Define sn to be the following sum:
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sn=2n-1∑m=12-nf(a+m(b-a)/2n) |
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Making the substitution m′=2m and using the fact that 1+(-1)m′=0 when m′ is odd to express the sum over even values of m′ as a sum over all values of m′, this becomes
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sn=2n+1-1∑m=1(1+(-1)m′)2-1-nf(a+m(b-a)/2n) |
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Subtracting this sum from sn+1 and simplifying gives
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sn+1-sn=2n+1-1∑m=1(-1)m+12-nf(a+m(b-a)/2n) |
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Using the telescoping sum trick, we may write
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sk=k∑n=1(sn-sn-1)=k∑n=12n-1∑m=1(-1)m+12-nf(a+m(b-a)/2n) |
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To complete




the proof, we must investigate the limit as k→∞. Since f is assumed continuous

and the interval [a,b] is compact
, f is uniformly continuous
. This means that, for every ϵ>0, there exists a δ>0 such that |x-y|<δ implies |f(x)-f(y)|<ϵ. By the Archimedean property, there exists an integer k>0 such that 2kδ>|a-b|. Hence, |f(x)-f(a+m(b-a)/2n)|≤ϵ when x lies in the interval [a+(m-1)(b-a)/2n,a+(m+1)(b-a)/2n]. Thus, (a-b)sk+|a-b|ϵ is a Darboux upper sum for the integral
and (b-a)sk-|a-b|ϵ is a Darboux lower sum. (Darboux’s definition of the integral may be thought of as a modern incarnation of the ancient method of exhaustion.) Hence
Taking the limit ϵ→0, we see that
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b∫af(x)𝑑x=∞∑n=1An=(b-a)∞∑n=12n-1∑m=1(-1)m+12-nf(a+m(b-a)/2n) |
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