Alternative proof for the 1st mean-value theorem for integrals

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SUMMARY

The discussion focuses on proving the first mean-value theorem for integrals, specifically the equation \(\int^{b}_{a}f(x)g(x)dx=f(\xi)\int^{b}_{a}g(x)dx\). Elucidus suggests using the Lagrange mean-value theorem applied to an integral with a variable upper limit. The proof involves defining two functions, \(F(t) = \int_{a}^{t}f(x)g(x)dx\) and \(G(t) = \int_{a}^{t}g(x)dx\), and establishing the existence of \(\xi\) such that \(\frac{F'(\xi)}{G'(\xi)} = \frac{F(b)-F(a)}{G(b)-G(a)}\).

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can anyone tell me how to prove the 1st mean-value theorem for integral
\int^{b}_{a}f(x)g(x)dx=f(\xi)\int^{b}_{a}g(x)dx
by applying Lagrange mean-value theorem to an integral with variable upper limit?
thanks a lot.
 
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I think I've seen a proof that uses the Cauchy Mean Value Theorem using

F(t) = \int_{a}^{t}f(x)g(x)\;dx

G(t) = \int_{a}^{t}g(x)\;dx

So there exists \xi so that

\frac{F'(\xi)}{G'(\xi)} = \frac{F(b)-F(a)}{G(b)-G(a)}

or at least that is what I remember.

--Elucidus
 

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