16 Riemann Sums and Integrals

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1 16 Riema Sums ad Itegrals Defiitio: A partitio P of a closed iterval [a, b], (b >a)isasetof 1 distict poits x i (a, b) togetherwitha = x 0 ad b = x, together with the covetio that i>j x i >x j. Defiitio: ApartitioQ is said to be a refiemet of a partitio P if both are partitios of the same closed iterval [a, b], ad P Q, i.e., P is a proper subset of Q. If Q is a refiemet of P, the every poit i P is also i Q, butq has poits (a, b) thatareotip. We will write Q P to mea Q is a refiemet of P. Defiitio: If f(x) isafuctiodefiedo[a, b] adp is ay partitio of [a, b], a Riema Sum S(f,P) isaysumoftheform S(f,P) = f(x k )(x k x k 1 ), where x k [x k 1,x k ], ad is the umber of subitervals i P. Because the rule for pickig the x k i each iterval of the Riema Sum is ot specified, Riema sums that use differet rules for pickig x k will i geeral have differet values, eve with the same partitio P. Thus, it does t seem that Riema Sums could have much value, give the variability i the partitio P ad the rule for pickig x k. The goal of this sectio is to show that, regardless of the rule for x k, all Riema Sums of a particular fuctio f o a particular iterval [a, b] give the same umber, i the limit that the size of each subiterval is dimiished to zero. For each iteger value of k i 1 k, let ad The, for all x k [x k 1,x k ], M k =max{f(x) x [x k 1,x k ]} m k =mi{f(x) x [x k 1,x k ]}. m k f(x k ) M k. 50

2 Multiply by x k =(x k x k 1 ) > 0, ad sum over k: or where m k x k f(x k ) x k M k x k, L(f,P) S(f,P) U(f,P) L(f,P) = m k x k (3) is the lower Riema sum of f for the partitio P,ad U(f,P) = M k x k (4) is the upper Riema sum of f for the partitio P. Thus we have proved Theorem 16.1 If S(f,P) is ay Riema sum for a fuctio f(x) over a partitio P of a closed iterval [a, b], the L(f,P) S(f,P) U(f,P) where L(f,P) ad U(f,P), defied i (3) ad (4) above, are the lower ad upper sums o the same partitio P. This is true for all choices of x k i S(f,P). Thus the Riema Sums of f o a specific partitio P ca oly vary so much as a result of varyig the choice of x k. I particular, they are bouded above by the upper sum, ad bouded below by the lower sum. So we ca separate the variability i Riema sums due to the choice of x k from the variability i Riema sums from the choice of partitio. I particular, it will be useful to ivestigate how L(f,P) ad U(f,P) varyasweicreasetheumberof poits i the partitio P. Let us add oe poit to partitio P to produce the partitio Q. ForQ to be a partitio distict from P, the ew poit x must lad i the iterior of a existig subiterval of P, say I j =[x j 1,x j ]. The arrival of x splits I j ito 51

3 two subitervals, Ij =[x j 1,x ]adij =[x,x j ]. These ew subitervals will appear i the Riema sum for Q, while I j will appear i the sum for P. Let m j ad Mj be the miimum ad maximum y-values i Ij, ad m j be the miimum ad maximum y-values for Ij. Now, imagiig ad Mj assemblig I j from Ijad I ad j, we see that M j =max(m j,m j ) m j =mi(m j,m j ). This meas that M j M j or M j,adm j m j or m j. Cosider the cotributio of I j to the Riema sum L(f,P). This is m j (x j x j 1 ) = m j (x j x )+m j (x x j 1 ) m j (x j x )+m j(x x j 1 ), which are the cotributios to the Riema sum L(f,Q) by I j ad I j. Addig the cotributios from all the other subitervals of L(f,P), which are the same as the cotributios of all other subitervals of L(f,Q), we have L(f,P) L(f,Q). Similarly, cosider the cotributio of I j to U(f,P). This is M j (x j x j 1 ) = M j (x j x )+M j (x x j 1 ) M j (x j x )+M j(x x j 1 ), which are the cotributios to the Riema sum U(f,Q) byi j ad I j. Thus, U(f,P) U(f,Q). Now ay partitio Q that is a refiemet of P ca be obtaied by addig a umber of poits oe at a time to P. Thus, these results are geeral ad ca be summarized as Theorem 16.2 If f is a fuctio, P ad Q are partitios of the closed iterval [a, b], ad Q is a refiemet of P (i.e., Q P ), the L(f,P) L(f,Q) U(f,Q) U(f,P). 52

4 The middle iequality is obtaied by usig Theorem Thus, as we add more poits to our partitios, the values of the upper ad lower sums move closer to oe aother (or at least ot farther apart). This is i accord with our ituitio. Theorem 16.3 If P ad Q are ay partitios of the closed iterval [a, b], ad f is a fuctio o that iterval, the L(f,P) U(f,Q). Proof: Note that P Q is a refiemet of both P ad Q, because The, by Theorem 16.2, Also, by Theorem Fially, by Theorem P (P Q) adq (P Q). L(f,P) L(f,P Q). L(f,P Q) U(f,P Q) U(f,P Q) U(f,Q) Theorem 16.4 Let f be a fuctio defied o the iterval [a, b]. Defie the set L as ad the set U as L = {x R apartitiop of [a, b] s. t. x = L(f,P)} U = {x R apartitiop of [a, b] s. t. x = U(f,P)}. The l =(l.u.b. L) ad u =(g.l.b. U) both exist, ad, L L ad U U, L l U ad L u U. Proof: Let P ad Q be ay partitios of [a, b]. The set L has the followig properties. 53

5 1. The set L has at least oe upper boud, because from Theorem 16.3 we kow that L(f,P) U(f,Q). But the L(f,P) aremembersofthesetl. Thus, each U(f,Q) isa upper boud of the set L. 2. L is ot empty, because it cotais L(f,[a, b]) = m (b a), where m =mi{f(x) x [a, b]}. The by the Least Upper Boud Priciple, l =l.u.b.l exists, ad has the followig property: L(f,P) l U(f,Q). The first iequality holds because L(f,P) L, adl is a upper boud of L. The secod holds because l is the least upper boud, ad each U(f,Q) is a upper boud, of L. The same properties hold, i reverse, for the set U, which has the followig properties. 1. The set U has at least oe lower boud, because from Theorem 16.3 we kow that L(f,P) U(f,Q). But the U(f,Q) are members of the set U. Thus, each L(f,P) isa lower boud of the set U. 2. U is ot empty, because it cotais U(f,[a, b]) = M (b a), where M =max{f(x) x [a, b]}. The by the Greatest Lower Boud Priciple, u =g.l.b.u exists, ad has the followig property: L(f,P) u U(f,Q). The first iequality holds because u is the greatest lower boud, ad each L(f,P)isalowerboud,ofU. The secod iequality holds because U(f,P) U, adu is a lower boud of U. Theorem 16.5 For l ad u defied i Theorem 16.4, l u. 54

6 Proof: 1. Let Q be ay partitio of [a, b]. The, by Theorem 16.4, l U(f,Q), because l is the least upper boud of L ad ay U(f,Q) U is a upper boud of L ad thus is greater tha the least upper boud. 2. But the, sice l ay member of U, l is a lower boud of U. 3. But the l must be less tha or equal to the greatest lower boud, u. Now, if l = u, the o matter how may poits we ad to the partitios of [a, b], we see that L(f,P) l<u U(f,P), so we ca t prove that the Riema sums S(f,P) will coverge o oe particular value. However, if l = u, the it is possible that the L(f,P) adthe U(F, P) cacoverge,adthereforealltheriemasumswillcovergeto the same value. This motivates the followig defiitio: Defiitio: Let f be a fuctio defied o a closed iterval [a, b], ad let l ad u be as defied i Theorem The, if l = u, wesaythatthe Riema Itegral R(f,[a, b]) exists, ad has the value l = u. The usual otatio, of course, for R(f,[a, b]) is b a f(x)dx. The questio ow arises, what has to be true about the upper ad lower sums for this itegral to exist? What must happe to the partitios to get the Riema sums to coverge? Is there some coditio o the fuctio f that is ecessary for covergece? Theorem 16.6 A ecessary coditio for the existece of a Riema Itegral R(f,[a, b]), that is, for l = u, is > 0, apartitiop of [a, b] such that U(f,P ) L(f,P ) <. 55

7 This theorem starts to aswer this questio by redirectig our search to particular properties of the partitio. This ecessary coditio is phrased more i the laguage of a limitig process, by itroducig a. But istead of provig the existece of a δ() for each, istead we must prove the existece of a partitio P of [a, b], oe for each. This is challegig, but the proof makes good use of the properties of the least upper boud ad greatest lower boud. Proof: (Of Theorem 16.6) Let s (for sum) = l = u. The s =l.u.b. L =g.l.b. U. First, use the properties of the least upper boud. 1. Because s is the least upper boud of L, ay umber less tha s is ot a upper boud of L. 2. Let > 0. The, sice s <s,the umber s 2 2 boud of L. is ot a upper 3. There must, therefore, L L >s But by the defiitio of the set L, this meas that apartitioq of [a, b] such that L = L(f,Q ), 5. By (3), L(f,Q ) >s 2. Now we use the properties of the greatest lower boud. 1. Because s is the greatest lower boud of U, ay umber greater tha s is ot alowerboudofu. 2. The, sice s + 2 >s,the umber s + 2 is ot a lower boud of U. 3. There must, therefore, U U <s But by the defiitio of the set U, this meas that apartitioq + of [a, b] such that U = U(f,Q + ), 5. By (3), U(f,Q + ) <s+ 2. Now we have two partitios Q together to get oe: ad Q +, ad we eed to combie them 56

8 1. Let P = Q Q +. Sice P is a refiemet of both Q ad Q +, Theorem 16.2 applies, ad U(f,P ) U(f,Q + ) <s For the same reaso, L(f,P ) L(f,Q ) >s But U(f,P ) <s+ 2 ad L(f,P ) < s + 2 mea that which was to be proved. U(f,P ) L(f,P ) <, Now we have a ecessary coditio for the existece of the itegral, but is it sufficiet? The ext theorem shows that it is. Theorem 16.7 If, for ay > 0, there exists a partitio P of [a, b] such that U(f,P ) L(f,P ) <, the the Riema itegral of f over [a, b] exists. Proof: 1. Let be ay positive umber. The by hypothesis there exists a partitio P of [a, b] such that 2. Now, U(f,P ) L(f,P ) <. u = g.l.b. {U R apartitiop of [a, b] such that U = U(f,P)} is a lower boud of the set U, ad so u U(f,P ) U. 3. Also, l = l.u.b. {L R apartitiop of [a, b] such that L = L(f,P)} is a upper boud of the set L, ad so l L(f,P ) L. 4. Sice u U(f,P )ad l L(f,P ), we have from (1) for all > 0. u l U(f,P ) L(f,P ) < 5. From Theorem 16.5, we kow that u l 0. 57

9 6. If u l were greater tha 0, the we could create a cotradictio with (4) by pickig 0 < < 1 (u l) Thus u = l, ad the Riema itegral of f over [a, b] exists. Now we are able to specify a coditio o the fuctio f that will guaratee that we ca costruct a partitio P obeyig the hypothesis of Theorem 16.7 for ay > 0. First, though, we eed the followig defiitio. Defiitio: For ay partitio P of a closed iterval [a, b], its orm is defied as the quatity P = max x k x k 1 > 0. That is, P is the legth of the logest subiterval i it. Now we are ready for the crow jewel: Theorem 16.8 If f is cotiuous o a closed iterval [a, b], the the Riema itegral of f o [a, b] exists. Proof: 1. Sice f is cotiuous o [a, b], for all > 0thereexistsaδ() > 0such that for all x,x i [a, b], x x < δ()impliesthat f(x ) f(x ) <. 2. Now, costruct ay partitio P with the requiremet that P < δ. b a This ca be doe for ay > 0, by addig eough poits to the partitio. 3. The for all k such that 1 k, we have x k x k 1 < δ. b a 4. I each subiterval [x k 1,x k ]ofp, let x k ad x k be the values of x where f attais its maximum value M k ad its miimum value m k, respectively, o the subiterval [x k 1,x k ]. Note that by the Extreme Value Theorem, Theorem 13.3, x k ad x k must exist for each subiterval of P. 58

10 5. The, sice x k,x k [x k 1,x k ], x k x k < x k x k 1 < δ b a. 6. Sice f is cotiuous, this implies that M k m k = f(x k) f(x k) < b a. 7. But the U(f,P ) L(f,P ) = < = = (M k m k ) (x k x k 1 ) b a (x k x k 1 ) b a (x x 0 ) (b a) =. b a 8. Thus, by Theorem 16.7, the Riema itegral of f o [a, b] exists. Now that we kow that the itegral exists whe f is cotiuous, we ca also show that ay Riema sum, ot just L(f,P) oru(f,p), ca be used to calculate it. Theorem 16.9 If f is a cotiuous fuctio o the closed iterval [a, b], the ay Riema sum S(f,P) will coverge to the Riema itegral i the limit P 0. Proof: Let f be cotiuous o the closed iterval [a, b]. 1. By Theorem 16.8, the Riema itegral I = b a b a f(x)dx will exist. f(x)dx 2. Also, because f is cotiuous o [a, b] we kow that for all > 0there exists a δ() > 0suchthatforallx,x i [a, b], x x < δ() implies that f(x ) f(x ) <. 59

11 3. By Theorem 16.8, for all > 0, there exists a partitio P such that U(f,P ) L(f,P ) < /2 with P < δ. 2(b a) 4. By Theorems 16.4 ad 16.5, L(f,P ) I U(f,P ). 5. Now let S(f,P) = partitio P of [a, b]. f(x k ) x k be a Riema sum of f over ay 6. By Theorem 16.1, L(f,P ) S(f,P ) U(f,P ). 7. The S(f,P ) I = S(f,P ) L(f,P )+L(f,P ) I S(f,P ) L(f,P ) + L(f,P ) I (by Triagle Iequality) = (S(f,P ) L(f,P )) + (I L(f,P )) (by (6) ad (4)) (U(f,P ) L(f,P )) + (U(f,P ) L(f,P )) (also by (6) ad (4)) < /2+/2 = (by (3)) 8. Sice this is true for all positive, we have lim S(f,P) =I = P 0 b a f(x)dx. 60