In this problem, I need to use Beta density function to solve the integration.
$$ \int_{0}^{100}x^{2}\left(\,100 - x\,\right)^{2}\,{\rm d}x $$
After applying $\,{\rm Beta}\left(\, 3,3\,\right)$ I got $2!\,2!/{5!}$. However there should be a factor of $100^{5}$ which I did not get in my answer.
I don't understand where this factor came from.
This factor comes from the change of variable $t = x/100$ (recall that the Beta distribution is defined on $[0,1]$, not $[0,100]$ — you have to scale things down) $$\begin{align} \int_0^{100} x^2(100-x)^2 dx &= \int_0^1 (100t)^2(100-100t)^2 (100dt) \\ &= (100)^2\cdot(100)^2 \cdot 100\int_0^1 t^2(1-t)^2 dt \\ &= 100^5 \int_0^1 B(3,3)\beta_{3,3}(t) dt = {10^5}{B(3,3)} \end{align}$$ with $\frac{1}{B(3,3)} = \frac{\Gamma(3+3)}{\Gamma(3)\Gamma(3)} = \frac{5!}{2!\cdot 2!}$.