Chapter 5. The Riemann Integral. 5.1 The Riemann integral Partitions and lower and upper integrals. Note: 1.5 lectures

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1 Chpter 5 The Riem Itegrl 5.1 The Riem itegrl Note: 1.5 lectures We ow get to the fudmetl cocept of itegrtio. There is ofte cofusio mog studets of clculus betwee itegrl d tiderivtive. The itegrl is (iformlly) the re uder the curve, othig else. Tht we c compute tiderivtive usig the itegrl is otrivil result we hve to prove. I this chpter we defie the Riem itegrl usig the Drboux itegrl, which is techiclly simpler th (but equivlet to) the trditiol defiitio s doe by Riem Prtitios d lower d upper itegrls We wt to itegrte bouded fuctio defied o itervl [,b]. We first defie two uxiliry itegrls tht c be defied for ll bouded fuctios. Oly the c we tlk bout the Riem itegrl d the Riem itegrble fuctios. Defiitio A prtitio P of the itervl [,b] is fiite set of umbers {x,x 1,x 2,...,x } such tht = x < x 1 < x 2 < < x 1 < x = b. We write x i := x i x i 1. Nmed fter the Germ mthemtici Georg Friedrich Berhrd Riem ( ). Nmed fter the Frech mthemtici Je-Gsto Drboux ( ). 147

2 148 CHAPTER 5. THE RIEMANN INTEGRAL Let f : [,b] R be bouded fuctio. Let P be prtitio of [,b]. Defie m i := if{ f (x) : x i 1 x x i }, M i := sup{ f (x) : x i 1 x x i }, L(P, f ) := U(P, f ) := m i x i, M i x i. We cll L(P, f ) the lower Drboux sum d U(P, f ) the upper Drboux sum. The geometric ide of Drboux sums is idicted i Figure 5.1. The lower sum is the re of the shded rectgles, d the upper sum is the re of the etire rectgles. The width of the ith rectgle is x i, the height of the shded rectgle is m i d the height of the etire rectgle is M i. Figure 5.1: Smple Drboux sums. Propositio Let f : [,b] R be bouded fuctio. Let m,m R be such tht for ll x we hve m f (x) M. For y prtitio P of [,b] we hve m(b ) L(P, f ) U(P, f ) M(b ). (5.1) Proof. Let P be prtitio. The ote tht m m i for ll i d M i M for ll i. Also m i M i for ll i. Filly x i = (b ). Therefore, m(b ) = m ( x i ) = m x i m i x i M i x i M x i = M ( x i ) = M(b ).

3 5.1. THE RIEMANN INTEGRAL 149 Hece we get (5.1). I other words, the set of lower d upper sums re bouded sets. Defiitio As the sets of lower d upper Drboux sums re bouded, we defie f (x) dx := sup{l(p, f ) : P prtitio of [,b]}, f (x) dx := if{u(p, f ) : P prtitio of [,b]}. We cll the lower Drboux itegrl d the upper Drboux itegrl. To void worryig bout the vrible of itegrtio, we ofte simply write f := f (x) dx d f := If itegrtio is to mke sese, the the lower d upper Drboux itegrls should be the sme umber, s we wt sigle umber to cll the itegrl. However, these two itegrls my i fct differ for some fuctios. Exmple 5.1.4: Tke the Dirichlet fuctio f : [,1] R, where f (x) := 1 if x Q d f (x) := if x / Q. The 1 f = d 1 f = 1. The reso is tht for every i we hve tht m i = if{ f (x) : x [x i 1,x i ]} = d M i = sup{ f (x) : x [x i 1,x i ]} = 1. Thus L(P, f ) = U(P, f ) = x i =, 1 x i = x i = 1. Remrk The sme defiitio of f d f is used whe f is defied o lrger set S such tht [,b] S. I tht cse, we use the restrictio of f to [,b] d we must esure tht the restrictio is bouded o [,b]. To compute the itegrl we ofte tke prtitio P d mke it fier. Tht is, we cut itervls i the prtitio ito yet smller pieces. Defiitio Let P = {x,x 1,...,x } d P = { x, x 1,..., x m } be prtitios of [,b]. We sy P is refiemet of P if s sets P P.

4 15 CHAPTER 5. THE RIEMANN INTEGRAL Tht is, P is refiemet of prtitio if it cotis ll the umbers i P d perhps some other umbers i betwee. For exmple, {,.5,1,2} is prtitio of [,2] d {,.2,.5,1,1.5,1.75,2} is refiemet. The mi reso for itroducig refiemets is the followig propositio. Propositio Let f : [,b] R be bouded fuctio, d let P be prtitio of [,b]. Let P be refiemet of P. The L(P, f ) L( P, f ) d U( P, f ) U(P, f ). Proof. The tricky prt of this proof is to get the ottio correct. Let P := { x, x 1,..., x m } be refiemet of P := {x,x 1,...,x }. The x = x d x = x m. I fct, we c fid itegers k < k 1 < < k such tht x j = x for j =,1,2,...,. Let x j = x j 1 x j. We get tht x j = x p. p= 1 +1 Let m j be s before d correspod to the prtitio P. Let m j := if{ f (x) : x j 1 x x j }. Now, m j m p for 1 < p. Therefore, m j x j = m j x p = p= 1 +1 m j x p p= 1 +1 m p x p. p= 1 +1 So L(P, f ) = m j x j j=1 j=1 p= 1 +1 m p x p = m m j x j = L( P, f ). j=1 The proof of U( P, f ) U(P, f ) is left s exercise. Armed with refiemets we prove the followig. The key poit of this ext propositio is tht the lower Drboux itegrl is less th or equl to the upper Drboux itegrl. Propositio Let f : [,b] R be bouded fuctio. Let m,m R be such tht for ll x we hve m f (x) M. The m(b ) f Proof. By Propositio we hve for y prtitio P m(b ) L(P, f ) U(P, f ) M(b ). f M(b ). (5.2)

5 5.1. THE RIEMANN INTEGRAL 151 The iequlity m(b ) L(P, f ) implies m(b ) f. Also U(P, f ) M(b ) implies f M(b ). The key poit of this propositio is the middle iequlity i (5.2). Let P 1,P 2 be prtitios of [,b]. Defie P := P 1 P 2. The set P is prtitio of [,b]. Furthermore, P is refiemet of P 1 d it is lso refiemet of P 2. By Propositio we hve L(P 1, f ) L( P, f ) d U( P, f ) U(P 2, f ). Puttig it ll together we hve L(P 1, f ) L( P, f ) U( P, f ) U(P 2, f ). I other words, for two rbitrry prtitios P 1 d P 2 we hve L(P 1, f ) U(P 2, f ). Now we recll Propositio Tkig the supremum d ifimum over ll prtitios we get I other words f f Riem itegrl sup{l(p, f ) : P prtitio} if{u(p, f ) : P prtitio}. We c filly defie the Riem itegrl. However, the Riem itegrl is oly defied o certi clss of fuctios, clled the Riem itegrble fuctios. Defiitio Let f : [,b] R be bouded fuctio such tht f (x) dx = The f is sid to be Riem itegrble. The set of Riem itegrble fuctios o [,b] is deoted by R[,b]. Whe f R[,b] we defie As before, we ofte simply write f (x) dx := f := f (x) dx = The umber f is clled the Riem itegrl of f, or sometimes simply the itegrl of f. By defiitio, y Riem itegrble fuctio is bouded. By ppelig to Propositio we immeditely obti the followig propositio. Propositio Let f : [,b] R be Riem itegrble fuctio. Let m,m R be such tht m f (x) M. The m(b ) f M(b ).

6 152 CHAPTER 5. THE RIEMANN INTEGRAL Ofte we use weker form of this propositio. Tht is, if f (x) M for ll x [,b], the f M(b ). Exmple : We itegrte costt fuctios usig Propositio If f (x) := c for some costt c, the we tke m = M = c. I iequlity (5.2) ll the iequlities must be equlities. Thus f is itegrble o [,b] d f = c(b ). Exmple : Let f : [, 2] R be defied by 1 if x < 1, f (x) := 1/2 if x = 1, if x > 1. We clim tht f is Riem itegrble d tht 2 f = 1. Proof: Let < ε < 1 be rbitrry. Let P := {,1 ε,1+ε,2} be prtitio. We use the ottio from the defiitio of the Drboux sums. The m 1 = if{ f (x) : x [,1 ε]} = 1, M 1 = sup{ f (x) : x [,1 ε]} = 1, m 2 = if{ f (x) : x [1 ε,1 + ε]} =, M 2 = sup{ f (x) : x [1 ε,1 + ε]} = 1, m 3 = if{ f (x) : x [1 + ε,2]} =, M 3 = sup{ f (x) : x [1 + ε,2]} =. Furthermore, x 1 = 1 ε, x 2 = 2ε d x 3 = 1 ε. We compute L(P, f ) = U(P, f ) = 3 3 m i x i = 1 (1 ε) + 2ε + (1 ε) = 1 ε, M i x i = 1 (1 ε) + 1 2ε + (1 ε) = 1 + ε. Thus, 2 2 f f U(P, f ) L(P, f ) = (1 + ε) (1 ε) = 2ε. By Propositio we hve 2 f 2 f. As ε ws rbitrry we see tht 2 f = 2 f. So f is Riem itegrble. Filly, 1 ε = L(P, f ) 2 f U(P, f ) = 1 + ε. Hece, 2 f 1 ε. As ε ws rbitrry, we hve tht 2 f = 1.

7 5.1. THE RIEMANN INTEGRAL 153 It my be worthwhile to sum up prt of the techique of the exmple i propositio. Propositio Let f : [,b] R be bouded fuctio. The f is Riem itegrble if for every ε >, there exists prtitio P such tht Proof. If for every ε >, P exists we hve: Therefore, f = f, d f is itegrble. U(P, f ) L(P, f ) < ε. f f U(P, f ) L(P, f ) < ε. 1 Exmple : Let us show tht 1+x is itegrble o [,b] for y b >. We will see lter tht ll cotiuous fuctios re itegrble, but let us demostrte how we c do it directly. Let ε > be give. Tke N d pick x j := ib/, to form the prtitio P := {x,x 1,...,x } of [,b]. We hve x j = b/ for ll j. For for y subitervl [x j 1,x j ] we obti { } 1 m j = if 1 + x : x [x j 1,x j ] = 1 { } 1, M j = sup 1 + x j 1 + x : x [x j 1,x j ] = 1. x j 1 The we hve U(P, f ) L(P, f ) = x j j=1(m j m j ) = = b ( 1 j= ) ( j 1)b/ 1 + ib/ = b ( ) = b2 b/ 1 + b/ (b + 1). The sum telescopes, the terms successively ccel ech other, somethig we hve see before. b Pickig to be such tht 2 (b+1) < ε the propositio is stisfied d the fuctio is itegrble More ottio Whe f : S R is defied o lrger set S d [,b] S, we sy tht f is Riem itegrble o [,b] if the restrictio of f to [,b] is Riem itegrble. I this cse, we sy f R[,b], d we write f to me the Riem itegrl of the restrictio of f to [,b]. It is useful to defie the itegrl f eve if b. Suppose tht b < d tht f R[b,], the defie f := f. b

8 154 CHAPTER 5. THE RIEMANN INTEGRAL For y fuctio f we defie f :=. At times, the vrible x my lredy hve some other meig. Whe we eed to write dow the vrible of itegrtio, we my simply use differet letter. For exmple, Exercises f (s) ds := Exercise 5.1.1: Let f : [,1] R be defied by f (x) := x 3 d let P := {,.1,.4,1}. Compute L(P, f ) d U(P, f ). Exercise 5.1.2: Let f : [,1] R be defied by f (x) := x. Show tht f R[,1] d compute 1 f usig the defiitio of the itegrl (but feel free to use the propositios of this sectio). Exercise 5.1.3: Let f : [,b] R be bouded fuctio. Suppose tht there exists sequece of prtitios {P k } of [,b] such tht ( lim U(Pk, f ) L(P k, f ) ) =. k Show tht f is Riem itegrble d tht Exercise 5.1.4: Fiish the proof of Propositio f = lim k U(P k, f ) = lim k L(P k, f ). Exercise 5.1.5: Suppose tht f : [ 1,1] R is defied s Prove tht f R[ 1,1] d compute 1 propositios of this sectio). 1 f (x) := { 1 if x >, if x. Exercise 5.1.6: Let c (,b) d let d R. Defie f : [,b] R s f usig the defiitio of the itegrl (but feel free to use the f (x) := { d if x = c, if x c. Prove tht f R[,b] d compute f usig the defiitio of the itegrl (but feel free to use the propositios of this sectio).