Suppose that f : [a, b] → R is an integrable function Then f is also integrable on [a, b] Proof Let ϵ > 0 be given Since f is integrable, there exists a partition P
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[PDF] Chapter 5 Integration §1 The Riemann Integral Let a and b be two
Theorem 1 1 A bounded function f on [a, b] is integrable if and only if for each ε > 0 there exists a partition P of [a, b] such that U(f,P) − L(f,P) < ε Proof Suppose
[PDF] 1 f and f 2 are integrable when f is integrable
Suppose that f : [a, b] → R is an integrable function Then f is also integrable on [a, b] Proof Let ϵ > 0 be given Since f is integrable, there exists a partition P
[PDF] MATH 104, HOMEWORK - Math Berkeley
You may also find that things are clearer if you reorganize the argument a bit ) Solution Suppose f is integrable on [a, b] For Exercise 33 5, we need to show that
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Show that the converse is not true by finding a function f that is not integrable on [ a, b] but that f is integrable on [a, b] Solution 2 Consider the function f(x) = { 1 if x
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Compute the upper and lower Darboux sums associated to the function f(x) 4) Show that if f is integrable on [a, b] and if [c, d] c [a, b] then f is integrable on [c,
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Theorem If f is continuous on [a, b], then f is Riemann-integrable on [a, b] Proof We use the condition (*) to prove that f is Riemann-integrable If ϵ > 0, we set
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the next chapter In this chapter, we define the Riemann integral and prove some Proof If sup g = ∞, then sup f ≤ sup g Otherwise, if f ≤ g and g is bounded
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(b) Find f : [0,1] → R such that f2 is integrable but f is not integrable 4 Let f and g be two integrable functions on [a, b] (a) If f(x) ≤ g(x) for all x ∈ [a, b], show that
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If fi [a,b] R is a bounded function, then ffon da e § fonda 2 A bounded If f is Riemann integrable on [a, o and Proof f is continuous on [a, b] = f is bounded
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Suppose there exist sequences (Un) and (Ln) of upper and lower Darboux sums for f such that Un − Ln → 0 Show that f is integrable and that ∫ b a f = lim
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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 haveU(|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?BThis 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 whenn > 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.