Proof with Darboux integral - question.

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To prove that the integral of a Riemann integrable function f(x) bounded above by M satisfies ∫[a,b] f(x) < M(b-a), one must recognize that the integral equals both the lower and upper Darboux sums. The relationship can be expressed as ∫[a,b] f(x) = sup{f(x)} * (b-a), which leads to the conclusion that this value is less than or equal to M(b-a). The discussion indicates uncertainty about the rigor of this proof and suggests that it may relate to the integral mean value theorem. Further clarification and guidance on the proof's approach are requested. The conversation highlights the need for a solid understanding of the Darboux integral properties.
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.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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