[PDF] Chapter 5. Integration §1. The Riemann Integral Let a and b be two

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Chapter 5. Integration

§1. The Riemann Integral

Letaandbbe two real numbers witha < b. Then [a;b] is a closed and bounded interval in IR. By apartitionPof [a;b] we mean a finite ordered set{t0;t1;:::;tn}such that a=t0< t1<···< tn=b: ThenormofPis defined by∥P∥:= max{ti-ti-1:i= 1;2;:::;n}. Supposefis a bounded real-valued function on [a;b]. Given a partition{t0;t1;:::;tn} of [a;b], for eachi= 1;2;:::;n, let m Theupper sumU(f;P) and thelower sumL(f;P) for the functionfand the partition

Pare defined by

U(f;P) :=n∑

i=1M i(ti-ti-1) andL(f;P) :=n∑ i=1m i(ti-ti-1):

Theupper integralU(f) offover [a;b] is defined by

U(f) := inf{U(f;P) :Pis a partition of [a;b]}

and thelower integralL(f) offover [a;b] is defined by

L(f) := sup{L(f;P) :Pis a partition of [a;b]}:

A bounded functionfon [a;b] is said to be (Riemann)integrableifL(f) =U(f). In this case, we write∫b a f(x)dx=L(f) =U(f):

By convention we define

a b f(x)dx:=-∫ b a f(x)dxand∫ a a f(x)dx:= 0: A constant function on [a;b] is integrable. Indeed, iff(x) =cfor allx∈[a;b], then L(f;P) =c(b-a) andU(f;P) =c(b-a) for any partitionPof [a;b]. It follows that b a cdx=c(b-a): 1 Suppose thatP={t0;t1;:::;tn}is a partition of [a;b], and thatP1is a partition obtained fromPby adding one more pointt∗∈(ti-1;ti) for somei. The lower sums forPandP1 m L(f;P1)-L(f;P) =m′(t∗-ti-1) +m′′(ti-t∗)-mi(ti-ti-1): m m

Consequently,

Now suppose thatPNis a mesh obtained fromPby addingNpoints. An induction argument shows that

Similarly we have

By the definition ofL(f) andU(f), for eachn∈IN there exist partitionsPandQof [a;b] such that Consider the partitionP∪Qof [a;b]. SinceP⊆P∪QandQ⊆P∪Q, by (1) and (2) we get We are in a position to establish the following criterion for a bounded function to be integrable. 2 Theorem 1.1.A bounded functionfon[a;b]is integrable if and only if for each" >0 there exists a partitionPof[a;b]such that

U(f;P)-L(f;P)< ":

Proof.Suppose thatfis integrable on [a;b]. For" >0, there exist partitionsP1andP2 such that

L(f;P1)> L(f)-"

2 andU(f;P2)< U(f) +" 2

ForP:=P1∪P2we have

L(f)-"

2 2

SinceL(f) =U(f), it follows thatU(f;P)-L(f;P)< ".

Conversely, suppose that for each" >0 there exists a partitionPof [a;b] such that U(f;P)-L(f;P)< ". ThenU(f;P)< L(f;P) +". It follows that that is,fis integrable. Letfbe a bounded real-valued function on [a;b] and letP={t0;t1;:::;tn}be a partition of [a;b]. For eachi= 1;2;:::;n, choosei∈[xi-1;xi]. The sum n i=1f(i)(ti-ti-1) is called aRiemann sumoffwith respect to the partitionPand points{1;:::;n}. Theorem 1.2.Letfbe a bounded real-valued function on[a;b]. Thenfis integrable on [a;b]if and only if there exists a real numberIwith the following property: For any" >0 there exists some >0such that n wheneverP={t0;t1;:::;tn}is a partition of[a;b]with∥P∥< andi∈[ti-1;ti]for i= 1;2;:::;n. If this is the case, then b a f(x)dx=I: 3 Proof.Let"be an arbitrary positive number. Suppose that (3) is true for some partition P={t0;t1;:::;tn}of [a;b] and pointsi∈[ti-1;ti],i= 1;2;:::;n. Then

L(f;P) = inf{

n∑ i=1f(i)(ti-ti-1) :i∈[xi-1;xi] fori= 1;2;:::;n} ≥I-" and

U(f;P) = sup{

n∑ i=1f(i)(ti-ti-1) :i∈[xi-1;xi] fori= 1;2;:::;n} on [a;b]. Moreover,L(f) =U(f) =I. Conversely, suppose thatfis integrable on [a;b]. LetM:= sup{|f(x)|:x∈[a;b]} andI:=L(f) =U(f). Given an arbitrary" >0, there exists a partitionQof [a;b] such thatL(f;Q)> I-"=2 andU(f;Q)< I+"=2. Suppose thatQhasNpoints. Let P={t0;t1;:::;tn}be a partition of [a;b] with∥P∥< . Consider the partitionP∪Qof [a;b]. By (1) and (2) we have :="=(4MN). Since∥P∥< , we deduce from the foregoing inequalities that Thus, withi∈[ti-1;ti] fori= 1;2;:::;nwe obtain

This completes the proof.

Theorem 1.3.Letfbe a bounded function from a bounded closed interval[a;b]toIR. If the set of discontinuities offis finite, thenfis integrable on[a;b]. Proof.LetDbe the set of discontinuities off. By our assumption,Dis finite. So the setD∪ {a;b}can be expressed as{d0;d1;:::;dN}witha=d0< d1<···< dN=b. Let M:= sup{|f(x)|:x∈[a;b]}. For an arbitrary positive number", we choose >0 such 4 that < "=(8MN) and <(dj-dj-1)=3 for allj= 1;:::;N. Forj= 0;1;:::;N, let x j:=dj-andyj:=dj+. Then we have a=d0< y0< x1< d1< y1<···< xN< dN=b: LetEbe the union of the intervals [d0;y0], [x1;d1];[d1;y1];:::;[xN-1;dN-1];[dN-1;yN-1], and [xN;dN]. There are 2Nintervals in total. Forj= 1;:::;N, letFj:= [yj-1;xj]. Further, letF:=∪Nj=1Fj. The functionfis continuous onF, which is a finite union of bounded closed intervals. Hencefis uniformly continuous onF. There exists some >0 such that|f(x)-f(y)|< "=(2(b-a)) wheneverx;y∈Fsatisfying|x-y|< . For each j∈ {1;:::;N}, letPjbe a partition ofFjsuch that∥Pj∥< . Let

P:={a;b} ∪D∪(∪Nj=1Pj):

The setPcan be arranged as{t0;t1;:::;tn}witha=t0< t1<···< tn=b. Consider

U(f;P)-L(f;P) =n∑

i=1(Mi-mi)(ti-ti-1); [ti-1;ti] is either contained inEor inF, but not in both. Hence n i=1(Mi-mi)(ti-ti-1) =∑ [ti1;ti]⊆E(Mi-mi)(ti-ti-1) +∑ [ti1;ti]⊆F(Mi-mi)(ti-ti-1): There are 2Nintervals [ti-1;ti] contained inE. Each interval has length < "=(8MN). 2 If [ti-1;ti]⊆F, thenti-ti-1< ; henceMi-mi< "=(2(b-a)). Therefore,

2(b-a)∑

[ti1;ti]⊆F(ti-ti-1)<"

2(b-a)(b-a) ="

2 From the above estimates we conclude thatU(f;P)-L(f;P)< ". By Theorem 1.1, the functionfis integrable on [a;b]. Example 1.Let [a;b] be a closed interval witha < b, and letfbe the function on [a;b] given byf(x) =x. By Theorem 1.3,fis integrable on [a;b]. LetP={t0;t1;:::;tn}be a partition of [a;b] and choosei:= (ti-1+ti)=2∈[ti-1;ti] fori= 1;2;:::;n. Then n i=1f(i)(ti-ti-1) =1 2 n i=1(ti+ti-1)(ti-ti-1) =1 2 n i=1(t2i-t2i-1) =1 2 (t2n-t20) =1 2 (b2-a2): 5

By Theorem 1.2 we have∫b

a xdx=1 2 (b2-a2): More generally, for a positive integerk, letfkbe the function given byfk(x) =xkfor x∈[a;b]. Choose i:=(tki-1+tk-1 i-1ti+···+tki k+ 1) 1=k ; i= 1;2;:::;n: n i=1f k(i)(ti-ti-1) =1 k+ 1n i=1(tk+1 i-tk+1 i-1) =1 k+ 1(tk+1n-tk+10) =1 k+ 1(bk+1-ak+1):

By Theorem 1.2 we conclude that

b a xkdx=1 k+ 1(bk+1-ak+1): andg(0) := 0. The only discontinuity point ofgis 0. By Theorem 1.3,gis integrable on [0;1]. Note thatgis not uniformly continuous on (0;1). Indeed, letxn:= 1=(2n) and y n:= 1=(2n+=2) forn∈IN. Then limn→∞(xn-yn) = 0. But |f(xn)-f(yn)|=|cos(2n)-cos(2n+=2)|= 1∀n∈IN: Hencegis not uniformly continuous on (0;1). On the other hand, the functionugiven continuous on (0;1]. Theorem 1.3 is not applicable tou, becauseuis unbounded. Example 3.Lethbe the function on [0;1] defined byh(x) := 1 ifxis a rational number in [0;1] andh(x) := 0 ifxis an irrational number in [0;1]. LetP={t0;t1;:::;tn}be a partition of [0;1]. Fori= 1;:::;nwe have m i:= inf{h(x) :x∈[ti-1;ti]}= 0 andMi:= sup{h(x) :x∈[ti-1;ti]}= 1: HenceL(h;P) = 0 andU(h;P) = 1 for every partitionPof [0;1]. Consequently,L(h) = 0 andU(h) = 1. This shows thathis not Riemann integrable on [0;1]. 6

§2. Properties of the Riemann Integral

In this section we establish some basic properties of the Riemann integral. Theorem 2.1.Letfandgbe integrable functions from a bounded closed interval[a;b] toIR. Then (1)For any real numberc,cfis integrable on[a;b]and∫b a(cf)(x)dx=c∫b af(x)dx; (2)f+gis integrable on[a;b]and∫b a(f+g)(x)dx=∫b af(x)dx+∫b ag(x)dx. Proof.Suppose thatfandgare integrable functions on [a;b]. WriteI(f) :=∫b af(x)dx andI(g) :=∫b ag(x)dx. Let"be an arbitrary positive number. By Theorem 1.2, there exists some >0 such that n wheneverP={t0;t1;:::;tn}is a partition of [a;b] with∥P∥< andi∈[ti-1;ti] for i= 1;2;:::;n. It follows that n i=1(cf)(i)(ti-ti-1)-cI(f)=|c|n

Hencecfis integrable on [a;b] and∫b

a(cf)(x)dx=c∫b af(x)dx. Moreover, n i=1(f+g)(i)(ti-ti-1)-[I(f) +I(g)] n i=1f(i)(ti-ti-1)-I(f)+n

Thereforef+gis integrable on [a;b] and∫b

a(f+g)(x)dx=∫b af(x)dx+∫b ag(x)dx. Theorem 2.2.Letfandgbe integrable functions on[a;b]. Thenfgis an integrable function on[a;b]. Proof.Let us first show thatf2is integrable on [a;b]. Sincefis bounded, there exists for any partitionPof [a;b]. Let" >0. Sincefis integrable on [a;b], by Theorem 1.1 7 there exists a partitionPof [a;b] such thatU(f;P)-L(f;P)< "=(2M). Consequently, U(f2;P)-L(f2;P)< ". By Theorem 1.1 again we conclude thatf2is integrable on [a;b]. Note thatfg= [(f+g)2-(f-g)2]=4. By Theorem 2.1,f+gandf-gare integrable on [a;b]. By what has been proved, both (f+g)2and (f-g)2are integrable on [a;b]. Using Theorem 2.1 again, we conclude thatfgis integrable on [a;b]. functionfis integrable on[a;b], thenf|[c;d]is integrable on[c;d]. Proof.Suppose thatfis integrable on [a;b]. Let"be an arbitrary positive number. By Theorem 1.1, there exists a partitionPof [a;b] such thatU(f;P)-L(f;P)< ". It follows thatU(f;P∪ {c;d})-L(f;P∪ {c;d})< ". LetQ:= (P∪ {c;d})∩[c;d]. ThenQis a partition of [c;d]. We have

Hencef|[c;d]is integrable on [c;d].

Theorem 2.4.Letfbe a bounded real-valued function on[a;b]. Ifa < c < b, and iffis integrable on[a;c]and[c;b], thenfis integrable on[a;b]and b a f(x)dx=∫ c a f(x)dx+∫ b c f(x)dx: Proof.Suppose thatfis integrable on [a;c] and [c;b]. We writeI1:=∫c af(x)dxand I

2:=∫b

cf(x)dx. Let" >0. By Theorem 1.1, there exist a partitionP1={s0;s1;:::;sm} of [a;c] and a partitionP2={t0;t1;:::;tn}of [c;b] such that

U(f;P1)-L(f;P1)<"

2 andU(f;P2)-L(f;P2)<" 2 LetP:=P1∪P2={s0;:::;sm-1;t0;:::;tn}. ThenPis a partition of [a;b]. We have L(f)≥L(f;P) =L(f;P1) +L(f;P2)> U(f;P1) +U(f;P2)-"≥I1+I2-" and

It follows that

c a f(x)dx+∫ b c c a f(x)dx+∫ b c f(x)dx+": 8 Since the above inequalities are valid for all" >0, we conclude thatfis integrable on [a;b] and∫b af(x)dx=∫c af(x)dx+∫b cf(x)dx. Leta;b;cbe real numbers in any order, and letJbe a bounded closed interval con- taininga,b, andc. Iffis integrable onJ, then by Theorems 2.3 and 2.4 we have∫b a f(x)dx=∫ c a f(x)dx+∫ b c f(x)dx: then∫b ag(x)dx. Proof.By Theorem 2.1,h:=g-fis integrable on [a;b]. Sinceh(x)≥0 for allx∈[a;b], it is clear thatL(h;P)≥0 for any partitionPof [a;b]. Hence,∫b ah(x)dx=L(h)≥0.

Applying Theorem 2.1 again, we see that∫b

a g(x)dx-∫ b a f(x)dx=∫ b a h(x)dx≥0: Theorem 2.6.Iffis an integrable function on[a;b], then|f|is integrable on[a;b]and∫ b a b a |f(x)|dx: Proof.LetP={t0;t1;:::;tn}be a partition of [a;b]. For eachi∈ {1;:::;n}, letMiand m idenote the supremum and infimun respectively offon [ti-1;ti], and letM∗iandm∗i denote the supremum and infimun respectively of|f|on [ti-1;ti]. Then M i-mi= sup{f(x)-f(y) :x;y∈[ti-1;ti]} and M ∗i-m∗i= sup{|f(x)| - |f(y)|:x;y∈[ti-1;ti]}:

By the triangle inequality,

and n∑ i=1(Mi-mi)(ti-ti-1): number. By our assumption,fis integrable on [a;b]. By Theorem 1.1, there exists a partitionPsuch thatU(f;P)-L(f;P)< ". HenceU(|f|;P)-L(|f|;P)< ". By using Theorem 1.1 again we conclude that|f|is integrable on [a;b]. Furthermore, since a b a |f(x)|dxand-∫ b a b a |f(x)|dx:

Therefore

∫b a|f(x)|dx. 9

§3. Fundamental Theorem of Calculus

In this section we give two versions of the Fundamental Theorem of Calculus and their applications. Letfbe a real-valued function on an intervalI. A functionFonIis called an antiderivativeoffonIifF′(x) =f(x) for allx∈I. IfFis an antiderivative off, then so isF+Cfor any constantC. Conversely, ifFandGare antiderivatives offonI, then G ′(x)-F′(x) = 0 for allx∈I. Thus, there exists a constantCsuch thatG(x)-F(x) =C for allx∈I. Consequently,G=F+C. The following is the first version of the Fundamental Theorem of Calculus. Theorem 3.1.Letfbe an integrable function on[a;b]. IfFis a continuous function on [a;b]and ifFis an antiderivative offon(a;b), then b a f(x)dx=F(x)b a :=F(b)-F(a): Proof.Let" >0. By Theorem 1.1, there exists a partitionP={t0;t1;:::;tn}of [a;b] such thatU(f;P)-L(f;P)< ". Sincet0=aandtn=bwe have

F(b)-F(a) =n∑

i=1[F(ti)-F(ti-1)]: By the Mean Value Theorem, for eachi∈ {1;:::;n}there existsxi∈(ti-1;ti) such that F(ti)-F(ti-1) =F′(xi)(ti-ti-1) =f(xi)(ti-ti-1):

Consequently,

On the other hand,

b a

Thus bothF(b)-F(a) and∫b

af(x)dxlie in [L(f;P);U(f;P)] withU(f;P)-L(f;P)< ". Hence [F(b)-F(a)]-∫ b a f(x)dx< ": Since the above inequality is valid for all" >0, we obtain∫b af(x)dx=F(b)-F(a). 10

Example 1.Letkbe a positive integer. Find∫b

axkdx. Solution. We know that the functiongk:x7→xk+1=(k+ 1) is an antiderivative of the functionfk:x7→xk. By the Fundamental Theorem of Calculus we obtain b a xkdx=xk+1 k+ 1 b a =bk+1-ak+1 k+ 1:

Example 2.Find the integral∫2

11=xdx.

Solution. On the interval (0;∞), the functionx7→lnxis an antiderivative the function x7→1=x. By the Fundamental Theorem of Calculus we obtain 2 11 x dx= lnx2

1= ln2-ln1 = ln2:

This integral can be used to find the limit

lim n→∞( 1 n+ 1+1 n+ 2+···+1 2n) Indeed, letf(x) := 1=xforx= [1;2], and letti= 1 +i=nfori= 0;1;:::;n. Then

P:={t0;t1;:::;tn}is a partition of [1;2] and

1 n+ 1+1 n+ 2+···+1

2n=n∑

i=1f(ti)(ti-ti-1) is a Riemann sum offwith respect toPand points{t1;:::;tn}. By Theorem 1.2 we get lim n→∞( 1 n+ 1+1 n+ 2+···+1 2n) = lim n→∞n i=1f(ti)(ti-ti-1) =∫ 2 11 x dx= ln2: Example 3.A curve in plane is represented by a continuous mappingu= (u1;u2) from [a;b] to IR2. We useL(u) to denote the length of the curve. Suppose thatu′1andu′2are continuous on [a;b]. Thenuis rectifiable. Fort∈[a;b], lets(t) denote the length of the curveu|[a;t]. It was proved in Theorem 7.1 of Chapter 4 that s [u′1(t)]2+ [u′2(t)]2; t∈[a;b]: By Theorem 3.1 (the Fundamental Theorem of Calculus), we obtain

L(u) =s(b)-s(a) =∫

b a s′(t)dt=∫ b [u′1(t)]2+ [u′2(t)]2dt: The following is the second version of the Fundamental Theorem of Calculus. 11 Theorem 3.2.Letfbe an integrable function on[a;b]. Define

F(x) :=∫

x a f(t)dt; x∈[a;b]: ThenFis a continuous function on[a;b]. Furthermore, iffis continuous at a point c∈[a;b], thenFis differentiable atcand F ′(c) =f(c): for allx∈[a;b]. Ifx;y∈[a;b] andx < y, then

F(y)-F(x) =∫

y a f(t)dt-∫ x a f(t)dt=∫quotesdbs_dbs21.pdfusesText_27

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