[PDF] 1 f and f 2 are integrable when f is integrable PDF integrals.pdf

1 f and f2 are integrable when f is integrable Lemma 1 1 Let f : [a, b] → R be a bounded function and let P = {x0,x1, ,xn} be a partition of [a, b] Then for each i

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

L(f) := sup{L(f,P) : P is a partition of [a, b]} A bounded function f on [a, b] is said to be (Riemann) integrable if L(f) = U(f) In this case, we write ∫ b a f(x)dx = L(f)

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1|f|andf2are integrable whenfis integrable

Lemma 1.1.Letf: [a,b]→Rbe a bounded function and letP={x0,x1,...,xn}be a partition of [a,b].Then for eachi? {1,2,...,n},Mi(f)-mi(f) = sup{|f(x)-f(y)|:x,y?[xi-1,xi]}. Proof.Letx,y?[xi-1,xi].Without loss of generality assume thatf(x)≥f(y) and observe that M follows that M Let? >0 be given. There existx,y?[xi-1,xi] such thatf(x)> Mi(f)-?2 andf(y)< mi(f)+?2 Sof(x)-f(y)> Mi(f)-mi(f)-?, and therefore,|f(x)-f(y)|> Mi(f)-mi(f)-?.It now follows that sup{|f(x)-f(y)|:x,y?[xi-1,xi]}> Mi(f)-mi(f)-?.Since this holds for any? >0, we have sup{|f(x)-f(y)|:x,y?[xi-1,xi]} ≥Mi(f)-mi(f).(2)

The inequalities (1) and (2) imply the desired equality.Theorem 1.2.Suppose thatf: [a,b]→Ris an integrable function. Then|f|is also integrable on

[a,b]. Proof.Let? >0 be given. Sincefis integrable, there exists a partitionP={x0,x1,...,xn}of [a,b] such thatU(f,P)-L(f,P)< ?.For anyi? {1,2,...,n}and allx,y?[xi-1,xi], we have

U(|f|,P)-L(|f|,P) =n?

i=1(Mi(f)-mi(f))Δxi =U(f,P)-L(f,P)< ?.

This shows that|f|is integrable on [a,b].Theorem 1.3.Suppose thatf: [a,b]→Ris an integrable function. Thenf2is also integrable on

[a,b]. Proof.Sincefis bounded on [a,b], there exists aB >0 such that|f(x)+f(y)|< Bfor allx,y?[a,b.] Now let? >0 be given. Sincefis integrable, there exists a partitionP={x0,x1,...,xn}of [a,b] such thatU(f,P)-L(f,P)B(Mi(f)-mi(f)).Now,

U(f2,P)-L(f2,P) =n?

i=1B(Mi(f)-mi(f))Δxi =B(U(f,P)-L(f,P))< B?B

This shows thatf2is integrable on [a,b].

2 Integration for continuous function

Theorem 2.1.Letf: [a,b]→Rbe continuous on[a,b]and letPn={x0=a,x1=a+(b-a)n ,x2= a+ 2(b-a)n ,...,xn=b}.Then? b a f= limn→∞U(f,Pn) = limn→∞L(f,Pn).

Proof.It suffices to show that limn→∞(U(f,Pn)-L(f,Pn)) = 0 since exercise 29.5 in [1] will then imply

the result. Let? >0 be given. Sincefis uniformly continuous on [a,b], there exists aδ >0 such that

when|x-y|< δ,|f(x)-f(y)|U(f,Pn)-L(f,Pn) =n? i=1(Mi-mi)Δxi=n? i=1(f(ti)-f(si))ΔxiSince lim n→∞(b-a)n = 0, there exists aN?Rsuch that whenn > N, we have(b-a)n < δ.So when

n > N, we getU(f,Pn)-L(f,Pn)< ?, which implies that limn→∞(U(f,Pn)-L(f,Pn)) = 0.Corollary 2.2.Suppose thatf: [a,b]→Ris continuous on[a,b].LetPn={x0=a,x1=a+

(b-a)n ,x2=a+2(b-a)n ,...,xn=b}and for eachi? {1,2,...,n}, letx?i?[xi-1,xi]be sample points. Then? b a f= limn→∞n i=1f(x?i)Δxi.

L(f,Pn) =n?