Proof regarding an integral function I want to prove that $\exists c\in (-\infty,2]$ s.t let $h(x)=\int_0^x \frac{t^3e^{-2t}}{\sqrt{2-t}}$ then $h(2)=\frac{2c^3e^{-2c}}{\sqrt{2-c}}$.
The integral function $f$ is defined and continous in $(-\infty,2]$ since I can continously extends the function in $2$ for the reason why for $x\to 2^-$ then $\frac{t^3e^{-2t}}{\sqrt{2-t}}\sim \frac{8e^{-4}}{\sqrt{2-t}}$ that it is integrable.
The only thing thta I have thought is that by mean value theorem I would have $h(2)=f(c)(2-0)$ (f is the integrand) with $c\in [0,2]$, but what's about $(-\infty,2]$?
I have thought I can apply the mean integral value theorem, but I don't have a bounded interval so it is not possible. How can I prove this?
$\textbf{Edit with my work (thanks to the suggestions of André Armatowski)}$.
I know that the integrand function is continous and differentiable on $(-\infty, 2)$. Now so surely in $[0+\epsilon, 2-\epsilon]$ ( $\forall\epsilon&gt;0$)the integrand will be continous and differentiable and here I will apply the mean value theorem:
$$\exists c\in [0+\epsilon, 2-\epsilon]: \int_{0+\epsilon}^{2-\epsilon}f(t)\, dt=2\frac{c^3e^{-2c}}{\sqrt{2-c}} $$. Now since $[0+\epsilon, 2-\epsilon]\subseteq (-\infty,2]$ $\forall \epsilon&gt;0$ then this means that I have completed the proof.
We consider as in the question the function $h$:
$$h(x)=\int_{0}^{x}\frac{t^3e^{-2t}}{\sqrt{2-t}}\,dt$$
Because our goal is to use the Mean Value Theorem on $[0,2]$, we must first confirm that $h$ satisfies the assumptions of being continuous on $[0,2]$ and differentiable on $(0,2)$.
The differentiability of $h$ on $(0,2)$ is an immediate consequence of The Second Fundemental Theorem Of Calculus. The differentiability on $(0,2)$ for $h$ implies continuity of $h$ on $(0,2)$. To make $h$ continuous at $x=2$ we can define $h(2)=\lim_{x\to 2^{-}}h(x)$, provided it exists:
$$\lim_{x\to 2^{-}}h(x)=\lim_{x\to 2^{-}}\int_{0}^{x}\frac{t^{3}e^{-2t}}{\sqrt{2-t}}\,dt \leq \lim_{x\to 2^{-}}K\int_{0}^{x}\frac{1}{\sqrt{2-t}}\,dt=K2^{\frac{3}{2}}.$$
We can therefore extend $h$ to be a continuous function on $[0,2]$ by taking $h(0)=0$ and noting that $h(2) = \lim_{x\to 2^{-}}h(x)$ exists by monotonicity of $h$ and that it is bounded above by the above estimate for some $K&gt;0$.
Applying the Mean Value Theorem to $h$ on $[0,2]$ there exists a $c\in(0,2)$ such that:
$$h(2)=\frac{2c^{3}e^{-2c}}{\sqrt{2-c}}$$
The question, as stated, makes no sense, but a data point:
$$h(2) = \frac{444 e^4-791 \sqrt{\pi }
   \text{erfi}(2)}{64 \sqrt{2} e^4}.$$