Second Mean Value Theorem for Integrals Meaning The Second Mean Value Theorem for Integrals says that for $f (x)$ and $g(x)$ continuous on $[a, b]$  and $g(x)\ge 0$
$$\int_a^bf(x)g(x)\,dx=f(a)\int_a^cg(x)\,dx+f(b)\int_c^bg(x)\,dx$$
I have a difficult time understanding what this means, as opposed to the first mean value theorem for integrals, which is easy to conceptualize. Is there a graphical or 'in words' interpretation of this theorem that I may use to understand it better?
 A: I mentioned in a comment that you need more requirements on $f$ than just that is continuous.  To give you a verbal explanation of the theorem I will assume it is non-decreasing.  Then you can look at it as follows:
Since $f$ is non decreasing, $f(a)$ must be the minimum of $f$ over the interval, and $f(b)$ must be the maximum.  Now it must be true that:
$$\int_a^b f(x)g(x)dx \geq f(a) \int_a^b g(x) dx$$
and 
$$\int_a^b f(x)g(x)dx \leq f(b) \int_a^b g(x) dx$$
Now consider the function $F$ of $c$ given by
$$F(c) = f(a)\int_a^c g(x)dx + f(b)\int_c^b g(x) dx$$
This function must satisfy $F(b) \leq \int_a^b f(x)g(x)dx$ and also
$F(a) \geq \int_a^b f(x)g(x)dx$.  Since it is continuous there must be a $c^*$ where equality holds. (By the intermediate value theorem).
So to put it in words.  If you integrate a function $g$ from $a$ to $b$ and weight it by an increasing function $f$, then the weighted integral must be greater than the integral of $g$ times $f$'s min and less than the integral times $f$'s max.  So there must be a point in between where $f$s min times some of $g$'s integral plus $f$'s max times the rest of $g$'s integral equals the total weighted integral.
