The mean value theorems you cite apply to integrals over bounded intervals, so some additional analysis is required.
Note that the integral $\int_0^\infty f(x) e^{-x} \, dx$ is convergent by the Weierstrass test.
Let $m = \inf_{x \in[0,\infty)} f(x)$ and $M = \sup_{x \in[0,\infty)} f(x)$. Since $me^{-x} \leqslant f(x)e^{-x} \leqslant Me^{-x}$, we have
$$m = \int_0^\infty me^{-x} \, dx \leqslant I = \int_0^{-\infty}f(x) e^{-x} \, dx \leqslant \int_0^\infty Me^{-x} \, dx = M,$$
and $m \leqslant I \leqslant M$.
If $m < I < M$ then there exist $a$ and $b$ such that $m \leqslant f(a) < I < f(b) \leqslant M$. (This is a basic property of the infimum and supremum.) By the familiar intermediate value theorem -- which applies to continuous functions on compact intervals -- there exists $c \in (a,b) \subset [0,\infty)$ such that
$$f(c) = I = \int_0^\infty f(x) e^{-x} \, dx.$$
See if you can handle the cases where $I = m$ and $I = M$.
Hint: If $I = m$ then $\int_0^\infty [f(x) - m]e^{-x} \, dx = 0$ where $f(x) - m \geqslant 0$.
Best Answer
The mean value theorem for integrals says that there exists a $c$ for which $f$ agrees with its average; that is,
$$f(c) = \frac1{5 - 0} \int_0^5 f(t) dt$$
So compute the integral
$$\int_0^5 t^5 dt$$ and get a number; you should get $5^6/6$; then solve
$$c^5 = \frac{5^5}{6}$$