How does one prove that $\int_0^4xe^{(x-2)^4}dx=2k$ if $\int_0^4e^{(x-2)^4}dx=k$? 
Suppose $\int_0^4e^{(x-2)^4}dx=k$. Prove that $\int_0^4xe^{(x-2)^4}dx=2k$.

I got stuck doing by parts:
$$  u=x \ \ \ \ dv=e^{(x-2)^4}dx $$
$$  du=dx \ \ \ \ \ \ \ v=? $$
What can be $v$. I can't say $v=k$, can I? 
 A: By the mean value theorem for integrals, it follows that there is a point $x_0$ in the interval $(0,4)$ so that $$\int_0^4xe^{(x-2)^4}\mathrm dx=x_0\int_0^4e^{(x-2)^4}\mathrm dx=kx_0.$$
It then remains to show that $x_0=2.$ I have not been able to see how to do this more rigorously, but a heuristic argument for why this must be so is as follows: You may view the calculations above as trying to determine the $x$-coordinate $x_0$ of the centre of gravity of the region defined by the equations $y=0,\,y=e^{(x-2)^4},\,x=0,\,x=4,$ so that the first integral above is the moment of inertia of the region about the $y$-axis. Then since the region is symmetric about the line $x=2,$ it follows that the $x$-coordinate of the centre of gravity, $x_0,$ must be equal to $2.$
A: A substitution $x = t +2$ yields:
$$\begin{align*}\int_0^4 x \exp((x-2)^4) \, dx &= \int_{-2}^2 (t + 2) \exp(t^4) \, dt = \color{red}{\int_{-2}^2 t \exp(t^4) \, dt} + 2\int_{-2}^2 \exp(t^4) \, dt \\
&= 2 \int_0^4 \exp((x - 2)^4) \, dx\end{align*}$$
Herein, the red integral vanishes as the function is symmetric around the origin.
