Proof with Darboux integral - question.

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The discussion centers on proving that for a Riemann integrable function f(x) bounded above by a real number M, the integral ∫[a,b] f(x) is less than M(b-a). The user correctly identifies that since f(x) is Riemann integrable, the integral equals both the lower and upper Darboux sums. The conclusion drawn is that ∫[a,b] f(x) = sup f(x) [a,b] Σ[i=1,n] Δxi ≤ M(b-a), establishing the required inequality.

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peripatein
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Hello,

Homework Statement


I was asked to prove that for f(x), Riemann integrable and bounded from above by real number M:
∫[a,b] f(x) < M(b-a)


Homework Equations





The Attempt at a Solution


Since f(x) is Riemann integrable, ∫[a,b] f(x) must be equal to both the lower and upper Darboux sums. Therefore: ∫[a,b] f(x) = supf(x) [a,b]Ʃ[i=1,n]Δxi = supf(x) [a,b](b-a) <= M(b-a)
I am really not sure this is rigorous enough, or even how it ought to be approached. I'd appreciate some guidance please.
 
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I think this can be seen as a consequence of the integral mean value theorem.
 

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