Use the weighted mean value theorem (Theorem 3.16 in Apostol) to prove:
Recall the weighted mean value theorem:
For functions and
continuous on
, if
never changes sign in
then there exists
such that
Proof. Let
Then substituting our definitions of and
,
Since and
are continuous and
does not change sign on
we may apply Theorem 3.16,
Since is strictly decreasing on
, we have
Then,
It seems safer to do integration on the product of g and f, as in the proof of 3.16 (and not like done here).
Sorry, the way you proved it seems valid (apart from the mentioned error in the last row) and probably the intended way, because it uses the required theorem.
The inequality you end up with is not the one the question is asking for; there is an extra factor of f(c).
remember that f(c)(integral of g(x)) = (integral of f(x)g(x)) which is the original equation. In other words both the function he ended up with and the one being asked for are equivalent.
*edit although he did make a typo and rewrote f(c) in the final line when he had already factored it back in.