How to find the limit that involves the factorials? $$ \lim_{n\to\infty}\left(\int_a^b((x-a)(b-x))^n\mathrm{d}x\right)^{\frac{1}{n}}$$
I have found the form of the integrand to be $$\frac{(n!)^2}{(2n+1)!}(b-a)^{2n+1}$$ Now, splitting the limit into two different parts i would need to solve $$\lim_{n\to\infty}(\frac{(n!)^2}{(2n+1)!} )^{\frac{1}{n}}$$ My mind went to Stirling's approximation formula but i don't think it could help.
 A: You want to find
$$\lim_{n\to\infty} \left(\frac{(n!)^2}{(2n+1)!} (b-a)^{2n+1}\right)^{\frac{1}{n}}$$
Clearly, $\lim_{n\to\infty}(b-a)^{\frac{2n+1}{n}} = (b-a)^2$. I claim that
$$\lim_{n\to\infty} \left(\frac{(n!)^2}{(2n+1)!}\right)^\frac{1}{n} = \frac{1}{4}$$
One way to do this is, as you say, to use Stirling's approximation:
$$ n! \sim \frac{1}{\sqrt{2\pi n}}\left(\frac{n}{e}\right)^n$$
$$ (2n+1)! \sim \frac{1}{\sqrt{2\pi (2n+1)}}\left(\frac{2n+1}{e}\right)^{2n+1}$$
Note that after taking $n$-th roots, the terms $\frac{1}{\sqrt{2\pi n}},\frac{1}{\sqrt{2\pi (2n+1)}}$ and the dangling factor of $\frac{2n+1}{e}$ don't really matter, so
$$\lim_{n\to\infty} \left(\frac{(n!)^2}{(2n+1)!}\right)^\frac{1}{n} = \lim_{n\to\infty} \left(\frac{n}{e}\right)^2\left(\frac{e}{2n+1}\right)^2 = \frac{1}{4}$$
as advertised.
More generally, it is a fact that $\lim_{p\to\infty}\left\lVert f\right\rVert_p = \left\lVert f\right\rVert_\infty$, where $\left\lVert f\right\rVert_p = \left(\int_{[a,b]} \left|f\right|^p\right)^{\frac{1}{p}}$, and for continuous functions, $\left\lVert f\right\rVert_\infty$ is equal to $\max_{[a,b]}|f|$. Indeed, in this example, the function $f(x)=(x-a)(b-x)$ attains a maximum value $\frac{(b-a)^2}{4}$ at $x=\frac{a+b}{2}$.
A: $$y=\left(\frac{(n!)^2}{(2n+1)!} (b-a)^{2n+1}\right)^{\frac{1}{n}}$$
$$\log(y)=\frac{1}{n}\Bigg[2\log(n!)+(2n+1)\log(b-a)-\log((2n+1)!) \Bigg]$$ Using Stirling approximation and continuing with Taylor expansions
$$n\,\log(y)=\log\Big[\frac 14{(b-a)^2} \Big]n+\frac{1}{2} \log \left(\frac{\pi  (b-a)^2}{4 n}\right)-\frac 3 {8n}+O\left(\frac{1}{n^2}\right)$$
$$\log(y)=\log\Big[\frac 14{(b-a)^2} \Big]+\frac{1}{2n} \log \left(\frac{\pi  (b-a)^2}{4 n}\right)+O\left(\frac{1}{n^2}\right)$$
$$y=e^{\log(y)}\sim\frac 14{(b-a)^2}\exp\Bigg[\frac{1}{2n} \log \left(\frac{\pi  (b-a)^2}{4 n}\right) \Bigg]$$
