I know a bunch of different versions of the mean value theorem for integrals, and yet none of them are able to solve my problem, but it sure as heck looks like one of them should.
I have attempted the following versions:
$1)$ let $f$ be a continuous function on $[a,b]$. Then there is $c \in [a,b]$ such that $$\int_a^b f(x) dx = f(c)(b-a)\,.$$
$2)$ Let $f$ be a continuous function and $\alpha$ be a continuous function of bounded variation. Then there is $c \in [a,b]$ such that $$\int_a^b f(x) d \alpha (x) = f(a)\int_a^c f(x) d\alpha (x) + f(b)\int_c^bf(x) d \alpha(x)\,.$$
$3)$ the two theorems stated in this question here.
Yet, unless I'm crazy, none of these theorems are able to prove the following fact, which certainly feels like some kind of mean value theorem problem.
Problem: Suppose $f$ is bounded and continuous. Show that there is some $c \in (0, \infty)$ such that $$\int_0^\infty f(x) e^{-x} dx = f(c)\,.$$
Do I need a different version? Something else all together? Did I just miss something obvious? This was an old test problem, and I need someone to put me out of my misery over this question.