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

[y_lindsay]

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.

 

[Elucidus]

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