# 4: RIEMANN SUMS, RIEMANN INTEGRALS, FUNDAMENTAL THEOREM OF CALCULUS

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1 4: RIEMA SUMS, RIEMA ITEGRALS, FUDAMETAL THEOREM OF CALCULUS STEVE HEILMA Contents 1. Review 1 2. Riemnn Sums 2 3. Riemnn Integrl 3 4. Fundmentl Theorem of Clculus 7 5. Appendix: ottion Review Theorem 1.1 (Lest Upper Bound Property). Let E be nonempty subset of R. If E hs some upper bound, then E hs exctly one lest upper bound. Theorem 1.2. Let < b be rel numbers, nd let f : [, b] R be function which is continuous on [, b]. Then f is lso uniformly continuous on [, b]. Proposition 1.3. Let X be subset of R, let x 0 be limit point of X, let f : X R be function, nd let L be rel number. Then the following two sttements re equivlent. f is differentible t x 0 on X with derivtive L. For every ε > 0, there exists δ = δ(ε) > 0 such tht, if x X stisfies x x 0 < δ, then f(x) [f(x 0 ) + L(x x 0 )] ε x x 0. Corollry 1.4 (Men Vlue Theorem). Let < b be rel numbers, nd let f : [, b] R be continuous function which is differentible on (, b). Then there exists x (, b) such tht f f(b) f() (x) =. b Proposition 1.5. Let X be subset of R, let x 0 be limit point of X, nd let f : X R be function. If f is differentible t x 0, then f is lso continuous t x 0. Theorem 1.6 (Properties of Derivtives). Let X be subset of R, let x 0 be limit point of X, nd let f : X R nd g : X R be functions. (iv) If f, g re differentible t x 0, then fg is differentible t x 0, nd (fg) (x 0 ) = f (x 0 )g(x 0 ) + g (x 0 )f(x 0 ). (Product Rule) Dte: December 3,

2 2. Riemnn Sums Within clculus, the two most fundmentl concepts re differentition nd integrtion. We hve covered differentition lredy, nd we now move on to integrtion. Defining n integrl is firly delicte. In the cse of the derivtive, we creted one limit, nd the existence of this limit dictted whether or not the function in question ws differentible. In the cse of the Riemnn integrl, there is lso limit to discuss, but it is much more complicted thn in the cse of differentition. We should mention tht there is more thn one wy to construct n integrl, nd the Riemnn integrl is only one such exmple. Within this course, we will only be discussing the Riemnn integrl. The Riemnn integrl hs some deficiencies which re improved upon by other integrtion theories. However, those other integrtion theories re more involved, so we focus for now only on the Riemnn integrl. Our strting point will be prtitions of intervls into smller intervls, which will form the bckbone of the Riemnn sum. The Riemnn sum will then be used to crete the Riemnn integrl through limiting constructing. Definition 2.1 (Prtition). Let < b be rel numbers. A prtition P of the intervl [, b] is finite subset of rel numbers x 0,..., x n such tht We write P = {x 0, x 1,..., x n }. = x 0 < x 1 < < x n 1 < x n = b. Remrk 2.2. Let P, P be prtitions of [, b]. Then the union P P of P nd P is lso prtition of [, b]. Definition 2.3 (Upper nd Lower Riemnn Sums). Let < b be rel numbers, let f : [, b] R be bounded function, nd let P = {x 0,..., x n } be prtition of [, b]. For every integer 1 i n, the function f [xi 1,x i ] is lso bounded function. So, sup x [xi 1,x i ] f(x) nd inf x [xi 1,x i ] f(x) exist by the Lest Upper Bound property (Theorem 1.1). We therefore define the upper Riemnn sum U(f, P ) by U(f, P ) := n ( sup ) f(x) (x i x i 1 ). We lso define the lower Riemnn sum L(f, P ) by n ( ) L(f, P ) := inf f(x) (x i x i 1 ). Remrk 2.4. For ech integer 1 i n, we define function g : [, b] R such tht g(x) := sup y [xi 1,x i ] f(y) for ll x i 1 x < x i, with g(b) := f(b). Then g is constnt on [x i 1, x i ) for ll 1 i n, nd f(x) g(x) for ll x [, b]. The upper Riemnn sum U(f, P ) then represents the re under the function g, which is ment to upper bound the re under the function Similrly, for ech integer 1 i n, we define function h: [, b] R such tht h(x) := inf y [xi 1,x i ] f(y) for ll x i 1 x < x i, with h(b) := f(b). Then h is constnt on [x i 1, x i ) for ll 1 i n, nd h(x) g(x) for ll x [, b]. The lower Riemnn sum L(f, P ) then represents the re under the function g, which is ment to lower bound the re under the function 2

3 Definition 2.5 (Upper nd Lower Integrls). Let < b be rel numbers, let f : [, b] R be bounded function. We define the upper Riemnn integrl f of f on [, b] by f := inf{u(f, P ): P is prtition of [, b]}. We lso define the lower Riemnn integrl f of f on [, b] by f := sup{l(f, P ): P is prtition of [, b]}. Lemm 2.6. Let f : [, b] R be bounded function, so tht there exists rel number M such tht M f(x) M for ll x [, b]. Then In prticulr, nd M(b ) f exist s rel numbers. f M(b ) Proo If we choose P to be the prtition P = {, b}, then U(f, P ) = (b ) sup x [,b] f(x) nd L(f, P ) = (b ) inf x [,b] f(x). So, U(f, P ) (b )M nd L(f, P ) (b )( M). So, M(b ) f nd f M(b ) by the definition of supremum nd infimum, respectively. We now show tht f Let P be ny prtition of [, b]. By the definition of L(f, P ) nd U(f, P ), we hve < L(f, P ) U(f, P ) < +. So, we know tht the set {U(f, P ): P is prtition of [, b]} is nonempty nd bounded from below. Similrly, the set {L(f, P ): P is prtition of [, b]} is nonempty nd bounded from bove. Then, by the lest upper bound property (Theorem 1.1), f nd f exist s rel numbers. So, given ny ε > 0, choose prtition P such tht L(f, P ) f ε. (Such prtition P exists by the definition of the supremum.) We then hve f L(f, P ) + ε U(f, P ) + ε. Tking the infimum over prtitions P of [, b] of both sides of this inequlity, we get f f + ε. Since ε > 0 is rbitrry, we conclude tht f f, s desired. 3. Riemnn Integrl Definition 3.1 (Riemnn Integrl). Let < b be rel numbers, let f : [, b] R be bounded function. If f = f we sy tht f is Riemnn integrble on [, b], nd we define f := Remrk 3.2. Defining the Riemnn integrl of n unbounded function tkes more cre, nd we defer this issue to lter courses. f = 3

4 Theorem 3.3 (Lws of integrtion). Let < b be rel numbers, nd let f, g : [, b] R be Riemnn integrble functions on [, b]. Then (i) The function f + g is Riemnn integrble on [, b], nd (f + g) = ( f) + ( g). (ii) For ny rel number c, cf is Riemnn integrble on [, b], nd (cf) = c( f). (iii) The function f g is Riemnn integrble on [, b], nd (f g) = ( f) ( g). (iv) If f(x) 0 for ll x [, b], then f 0. (v) If f(x) g(x) for ll x [, b], then f g. (vi) If there exists rel number c such tht f(x) = c for x [, b], then f = c(b ). (vii) Let c, d be rel numbers such tht c < b d. Then [c, d] contins [, b]. Define F (x) := f(x) for ll x [, b] nd F (x) := 0 otherwise. Then F is Riemnn integrble on [c, d], nd d F = c (viii) Let c be rel number such tht < c < b. Then f [,c] nd f [c,b] re Riemnn integrble on [, c] nd [c, b] respectively, nd Exercise 3.4. Prove Theorem 3.3. f = c f [,c] + c f [c,b]. Remrk 3.5. Concerning Theorem 3.3(viii), we often write c f insted of c f [,c] Riemnn integrbility of continuous functions. So fr we hve discussed some properties of Riemnn integrble functions, but we hve not shown mny functions tht re ctully Riemnn integrble. In this section, we show tht continuous function on closed intervl is Riemnn integrble. Theorem 3.6. Let < b be rel numbers, nd let f : [, b] R be continuous function on [, b]. Then f is Riemnn integrble. Proo We will produce fmily of prtitions of the intervl [, b] such tht the upper nd lower Riemnn integrls of f re rbitrrily close to ech other. From Theorem 1.2, f is uniformly continuous on [, b]. Let ε > 0. Then, by uniform continuity of f, there exists δ = δ(ε) > 0 such tht, if x, y [, b] stisfy x y < δ, then f(x) f(y) < ε. By the Archimeden property, there exists positive integer such tht (b )/ < δ. Consider the prtition P of the intervl [, b] of the form P = {x 0,..., x } = {, +(b )/, +2(b )/, +3(b )/,..., +( 1)(b )/, b}. ote tht x i x i 1 = (b )/ for ll 1 i. Since f is continuous on [, b], f is lso continuous on [x i 1, x i ] for ech 1 i. In prticulr, f [xi 1,x i ] chieves its mximum nd minimum for ll 1 i. So, for ech 1 i, there exist m i, M i [x i 1, x i ] such tht inf f(x) = f(m i ), nd sup f(x) = f(m i ). Since x i x i 1 = (b )/ < δ, we hve m i M i < δ for ech 1 i n. Since f is uniformly continuous, we conclude tht inf f(x) = f(m i ) > f(m i ) ε = ( sup f(x)) ε, 1 i n. ( ) 4

5 We now estimte U(f, P ) nd L(f, P ). By the definition of U(f, P ) nd L(f, P ), we hve L(f, P ) U(f, P ). However, L(f, P ) is lso close to U(f, P ) by ( ): L(f, P ) = b ( inf f(x)) > b By the definition of f, we conclude tht By the definition of f, we conclude tht ( ) [( sup f(x)) ε] = (b )ε + U(f, P ). f > (b )ε + U(f, P ). Since ε > 0 is rbitrry, we get f > (b )ε + f Combining this inequlity with Lemm 2.6, we conclude tht f = Riemnn integrble. Tht is, f is Exercise 3.7. Let < b be rel numbers. Let f : [, b] R be bounded function. Let c [, b]. Assume tht, for ech δ > 0, we know tht f is Riemnn integrble on the set {x [, b]: x c δ}. Then f is Riemnn integrble on [, b] Piecewise Continuous Functions. We cn now expnd bit more the fmily of functions tht re Riemnn integrble. Proposition 3.8. Let < b be rel numbers. Assume tht f : [, b] R is continuous t every point of [, b], except for finite number of points. Then f is Riemnn integrble. Proo By Theorem 3.3(viii) nd n inductive rgument, it suffices to consider the cse tht f is discontinuous t single point c [, b]. Let δ > 0. Then f is continuous on the set E := {x [, b]: x c δ}. ote tht E consists of either one or two closed intervls. Since f E is continuous, we then conclude tht f E Riemnn integrble by Theorem 3.6. Then Exercise 3.7 sys tht f is Riemnn integrble on [, b], s desired Monotone Functions. It turns out tht monotone functions re Riemnn integrble s well. There exist monotone functions tht re not piecewise continuous, so the current section is not subsumed by the previous one. Proposition 3.9. Let < b be rel numbers, nd let f : [, b] R be monotone function. Then f is Riemnn integrble. 5

6 Proo Let ε > 0. Without loss of generlity, f is monotone incresing. Then f() f(x) f(b) for ll x [, b], so f is bounded. By the Archimeden property, there exists positive integer such tht (b )(f(b) f())/ < ε. Consider the prtition P of the intervl [, b] of the form P = {x 0,..., x } = {, +(b )/, +2(b )/, +3(b )/,..., +( 1)(b )/, b}. ote tht x i x i 1 = (b )/ for ll 1 i. We now estimte U(f, P ) nd L(f, P ). By the definition of U(f, P ) nd L(f, P ), we hve L(f, P ) U(f, P ). However, since f is monotoniclly incresing, L(f, P ) = b b ( inf f(x)) b ( 1 f(x 0 ) + ( sup f(x)) ) ( ) f(x i 1 ) = b b (f() f(b)) + U(f, P ) ε + U(f, P ). By the definition of f, we conclude tht By the definition of f, we conclude tht f ε + U(f, P ). ( f(x 0 ) + 1 f(x i ) = b (f(x 0) sup f(x)) + U(f, P ) x [x 1,x ] ) Since ε > 0 is rbitrry, we get f ε + f Combining this inequlity with Lemm 2.6, we conclude tht f = Riemnn integrble. Tht is, f is 3.4. A on-riemnn Integrble Function. Unfortuntely, not every function is Riemnn integrble. We hve seen tht unbounded functions cuse some difficulty in our definition of the Riemnn integrl, since their Riemnn sums cn be + or. However, there re even bounded functions tht re not Riemnn integrble. Consider the following function f : R [0, 1], which we encountered in our investigtion of limits. { 1, if x Q f(x) := 0, if x / Q. 6

7 For ny prtition P of [0, 1], we utomticlly hve L(f, P ) = 0 nd U(f, P ) = 1. (Justify this sttement.) Therefore, 1 f = 0 nd 1 f = 1, so tht this function f is not Riemnn 0 0 integrble on [0, 1]. 4. Fundmentl Theorem of Clculus The Fundmentl Theorem of Clculus sys, roughly speking, tht differentition nd integrtion negte ech other. This fct is remrkble on its own, but it will lso llow us to ctully compute wide rnge of integrls. (ote tht we hve not yet been ble to compute ny integrls.) Theorem 4.1 (First Fundmentl Theorem of Clculus). Let < b be rel numbers. Let f : [, b] R be continuous function on [, b]. Assume tht f is lso differentible on [, b], nd f is Riemnn integrble on [, b]. Then f = f(b) f(). Proo Let P = {x 0,..., x n } be prtition of [, b]. Then n f(b) f() = f(x n ) f(x 0 ) = (f(x i ) f(x i 1 )). By the Men Vlue Theorem (Corollry 1.4), for ech 1 i n there exists y i [x i 1, x i ] such tht (x i x i 1 )f (y i ) = f(x i ) f(x i 1 ). Substituting these equlities into ( ), we get n f(b) f() = (x i x i 1 )f (y i ). Applying the definitions of L(f, P ) nd U(f, P ), we hve From Definition 2.5, we get L(f, P ) f(b) f() U(f, P ). ( ) f f(b) f() f. ( ) Since f is Riemnn integrble, f = f = f. So, ( ) implies tht f = f(b) f(), s desired. Theorem 4.2 (Second Fundmentl Theorem of Clculus). Let < b be rel numbers. Let f : [, b] R be Riemnn integrble function. Define function F : [, b] R by F (x) := Then F is continuous. Moreover, if x 0 [, b] nd if f is continuous t x 0, then F is differentible t x 0 nd F (x 0 ) = f(x 0 ). x 7

8 Proo Since f : [, b] R is Riemnn integrble, f is bounded by the definition of Riemnn integrbility. So, there exists rel number M such tht M f(x) M for ll x [, b]. Let x, y [, b]. Without loss of generlity, x y. Then, by Theorem 3.3(viii) So, by Theorem 3.3(v), F (y) F (x) = M(y x) x f x f = f = F (y) F (x) = x x ( ) f M(y x). Tht is, F (y) F (x) M y x. Interchnging the roles of x nd y leves this sttement unchnged, so for ny x, y [, b], we hve F (y) F (x) M y x. In prticulr, F is uniformly continuous, so F is continuous. ow, suppose f is continuous t x 0. Using Proposition 1.3, it suffices to show: there exists rel number L such tht, for ny ε > 0, there exists δ > 0 such tht, if y [, b] stisfies y x 0 < δ, then F (y) [F (x 0 ) + L(y x 0 )] ε y x 0. ( ) We set L := f(x 0 ). Let ε > 0. Applying the continuity of f t x 0, there exists δ > 0 such tht if y [, b] stisfies y x 0 < δ, then f(x 0 ) ε f(y) f(x 0 ) + ε. Assume first tht y stisfies y > x 0. Then integrting nd pplying Theorem 3.3(v), (f(x 0 ) ε)(y x 0 ) x 0 f (f(x 0 ) + ε)(y x 0 ). So, using ( ), F (y) F (x 0 ) f(x 0 )(y x 0 ) = ( f) f(x 0 )(y x 0 ) ε y x 0. x 0 Tht is, we proved ( ) holds for y > x 0. The cse y < x 0 is proven similrly, nd the cse y = x 0 follows since then both sides of ( ) re zero Consequences of the Fundmentl Theorem. One of the consequences of the Fundmentl Theorem of Clculus is tht we cn now ctully compute some integrls. For exmple, if α Q, α 1, nd if 0 < < b re rel numbers, then f(x) := (α + 1) 1 x α+1 stisfies f (x) = x α. So, by Theorem 4.1, x α = 1 α + 1 (bα+1 α+1 ). Remrk 4.3. Let β Q, let x > 0 nd let f(x) := x β. Let s justify the formul f (x) = βx β 1. Write β = p/q with p Z nd q with q 0. Then f(x) = (x p ) 1/q. Recll tht the function h(x) = x 1/q is differentible for x > 0 by the Inverse Function Theorem. If p 0, then we hve lredy verified by explicit clcultion tht g(x) := x p is differentible. If p < 0, then g(x) := x p = 1/x p is differentible by the quotient rule. In summry, we 8

9 cn write f(x) = h(g(x)), where h is differentible when g(x) > 0, g is differentible when x > 0, nd g(x) > 0 when x > 0. So, f is differentible when x > 0, by the chin rule. Theorem 4.4. Let < b be rel numbers. Let f, g : [, b] R be Riemnn integrble functions. Then the product f g is Riemnn integrble. Theorem 4.5 (Integrtion by Prts). Let < b be rel numbers. Let f, g : [, b] R be differentible functions such tht f nd g re Riemnn integrble. Then fg = f(b)g(b) f()g() Proo Since f is differentible on [, b] it is continuous on [, b] by Proposition 1.5. So, f is Riemnn integrble by Theorem 3.6, nd then fg is Riemnn integrble by Theorem 4.4. Similrly, f g is Riemnn integrble. Since f, g re differentible, Theorem 1.6(iv) sys tht fg is differentible nd (fg) = f g + fg. Since f g nd fg re Riemnn integrble, f g + fg is Riemnn integrble by Theorem 3.3(i). So, pplying Theorem 4.1, (f g + g f) = f g (fg) = f(b)g(b) f()g(). Theorem 4.6 (Chnge of Vribles, version 1). Let < b be rel numbers. Let φ: [, b] [φ(), φ(b)] be differentible function such tht φ() < φ(b) nd such tht φ is Riemnn integrble. Let f : [φ(), φ(b)] R be continuous on [φ(), φ(b)]. Then (f φ)φ is Riemnn integrble on [, b], nd (f φ)φ = φ(b) Proo Since φ is differentible, φ is continuous. Then f φ is continuous, since it is the composition of two continuous functions. For t [φ(), φ(b)], define F (t) := t Recll φ() tht f is Riemnn integrble by Theorem 3.6. ow, F (t) = f(t) for ll t [φ(), φ(b)] by the second fundmentl theorem of clculus, Theorem 4.2. For ny x [, b], define g(x) := F φ(x). Then, by the Chin Rule, we hve φ() g (x) = F (φ(x))φ (x) = f(φ(x))φ (x). ote tht g is the product of two Riemnn integrble functions, so g is Riemnn integrble (from Theorem 4.4). So, pplying the first fundmentl theorem of clculus, Theorem 4.1, we get (f φ)φ = g = g(b) g() = F (φ(b)) F (φ()) = F (φ(b)) = φ(b) The following theorem is more difficult to prove, but it llows chnge of vribles for ny Riemnn integrble function φ() 9

10 Theorem 4.7 (Chnge of Vribles, version 2). Let < b be rel numbers. Let φ: [, b] [φ(), φ(b)] be differentible, strictly monotone incresing function. Assume tht φ is Riemnn integrble on [, b]. Let f : [φ(), φ(b)] R be Riemnn integrble on [φ(), φ(b)]. Then (f φ)φ is Riemnn integrble on [, b], nd (f φ)φ = φ(b) φ() 5. Appendix: ottion Let A, B be sets in spce X. Let m, n be nonnegtive integers. Z := {..., 3, 2, 1, 0, 1, 2, 3,...}, the integers := {0, 1, 2, 3, 4, 5,...}, the nturl numbers Z + := {1, 2, 3, 4,...}, the positive integers Q := {m/n: m, n Z, n 0}, the rtionls R denotes the set of rel numbers R = R { } {+ } denotes the set of extended rel numbers C := {x + y 1 : x, y R}, the complex numbers denotes the empty set, the set consisting of zero elements mens is n element o For exmple, 2 Z is red s 2 is n element of Z. mens for ll mens there exists F n := {(x 1,..., x n ): x i F, i {1,..., n}} A B mens A, we hve B, so A is contined in B A B := {x A: x / B} A c := X A, the complement of A A B denotes the intersection of A nd B A B denotes the union of A nd B Let E be subset of R { } {+ }. Let ( n ) n=0 be sequence of rel numbers. sup(e) denotes the smllest upper bound of E inf(e) denotes the lrgest lower bound of E lim sup( n ) n=0 := lim lim inf( n ) n=0 sup n m n ( n ) n=m := lim n inf m n ( n) n=m 10

11 5.1. Set Theory. Let X, Y be sets, nd let f : X Y be function. The function f : X Y is sid to be injective (or one-to-one) if nd only if: for every x, x V, if f(x) = f(x ), then x = x. The function f : X Y is sid to be surjective (or onto) if nd only if: for every y Y, there exists x X such tht f(x) = y. The function f : X Y is sid to be bijective (or one-to-one correspondence) if nd only if: for every y Y, there exists exctly one x X such tht f(x) = y. A function f : X Y is bijective if nd only if it is both injective nd surjective. Two sets X, Y re sid to hve the sme crdinlity if nd only if there exists bijection from X onto Y. UCLA Deprtment of Mthemtics, Los Angeles, CA E-mil ddress: 11

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