Calculus Problem by LOGA 3

$I = \int_{0}^{10} x^n(10-x)^n \ dx$

Taking $x = 10z$ transforms the integral to:

$I = 10^{2n+1}\int_{0}^{1} z^n(1-z)^n \ dz$ $I = 10^{2n+1} \Beta(n+1,n+1)$

Where $\Beta(x,y)$ denotes the Beta function. Knowing that:

$\Beta(x,y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$

$\implies I =10^{2n+1} \frac{(n!)^2}{(2n+1)!}$ $\because \Gamma(n+1) =n!$

Where $n$ is a positive integer.

$I^{1/n} = \left(10^{2n+1} \frac{(n!)^2}{(2n+1)!}\right)^{1/n}$ $I^{1/n} = 10^{2+1/n} \left( \frac{(n!)^2}{(2n+1)!}\right)^{1/n}$ $\lim_{n \to \infty} I^{1/n} = \lim_{n \to \infty} 10^{2+1/n} \left( \frac{(n!)^2}{(2n+1)!}\right)^{1/n}$ $\lim_{n \to \infty} I^{1/n} = 100\lim_{n \to \infty}\left( \frac{(n!)^2}{(2n+1)!}\right)^{1/n}$

Using the Stirling approximation of the factorial:

$n! \approx \sqrt{2 \pi n} \left(\frac{n}{e}\right)^n$

Plugging this into the expression above gives:

$\lim_{n \to \infty} I^{1/n} = 100\lim_{n \to \infty}\left( \frac{\left( \sqrt{2 \pi n} \left(\frac{n}{e}\right)^n \right)^2}{\sqrt{2 \pi (2n+1)} \left(\frac{2n+1}{e}\right)^{2n+1}}\right)^{1/n}$

Simplifying the above expression leads to the following trivial limit:

$\lim_{n \to \infty} I^{1/n} =100\lim_{n \to \infty} \frac{n^2}{(2n+1)^2}$ $\implies \lim_{n \to \infty} I^{1/n} = 25$

Some intermediate steps have been skipped in the solution.