Proof F' : (, ) R is otiuous d ouded implies tht F' is itegrle y Theorem 34 d tht the itegrl does ot deped o the vlues of the etesio of F' t the ed poits Hee, the theorem follows from Theorem 4 Remr Suppose f : [, ] R is itegrle d hs "ti-derivtive" F : [, ] R, whih is otiuous o [, ] d F'() = f () for ll i (, ) Note tht F eed ot e differetile t the ed poits of the itervl The y Theorem 4, f F F The sigifie of Theorem 43 is tht we use y "ti-derivtive" to ompute defiite itegrl Let f e futio otiuous o [, ] d so itegrle o [, ] For y otiuous futio F defied o [, ] suh tht F'() = f () for ll i (, ), (ie F is "ti-derivtive" of f i the sese give ove), f F F F This is the usul form tht the theorem is used Emple d Let f () = The f is otiuous o [0, ] We reogize immeditely tht F is ti-derivtive of f Thus 3 6 d Remr Not ll otiuous futios hve ti-derivtives epressile i terms of elemetry futios By elemetry futio, we me y futio formed from rel ostt, the idetity futio, the epoetil futio, the logrithmi futio, the trigoometri futios d the iverse trigoometri futios, y ddig, multiplyig, dividig or formig ompositio Not ll itegrle futios hve ti-derivtives For emple, the futio does ot hve /4 ti-derivtive i terms of elemetry futios o [-, ] d the futio f, 0 does ot hve ti-derivtive o [-, ] ut is itegrle We, 0 still ompute whih is 0 f Theorem 45 (Seod Fudmetl Theorem of Clullus) Suppose f : [, ] R is itegrle Defie F: [, ] R y F f for i [, ] The F is otiuous o [, ] If f is otiuous t i [, ], the F is differetile t d F'() = f () Proof Sie f is itegrle, f is ouded o [, ] Thus, there eists positive rel umer K suh tht K f () K for ll i [, ] We shll show tht F is Ng Tze Beg 9

30 Chpter 5 Itegrtio uiformly otiuous d hee otiuous For y < y suh tht y we hve tht K f (t) K for ll t i [, y] Hee, tig itegrls, y Theorem 3, y K y f K y d so y K( y ) = K y () f For y < y suh tht y, y y y F y f f f F f Therefore, y F y F f () Similrly, if y, y iterhgig the role of d y ove, we get, F F y y f (3) It follows from (), () d (3) tht, for y y, F F y K y Hee for y > 0, te to e y rel umer suh tht 0 < < /K The for ll, y i [, ], y < F() F(y) K y < K /K = This mes F is uiformly otiuous o [, ] d so otiuous o [, ] Prt () Suppose f is otiuous t 0 [, ] We shll show tht F f t dt is differetile t 0 For > 0, F F 0 f 0 f 0 f 0 f 0 f 0 f d for < 0, F F 0 f 0 f f f 0 f 0 f 0 f where we use the ovetio tht whe < d i [, ], d f d f For 0, let H F F 0 0 f 0 0 The y the defiitio of the derivtive, lim, where the limit is te to e the pproprite left or 0 H F 0 right limit whe 0 = or We shll show tht H() teds to f ( 0 ) s teds to 0 Sie f is otiuous t 0, give > 0 there eists > 0 suh tht 0 f f 0 f 0 f f 0 Tht is to sy, for ll i [, ] with 0, 0, we hve f 0 f f (4) Now we osider the right limit first For i ( 0, 0 + ), usig (4) d Theorem 3, we oti 0 f 0 0 Tht is to sy, for y i ( 0, 0 + ), f f 0 f 0 f t dt f 0 0 Dividig the ove iequlity y ( 0 ) > 0, we oti f 0 0 f t dt 0 H f 0 Therefore, H f 0 This shows tht lim H f 0 Similrly, for the left limit, we te i ( 0, 0 ) The from (4) d Theorem 3, we oti for y i ( 0, 0 ), 0 0 f 0 0 f 0 f 0 Ng Tze Beg 30

31 Chpter 5 Itegrtio 0 f t dt As efore we get f 0 0 f t dt 0 0 H f 0 This shows tht the left limit lim Therefore, if 0, is i (, ), the H f 0 0 lim H d so By the ove rgumet 0 lim 0 H f 0 F 0 lim 0 H f 0 F lim H f d F lim H f This shows tht for ll i [, ], F'() = f () This ompletes the proof Corollry 46 Suppose f : [, ] R is otiuous Defie F: [, ] R y F f for i [, ] The F is differetile d F'() = f () for ll i [, ], ie, F is ti-derivtive of f Defiitio 47 Suppose f : [, ] R is itegrle Suppose d d re two poits d i [, ] suh tht < d We defie With this defiitio, we hve the d f f y for y, y i [, ], f y f I prtiulr, for y, y d z i [, ], z y z f f y f Corollry 48 Suppose f : [, ] R is otiuous Let e i (, ] Defie F: [, ] R y F f for i [, ] The F is differetile d F'() = f () for ll i [, ], ie, F is ti-derivtive of f Proof Note tht F f f f Therefore, y Corollry 46, F d d f d d f f for ll i [, ] Hee F' = f Corollry 49 Suppose I is ope itervl d f : I R is otiuous Let 0 e i I d defie F: I R y F f for i I The F'() = f () for eh i I 0 Emple 50 () Let g: R R e defied y g t dt Let f t t The g f t dt F Thus y Corollry 49, g F f () Let h: R R e defied y h t dt Therefore, h F, where F is defied i emple () By the Chi Rule for differetitio, h F f, where the futio f is defied i Emple () Hee h 4 Corollry 49 provides wy to fid the ti-derivtive of otiuous futio It is the turl questio to s whether if we rel the otiuity oditio to oe of itegrility, the the olusio of Corollry 49 holds The swer is o i geerl A outeremple is is provided y the futio i Emple 5 Emple 5 Let g : [0, ] R, e futio defied y Ng Tze Beg 3

32 Chpter 5 Itegrtio 0 if is irrtiol g q if 0 is rtiol d p q i its lowest terms if 0 or The g is ouded, itegrle d 0 g 0 We shll ow show tht g is itegrle y usig oe of the equivlet oditios for itegrility i Theorem Give y > 0, y the Arhimede property of R, there eists positive iteger m > suh tht /m < Oserve tht there oly e fiite umer of reiprols of itegers tht re greter th or equl to /m: Te y rtiol umer p/q with p/q i its lowest terms d 0 < p/q This mes tht p q d the gretest ommo divisor of p d q is g(p/q) /m if d oly if /q /m if d oly if q m Thus, the umer of rtiol umers i [0, ] tht hve vlues greter th or equl to /m is fiite d this set iludes the poit 0 sie g(0) = Tht is the fiite set S m = { p/q: q =,, m; p =,, q} {0} is preisely the set o whih the vlues of g re greter th or equl to /m We shll show tht g stisfies oditio (3) of Theorem Rell tht the fiite set S m is preisely the set {y [0, ]: g(y) /m} Note tht 0, S m Let the umer of poits of S m e + Order the elemets y 0, y, y,, y of S m s follows: 0 = y 0 < y < y < y < y = Choose pir of poits, eh pir ostitute itervl otiig eh y i i its iterior d of legth < / for i =,,, d suh tht they re ll mutully disjoit Tht is, we hoose < < < suh tht 0 = y 0 y 3 y y y = We hoose further two more poits, 0 d d me y 0 s, y s suh tht 0 = = y 0 0 d = y = We further require tht j j () j 0 Oviously, P: 0 = = forms prtitio for [0, ] Now y the desity of the irrtiol umers i y itervl, for i = 0,,, m i( g, P ) = if{g(): [ i -, i ]}= () Now sie for eh j = 0,,,,, y j [ j j ] d so y the defiitio of S m, M j (g, P) = sup{g(): [ j -, j ]}= g (y j ) (3) for j = 0,,,, Now euse for j =,,,, [ j -, j ] S m =, M j - (g, P ) = sup{ g (): [ j -, j ]} /m (4) for j =,,, U(g, P) L(g, P) = M i g, P i m i g, P i, i 0 i 0 = M i g, P i y () i 0 Ng Tze Beg 3

33 Chpter 5 Itegrtio = j 0 M j g, P j j 0 M j g, P j g y j j m j y (3) d (4) j 0 j 0 j j m sie g (y j ) j 0 j 0 < / + y () m j 0 j < / + /m sie j 0 j < / + / = Therefore, give y we fid prtitio P suh tht, U(g, P) L(g, P) < By Theorem (), g is itegrle Furthermore y Theorem 39, for y sequee of prtitios ( P ) of [0, ] with P 0 (te ( P ) to e the sequee of regulr prtitios), L g, P But for eh positive iteger, y the desity of the 0 g irrtiol umers, L(g, P ) = 0 Therefore, 0 g 0 Remr Sie g is o-egtive d, the futio G :[0,] R defied y 0 g 0 G 0 g = 0 for eh i [0,] Hee, G'() = 0 d so G'() g() for ll rtiol i [0,] Thus G is ot ti-derivtive of g This gives outeremple to relig the otiuity oditio i Corollry 50 The futio g i Emple 5 is otiuous t every irrtiol poits d disotiuous t every rtiol poits i [0, ] Therefore, we pply Leesgue Theorem (referee: Theorem 33 Chpter 4) to olude tht g is itegrle euse the rtiol poits ostitute set of mesure zero We show elow tht g is otiuous t irrtiol poits Give y > 0, hoose positive iteger m suh tht /m < Te irrtiol umer i [0, ] Let = mi{ y : y S m } The > 0 sie is irrtiol Oviously the ope itervl (, + ) do ot meet S m This is, euse if (, + ) S m, the there eists p/q i S m suh tht - p/q < otrditig tht - p/q mi{ y : y S m } = Tht mes for ll y i (, + ) [0, ], y S m d osequetly, g(y) < /m < Therefore, g(y) g() = g(y) 0 = g(y) < We hve thus show tht for ll y i [0, ] suh tht y < we hve g(y) g() < It follows y defiitio tht g is otiuous t Sie is ritrry, g is otiuous t every irrtiol poit i [0, ] Now we shll show tht g is disotiuous t y rtiol poit For y rtiol poit i [0, ], y defiitio g() > 0 Let = g()/ > 0 For y > 0, y the desity of the irrtiol umers i y itervl, there eists irrtiol umer y i (, + ) [0, ] The g( y ) g() = 0 g() = g() > g()/ = We hve thus show tht for y > 0, we fid y i (, + ) [0, ] suh tht g( y ) g() > g()/ = This mes g is ot otiuous t, if is rtiol Hee g Ng Tze Beg 33