# (a) A partition P of [a, b] is a finite subset of [a, b] containing a and b. If Q is another partition and P Q, then Q is a refinement of P.

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1 Chpter 7: The Riemnn Integrl When the derivtive is introdued, it is not hrd to see tht the it of the differene quotient should be equl to the slope of the tngent line, or when the horizontl xis is time nd the vertil is distne, equl to the instntneous veloity. But the integrl is not so esily interpreted: Why is the re under the urve in ny wy relted to the ntiderivtive? It relly ws stroke of genius by Newton nd Leibnitz to see tht onnetion. In ft, integrtion, in the sense of re or volume, ws invented first, mny enturies erlier: Arhimedes ws involved in trying to ompute the volume of wine sk. The derivtive ws muh hrder onept. In ft, Zeno s prdoxes show how hrd the Greeks found the onept of it in generl. Differentition ould relly only be reognized nd investigted fter Desrtes invented oordinte geometry. As result, few lulus texts, seeking historil pproh to the subjet, over integrtion before differentition. Definition. Let f be bounded funtion on ompt intervl [, b]. (We ll see lter why the underlined words re importnt. For now, just note tht we re not ssuming tht f is even ontinuous. ( A prtition P of [, b] is finite subset of [, b] ontining nd b. If Q is nother prtition nd P Q, then Q is refinement of P. (b Write P = {x 0, x, x 2,..., x n } where = x 0 < x < x 2 < < x n = b. Then U(f, P = L(f, P = n sup{f(x : x i x x i }(x i x i i= n inf{f(x : x i x x i }(x i x i i= re the upper nd lower sums with respet to f nd P. ( The upper nd lower integrls of f on [, b] re U(f = inf{u(f, P } nd L(f = sup{l(f, P }, where the inf nd sup re tken over ll prtitions P of [, b]. (d If U(f = L(f, then f is integrble on [, b], nd their ommon vlue is denoted b f or f(x dx. b Exmple. On the intervl [, b] with prtition P s shown, onsider the funtion f. We hve U(f, P = N(x x 0 + M(x 2 x + N(x 3 x 2 + M(x 4 x 3 + M(x 5 x 4 L(f, P = 0(x x 0 + N(x 2 x + 0(x 3 x 2 + N(x 4 x 3 + 0(x 5 x 4 The blue region shows the re representing the upper sum, nd the pink region the lower sum.

2 With prtition Q tht refines P, the upper sum dereses nd the lower sum inreses: Repling the prtitions with ever finer ones, we see tht the upper sums nd lower sums pproh eh other, with the re of the three tringles s the ommon it: U(f = L(f = b f.

3 We see now why we restrited to bounded funtions: So tht the sup nd inf of f on eh subintervl exists (nd is finite. And similrly we restrited to bounded intervls so tht we do not hve infinitely long subintervls or infinitely mny subintervls, mking the upper nd lower sums muh hrder or impossible to interpret. (We restrited to losed intervls so tht the funtion hs vlues t the endpoints. Lter in lulus ourse integrls of unbounded funtions or over unbounded or unlosed intervls might be llowed, but suh things re lwys lled improper integrls nd interpreted s its of proper integrls, its tht my or my not exist: Exmple. Improper integrls: 0 dx = x ε 0 + x dx = M sin x dx =? M ε M x dx = M The orret version of the lst one is M x dx = sin x dx = [ 2x /2] ε 0 + M [2x /2] M = 2( ε = 2 ε ε 0 + = M 2( M = [ os M x]m M = 0 = 0 M sin x dx = N M M N sin x dx = N [ os M x]m N does not exist The it where the left nd right endpoint pproh nd t the sme rte is the lled the prinipl vlue of the improper integrl. Exmple. A funtion tht is not integrble: The Dirihlet funtion χ Q on [0, ]. Every subintervl in every prtition ontins rtionl numbers, so the supremum of the χ Q -vlues on the subintervl is, so the upper sum for every prtition is, so the upper integrl is. But every subintervl in every prtition lso ontins irrtionl numbers, so the infimum of the χ Q -vlues on the subintervl is 0, so the lower sum for every prtition is 0, so the lower integrl is 0. Lemm. For ny bounded funtion f on the ompt intervl [, b]: ( For ny prtition P of [, b], L(f, P U(f, P. (b For ny prtitions P, Q of [, b], if P Q, then L(f, P L(f, Q nd U(f, P U(f, Q. ( For ny prtition P of [, b], L(f, P U(f Therefore, L(f U(f. Proof. ( On ny subintervl [x i, x i ], inf{f(x : x i x x i } sup{f(x : x i x x i }. (b It is enough to ssume tht Q hs only one more element thn P, sy P = {x 0, x,..., x n } nd Q hs the dditionl division point x between x j nd x j. Then p = inf{f(x : x j x

4 x j } is less thn or equl to eh of q = inf{f(x : x j x x } nd q 2 = inf{f(x : x x x j } it is equl to t lest one, but no lrger thn either. So the term p(x j x j in L(f, P tht orresponds to [x j, x j ] is t most the sum q (x x j + q 2 (x j x of the two terms in L(f, Q tht orrespond to [x j, x ] nd [x, x j ] beuse x j x j = (x x j + (x j x. Adding up ll the terms for ll the subintervls in eh prtition, we see tht L(f, P L(f, Q. Similrly, U(f, P U(f, Q. ( Assume BWOC tht there is prtition P for whih L(f, P > U(f. Then L(f, P is not lower bound for the set of ll upper sums, so there is prtition Q for whih L(f, P > U(f, Q. But then P Q is refinement of both P nd Q, so by (b nd (, we hve L(f, P L(f, P Q U(f, P Q U(f, Q, ontrdition. Thus, U(f is n upper bound on the set of ll lower sums, it is t lest s lrge s the lest upper bound, nd we hve the Therefore sentene. Here is nother wy to sy integrble, in whih, insted of finding one prtition P tht mkes L(f, P lmost s lrge s possible nd nother, Q, tht mkes U(f, P lmost s smll s possible, we sy tht it is enough to find single prtition tht mkes the upper nd lower sums lose to eh other: Proposition. A bounded funtion f : [, b] R is integrble iff, ε > 0, P prtition of [, b] s.t. U(f, P L(f, P < ε. Proof. ( Assume BWOC tht U(f L(f. Then then U(f L(f = ε > 0. By hypothesis, P s.t. U(f, P L(f, P < ε. But beuse L(f, P L(f nd U(f, P U(f, we hve U(f L(f U(f, P L(f, P < ε = U(f L(f, /\. ( Let ε > 0 be given, nd pik P, Q s.t. U(f, P b f < ε/2 nd b f L(f, Q < ε/2 Then beuse P Q is ommon refinement of P, Q, we hve so L(f, Q L(f, P Q U(f, P Q U(f, P, U(f, P Q L(f, P Q U(f, P L(f, Q ( b = U(f, P f + ( b f L(f, Q < ε. Now we show tht ontinuous funtion (on ompt intervl is integrble. Beuse not ll ontinuous funtions re differentible, we see tht it is hrder for funtion to be differentible thn for it to be integrble. Theorem. If f : [, b] R is ontinuous, then it is integrble. Proof. Let ε > 0 be given. Beuse f is uniformly ontinuous, there is δ > 0 for whih x i x j < δ implies f(x i f(x j < ε/(b. Pik prtition P for whih ll the subintervls [x i, x i ] hve length less thn δ. (For exmple, we might hoose n in N so lrge tht (b /n < δ nd set P = {x i = +i(b /n : i = 0,,..., n}. Then beuse f ttins its inf nd sup on eh [x i, x i ], sy t t i nd u i respetively, we hve t i u i x i x i < δ, so f(t i f(u i < ε/(b, so U(f, P L(f, P = n (f(t i f(u i (x i x i < i= n i= ε b (x i x i = ε b (x n x 0 = ε,

5 the seond lst equlity beuse the sum (x x 0 + (x 2 x + + (x n x n telesopes, i.e., ll the terms nel exept the seond nd seond lst. But there re mny disontinuous integrble funtions; our first exmple (the three tringles ws disontinuous t two points but still integrble. We hve seen tht the Dirihlet funtion on [0, ] is not integrble. But we tht the Thome funtion is integrble on this intervl: Exmple. Rell tht the Thome funtion is given by { 0 if x is irrtionl t(x = n if x = m n in lowest terms Beuse every subintervl of every prtition of [0, ] ontins irrtionl numbers, the lower sum of t with respet to every prtition is 0, so the lower integrl of t is 0. Thus, to see tht t is integrble, we need to show tht the upper integrl is 0, i.e., tht, given ε > 0, we n find prtition P with respet to whih U(t, P < ε. We n do tht: There re only finitely mny rtionls x in [0, ] for whih f(x > ε/2. Pik prtition P of [0, ] so tht these finitely mny rtionls re the enters (or, for x = 0 nd x =, the ends of subintervls with totl length ε/2. Then in U(f, P, the terms orresponding to those subintervls dd up to t most times the totl length of these subintervls nd hene less thn ε/2. In the other subintervls the vlue of the funtion is t most ε/2, so those subintervls ontribute to U(f, P totl of less thn (ε/2. It follows tht U(f, P < ε. In ft, it is shown in the text tht bounded funtion on ompt intervl is integrble iff its set of disontinuities hs mesure zero. We won t try to explin this term, beuse the right wy to do tht is to tke totlly different pproh to the ourse, in terms of Lebesgue integrtion. So we will only sy tht ll ountble sets, inluding finite sets, nd the Cntor set hve mesure zero, so tht funtions tht re disontinuous only on these sets re integrble. ( Mesure is onept tht extends the ide of length, so set of length l hs mesure l; but it is hrd to ssign length to set like the rtionls, whih hs mesure 0. Remember tht the Dirihlet funtion is disontinuous everywhere, so its set of disontinuities in [0, ] hs mesure one; but the Thome funtion is disontinuous only on the rtionls, whih hve mesure 0. And sure enough, the Dirihlet funtion is not integrble, while the Thome funtion is. [At this point students re redy to do the eleventh problem set.] Proposition. (Properties of the definite integrl Suppose f, g re integrble on [, b] nd k is onstnt. Then: (0 If < < b, then b f = f + b f. ( b (f + g = b f + b g. (2 b kf = k b f. (3 If f g on [, b], then b f b g. Corollry. (of (3: (3 If m f M on [, b], then m(b b f M(b. (3b f is integrble nd b f b f.

6 The ft tht f is integrble is new informtion the proof is n exerise. To do it, you should verify tht, if m i, M i re the inf nd sup of f on [x i, x i ] nd m i, M i re the sme for f, then M i m i M i m i. To get the seond ssertion in (3b, use f f f on [.b]. Remrks on the proposition: (0 Prt of wht is to be proved here is tht f is integrble on [, b] iff it is integrble on both [, ] nd [, b]. But if we define f = 0 nd b f = b f, then we don t need to dd the If prt of this sttement s long s f is integrble on the intervls determined by, b,. ( The proof is n exerise. (2 Here is proof of the k < 0 se: Let ε > 0 be given, nd pik prtition P of [, b] for whih U(f, P b f < ε/(2 k nd b f L(f, P < ε/(2 k. Then beuse, on eh subintervl [x i, x i ] in P we hve sup{kf(x : x i x x i } = k inf{f(x : x i x x i } nd similrly with sup nd inf reversed, it follows tht U(kf, P L(kf, P = kl(f, P ku(f, P = k (U(f, P L(f, P ( b b k U(f, P f + f L(f, P < ε 2 + ε 2 = ε, so kf is integrble, nd U(kf, P k so k b f = b kf.// b ( f = k L(f, P b ( b f = k f L(f, P < ε 2 < ε, (3 Beuse f g, for ny subintervl [x i, x i ] of ny prtition P of [, b] we hve sup{f(x : x i x x i } sup{g(x : x i x x i } so U(f, P U(g, P, so b f = inf P U(f, P inf P U(g, P = b g. [At this point students re redy to do the twelfth problem set.] Theorem. (Fundmentl Theorem of Clulus ( If f is the derivtive of F on [, b] nd f is integrble, then b f = F (b F (. (2 Let g : [, b] R be integrble, nd define G(t = t g(x dx. Then G : [, b] R is ontinuous. If g is ontinuous t in [, b], then G is differentible t nd G ( = g(. Proof. ( For ny subintervl [x i, x i ] in ny prtition P of [, b], by the Men Vlue Theorem x i [x i, x i ] s.t. F (x i F (x i = f(x i (x i x i, so n i= (F (x i F (x i is between L(f, P nd U(f, P. But the sum is telesoping, to F (b F (, independent of the prtition P, so F (b F ( is between L(f nd U(f, whih re both equl to b f.

7 (2 In ft, we n show tht G is Lipshitz, whih is stronger thn ontinuous: Let M be n upper bound on g in [, b]. Then for ll x, x 2 [, b], we hve 2 x G(x G(x 2 = g g = g M x x 2. Now suppose g is ontinuous t. Then G G(x G( ( = = g x x x x, so we wnt to show, given ε > 0, δ > 0 s.t. 0 < x < δ = g/(x g( < ε. Now there is δ > 0 s.t., if x < δ, then g(x g( < ε, nd for tht δ, if 0 < x < δ, then g ( x g( = x (g(t g(dt x g(t g( dt ε dt = ε. x x (In the middle of this, note tht, if x < 0, then ny x 2 of nonnegtive funtion is lso 0. Here re some exmples of funtions g nd the funtions G defined by their integrls, G(x = g(t dt. Exmple. In these two exmples, = 2, so tht, no mtter wht g is, G( 2 = 2 2 g = 0. We re interested espeilly in the points t whih the definitions of g hnge from one formul to nother. First: { if x < 0 g(x =, if x 0 { G(x = g(t dt = 2 dt = x 2 if x < 0 2 G(0 + 0 dt = 2 + x if x 0 Here, G is differentible exept t the disontinuity x = 0 of g. Seond: { if x g(x =, 2x 3 if x > { x 2 if x G(x = g(t dt = 2 G( + (2t 3dt = 3 + (x2 3x ( 2 = x 2 3x if x > Beuse this g is ontinuous t the point where its definition hnges, the orresponding G is differentible there. Here re the orresponding grphs, with the g s in green nd the G s in red:

8 Though our text doesn t seem to inlude them, I hve seen in other texts exmples of funtions defined by integrls where the upper, nd mybe lso the lower, it(s of integrtion re themselves funtions. In order to find their derivtives, we only need to throw in the Chin Rule: Suppose we re given H(x = f(x g(t dt. We n isolte n intervening funtion: G(u = u g(t dt. Then H(x = G(f(x, so (d/dx(h(x = G (f(x f (x = g(f(xf (x. Exmple. d dx 5x x ( exp( t 2 dt = d 5x 2 +3 exp( t 2 dt dx 0 2x 0 exp( t 2 dt = exp( (5x (0x exp( (2x 2 2 = 0x exp( (5x exp( (2x 2. Here, the intervening funtion is G(u = u 0 exp( t2 dt, so tht G (u = exp( u 2. There is nothing speil bout 0; beuse exp( t 2 is ontinuous everywhere, ny onstnt would hve worked. [At this point students re redy to do the thirteenth problem set.]

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