The Riemann Integral. Chapter 1

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1 Chpter The Riemnn Integrl now of some universities in Englnd where the Lebesgue integrl is tught in the first yer of mthemtics degree insted of the Riemnn integrl, but now of no universities in Englnd where students lern the Lebesgue integrl in the first yer of mthemtics degree. (Approximte quottion ttributed to T. W. Körner) Let f : [,b] R be bounded (not necessrily continuous) function on compct (closed, bounded) intervl. We will define wht it mens for f to be Riemnn integrble on [,b] nd, in tht cse, define its Riemnn integrl f. The integrl of f on [,b] is rel number whose geometricl interprettion is the signed re under the grph y = f(x) for x b. This number is lso clled the definite integrl of f. By integrting f over n intervl [,x] with vrying right end-point, we get function of x, clled the indefinite integrl of f. The most importnt result bout integrtion is the fundmentl theorem of clculus, which sttes tht integrtion nd differentition re inverse opertions in n ppropritely understood sense. Among other things, this connection enbles us to compute mny integrls explicitly. Integrbility is less restrictive condition on function thn differentibility. Roughly speking, integrtion mkes functions smoother, while differentition mkes functions rougher. For exmple, the indefinite integrl of every continuous function exists nd is differentible, wheres the derivtive of continuous function need not exist (nd generlly doesn t). The Riemnn integrl is the simplest integrl to define, nd it llows one to integrte every continuous function s well s some not-too-bdly discontinuous functions. There re, however, mny other types of integrls, the most importnt of which is the Lebesgue integrl. The Lebesgue integrl llows one to integrte unbounded or highly discontinuous functions whose Riemnn integrls do not exist, nd it hs better mthemticl properties thn the Riemnn integrl. The definition of the Lebesgue integrl requires the use of mesure theory, which we will not

2 2. The Riemnn Integrl describe here. In ny event, the Riemnn integrl is dequte for mny purposes, nd even if one needs the Lebesgue integrl, it s better to understnd the Riemnn integrl first... Definition of the Riemnn integrl We sy tht two intervls re lmost disjoint if they re disjoint or intersect only t common endpoint. For exmple, the intervls [,] nd [,3] re lmost disjoint, wheres the intervls [,2] nd [,3] re not. Definition.. Let I be nonempty, compct intervl. A prtition of I is finite collection {I,I 2,...,I n } of lmost disjoint, nonempty, compct subintervls whose union is I. A prtition of [,b] with subintervls = [x k,x k ] is determined by the set of endpoints of the intervls = x < x < x 2 < < x n < x n = b. Abusing nottion, we will denote prtition P either by its intervls P = {I,I 2,...,I n } or by the set of endpoints of the intervls P = {x,x,x 2,...,x n,x n }. We ll dopt either nottion s convenient; the context should mke it cler which one is being used. There is lwys one more endpoint thn intervl. Exmple.2. The set of intervls {[,/5],[/5,/4],[/4,/3],[/3,/2],[/2,]} is prtition of [,]. The corresponding set of endpoints is {,/5,/4,/3,/2,}. We denote the length of n intervl I = [,b] by I = b. Note tht the sum of the lengths = x k x k of the lmost disjoint subintervls in prtition {I,I 2,...,I n } ofn intervl I is equl to length of the whole intervl. This is obvious geometriclly; lgebriclly, it follows from the telescoping series n n = (x k x k ) k= k= = x n x n +x n x n 2 + +x 2 x +x x = x n x = I. Suppose tht f : [,b] R is bounded function on the compct intervl I = [,b] with M = supf, m = inf f. I I

3 .. Definition of the Riemnn integrl 3 If P = {I,I 2,...,I n } is prtition of I, let M k = sup f, m k = inf f. These suprem nd infim re well-defined, finite rel numbers since f is bounded. Moreover, m m k M k M. If f is continuous on the intervl I, then it is bounded nd ttins its mximum nd minimum vlues on ech subintervl, but bounded discontinuous function need not ttin its supremum or infimum. We define the upper Riemnn sum of f with respect to the prtition P by n n U(f;P) = M k = M k (x k x k ), k= nd the lower Riemnn sum of f with respect to the prtition P by n n L(f;P) = m k = m k (x k x k ). k= Geometriclly, U(f;P) is the sum of the res of rectngles bsed on the intervls tht lie bove the grph of f, nd L(f;P) is the sum of the res of rectngles tht lie below the grph of f. Note tht k= k= m(b ) L(f;P) U(f;P) M(b ). Let Π(,b), or Π for short, denote the collection of ll prtitions of [,b]. We define the upper Riemnn integrl of f on [,b] by U(f) = inf P Π U(f;P). The set {U(f;P) : P Π} of ll upper Riemnn sums of f is bounded from below by m(b ), so this infimum is well-defined nd finite. Similrly, the set {L(f;P) : P Π} of ll lower Riemnn sums is bounded from bove by M(b ), nd we define the lower Riemnn integrl of f on [,b] by L(f) = sup L(f;P). P Π Theseupper nd lowersums nd integrlsdepend onthe intervl [,b]s well sthe function f, but to simplify the nottion we won t show this explicitly. A commonly used lterntive nottion for the upper nd lower integrls is U(f) = f, L(f) = Note the use of lower-upper nd upper-lower pproximtions for the integrls: we tke the infimum of the upper sums nd the supremum of the lower sums. As we show in Proposition.3 below, we lwys hve L(f) U(f), but in generl the upper nd lower integrls need not be equl. We define Riemnn integrbility by their equlity. f.

4 4. The Riemnn Integrl Definition.3. A bounded function f : [,b] R is Riemnn integrble on [,b] if its upper integrl U(f) nd lower integrl L(f) re equl. In tht cse, the Riemnn integrl of f on [,b], denoted by f(x)dx, f, f [,b] or similr nottions, is the common vlue of U(f) nd L(f). An unbounded function is not Riemnn integrble. In the following, integrble will men Riemnn integrble, nd integrl will men Riemnn integrl unless stted explicitly otherwise..2. Exmples of the Riemnn integrl Let us illustrte the definition of Riemnn integrbility with number of exmples. Exmple.4. Define f : [,] R by { /x if < x, f(x) = if x =. Then x dx isn t defined s Riemnn integrl becuse f is unbounded. In fct, if < x < x 2 < < x n < is prtition of [,], then sup f =, [,x ] so the upper Riemnn sums of f re not well-defined. An integrl with n unbounded intervl of integrtion, such s x dx, lso isn t defined s Riemnn integrl. In this cse, prtition of [, ) into finitely mny intervls contins t lest one unbounded intervl, so the corresponding Riemnn sum is not well-defined. A prtition of [, ) into bounded intervls (for exmple, = [k,k+] with k N) gives n infinite series rther thn finite Riemnn sum, leding to questions of convergence. One cn interpret the integrls in this exmple s limits of Riemnn integrls, or improper Riemnn integrls, dx = lim x ǫ + ǫ x dx, r dx = lim x r x dx, but these re not proper Riemnn integrls in the sense of Definition.3. Such improper Riemnn integrls involve two limits limit of Riemnn sums to define the Riemnn integrls, followed by limit of Riemnn integrls. Both of the improper integrls in this exmple diverge to infinity. (See Section..)

5 .2. Exmples of the Riemnn integrl 5 Next, we consider some exmples of bounded functions on compct intervls. Exmple.5. The constnt function f(x) = on [,] is Riemnn integrble, nd dx =. To show this, let P = {I,I 2,...,I n } be ny prtition of [,] with endpoints Since f is constnt, nd therefore {,x,x 2,...,x n,}. M k = sup f =, m k = inf f = U(f;P) = L(f;P) = for k =,...,n, n (x k x k ) = x n x =. k= Geometriclly, this eqution is the obvious fct tht the sum of the res of the rectngles over (or, equivlently, under) the grph of constnt function is exctly equl to the re under the grph. Thus, every upper nd lower sum of f on [,] is equl to, which implies tht the upper nd lower integrls U(f) = inf U(f;P) = inf{} =, P Π re equl, nd the integrl of f is. L(f) = sup L(f;P) = sup{} = P Π Moregenerlly, thesmergumentshowsthteveryconstntfunction f(x) = c is integrble nd cdx = c(b ). The following is n exmple of discontinuous function tht is Riemnn integrble. Exmple.6. The function is Riemnn integrble, nd f(x) = { if < x if x = f dx =. To show this, let P = {I,I 2,...,I n } be prtition of [,]. Then, since f(x) = for x >, M k = sup f =, m k = inf f = for k = 2,...,n. The first intervl in the prtition is I = [,x ], where < x, nd M =, m =, since f() = nd f(x) = for < x x. It follows tht Thus, L(f) = nd U(f;P) = x, L(f;P) =. U(f) = inf{x : < x } =,

6 6. The Riemnn Integrl so U(f) = L(f) = re equl, nd the integrl of f is. In this exmple, the infimum of the upper Riemnn sums is not ttined nd U(f;P) > U(f) for every prtition P. A similr rgument shows tht function f : [,b] R tht is zero except t finitely mny points in [, b] is Riemnn integrble with integrl. The next exmple is bounded function on compct intervl whose Riemnn integrl doesn t exist. Exmple.7. The Dirichlet function f : [,] R is defined by { if x [,] Q, f(x) = if x [,]\Q. Tht is, f is one t every rtionl number nd zero t every irrtionl number. This function is not Riemnn integrble. If P = {I,I 2,...,I n } is prtition of [,], then M k = supf =, m k = inf =, since every intervl of non-zero length contins both rtionl nd irrtionl numbers. It follows tht U(f;P) =, L(f;P) = for every prtition P of [,], so U(f) = nd L(f) = re not equl. The Dirichlet function is discontinuous t every point of [, ], nd the morl of the lst exmple is tht the Riemnn integrl of highly discontinuous function need not exist..3. Refinements of prtitions As the previous exmples illustrte, direct verifiction of integrbility from Definition.3 is unwieldy even for the simplest functions becuse we hve to consider ll possible prtitions of the intervl of integrtion. To give n effective nlysis of Riemnn integrbility, we need to study how upper nd lower sums behve under the refinement of prtitions. Definition.8. A prtition Q = {J,J 2,...,J m } is refinement of prtition P = {I,I 2,...,I n } if every intervl in P is n lmost disjoint union of one or more intervls J l in Q. Equivlently, if we represent prtitions by their endpoints, then Q is refinement of P if Q P, mening tht every endpoint of P is n endpoint of Q. We don t require tht every intervl or even ny intervl in prtition hs to be split into smller intervls to obtin refinement; for exmple, every prtition is refinement of itself. Exmple.9. Consider the prtitions of [, ] with endpoints P = {,/2,}, Q = {,/3,2/3,}, R = {,/4,/2,3/4,}. Thus, P, Q, nd R prtition [,] into intervls of equl length /2, /3, nd /4, respectively. Then Q is not refinement of P but R is refinement of P.

7 .3. Refinements of prtitions 7 Given two prtitions, neither one need be refinement of the other. However, two prtitions P, Q lwys hve common refinement; the smllest one is R = P Q, mening tht the endpoints of R re exctly the endpoints of P or Q (or both). Exmple.. Let P = {,/2,} nd Q = {,/3,2/3,}, s in Exmple.9. Then Q isn t refinement of P nd P isn t refinement of Q. The prtition S = P Q, or S = {,/3,/2,2/3,}, is refinement of both P nd Q. The prtition S is not refinement of R, but T = R S, or T = {,/4,/3,/2,2/3,3/4,}, is common refinement of ll of the prtitions {P,Q,R,S}. As we show next, refining prtitions decreses upper sums nd increses lower sums. (The proof is esier to understnd thn it is to write out drw picture!) Theorem.. Suppose tht f : [,b] R is bounded, P is prtitions of [,b], nd Q is refinement of P. Then Proof. Let U(f;Q) U(f;P), L(f;P) L(f;Q). P = {I,I 2,...,I n }, Q = {J,J 2,...,J m } be prtitions of [,b], where Q is refinement of P, so m n. We list the intervls in incresing order of their endpoints. Define M k = sup f, m k = inf f, M l = sup J l f, m l = inf J l f. Since Q is refinement of P, ech intervl in P is n lmost disjoint union of intervls in Q, which we cn write s = q k l=p k J l for some indices p k q k. If p k < q k, then is split into two or more smller intervls in Q, nd if p k = q k, then belongs to both P nd Q. Since the intervls re listed in order, we hve If p k l q k, then J l, so p =, p k+ = q k +, q n = m. M l M k, m k m l for p k l q k. Using the fct tht the sum of the lengths of the J-intervls is the length of the corresponding I-intervl, we get tht q k l=p k M l J l q k q k M k J l = M k J l = M k. l=p k l=p k

8 8. The Riemnn Integrl It follows tht m n q k n U(f;Q) = M l J l = M l J l M k = U(f;P) l= k=l=p k k= Similrly, q k q k m l J l m k J l = m k, l=p k l=p k nd n q k n L(f;Q) = m l J l m k = L(f;P), k= l=p k k= which proves the result. It follows from this theorem tht ll lower sums re less thn or equl to ll upper sums, not just the lower nd upper sums ssocited with the sme prtition. Proposition.2. If f : [,b] R is bounded nd P, Q re prtitions of [,b], then L(f;P) U(f;Q). Proof. Let R be common refinement of P nd Q. Then, by Theorem., L(f;P) L(f;R), U(f;R) U(f;Q). It follows tht L(f;P) L(f;R) U(f;R) U(f;Q). An immedite consequence of this result is tht the lower integrl is lwys less thn or equl to the upper integrl. Proposition.3. If f : [,b] R is bounded, then Proof. Let A = {L(f;P) : P Π}, L(f) U(f). B = {U(f;P) : P Π}. From Proposition.2, b for every A nd b B, so Proposition 2.9 implies tht supa infb, or L(f) U(f)..4. The Cuchy criterion for integrbility The following theorem gives criterion for integrbility tht is nlogous to the Cuchy condition for the convergence of sequence. Theorem.4. A bounded function f : [,b] R is Riemnn integrble if nd only if for every ǫ > there exists prtition P of [,b], which my depend on ǫ, such tht U(f;P) L(f;P) < ǫ.

9 .4. The Cuchy criterion for integrbility 9 Proof. First, suppose tht the condition holds. Let ǫ > nd choose prtition P tht stisfies the condition. Then, since U(f) U(f;P) nd L(f;P) L(f), we hve U(f) L(f) U(f;P) L(f;P) < ǫ. Since this inequlity holds for every ǫ >, we must hve U(f) L(f) =, nd f is integrble. Conversely, suppose tht f is integrble. Given ny ǫ >, there re prtitions Q, R such tht U(f;Q) < U(f)+ ǫ 2, L(f;R) > L(f) ǫ 2. Let P be common refinement of Q nd R. Then, by Theorem., U(f;P) L(f;P) U(f;Q) L(f;R) < U(f) L(f)+ǫ. Since U(f) = L(f), the condition follows. If U(f;P) L(f;P) < ǫ, then U(f;Q) L(f;Q) < ǫ for every refinement Q of P, so the Cuchy condition mens tht function is integrble if nd only if its upper nd lower sums get rbitrrily close together for ll sufficiently refined prtitions. It is worth considering in more detil wht the Cuchy condition in Theorem.4 implies bout the behvior of Riemnn integrble function. Definition.5. The oscilltion of bounded function f on set A is k= osc f = sup f inf f. A A A If f : [,b] R is bounded nd P = {I,I 2,...,I n } is prtition of [,b], then n n n U(f;P) L(f;P) = supf inf f = oscf. A function f is Riemnn integrble if we cn mke U(f;P) L(f;P) s smll s we wish. This is the cse if we cn find sufficiently refined prtition P such tht the oscilltion of f on most intervls is rbitrrily smll, nd the sum of the lengths of the remining intervls (where the oscilltion of f is lrge) is rbitrrily smll. For exmple, the discontinuous function in Exmple.6 hs zero oscilltion on every intervl except the first one, where the function hs oscilltion one, but the length of tht intervl cn be mde s smll s we wish. Thus, roughly speking, function is Riemnn integrble if it oscilltes by n rbitrry smll mount except on finite collection of intervls whose totl length is rbitrrily smll. Theorem.87 gives precise sttement. One direct consequence of the Cuchy criterion is tht function is integrble if we cn estimte its oscilltion by the oscilltion of n integrble function. Proposition.6. Suppose tht f,g : [,b] R re bounded functions nd g is integrble on [,b]. If there exists constnt C such tht k= osc I f Cosc I g on every intervl I [,b], then f is integrble. k=

10 . The Riemnn Integrl Proof. If P = {I,I 2,...,I n } is prtition of [,b], then n [ ] U (f;p) L(f;P) = supf inff = C k= n k= osc f n k= osc g C[U(g;P) L(g;P)]. Thus, f stisfies the Cuchy criterion in Theorem.4 if g does, which proves tht f is integrble if g is integrble. We cn lso give sequentil chrcteriztion of integrbility. Theorem.7. A bounded function f : [,b] R is Riemnn integrble if nd only if there is sequence (P n ) of prtitions such tht In tht cse, lim [U(f;P n) L(f;P n )] =. n f = lim n U(f;P n) = lim n L(f;P n). Proof. First, suppose tht the condition holds. Then, given ǫ >, there is n n N such tht U(f;P n ) L(f;P n ) < ǫ, so Theorem.4 implies tht f is integrble nd U(f) = L(f). Furthermore, since U(f) U(f;P n ) nd L(f;P n ) L(f), we hve U(f;P n ) U(f) = U(f;P n ) L(f) U(f;P n ) L(f;P n ). Since the limit of the right-hnd side is zero, the squeeze theorem implies tht It lso follows tht lim U(f;P n) = U(f) = n lim L(f;P n) = lim U(f;P n) lim [U(f;P n) L(f;P n )] = n n n Conversely, if f is integrble then, by Theorem.4, for every n N there exists prtition P n such tht U(f;P n ) L(f;P n ) < n, f f. nd U(f;P n ) L(f;P n ) s n. Note tht if the limits of U(f;P n ) nd L(f;P n ) both exist nd re equl, then lim [U(f;P n) L(f;P n )] = lim U(f;P n) lim L(f;P n), n n n so the conditions of the theorem re stisfied. Conversely, the proof of the theorem shows tht if the limit of U(f;P n ) L(f;P n ) is zero, then the limits of U(f;P n )

11 .5. Integrbility of continuous nd monotonic functions nd L(f;P n ) both exist nd re equl. This isn t true for generl sequences, where one my hve lim( n b n ) = even though lim n nd limb n don t exist. Theorem.7 provides one wy to prove the existence of n integrl nd, in some cses, evlute it. Exmple.8. Let P n be the prtition of [,] into n-intervls of equl length /n with endpoints x k = k/n for k =,,2,...,n. If = [(k )/n,k/n] is the kth intervl, then sup f = x 2 k, inf = x 2 k since f is incresing. Using the formul for the sum of squres n k 2 = 6 n(n+)(2n+), we get nd U(f;P n ) = L(f;P n ) = n k= n k= k= (See Figure.8.) It follows tht x 2 k n = n 3 n k 2 = 6 k= x 2 k n = n n 3 k 2 = 6 k= lim U(f;P n) = lim L(f;P n) = n n 3, nd Theorem.7 implies tht x 2 is integrble on [,] with x 2 dx = 3. ( + )( 2+ ) n n ( )( 2 ). n n The fundmentl theorem of clculus, Theorem.45 below, provides much esier wy to evlute this integrl..5. Integrbility of continuous nd monotonic functions The Cuchy criterion leds to the following fundmentl result tht every continuous function is Riemnn integrble. To provethis, we use the fct tht continuous function oscilltes by n rbitrrily smll mount on every intervl of sufficiently refined prtition. Theorem.9. A continuous function f : [,b] R on compct intervl is Riemnn integrble. Proof. A continuous function on compct set is bounded, so we just need to verify the Cuchy condition in Theorem.4. Let ǫ >. A continuous function on compct set is uniformly continuous, so there exists δ > such tht f(x) f(y) < ǫ for ll x,y [,b] such tht x y < δ. b

12 2. The Riemnn Integrl Upper Riemnn Sum = y x Lower Riemnn Sum = y x Upper Riemnn Sum = y x Lower Riemnn Sum = y x Upper Riemnn Sum = y x Lower Riemnn Sum = y x Figure. Upper nd lower Riemnn sumsfor Exmple.8 with n = 5,,5 subintervls of equl length.

13 .5. Integrbility of continuous nd monotonic functions 3 Choose prtition P = {I,I 2,...,I n } of [,b] such tht < δ for every k; for exmple, we cn tke n intervls of equl length (b )/n with n > (b )/δ. Since f is continuous, it ttins its mximum nd minimum vlues M k nd m k on the compct intervl t points x k nd y k in. These points stisfy x k y k < δ, so M k m k = f(x k ) f(y k ) < ǫ b. The upper nd lower sums of f therefore stisfy n n U(f;P) L(f;P) = M k m k = k= k= n (M k m k ) k= < ǫ b < ǫ, nd Theorem.4 implies tht f is integrble. n Exmple.2. The function f(x) = x 2 on [,] considered in Exmple.8 is integrble since it is continuous. Another clss of integrble functions consists of monotonic (incresing or decresing) functions. Theorem.2. A monotonic function f : [,b] R on compct intervl is Riemnn integrble. Proof. Suppose tht f is monotonic incresing, mening tht f(x) f(y) for x y. Let P n = {I,I 2,...,I n } be prtition of [,b] into n intervls = [x k,x k ], of equl length (b )/n, with endpoints k= x k = +(b ) k n, k =,,...,n,n. Since f is incresing, M k = sup f = f(x k ), m k = inf f = f(x k ). Hence, summing telescoping series, we get n U(f;P n ) L(U;P n ) = (M k m k )(x k x k ) k= = b n n [f(x k ) f(x k )] = b n [f(b) f()]. It follows tht U(f;P n ) L(U;P n ) s n, nd Theorem.7 implies tht f is integrble. k=

14 4. The Riemnn Integrl.2.8 y x Figure 2. The grph of the monotonic function in Exmple.22 with countbly infinite, dense set of jump discontinuities. The proof for monotonic decresing function f is similr, with sup f = f(x k ), inf f = f(x k ), or we cn pply the result for incresing functions to f nd use Theorem.23 below. Monotonic functions needn t be continuous, nd they my be discontinuous t countbly infinite number of points. Exmple.22. Let {q k : k N} be n enumertion of the rtionl numbers in [,) nd let ( k ) be sequence of strictly positive rel numbers such tht k =. Define f : [,] R by f(x) = k Q(x) k, k= Q(x) = {k N : q k [,x)}. for x >, nd f() =. Tht is, f(x) is obtined by summing the terms in the series whose indices k correspond to the rtionl numbers q k < x. For x =, this sum includes ll the terms in the series, so f() =. For every < x <, there re infinitely mny terms in the sum, since the rtionls re dense in [,x), nd f is incresing, since the number of terms increses with x. By Theorem.2, f is Riemnn integrble on [, ]. Although f is integrble, it hs countbly infinite number of jump discontinuities t every rtionl number in [,), which re dense in [,], The function is continuous elsewhere (the proof is left s n exercise).

15 .6. Properties of the Riemnn integrl 5 Figure 2 shows the grph of f corresponding to the enumertion {,/2,/3,2/3,/4,3/4,/5,2/5,3/5,4/5,/6,5/6,/7,...} of the rtionl numbers in [,) nd k = 6 π 2 k Properties of the Riemnn integrl The integrl hs the following three bsic properties. () Linerity: cf = c f, (2) Monotonicity: if f g, then (3) Additivity: if < c < b, then c f + f c (f +g) = In this section, we prove these properties nd derive few of their consequences. These properties re nlogous to the corresponding properties of sums (or convergent series): n n n n n c k = c k, ( k +b k ) = k + b k ; k= n k k= m k + k= k= k= f = n b k if k b k ; k= n k=m+ k = n k. k=.6.. Linerity. We begin by proving the linerity. First we prove linerity with respect to sclr multipliction nd then linerity with respect to sums. Theorem.23. If f : [,b] R is integrble nd c R, then cf is integrble nd cf = c g. f. k= f + Proof. Suppose tht c. Then for ny set A [,b], we hve sup A cf = csup A f. f, inf cf = cinf f, A A so U(cf;P) = cu(f;p) for every prtition P. Tking the infimum over the set Π of ll prtitions of [,b], we get g. k= U(cf) = inf U(cf;P) = inf cu(f;p) = c inf U(f;P) = cu(f). P Π P Π P Π

16 6. The Riemnn Integrl Similrly, L(cf;P) = cl(f;p) nd L(cf) = cl(f). If f is integrble, then which shows tht cf is integrble nd we hve Now consider f. Since Therefore U(cf) = cu(f) = cl(f) = L(cf), sup( f) = inf f, A A U( f;p) = L(f;P), cf = c inf A f. ( f) = supf, A L( f;p) = U(f;P). U( f) = inf U( f;p) = inf [ L(f;P)] = sup L(f;P) = L(f), P Π P Π P Π U(f;P) = U(f). L( f) = sup L( f;p) = sup P Π P Π Hence, f is integrble if f is integrble nd [ U(f;P)] = inf P Π ( f) = Finlly, if c <, then c = c, nd successive ppliction of the previous results shows tht cf is integrble with cf = c f. Next, we prove the linerity of the integrl with respect to sums. If f, g re bounded, then f +g is bounded nd It follows tht sup I (f +g) sup I f +sup I f. g, inf(f +g) inf f +infg. I I I osc(f +g) osc f +oscg, I I I so f+g is integrble if f, g re integrble. In generl, however, the upper (or lower) sum of f +g needn t be the sum of the corresponding upper (or lower) sums of f nd g. As result, we don t get (f +g) = simply by dding upper nd lower sums. Insted, we prove this equlity by estimting the upper nd lower integrls of f +g from bove nd below by those of f nd g. Theorem.24. If f,g : [,b] R re integrblefunctions, then f+g is integrble, nd (f +g) = f + f + g g.

17 .6. Properties of the Riemnn integrl 7 Proof. We first prove tht if f,g : [,b] R re bounded, but not necessrily integrble, then U(f +g) U(f)+U(g), L(f +g) L(f)+L(g). Suppose tht P = {I,I 2,...,I n } is prtition of [,b]. Then n U(f +g;p) = sup(f +g) k= n k= sup f + n k= U(f;P)+U(g;P). sup g Let ǫ >. Since the upper integrl is the infimum of the upper sums, there re prtitions Q, R such tht U(f;Q) < U(f)+ ǫ 2, U(g;R) < U(g)+ ǫ 2, nd if P is common refinement of Q nd R, then It follows tht U(f;P) < U(f)+ ǫ 2, U(g;P) < U(g)+ ǫ 2. U(f +g) U(f +g;p) U(f;P)+U(g;P) < U(f)+U(g)+ǫ. Sincethisinequlityholdsforrbitrryǫ >, wemusthveu(f+g) U(f)+U(g). Similrly, we hve L(f +g;p) L(f;P)+L(g;P) for ll prtitions P, nd for every ǫ >, we get L(f +g) > L(f)+L(g) ǫ, so L(f +g) L(f)+L(g). For integrble functions f nd g, it follows tht U(f +g) U(f)+U(g) = L(f)+L(g) L(f +g). Since U(f +g) L(f +g), we hve U(f +g) = L(f +g) nd f +g is integrble. Moreover, there is equlity throughout the previous inequlity, which proves the result. Although the integrl is liner, the upper nd lower integrls of non-integrble functions re not, in generl, liner. Exmple.25. Define f,g : [,] R by { if x [,] Q, f(x) = g(x) = if x [,]\Q, Tht is, f is the Dirichlet function nd g = f. Then so { if x [,] Q, if x [,]\Q. U(f) = U(g) =, L(f) = L(g) =, U(f +g) = L(f +g) =, U(f +g) < U(f)+U(g), L(f +g) > L(f)+L(g). The product of integrble functions is lso integrble, s is the quotient provided it remins bounded. Unlike the integrl of the sum, however, there is no wy to express the integrl of the product fg in terms of f nd g.

18 8. The Riemnn Integrl Theorem.26. If f,g : [,b] R re integrble, then fg : [,b] R is integrble. If, in ddition, g nd /g is bounded, then f/g : [,b] R is integrble. Proof. First, we show tht the squre of n integrble function is integrble. If f is integrble, then f is bounded, with f M for some M. For ll x,y [,b], we hve f 2 (x) f 2 (y) = f(x)+f(y) f(x) f(y) 2M f(x) f(y). Tking the supremum of this inequlity over x,y I [,b] nd using Proposition 2.9, we get tht [ ]. sup(f 2 ) inf I I (f2 ) 2M supf inf f I I mening tht osc(f 2 ) 2M osc f. I I If follows from Proposition.6 tht f 2 is integrble if f is integrble. Since the integrl is liner, we then see from the identity fg = [ (f +g) 2 (f g) 2] 4 tht fg is integrble if f, g re integrble. In similr wy, if g nd /g M, then g(x) g(y) = g(x) g(y) M 2 g(x) g(y). g(x)g(y) Tking the supremum of this eqution over x,y I [,b], we get ( ) ( ) [ ] sup inf M 2 supg inf I g I g g, I I mening tht osc I (/g) M 2 osc I g, nd Proposition.6 implies tht /g is integrble if g is integrble. Therefore f/g = f (/g) is integrble Monotonicity. Next, we prove the monotonicity of the integrl. Theorem.27. Suppose tht f,g : [,b] R re integrble nd f g. Then f Proof. First suppose tht f is integrble. Let P be the prtition consisting of the single intervl [, b]. Then so g. L(f;P) = inf f (b ), [,b] f L(f;P). If f g, then h = f g, nd the linerity of the integrl implies tht f g = h,

19 .6. Properties of the Riemnn integrl 9 which proves the theorem. One immedite consequence of this theorem is the following simple, but useful, estimte for integrls. Theorem.28. Suppose tht f : [,b] R is integrble nd Then M = supf, m = inf f. [,b] [,b] m(b ) f M(b ). Proof. Since m f M on [,b], Theorem.27 implies tht which gives the result. m f M, This estimte lso follows from the definition of the integrl in terms of upper nd lower sums, but once we ve estblished the monotonicity of the integrl, we don t need to go bck to the definition. A further consequence is the intermedite vlue theorem for integrls, which sttes tht continuous function on n intervl is equl to its vergevlue t some point. Theorem.29. If f : [,b] R is continuous, then there exists x [,b] such tht f(x) = f. b Proof. Since f is continuous function on compct intervl, it ttins its mximum vlue M nd its minimum vlue m. From Theorem.28, m b f M. By the intermedite vlue theorem, f tkes on every vlue between m nd M, nd the result follows. As shown in the proof of Theorem.27, given linerity, monotonicity is equivlent to positivity, f if f. We remrk tht even though the upper nd lower integrls ren t liner, they re monotone. Proposition.3. If f,g : [,b] R re bounded functions nd f g, then U(f) U(g), L(f) L(g).

20 2. The Riemnn Integrl Proof. From Proposition 2.2, we hve for every intervl I [,b] tht sup I f supg, inf f inf g. I I I It follows tht for every prtition P of [,b], we hve U(f;P) U(g;P), L(f;P) L(g;P). Tking the infimum of the upper inequlity nd the supremum of the lower inequlity over P, we get U(f) U(g) nd L(f) L(g). We cn estimte the bsolute vlue of n integrl by tking the bsolute vlue under the integrl sign. This is nlogous to the corresponding property of sums: n n n k. k= k= Theorem.3. If f is integrble, then f is integrble nd f f. Proof. First, suppose tht f is integrble. Since f f f, we get from Theorem.27 tht f f f, or f f. To complete the proof, we need to show tht f is integrble if f is integrble. For x,y [,b], the reverse tringle inequlity gives Using Proposition 2.9, we get tht f(x) f(y) f(x) f(y). sup I f inf I f supf inf f, I I mening tht osc I f osc I f. Proposition.6 then implies tht f is integrble if f is integrble. In prticulr, we immeditely get the following bsic estimte for n integrl. Corollry.32. If f : [,b] R is integrble then M = sup f, [,b] f M(b ).

21 .6. Properties of the Riemnn integrl Additivity. Finlly, we prove dditivity. This property refers to dditivity with respect to the intervl of integrtion, rther thn linerity with respect to the function being integrted. Theorem.33. Suppose tht f : [,b] R nd < c < b. Then f is Riemnn integrble on [,b] if nd only if it is Riemnn integrble on [,c] nd [c,b]. In tht cse, f = c Proof. Suppose tht f is integrbleon [,b]. Then, givenǫ >, there is prtition P of [,b] such tht U(f;P) L(f;P) < ǫ. Let P = P {c} be the refinement of P obtined by dding c to the endpoints of P. (If c P, then P = P.) Then P = Q R where Q = P [,c] nd R = P [c,b] re prtitions of [,c] nd [c,b] respectively. Moreover, It follows tht U(f;P ) = U(f;Q)+U(f;R), f + c f. L(f;P ) = L(f;Q)+L(f;R). U(f;Q) L(f;Q) = U(f;P ) L(f;P ) [U(f;R) L(f;R)] U(f;P) L(f;P) < ǫ, which proves tht f is integrble on [,c]. Exchnging Q nd R, we get the proof for [c,b]. Conversely, if f is integrble on [,c] nd [c,b], then there re prtitions Q of [,c] nd R of [c,b] such tht U(f;Q) L(f;Q) < ǫ 2, U(f;R) L(f;R) < ǫ 2. Let P = Q R. Then U(f;P) L(f;P) = U(f;Q) L(f;Q)+U(f;R) L(f;R) < ǫ, which proves tht f is integrble on [,b]. Finlly, with the prtitions P, Q, R s bove, we hve Similrly, f U(f;P) = U(f;Q)+U(f;R) < L(f;Q)+L(f;R)+ǫ < c f + c f +ǫ. f L(f;P) = L(f;Q)+L(f;R) > U(f;Q)+U(f;R) ǫ > c f + c f ǫ. Since ǫ > is rbitrry, we see tht f = c f + c f.

22 22. The Riemnn Integrl We cn extend the dditivity property of the integrl by defining n oriented Riemnn integrl. Definition.34. If f : [,b] R is integrble, where < b, nd c b, then f = b f, c c f =. With this definition, the dditivity property in Theorem.33 holds for ll,b,c R for which the oriented integrls exist. Moreover, if f M, then the estimte in Corollry.32 becomes f M b for ll,b R (even if b). The oriented Riemnn integrl is specil cse of the integrl of differentil form. It ssigns vlue to the integrl of one-form f dx on n oriented intervl..7. Further existence results for the Riemnn integrl In this section, we prove severl further useful conditions for the existences of the Riemnn integrl. First, we show tht chnging the vlues of function t finitely mny points doesn t chnge its integrbility of the vlue of its integrl. Proposition.35. Suppose tht f,g : [,b] R nd f(x) = g(x) except t finitely mny points x [,b]. Then f is integrble if nd only if g is integrble, nd in tht cse f = Proof. It is sufficient to prove the result for functions whose vlues differ t single point, sy c [,b]. The generl result then follows by induction. Since f, g differ t single point, f is bounded if nd only if g is bounded. If f, g re unbounded, then neither one is integrble. If f, g re bounded, we will show tht f, g hve the sme upper nd lower integrls becuse their upper nd lower sums differ by n rbitrrily smll mount with respect to prtition tht is sufficiently refined ner the point where the functions differ. Suppose tht f, g re bounded with f, g M on [,b] for some M >. Let ǫ >. Choose prtition P of [,b] such tht g. U(f;P) < U(f)+ ǫ 2. Let Q = {I,...,I n } be refinement of P such tht < δ for k =,...,n, where δ = ǫ 8M.

23 .7. Further existence results for the Riemnn integrl 23 Then g differs from f on t most two intervls in Q. (There could be two intervls if c is n endpoint of the prtition.) On such n intervl we hve sup g supf sup g +sup f 2M, nd on the remining intervls, sup Ik g sup Ik f =. It follows tht U(g;Q) U(f;Q) < 2M 2δ < ǫ 2. Using the properties of upper integrls nd refinements, we obtin U(g) U(g;Q) < U(f;Q)+ ǫ 2 U(f;P)+ ǫ 2 < U(f)+ǫ. Since this inequlity holds for rbitrry ǫ >, we get tht U(g) U(f). Exchnging f nd g, we see similrly tht U(f) U(g), so U(f) = U(g). An nlogous rgument for lower sums (or n ppliction of the result for upper sums to f, g) shows tht L(f) = L(g). Thus U(f) = L(f) if nd only if U(g) = L(g), in which cse f = g. Exmple.36. The function f in Exmple.6 differs from the -function t one point. It is integrble nd its integrl is equl to. The conclusion of Proposition.35 cn fil if the functions differ t countbly infinite number of points. One reson is tht we cn turn bounded function into n unbounded function by chnging its vlues t n infinite number of points. Exmple.37. Define f : [,] R by { n if x = /n for n N, f(x) = otherwise Then f is equl to the -function except on the countbly infinite set {/n : n N}, but f is unbounded nd therefore it s not Riemnn integrble. The result is still flse, however, for bounded functions tht differ t countbly infinite number of points. Exmple.38. The Dirichlet function in Exmple.7 is bounded nd differs from the -function on the countbly infinite set of rtionls, but it isn t Riemnn integrble. The Lebesgue integrl is better behved thn the Riemnn intgerl in this respect: two functions tht re equl lmost everywhere, mening tht they differ on set of Lebesgue mesure zero, hve the sme Lebesgue integrls. In prticulr, two functions tht differ on countble set re equl lmost everywhere (see Section.2). The next proposition llows us to deduce the integrbility of bounded function on n intervl from its integrbility on slightly smller intervls.

24 24. The Riemnn Integrl Proposition.39. Suppose tht f : [,b] R is bounded nd integrble on [,r] for every < r < b. Then f is integrble on [,b] nd f = lim r b Proof. Since f is bounded, f M on [,b] for some M >. Given ǫ >, let r = b ǫ 4M (where we ssume ǫ is sufficiently smll tht r > ). Since f is integrble on [,r], there is prtition Q of [,r] such tht r f. U(f;Q) L(f;Q) < ǫ 2. ThenP = Q {b}isprtitionof[,b]whoselstintervlis[r,b]. The boundedness of f implies tht supf inf f 2M. [r,b] [r,b] Therefore ( U(f;P) L(f;P) = U(f;Q) L(f;Q)+ sup [r,b] ) f inf f (b r) [r,b] < ǫ +2M (b r) = ǫ, 2 so f is integrble on [,b] by Theorem.4. Moreover, using the dditivity of the integrl, we get r f f = f M (b r) s r b. r An obvious nlogous result holds for the left endpoint. Exmple.4. Define f : [,] R by { sin(/x) if < x, f(x) = if x =. Then f is bounded on [, ]. Furthemore, f is continuous nd therefore integrble on [r,] for every < r <. It follows from Proposition.39 tht f is integrble on [,]. The ssumption in Proposition.39 tht f is bounded on [,b] is essentil. Exmple.4. The function f : [,] R defined by { /x for < x, f(x) = for x =, is continuous nd therefore integrble on [r,] for every < r <, but it s unbounded nd therefore not integrble on [, ].

25 .7. Further existence results for the Riemnn integrl 25.5 y x Figure 3. Grph of the Riemnn integrble function y = sin(/sinx) in Exmple.43. As corollry of this result nd the dditivity of the integrl, we prove generliztion of the integrbility of continuous functions to piecewise continuous functions. Theorem.42. If f : [,b] R is bounded function with finitely mny discontinuities, then f is Riemnn integrble. Proof. By splitting the intervl into subintervls with the discontinuities of f t n endpoint nd using Theorem.33, we see tht it is sufficient to prove the result if f is discontinuous only t one endpoint of [,b], sy t b. In tht cse, f is continuous nd therefore integrble on ny smller intervl [,r] with < r < b, nd Proposition.39 implies tht f is integrble on [, b]. Exmple.43. Define f : [,2π] R by { sin(/sinx) if x,π,2π, f(x) = if x =,π,2π. Thenf isboundedndcontinuousexcepttx =,π,2π,soitisintegrbleon[,2π] (see Figure 3). This function doesn t hve jump discontinuities, but Theorem.42 still pplies. Exmple.44. Define f : [,/π] R by { sgn[sin(/x)] if x /nπ for n N, f(x) = if x = or x /nπ for n N,

26 26. The Riemnn Integrl Figure 4. Grph of the Riemnn integrble function y = sgn(sin(/x)) in Exmple.44. where sgn is the sign function, if x >, sgnx = if x =, if x <. Then f oscilltes between nd countbly infinite number of times s x + (see Figure 4). It hs jump discontinuities t x = /(nπ) nd n essentil discontinuity t x =. Nevertheless, it is Riemnn integrble. To see this, note tht f is bounded on [, ] nd piecewise continuous with finitely mny discontinuities on [r,] for every < r <. Theorem.42 implies tht f is Riemnn integrble on [r,], nd then Theorem.39 implies tht f is integrble on [,]..8. The fundmentl theorem of clculus In the integrl clculus I find much less interesting the prts tht involve only substitutions, trnsformtions, nd the like, in short, the prts tht involve the known skillfully pplied mechnics of reducing integrls to lgebric, logrithmic, nd circulr functions, thn I find the creful nd profound study of trnscendentl functions tht cnnot be reduced to these functions. (Guss, 88) The fundmentl theorem of clculus sttes tht differentition nd integrtion re inverse opertions in n ppropritely understood sense. The theorem hs two prts: in one direction, it sys roughly tht the integrl of the derivtive is the originl function; in the other direction, it sys tht the derivtive of the integrl is the originl function.

27 .8. The fundmentl theorem of clculus 27 In more detil, the first prt sttes tht if F : [,b] R is differentible with integrble derivtive, then F (x)dx = F(b) F(). This result cn be thought of s continuous nlog of the corresponding identity for sums of differences, n (A k A k ) = A n A. k= The second prt sttes tht if f : [,b] R is continuous, then d dx x f(t)dt = f(x). This is continuous nlog of the corresponding identity for differences of sums, k k j j = k. j= j= The proof of the fundmentl theorem consists essentilly of pplying the identities for sums or differences to the pproprite Riemnn sums or difference quotients nd proving, under pproprite hypotheses, tht they converge to the corresponding integrls or derivtives. We ll split the sttement nd proof of the fundmentl theorem into two prts. (The numbering of the prts s I nd II is rbitrry.).8.. Fundmentl theorem I. First we prove the sttement bout the integrl of derivtive. Theorem.45 (FundmentltheoremofclculusI). IfF : [,b] Riscontinuous on [,b] nd differentible in (,b) with F = f where f : [,b] R is Riemnn integrble, then Proof. Let be prtition of [,b]. Then f(x)dx = F(b) F(). P = { = x,x,x 2,...,x n,x n = b} F(b) F() = n [F(x k ) F(x k )]. k= The function F is continuous on the closed intervl [x k,x k ] nd differentible in the open intervl (x k,x k ) with F = f. By the men vlue theorem, there exists x k < c k < x k such tht F(x k ) F(x k ) = f(c k )(x k x k ). Since f is Riemnn integrble, it is bounded, nd it follows tht m k (x k x k ) F(x k ) F(x k ) M k (x k x k ),

28 28. The Riemnn Integrl where M k = sup f, m k = inf f. [x k,x k ] [x k,x k ] Hence, L(f;P) F(b) F() U(f;P) for every prtition P of [,b], which implies tht L(f) F(b) F() U(f). Since f is integrble, L(f) = U(f) nd F(b) F() = f. In Theorem.45, we ssume tht F is continuous on the closed intervl [,b] nd differentible in the open intervl (, b) where its usul two-sided derivtive is defined nd is equl to f. It isn t necessry to ssume the existence of the right derivtive of F t or the left derivtive t b, so the vlues of f t the endpoints re rbitrry. By Proposition.35, however, the integrbility of f on [, b] nd the vlue of its integrl do not depend on these vlues, so the sttement of the theorem mkes sense. As result, we ll sometimes buse terminology, nd sy tht F is integrble on [,b] even if it s only defined on (,b). Theorem.45 imposes the integrbility of F s hypothesis. Every function F tht is continuously differentible on the closed intervl [, b] stisfies this condition, but the theorem remins true even if F is discontinuous, Riemnn integrble function. Exmple.46. Define F : [,] R by { x 2 sin(/x) if < x, F(x) = if x =. Then F is continuous on [, ] nd, by the product nd chin rules, differentible in (,]. It is lso differentible but not continuously differentible t, with F ( + ) =. Thus, { F cos(/x)+2xsin(/x) if < x, (x) = if x =. The derivtive F is bounded on [,] nd discontinuous only t one point (x = ), so Theorem.42 implies tht F is integrble on [,]. This verifies ll of the hypotheses in Theorem.45, nd we conclude tht F (x)dx = sin. There re, however, differentible functions whose derivtives re unbounded or so discontinuous tht they ren t Riemnn integrble. Exmple.47. Define F : [,] R by F(x) = x. Then F is continuous on [,] nd differentible in (,], with F (x) = 2 x for < x. This function is unbounded, so F is not Riemnn integrble on [,], however we define its vlue t, nd Theorem.45 does not pply.

29 .8. The fundmentl theorem of clculus 29 We cn, however, interpret the integrl of F on [,] s n improper Riemnn integrl. The function F is continuously differentible on [ǫ,] for every < ǫ <, so 2 x dx = ǫ. Thus, we get the improper integrl ǫ lim ǫ + ǫ 2 dx =. x The construction of function with bounded, non-integrble derivtive is more involved. It s not sufficient to give function with bounded derivtive tht is discontinuous t finitely mny points, s in Exmple.46, becuse such function is Riemnn integrble. Rther, one hs to construct differentible function whose derivtive is discontinuous on set of nonzero Lebesgue mesure; we won t give n exmple here. Finlly, we remrk tht Theorem.45 remins vlid for the oriented Riemnn integrl, since exchnging nd b reverses the sign of both sides Fundmentl theorem of clculus II. Next, we prove the other direction of the fundmentl theorem. We will use the following result, of independent interest, which sttes tht the verge of continuous function on n intervl pproches the vlue of the function s the length of the intervl shrinks to zero. The proof uses common trick of tking constnt inside n verge. Theorem.48. Suppose tht f : [,b] R is integrble on [,b] nd continuous t. Then +h lim f(x)dx = f(). h + h Proof. If k is constnt, we hve k = h +h kdx. (Tht is, the vergeof constnt is equl to the constnt.) We cn therefore write h +h f(x)dx f() = h +h [f(x) f()] dx. Let ǫ >. Since f is continuous t, there exists δ > such tht f(x) f() < ǫ for x < +δ. It follows tht if < h < δ, then +h f(x)dx f() h h sup +h f(x) f() h ǫ, which proves the result.

30 3. The Riemnn Integrl A similr proof shows tht if f is continuous t b, then b lim f = f(b), h + h b h nd if f is continuous t < c < b, then c+h lim f = f(c). h + 2h c h More generlly, if {I h : h > } is ny collection of intervls with c I h nd I h s h +, then lim f = f(c). h + I h I h The ssumption in Theorem.48 tht f is continuous t the point bout which we tke the verges is essentil. Exmple.49. Let f : R R be the sign function if x >, f(x) = if x =, if x <. Then h lim f(x)dx =, lim f(x)dx =, h + h h + h h nd neither limit is equl to f(). In this exmple, the limit of the symmetric verges h lim f(x)dx = h + 2h h is equl to f(), but this equlity doesn t hold if we chnge f() to nonzero vlue, since the limit of the symmetric verges is still. The second prt of the fundmentl theorem follows from this result nd the fct tht the difference quotients of F re verges of f. Theorem.5 (Fundmentl theorem of clculus II). Suppose tht f : [,b] R is integrble nd F : [,b] R is defined by F(x) = x f(t)dt. Then F is continuous on [,b]. Moreover, if f is continuous t c b, then F is differentible t c nd F (c) = f(c). Proof. First, note tht Theorem.33 implies tht f is integrble on [,x] for every x b, so F is well-defined. Since f is Riemnn integrble, it is bounded, nd f M for some M. It follows tht x+h F(x+h) F(x) = f(t)dt M h, which shows tht F is continuous on [, b] (in fct, Lipschitz continuous). x

31 .8. The fundmentl theorem of clculus 3 Moreover, we hve F(c+h) F(c) h = h c+h c f(t)dt. It follows from Theorem.48 tht if f is continuous t c, then F is differentible t c with [ ] F(c+h) F(c) F c+h (c) = lim = lim f(t)dt = f(c), h h h h where we use the pproprite right or left limit t n endpoint. The ssumption tht f is continuous is needed to ensure tht F is differentible. Exmple.5. If then F(x) = f(x) = x { for x, for x <, f(t)dt = c { x for x, for x <. The function F is continuous but not differentible t x =, where f is discontinuous, since the left nd right derivtives of F t, given by F ( ) = nd F ( + ) =, re different Consequences of the fundmentl theorem. The first prt of the fundmentl theorem, Theorem.45, is the bsic computtionl tool in integrtion. It llows us to compute the integrl of of function f if we cn find n ntiderivtive; tht is, function F such tht F = f. There is no systemtic procedure for finding ntiderivtives. Moreover, even if one exists, n ntiderivtive of n elementry function (constructed from power, trigonometric, nd exponentil functions nd their inverses) my not be nd often isn t expressible in terms of elementry functions. Exmple.52. For p =,,2,..., we hve [ ] d dx p+ xp+ = x p, nd it follows tht x p dx = p+. We remrk tht once we hve the fundmentl theorem, we cn use the definition of the integrl bckwrds to evlute limit such s [ ] n lim n n p+ k p = p+, k= since the sum is the upper sum of x p on prtition of [,] into n intervls of equl length. Exmple.8 illustrtes this result explicitly for p = 2.

32 32. The Riemnn Integrl Two importnt generl consequences of the first prt of the fundmentl theorem re integrtion by prts nd substitution (or chnge of vrible), which come from inverting the product rule nd chin rule for derivtives, respectively. Theorem.53 (Integrtion by prts). Suppose tht f,g : [,b] R re continuous on [,b] nd differentible in (,b), nd f, g re integrble on [,b]. Then fg dx = f(b)g(b) f()g() f gdx. Proof. The function fg is continuous on [,b] nd, by the product rule, differentible in (, b) with derivtive (fg) = fg +f g. Since f, g, f, g re integrble on [,b], Theorem.26 implies tht fg, f g, nd (fg), re integrble. From Theorem.45, we get tht fg dx+ which proves the result. f gdx = f gdx = f(b)g(b) f()g(), Integrtion by prts sys tht we cn move derivtive from one fctor in n integrl onto the other fctor, with chnge of sign nd the ppernce of boundry term. The product rule for derivtives expresses the derivtive of product in terms of the derivtives of the fctors. By contrst, integrtion by prts doesn t give n explicit expression for the integrl of product, it simply replces one integrl by nother. This cn sometimes be used to simplify n integrl nd evlute it, but the importnce of integrtion by prts goes fr beyond its use s n integrtion technique. Exmple.54. For n =,,2,3,..., let I n (x) = x t n e t dt. If n, integrtion by prts with f(t) = t n nd g (t) = e t gives I n (x) = x n e x +n x Also, by the fundmentl theorem, I (x) = It then follows by induction tht where, s usul,! =. x I n (x) = n! t n e t dt = x n e x +ni n (x). [ e t dt = e x. e x n k= ] x k, k! Since x k e x s x for every k =,,2,..., we get the improper integrl t n e t dt = lim r r t n e t dt = n!.

33 .8. The fundmentl theorem of clculus 33 This formul suggests n extension of the fctoril function to complex numbers z C, clled the Gmm function, which is defined for Rz > by the improper, complex-vlued integrl Γ(z) = t z e t dt. In prticulr, Γ(n) = (n )! for n N. The Gm function is n importnt specil function, which is studied further in complex nlysis. Next we consider the chnge of vrible formul for integrls. Theorem.55 (Chnge of vrible). Suppose tht g : I R differentible on n open intervl I nd g is integrble on I. Let J = g(i). If f : J R continuous, then for every,b I, Proof. Let f (g(x))g (x)dx = F(x) = x g(b) g() f(u)du. f(u)du. Since f is continuous, Theorem.5 implies tht F is differentible in J with F = f. The chin rule implies tht the composition F g : I R is differentible in I, with (F g) (x) = f (g(x))g (x). This derivtive is integrble on [,b] since f g is continuous nd g is integrble. Theorem.45, the definition of F, nd the dditivity of the integrl then imply tht which proves the result. f (g(x))g (x)dx = (F g) dx = F (g(b)) F (g()) = g(b) g() F (u)du, A continuous function mps n intervl to n intervl, nd it is one-to-one if nd only if it is strictly monotone. An incresing function preserves the orienttion of the intervl, while decresing function reverses it, in which cse the integrls re understood s pproprite oriented integrls. There is no ssumption in this theorem tht g is invertible, nd the result remins vlid if g is not monotone. Exmple.56. For every >, the incresing, differentible function g : R R defined by g(x) = x 3 mps (,) one-to-one nd onto ( 3, 3 ) nd preserves orienttion. Thus, if f : [,] R is continuous, f(x 3 ) 3x 2 dx = 3 3 f(u)du.

34 34. The Riemnn Integrl.5.5 y x Figure 5. Grphs of the error function y = F(x) (blue) nd its derivtive, the Gussin function y = f(x) (green), from Exmple.58. The decresing, differentible function g : R R defined by g(x) = x 3 mps (,) one-to-one nd onto ( 3, 3 ) nd reverses orienttion. Thus, f( x 3 ) ( 3x 2 )dx = 3 3 f(u)du = 3 3 f(u)du. The non-monotone, differentible function g : R R defined by g(x) = x 2 mps (,) onto [, 2 ). It is two-to-one, except t x =. The chnge of vribles formul gives f(x 2 ) 2xdx = f(u)du =. 2 The contributions to the originl integrl from [,] nd [,] cncel since the integrnd is n odd function of x. One consequence of the second prt of the fundmentl theorem, Theorem.5, is tht every continuous function hs n ntiderivtive, even if it cn t be expressed explicitly in terms of elementry functions. This provides wy to define trnscendentl functions s integrls of elementry functions. Exmple.57. One wy to define the logrithm ln : (, ) R in terms of lgebric functions is s the integrl lnx = x 2 t dt. The integrl is well-defined for every < x < since /t is continuous on the intervl [,x] (or [x,] if < x < ). The usul properties of the logrithm follow from this representtion. We hve (lnx) = /x by definition, nd, for exmple, mking the substitution s = xt in the second integrl in the following eqution,