[PDF] 1 fandf2are integrable when f is integrable





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

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1 fandf2are integrable when f is integrable

1 f and f2 are integrable when f is integrable Lemma 1 1 Let f : [ab] ? R be a bounded function and let P = {x 0x 1 x n} be a partition of [ab] Then for each i ? {12 n} M i(f)?m i(f) = sup{f(x)?f(y) : xy ? [x i?1x i]} Proof Let xy ? [x i?1x i] Without loss of generality assume that f(x) ? f(y) and



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We say that f is integrable on [ab] if there is a number V such that for every sequence of partitions {Pn} on [ab] such that {µ(Pn)} ? 0 and every sequence {Sn} where Sn is a sample for Pn {X (fPnSn)} ? V If f is integrable on [ab] then the number V just described is denoted by Z b a f and is called “the integral from a to b of f



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Therefore supL(f;P) over all partitions P (which is called the lower Darboux integral is less than or equal to inf U(f;P) (the upper Darboux integral) If these are equal then we say the function f is integrable and their common value is called the integral: b a f(x)dx D DeTurck Math 360 001 2017C: Integral/functions 5/28



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Theintegralof f is de?ned by ZcZf(x)dx+ Zc f(x)dx=f(x)dx In view of the previous theorem the integral does not dependonc It can also be computed as a repeated limit:Zbf(x)dx= limlim d?b? c?a+ Zf(x) dx = limc?a+ limd?b? Zf(x) c dx Finally the integral can be computed as a double limit (i e the limit of a function of two variables): Zf(x)dx=



Searches related to if f is integrable filetype:pdf

A MONOTONE FUNCTION IS INTEGRABLE Theorem Let f be a monotone function on [a;b] then f is integrable on [a;b] Proof We will prove it for monotonically decreasing functions The proof for increasing functions is similar First note that if f is monotonically decreasing then f(b) • f(x) • f(a) for all x 2 [a;b] so f is bounded on [a;b]

What is a definite integral off?

    The integral offon [a, b] is a real number whose geometrical interpretation is thesigned area under the graphy=f(x) fora?x?b. This number is also calledRb the de?nite integral off. By integratingfover aninterval [a, x] with varying rightend-point, we get a function ofx, called the inde?nite integral off.

Is F integrable on [a1a2],[a2a3],[an1an] andz an?

    If the restriction of f to each of the intervals[a1,a2],[a2,a3],···,[an?1,an]is integrable, then f is integrable on [a1,an]andZ an a1

Is 1/g integrable?

    meaning that oscI(1/g)?M2oscI g, and Proposition 1.16 implies that 1/gis grable ifgis integrable. Thereforef /g=f·(1/g) is integrable. 1.6.2. Monotonicity.Next, we prove the monotonicity of the integral. Theorem 1.27. Suppose thatf, g: [a, b]? are integrable andf?g. Then Zg. Proof. First suppose thatf?0 is integrable.

Are integrable functions bounded?

    The product of integrable functions is also integrable, as is the quotient pro-vided it remains bounded.

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?

i=1m i=1M iΔxi=U(f,Pn). Since limU(f,Pn) = limL(f,Pn), the Squeeze Theorem implies that b a f= limn→∞n i=1f(x?i)Δxi= limn→∞U(f,Pn) = limn→∞L(f,Pn).References [1] S. Lay,Analysis with an introduction to proof, Prentice Hall, Inc., Englewood Cliffs, NJ, 1986.quotesdbs_dbs20.pdfusesText_26
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