Part 4. Integration (with Proofs)


 Harriet Oliver
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1 Prt 4. Integrtion (with Proofs) 4.1 Definition Definition A prtition P of [, b] is finite set of points {x 0, x 1,..., x n } with = x 0 < x 1 <... < x n = b. Exmple { 0, 1, 1, 7, 1} nd { 1 0,, 3, 1} re prtitions of [0, 1] { For eh n 1 define P n = + }, (b ) i : 1 i n the rithmeti n prtition of [, b]. { Assume b > > 0. For eh n 1 define Q n = ( ) } b i/n : 1 i n, the geometri prtition of [, b]. A prtition of [, b] divides the intervl into n subintervls [x i 1, x i ], 1 i n. Let f be bounded funtion on [, b], so there exist m nd M suh tht m f (x) M for ll x [, b]. For eh 1 i n let nd M i = lub {f (x) : x [x i 1, x i ]}, m i = glb {f (x) : x [x i 1, x i ]}, whih exist by the Completeness of R. So m m i M i M for ll 1 i n. Definition For f bounded on [, b] the Upper Sum is nd the Lower Sum is U (P, f) = L (P, f) = n M i (x i x i 1 ) i=1 n m i (x i x i 1 ). i=1 1
2 Exmple An upper sum [ ] b A lower sum [ ] b It would pper from these digrms tht the lower nd upper sums re the res of regions whih either ontin or re ontined within the region under the grph between nd b. Be reful, I hve drwn nie, smooth funtion, the sitution might look different with more pthologil funtions. Also be reful beuse I hve drwn only nonnegtive funtions. Look t wht hppens if the funtion should be negtive for some x. If we ould ssign mesure of size to the region under grph we might expet it to be less thn the res mesured by U (P, f) for ll P, yet greter thn those mesured by L (P, f) for ll P. Exmple Let f : [0, 1] R be given by f (x) = x x 2. Find U (P, f) nd L (P, f) when (i) P = {0, 12 }, 1 nd (ii) P = {0, 13, 23 }, 1. Exmple Let f : [1, 2] R be given by f (x) = x. Find U (P n, f), L (P n, f), U (Q n, f) nd L (Q n, f) for the rithmeti nd geometri prtitions. Lemm For ll prtitions P of [, b] we hve 2
3 m (b ) L (P, f) U (P, f) M (b ). Proof Left to student. From this we see tht for f bounded on [, b], {U (P, f) : P prtition of [, b]} is nonempty set of rel numbers bounded below, by m (b ). So by Completeness of the Rels this set hs gretest lower bound. Similrly, {L (P, f) : P prtition of [, b]} hs lest upper bound. 3
4 Definition For f bounded on [, b], the Upper Integrl is nd the Lower Integrl is f = glb {U (P, f) : P prtition of [, b]} (1) f = lub {L (P, f) : P prtition of [, b]}. Definition If P nd D re two prtitions of set whih stisfy P D, we sy tht D is refinement of P. (We lso sy tht D is finer thn P nd, equivlently, P is orser thn D.) Lemm If f is bounded on [, b] nd D is refinement of P then L (P, f) L (D, f) U (D, f) U (P, f). [ ] b Corollry If f is ny bounded funtion on [, b] then f Definition A bounded funtion f on [, b] is Riemnn integrble over [, b] if f = The ommon vlue is lled the (Riemnn) integrl nd is denoted by f or f (x) dx. Note To sve time in letures I will often write tion nd le in ple of integrtion nd integrble. f. f. 4
5 ( ) Riemnn gve his definition of n integrl in Here we hve given n pproh due to Drboux from The upper nd lower sums bove should stritly be lled Upper nd Lower Drboux sums. They differ slightly from the Upper nd Lower Riemnn sums tht you might find in lterntive ounts of integrtion. But be reful! In book by Strihrtz the Drboux sums re lled Riemnn sums while, wht I would ll Riemnn sums, re lled Cuhy sums. Very onfusing! ( ) Exmple Let f : [0, 1] R, x x. Prove, by verifying the definition, tht f is Riemnn integrble over [0, 1]. Wht is the vlue of the integrl? Hint: Use sequene of Arithmeti Prtitions. Exmple Let f : [1, 2] R, x 1. Prove, by verifying the definition, tht x f is Riemnn integrble over [1, 2]. Wht is the vlue of the integrl? Hint: Use sequene of Geometri Prtitions. We now return to n exmple seen erlier in Question 1.2. Exmple (i) Let f : [0, 1] R be given by { 1 if x rtionl f (x) = 0 if x irrtionl. Show tht f is not Riemnn integrble over [0, 1]. ( ) Though erlier in the ourse I termed this funtion pthologil it is not, in the sheme of things, omplited funtion (imgine wht ontinuous nowheredifferentible funtion might look like). So it is wekness of the theory of Riemnn integrtion tht we n t integrte this funtion. In ourse 341 theory of integrtion due to Lebesgue, from 1902, is studied. With Lebesgue integrtion this funtion n be integrted. Cn you guess t the vlue of the integrl? ( ) (ii) Let f : [0, 1] R be given by { 1/q if x Q is written s p/q in redued form, f (x) = 0 if x irrtionl. (This ws seen in Question 2.17(ii), nd is disontinuous t every rtionl.) Show tht f is Riemnn integrble over [0, 1]. Question?? 5
6 4.2 Criteri for being Riemnn Integrble. Lemm If f : [, b] R is Riemnn integrble then, ε > 0, P : Proof f integrble mens f ε < L (P, f) U (P, f) < f = f = f + ε. f. (2) Let ε > 0 be given. Then f = glb {U (P, f)} P mens there exists P 1 suh tht U (P 1, f) < f + ε = f + ε (3) by (2). Similrly, f = lub {L (P, f)} P mens there exists P 2 suh tht L (P 2, f) > f + ε = f + ε (4) by (2). Set P = P 1 P 2. Then (3) nd (4) hold with P in ple of P 1 nd P 2, whih is the sttement of this lemm. The next result is of gret use when proving properties of integrtion. It is not so importnt when showing given funtion is integrble. Theorem Riemnn s Criteri. if, The bounded funtion f : [, b] R is Riemnn integrble if, nd only Proof ( =) Assume (5) holds. From ε > 0, P : U (P, f) L (P, f) < ε. (5) we see tht L (P, f) f f U (P, f) 0 f f < ε. 6
7 True for ll ε > 0 mens we must hve equlity throughout, nd so f is Riemnn integrble. (= ) Assume tht f is Riemnn integrble. Let ε > 0 be given. From Lemm (with ε/2 in ple ε) we n find P suh tht f ε 2 < L (P, f) U (P, f) < f + ε 2, i.e. U (P, f) L (P, f) < ε s required. Theorem If f : [, b] R is monotoni then f is Riemnn integrble. Theorem If f : [, b] R is ontinuous then f is Riemnn integrble. 4.3 Integrtion Rules For A, B R define A + B = { + b : A, b B}, the sumset of A nd B. Further, given A R write A = { : A} then the sumset A A = A + ( A) = { :, A}. (Not the empty set s might hve been thought!) Lemm If A, B R then lub (A + B) = luba + lubb nd glb (A + B) = glba + glbb Proof For ll A we hve luba, nd for ll b B we hve b lubb. Adding we get + b luba + lubb for ll + b A + B. (So luba + lubb is n upper bound for A + B for whih lub (A + B) is the lest of ll upper bounds.) Hene lub (A + B) luba + lubb. (6) Conversely, + b A + B mens tht + b lub (A + B) for ll + b A + B, i.e. for ll A nd ll b B. Let b B be given. So we hve lub (A + B) b for ll A. Tht is, lub (A + B) b is n upper bound for A while luba is the lest of ll suh upper bounds. Hene luba lub (A + B) b. 7
8 We thus hve b lub (A + B) luba for ll b B. prgrph this mens tht lubb lub (A + B) luba, or As in the lst Combine (6) nd (7) to get result. I leve the proof for glb to student. luba + lubb lub (A + B). (7) Applition The sumset A+B = A A = { :, A} hs the property tht if A A then A A. Alterntively η A A iff η A A, so we ould write A A = { η : η A A} { η : η A A} Strnge perhps, but it does llow us to sy tht lub (A A) = lub { η : η A A}. (8) Finlly note tht lubb = lub ( A) = glb (A). So the Lemm gives luba glba = luba + lub ( A) = lub (A + B) = lub { :, A}, by (8). Apply this with A i = {f (x) : x [x i 1, x i ]} to dedue M f i m f i = lub { f (x) f (y) : x, y [x i 1, x i ]} This is useful when ombined with Riemnn s Criteri. Theorem Assume tht the bounded funtions f nd g re Riemnn integrble on [, b]. Then (i) Sum Rule: f + g is integrble on [, b]. (ii) Produt Rule: fg is integrble on [, b]. (iii) Quotient Rule: f/g is integrble on [, b] if there exists C > 0 suh tht g (x) C for ll x [, b]. (iv) f, defined by f (x) = f (x) for ll x [, b], is integrble over [, b]. Proof (i) For x [x i 1, x i ] we hve m f i f (x) nd m g i g (x), nd so mf i + 8
9 m g i f (x)+g (x). Thus mf i +mg i is lower bound for {f (x) + g (x) : x [x i 1, x i ]}, for whih m f+g i is the gretest lower bound. Hene Similrly From these we get m f i + mg i mf+g i. M f i + M g i M f+g i. nd Hene U (P, f + g) U (P, f) + U (P, g), (9) L (P, f + g) L (P, f) + L (P, g). (10) U (P, f + g) L (P, f + g) U (P, f) L (P, f)+u (P, g) L (P, g), (11) for ll prtitions P. Let ε > 0 be given. The ft tht f is integrble implies, by Riemnn s Criteri, tht we n find P 1 suh tht U (P 1, f) L (P 1, f) < ε 2. (12) For the sme resons we n find P 2 suh tht U (P 2, g) L (P 2, g) < ε 2. (13) Tke P 0 = P 1 P 2. Then (12) nd (13) both hold for this prtition. From (11) we then get U (P 0, f + g) L (P 0, f + g) < ε 2 + ε 2 = ε. Thus, by Riemnn s Criteri, we find tht f + g is Riemnn integrble over [, b]. (ii) Let K > 0 nd N > 0 be bounds on f nd g, so f (x) K nd g (x) N for ll x [, b]. We use Applition nd exmine 9
10 M fg i m fg i = lub { f (x) g (x) f (y) g (y) : x, y [x i 1, x i ]}. In this we use ides seen when exmining limits of produts, f (x) g (x) f (y) g (y) = f (x) g (x) f (x) g (y) + f (x) g (y) f (y) g (y) f (x) g (x) g (y) + g (y) f (x) f (y) K g (x) g (y) + N f (x) f (y). So we get M fg i m fg i K ( M f i ) m f i + N (M g i m g i ). Thus U (P, fg) L (P, fg) K (U (P, f) L (P, f)) +N (U (P, g) L (P, g)) for ll prtitions P, gin by the Applition. Agin finish the proof s in (i) using Riemnn s riteri on f with ε/2k, on g with ε/2n nd then ombining the two prtitions found. (iii) It suffies to prove this result for 1/g; tht for f/g follows then by prt (ii). From the Applition, { M 1/g i m 1/g 1 i = lub g (x) 1 } g (y) : x, y [x i 1, x i ]. to get In this simply use 1 g (x) 1 g (y) g (x) g (y) C 2 nd U M 1/g i m 1/g i 1 C 2 (M g i m g i ), ( P, 1 ) ( L P, 1 ) 1 (U (P, g) L (P, g)) g g C2 10
11 for ll prtitions P. So finish off with Riemnn s Criteri with C 2 ε. (iv) For, b R we hve = b + b b + b nd b = b = b b + = b +. So b b b nd thus b b. Using the Applition we dedue nd M f i m f i M f i m f i, U (P, f ) L (P, f ) U (P, f) L (P, f) for ll prtitions P. Then n pplition of Riemnn s Criteri will give f integrble. Theorem Assume tht the bounded funtions f nd g re Riemnn integrble on [, b]. Then (i) Linerity. (f + g) = f + (ii) Additive Property. For < < b, f = f + (iii) If f (x) g (x) for ll x [, b] then (iv) f g. f f. Note It fits in with these results if we define f (t) dt = Proof (i) From (9) nd (10) we hve b f. g. f (t) dt. 11
12 L (P, f) + L (P, g) L (P, f + g) U (P, f + g) U (P, f) + U (P, g). (14) Let ε > 0 be given. From the Lemm bove we n find P f nd P g suh tht nd f ε 2 L ( P f, f ) U ( P f, f ) f + ε 2 (15) g ε 2 L (Pg, g) U (P g, g) g + ε 2. (16) Let P = P f P g, when both (15) nd (16) hold for this prtition. From (14), (15), (16) nd the ft tht f + g is integrble, we find tht f+ g ε L (P, f + g) True for ll ε > 0 gives the result. (f + g) U (P, f + g) f+ g+ε. (ii) I leve it to the student to hek tht if f is integrble over n intervl I then it is integrble over ny subintervl J I. So, sine f is now integrble over [, ], there exists P, prtition of [, ], suh tht f ε 2 L (P, f) U (P, f) Similrly, there exists P b, prtition of [, b], suh tht f ε 2 L ( P b, f ) U ( P b, f ) Set P = P P b, prtition of [, b]. Then f + ε 2. (17) f + ε 2. (18) U (P, f) = U (P, f) + U ( P b, f ) (19) L (P, f) = L (P, f) + L ( P b, f ). (20) Combining (17) (20) gives 12
13 f + f ε L (P, f) f f U (P, f) f + f + ε. True for ll ε > 0 mens we hve equlity in the entre, so f is integrble over [, b] with vlue f + f. (iii) Let ε > 0 be given. Sine g is integrble we n find P suh tht g ε L (P, g). The ssumption tht g f mens tht L (P, g) L (P, f). Finlly by definition, L (P, f) f. Combing these three fts gives g ε L (P, g) L (P, f) True for ll ε > 0 mens tht g f. (iii) We hve seen erlier tht f is integrble. Use prt (ii) on f f f to get the inequlity. Though not needed for our ourse, useful lterntive form of Riemnn s Criteri is Theorem A bounded funtion f : [, b] R is Riemnn integrble iff there exists sequene of prtitions {P n } n 1 of [, b] with P 1 P 2 P 3... suh tht lim n U (P n, f) = lim n L (P n, f). This ommon vlue is the vlue of the integrl. Proof ( ) Simply use L (P n, f) f for ll n 1, long with the Sndwih Theorem. f U (P n, f) ( ) Assume tht f is Riemnn integrble in whih se we hve Riemnn s Criteri. Let n 1 be given. Choose ε = 1/n in Riemnn s Criteri to find prtitionq n : U (Q n, f) L (Q n, f) < 1/n. The {Q n } my well not stisfy Q 1 Q 2 Q 3... so define f. 13
14 P n = Q j. 1 j n Then P n = Q j Q j = P n+1. 1 j n 1 j n+1 And Q n P n mens U (P n, f) U (Q n, f) nd L (P n, f) L (Q n, f) so U (P n, f) L (P n, f) U (Q n, f) L (Q n, f) < 1/n. Next, P 1 P 2 P 3... mens tht U (P 1, f) U (P 2, f) U (P 3, f)..., deresing sequene bounded below by m (b ). Thus it onverges, to α sy. Similrly, L (P 1, f) L (P 2, f) L (P 3, f)..., n inresing sequene bounded bove by M (b ). Thus this lso onverges, to β sy. Finlly, 0 α β = lim n (U (P n, f) L (P n, f)) lim n 1/n = 0. Hene α = β nd so lim n U (P n, f) = lim n L (P n, f) while L (P n, f) gives the vlue of the limit s f. f U (P n, f) ( ) For the sttement of our next riteri define P, of prtition P = { = x 0 < x 1 <... < x n = b}, to be P = mx 1 i n x i x i 1, i.e. the mximum prtition length. Theorem (Du BoisReymond 1875, Drboux 1875) A bounded funtion f : [, b] R is Riemnn Integrble if, nd only if, ε > 0, δ > 0, P D δ, U (P, f) L (P, f) < ε. Here D δ is the set of ll prtitions with P < δ. ( ) Exmple Use the rithmeti prtitions long with Riemnn s riteri to show tht f : [0, 3] R : x x is integrble nd find the vlue of 3 0 tdt. Theorem If f is ontinuous on [, b] then f is Riemnn integrble on [, b]. 14
15 Proof Not given in this ourse, but if interested, see ( ) The proof depends on property of ontinuous funtions tht we hve not disussed nd whih you will not study until ourse 251. This property is tht if f : [, b] R is ontinuous then, sine the domin [, b] is losed nd bounded, we hve ε > 0, δ > 0, x 1, x 2 [, b], x 1 x 2 < δ f (x 1 ) f (x 2 ) < ε. (21) Try to understnd why this sttement is different to the definition of ontinuity. A funtion tht stisfies (21) is sid to be uniformly ontinuous.( ) See Question?? Theorem If f is monotoni on [, b] then f is Riemnn integrble on [, b]. Exmple Let f : [0, 1] R be given by f (x) = Riemnn integrble over [0, 1]. 1. Prove tht f is 1 + x2 Exmple Let f : [0, 1] R be given by f (0) = 0 nd, for x (0, 1], f (x) = 1 n where n is the lrgest integer stisfying x 1 n. Drw the grph of f. Show tht f is Riemnn integrble on [0, 1]. How mny disontinuities does this funtion hve? Question?? 15
16 4.4 Integrtion s the inverse of differentition. Definition Let f : (, b) R be given. Assume there exists funtion F : [, b] R whih is ontinuous on [, b], differentible on (, b) nd stisfies F (x) = f (x) for ll x (, b). Then F is lled primitive (or ntiderivtive) of f. Note If primitive exists it will not be unique. If F is primitive for f then F +, where is onstnt, stisfies (F + ) = F = f. Thus F + is lso primitive for f. But if F 1 nd F 2 re two primitives for f then F 1 = f = F 2. Thus (F 1 F 2 ) = 0. By the InresingDeresing Theorem this mens tht F 1 F 2 =, for some onstnt. So, if F is primitive for f then F 0 is primitive if, nd only if, F 0 F is onstnt on [, b]. Theorem Fundmentl Theorem of Clulus. 1) If f is Riemnn Integrble on [, b] then is ontinuous on [, b]. F (x) = x f (t) dt 2) Further, if f is ontinuous on [, b] then F is differentible on (, b) nd F (x) = f (x) for ll x (, b). Proof 1) Sine f is bounded hoose N > 0 : f (x) N for ll x [, b]. Let ε > 0 be given. Let (, b) be given. Choose δ = ε/n. Assume x < δ. If x > onsider F (x) F () = = x x x f (t) dt f (t) dt f (t) dt = N (x ). x f (t) dt Ndt If < x onsider F (x) F () = F () F (x) N ( x) by bove. So, in ll ses, F (x) F () N x < Nδ = ε. 16
17 Hene F is ontinuous t nd thus on (, b). 2) Let ε > 0 be given. Let (, b) be given. We re told tht f is ontinuous t so there exists δ > 0 suh tht if x < δ then f (x) f () < ε. Assume x < δ. If x > onsider F (x) F () x f () = = 1 (F (x) F ()) (x ) f () x 1 x x x f (t) dt f () dt 1 x f (t) f () dt x 1 x εdt = ε. x Here we hve used t x < δ nd so f (t) f () < ε inside the integrl. If x < the sme result follows on writing F (x) F () f () x = F () F (x) f () x. So in ll ses Thus F (x) F () x f () < ε. F (x) F () lim x x exists, i.e. F is differentible t, with derivtive f (). Note: If f is integrble, this is not suffiient to sy tht F (x) = x f (t) dt is primitive for f. We need the extr ondition tht f is ontinuous. Corollry If g is ontinuous on [, b] nd G is primitive for g then x for ll y < x b. y g (t) dt = G (x) G (y) = [G (t)] x y Exmple Prove tht ln x, defined erlier s the inverse of e x, stisfies 17
18 x dt ln x = 1 t for ll x > 0. This is often tken s the definition of the nturl logrithm. Question?? Exmple (Cuhy) Prove tht if f (n+1) is ontinuous on [, b] then x (x t) n f (n+1) (t) dt = R n, f (x) n! for ll x b, where R n, is the reminder term for the nth Tylor Series. Proof Let x [, b]. By n erlier exmple we hve seen tht the Tylor polynomil, T n,t f (x) is primitive for (x t) n f (n+1) (t), n! s funtion of t. Apply the orollry with g (t) = (x t) n f (n+1) (t) /n!, ontinuous by ssumption on f, to get d (x t) n f (n+1) (t) dt = T n,d f (x) T n, f (x), n! for ll, d [, b]. Tke =, d = x, nd rell T n,x f (x) = f (x). Thus we get our result. Applitions We hve ln (1 + x) = x x2 2 + x3 3 x4 4 + x = vlid for 1 < x 1. r=1 ( 1) r 1 x r Note In prtiulr lim n R n,0 (ln (1 + x)) x=1 = 0. Putting x = 1 gives ln 2 = The Fundmentl Theorem tells us how to differentite n integrl. Exmple Let r Clulte G (t). x 3 G (x) = e t os tdt. x 2 18
19 Solution. Let F (y) = y 1 e t os tdt, so G (x) = F (x 3 ) F (x 2 ). Sine e t os t is ontinuous we hve, by the Fundmentl Theorem, tht F is differentible. Thus G (x) = 3x 2 F ( x 3) 2xF ( x 2) = 3x 2 e x3 os ( x 3) 2xe x2 os ( x 2). As further pplition, note tht the rule for differentiting produt of funtions g nd h n be restted s sying tht gh is primitive for gh +g h. So, if gh +g h is ontinuous, we n pply the Fundmentl Theorem, giving Thus we hve justified x y (gh + g h) (t) dt = [(gh) (t)] x y. Theorem Integrtion by prts. Assume tht g nd h hve ontinuous derivtives on [, b]. Then x y x g (t) h (t) dt = [g (t) h(t)] x y g (t) h (t) dt. ( )Exmple Assume tht f (n+1) is ontinuous on [, b]. Integrte 1 n! h 0 (h u) n f (n+1) (u) du by prts, repetedly. Question?? To use integrtion by prts in this exmple we need g (t) = (h t) n nd h (t) = f (n) (u) to both hve ontinuous derivtives. In prtiulr f (n+1) needs to be ontinuous on [0, h]. If you look bk t the vrious forms of the error term for Tylor s series you will see the integrl form required f (n+1) to be ontinuous while the other two forms did not.( ) I leve it to the student to justify the integrtion of the hin rule for differentition. This would give Theorem Chnge of Vrible or Integrtion by Substitution Assume f nd g re ontinuous on [, b]. Then y 19
20 f (x) dx = where g (β) = b nd g (α) =. β α (f g) (t) g (t) dt Proof Let F be primitive for f. Then, by the hin rule F (g (x)) hs derivtive f (g (x)) g (x), i.e. it is primitive for this. So, by the Corollry, β α f (g (t)) g (t) dt = F (g (β)) F (g (α)) = F (b) F () = f (x) dx.. Question?? Exmple With ln x expressed s n integrl, s we sw in n erlier exmple, prove tht for ll, b > 0 we hve ln b = ln + ln b. Question?? 4.5 Improper Integrls (We my well not over this setion in letures) We hve only defined the integrl for losed intervls nd funtions bounded on suh intervls. We finish with list of definitions tht try to extend the situtions in whih our definitions re meningful. Definition If f : [, ) R is Riemnn integrble on every intervl [, b] nd t lim t f (x) dx mkes sense with the limit existing, we define this limit to be f (x) dx nd we sy tht the integrl onverges. Otherwise is diverges. Similrly for f (x) dx. Exmple Show tht 1 1 x α dx onverges if, nd only if, α > 1. Question?? ( ) This my be reminisent of result in ourse 153 onerning those k for whih 1 r=1 onverges. But then there is often onnetion between the n k series 1 f (n) nd integrl f (t) dt. For instne, if f : [1, ) R is 1 20
21 positive deresing funtion then the series nd integrl either both diverge or both onverge. ( ) Definition We define f (x) dx s 0 f (x) dx + 0 f (x) dx, nd it onverges only if both of these do so seprtely. Definition (Guss, 1812) If f is defined on (, b] nd lim α f (x) dx α mkes sense nd the limit exists, then we define the limit to be f (x) dx nd sy tht the integrl onverges. Otherwise is diverges. Similrly for funtion defined on [, b). Exmple For wht α does 1 0 onverge? Question?? Definition Cuhy Prinipl Vlue. Assume tht f is not defined t in [, b] but, for ll η > 0, is bounded in [, η] [ + η, b]. Then the Prinipl Vlue Integrl is defined to be ( η ) P.V f (x) dx = lim f (x) dx + f (x) dx, η 0 +η provided tht this limit exists. dx x α 21
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