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I suppose I need to make use of Cauchy's mean value theorem to prove the statement but how am I able to do that?

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Let $F(x) = \int_{a}^{x}f(t) dt$ and $H(x) = \int_a^{x}f(t)g(t) dt$. Both of these functions are continuous, and they are differentiable on $(a,b)$. Furthermore, $f(x) > 0$ on $[a,b]$ implies $F(a) \neq F(b)$.

Thus we can use Cauchy's mean value theorem: there exists some $c \in (a,b)$ such that $$ \frac{H'(c)}{F'(c)} = \frac{H(b) - H(a)}{F(b) - F(a)}. $$

Using the facts that $H'(c) = f(c)g(c), F'(c) = f(c), H(a)=F(a)=0$ gives you the desired result (after multiplying both sides by $F(b) = \int_a^bf(t)dt$).

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  • $\begingroup$ Good job. But I wish you hadn't used $x$ as the dummy variable in your integrals when you're integrating to $x$. $\endgroup$ – Ted Shifrin Oct 19 '16 at 7:59
  • $\begingroup$ You are right, I was careless about that. $\endgroup$ – Dániel G. Oct 19 '16 at 8:54
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Just a variation on simmilar ideas:

From Mean Value Theorem for Integrals:

$$\tag{1}\int^b_af(x)dx=f(c)(b-a) \rightarrow f(c)=\frac{\int^b_af(x)dx}{(b-a)}$$

$$\tag{2}\int^b_af(x)g(x)dx=f(c)g(c)(b-a)$$

From $(1)$ and $(2)$ we get:

$$\int^b_af(x)g(x)dx=\frac{\int^b_af(x)dx}{(b-a)}g(c)(b-a)=g(c)\int^b_af(x)dx$$

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  • $\begingroup$ the values c in (1),(2) should be different values. $\endgroup$ – Detectives Aug 2 '17 at 6:16

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