Mth 554 Integrtion Hndout #9 4/12/96 Defn. A collection of n + 1 distinct points of the intervl [, b] P := {x 0 = < x 1 < < x i 1 < x i < < b =: x n } is clled prtition of the intervl. In this cse, we define the norm of the prtition by P := mx 1 i n x i. where x i := x i x i 1 is the length of the i-th subintervl [x i 1, x i ]. Defn. For given prtition P, we define the Riemnn upper sum of function f by U(P, f) := n M i x i where M i denotes the supremum of f over ech of the subintervls [x i 1, x i ]. Similrly, we define the Riemnn lower sum of function f by L(P, f) := n m i x i where m i denotes the infimum of f over ech of the subintervls [x i 1, x i ]. Since m i M i, we note tht L(P, f) U(P, f). for ny prtition P. Defn. Suppose P 1, P 2 re both prtitions of [, b], then P 2 is clled refinement of P 1, denoted by P 1 P 2, if s sets P 1 P 2. Note. If P 1 P 2, it follows tht P 2 P 1 since ech of the subintervls formed by P 2 is contined in subintervl which rises from P 1. Lemm. If P 1 P 2, then nd L(P 1, f) L(P 2, f). U(P 2, f) U(P 1, f).
Proof. Suppose first tht P 1 is prtition of [, b] nd tht P 2 is the prtition obtined from P 1 by dding n dditionl point z. The generl cse follows by induction, dding one point t t time. In prticulr, we let nd P 1 := {x 0 = < x 1 < < x i 1 < x i < < b =: x n } P 2 := {x 0 = < x 1 < < x i 1 < z < x i < < b =: x n } for some fixed i. We focus on the upper Riemnn sum for these two prtitions, noting tht the inequlity for the lower sums follows similrly. Observe tht nd U(P 2, f) := i 1 j=1 U(P 1, f) := n M j x j j=1 M j x j + M(z x i 1 ) + M(x i z) + n j=i+1 M j x j where M := sup [xi 1,z] f(x) nd M := sup [z,xi ] f(x). It then follows tht U(P 2, f) U(P 1, f) since M, M M i. Defn. If P 1 nd P 2 re rbitrry prtitions of [, b], then the common refinement of P 1 nd P 2 is the forml union of the two. Corollry. Suppose P 1 nd P 2 re rbitrry prtitions of [, b], then L(P 1, f) U(P 2, f). Proof. Let P be the common refinement of P 1 nd P 2, then L(P 1, f) L(P, f) U(P, f) U(P 2, f). Defn. The lower Riemnn integrl of f over [, b] is defined to be f(x)dx := sup ll prtitions P of [,b] L(P, f). Similrly, the upper Riemnn integrl of f over [, b] is defined to be f(x)dx := inf ll prtitions P of [,b] U(P, f).
By the definitions of lest upper bound nd gretest lower bound, it is evident tht for ny function f there holds f(x)dx f(x)dx. Defn. A function f is Riemnn integrble over [, b] if the upper nd lower Riemnn integrls coincide. We denote this common vlue by f(x) dx. Exmples: 1. k dx = k(b ). 2. x dx = 1 2 (b2 2 ). [Hint: Use n i = n(n + 1)/2.] Theorem. A necessry nd sufficient condition for f to be Riemnn integrble is given ɛ > 0, there exists prtition P of [, b] such tht ( ) U(P, f) L(P, f) < ɛ. Proof. First we show tht (*) is sufficient condition. This follows immeditely, since for ech ɛ > 0 tht there is prtition P such tht (*) holds, f(x)dx b f(x)dx U(P, f) L(P, f) < ɛ. Since ɛ > 0 ws rbitrry, then the upper nd lower Riemnn integrls of f must coincide. To prove tht (*) is necessry condition for f to be Riemnn integrble, we let ɛ > 0. By the definition of the upper Riemnn integrl s infimum of upper sums, we cn find prtition P 1 of [, b] such tht Similrly, we hve f(x)dx U(P 1, f) < f(x)dx + ɛ/2 f(x)dx ɛ/2 < L(P 2, f) f(x)dx. Let P be common refinement of P 1 nd P 2, then subtrcting the two previous inequlities implies, U(P, f) L(P, f) U(P 1, f) L(P 2, f) < ɛ. Defn. A Riemnn sum for f for prtition P of n intervl [, b] is defined by R(P, f, ξ) := n f(ξ j ) x j j=1
where the ξ j, stisfying x j 1 ξ j x j (1 j n), re rbitrry. Corollry. Suppose tht f is Riemnn integrble on [, b], then there is unique number γ ( = f(x)dx) such tht for every ɛ > 0 there exists prtition P of [, b] such tht if P P 1, P 2, then i.) ii.) iii.) 0 U(P 1, f) γ < ɛ 0 γ L(P 2, f) < ɛ γ R(P 1, f, ξ) < ɛ where R(P 1, f, ξ) is ny Riemnn sum of f for the prtition P 1. In this sense, we cn interpret f(x)dx = lim R(P, f, ξ). P 0 lthough we would ctully need to show little more to be entirely correct. Proof. Since L(P 2, f) γ U(P 1, f) for ll prtitions, we see tht prts i.) nd ii.) follow from the definition of the Riemnn integrl. To see prt iii.), we observe tht m j f(ξ j ) M j nd hence tht But we lso know tht both L(P 1, f) R(P 1, f, ξ) U(P 1, f). L(P 1, f) γ U(P 1, f) nd condition (*) hold, from which prt iii.) follows.
Theorem. If f is continuous on [, b], then f is Riemnn-integrble on [, b]. Proof. We use the condition (*) to prove tht f is Riemnn-integrble. If ɛ > 0, we set ɛ 0 := ɛ/(b ). Since f is continuous on [, b], f is uniformly continuous. Hence there is δ > 0 such tht f(y) f(x) < ɛ 0 if y x < δ. Suppose tht P < δ, then it follows tht M i m i ɛ 0 (1 i n). Hence U(P, f) L(P, f) = n (M i m i ) x i ɛ 0 (b ) = ɛ. Theorem. If f is monotone on [, b], then f is Riemnn-integrble on [, b]. Proof. If f is constnt, then we re done. We prove the cse for f monotone incresing. The cse for monotone decresing is similir. We gin use the condition (*) to prove tht f is Riemnn-integrble. If ɛ > 0, we set δ := ɛ/(f(b) f()) nd consider ny prtition P with P < δ. Since f is monotone incresing on [, b], then M i = f(x i ) nd m i = f(x i 1 ). Hence U(P, f) L(P, f) = n = n (M i m i ) x i (f(x i ) f(x i 1 )) x i P n (f(x i ) f(x i 1 )) < δ (f(b) f()) = ɛ. Theorem. (Properties of the Riemnn Integrl) Suppose tht f nd g re Riemnn integrble nd k is rel number, then i.) k f(x) dx = k f(x) dx ii.) f + g dx = f dx + g dx iii.) g f implies g dx f dx. iv.) f dx f dx Proof. To prove prt i.), we observe tht in cse k 0, then sup [xi 1,x i ] kf(x) = km i nd inf [xi 1,x i ] kf(x) = km i. Hence U(P, kf) = ku(p, f) nd L(P, kf) = kl(p, f). In the cse tht k < 0, then sup [xi 1,x i ] kf(x) = km i nd inf [xi 1,x i ] kf(x) = km i. It follows in this cse tht U(P, kf) = kl(p, f) nd L(P, kf) = ku(p, f) nd so k f(x)dx = k b f(x)dx
k f(x)dx = k f(x)dx. To prove property ii.) we notice tht sup I (f + g) sup I f + sup I g nd inf I f + inf I g inf I (f + g) for ny intervl I (for exmple, I = [x i 1, x i ]). Hence, (1) L(P, f) + L(P, g) L(P, f + g) U(P, f + g) U(P, f) + U(P, g). Let ɛ > 0, then since f nd g re Riemnn integrble, there exist prtitions P 1, P 2 such tht (2) U(P 1, f) L(P 1, f) < ɛ/2, U(P 2, g) L(P 2, g) < ɛ/2. If we let P be common refinement of P 1 nd P 2, then by combining inequlities (1) nd (2), we see tht see tht U(P, f + g) L(P, f + g) U(P, f) L(P, f) + U(P, g) L(P, g) U(P 1, f) L(P 1, f) + U(P 2, g) L(P 2, g) ɛ/2 + ɛ/2 = ɛ. Property iii.) follows directly from the definition of the upper nd lower integrls using, for exmple, the inequlity sup I g(x) sup I f(x). Property iv.) is proved by pplying property iii.) to the inequlity f f f, from which it follows tht f dx f dx f dx. But this inequlity implies property iv.). Defn. We extend the definition of the integrl to include generl limits of integrtion. These re consistent with our erlier definition. 1. f(x) dx = 0. 2. b f(x) dx = f(x) dx. Theorem. If f is Riemnn integrble on [, b], then it is Riemnn integrble on ech subintervl [c, d] [, b]. Moreover, if c [, b], then (3) f(x) dx = c f(x) dx + b f(x) dx. Proof. We show first tht condition (*) holds for the intervl [c, d]. Suppose ɛ > 0, then by (*) pplied to f over the intervl [, b], we hve tht there exists prtition P of [, b] such tht condition (*) holds. Let P be the refinement obtined from P c
which contins the points c nd d. Let P be the prtition obtined by restricting the prtition P to the intervl [c, d], then U(P, f) L(P, f) U( P, f) L( P, f) U(P, f) L(P, f) < ɛ nd so f is Riemnn integrble over [c, d]. To prove the identity (3), we use the fct tht condition (*) holds when f is Riemnn integrble. Let ɛ > 0, then for ɛ/3 > 0, we my pply (*) to ech of the intervls I = [, b], [, c] nd [c, b], respectively, to obtin prtitions P I which stisfy (4) 0 U I (P I, f) I f dx U I(P I, f) L I (P I, f) < ɛ/3. We let P be the prtition of [, b] formed by the union of the two prtitions P [,c], P [c,b], nd P be the common refinement of P nd P [,b]. Observing tht (5) U [,b] ( P, f) = U [,c] ( P 1, f) + U [c,b] ( P 2, f), we cn combine with inequlity (4) to obtin c f dx + c f dx f dx U [,c] ( P, f) c f dx + U [c,b] ( P, f) c f dx < 3ɛ 0 = ɛ. + U [,b] ( P, f) f dx Since ɛ > 0 ws rbitrry, then equlity (3) must hold. Theorem. (Intermedite Vlue Theorem for Integrls) If f is continuous on [, b], then there exists ξ between nd b such tht f(x) dx = f(ξ)(b ). Proof. Since f is continuous on [, b] nd for η := min [,b] f dx b f(x) η mx f(x), [,b] there holds then by the Intermedite Vlue Theorem for continuous functions, there exists ξ [, b] such tht f(ξ) = η. Theorem. (Fundmentl Theorem of Clculus, I. Derivtive of n Integrl) Suppose tht f is continuous on [, b] nd set F (x) := x f(y)dy, then F is differentible nd F (x) = f(x) for < x < b.
Proof. Notice tht F (x 0 + h) F (x 0 ) h = x0 +h x 0 f dx h = f(ξ) for some ξ between x 0 nd x 0 + h. Hence, s h 0, then ξ = ξ h converges to x 0 nd so the displyed difference quotient hs limit of f(x 0 ) s h 0. Theorem. (Fundmentl Theorem of Clculus, Prt II. Integrl of Derivtive) Suppose tht F is function with continuous derivtive on [, b], then F (y) dy = F (x) x=b x= := F (b) F (). Proof. Define G(x) := x F (y) dy, nd set H := F G. Since the derivtive of H is identiclly zero by Prt I of the Fundmentl Theorem of Clculus, then the Men Vlue Theorem implies tht H(b) = H(). Expressing this in terms of F nd G gives F (b) F (y) dy = F (), which estblishes the theorem.