If $I_n=\sqrt[n]{\int _a^b f^n(x) dx}$,for $n\ge 1$.Find with proof $\lim _{n \to \infty}I_n$ Let $a,b \in \mathbb R ,a<b$ and let  $f:[a,b] \to [0,\infty) $ a continuous and non constant.

attempt,using Reimann series

$I_n=\sqrt [n]{\int _a^b f^n(x) dx}$
$I_n=\sqrt [n]{\lim _{k \to \infty}\sum_{i=1}^k f^n(x_i^*) dx}$
for each $i=1,2,3...n$
$x_i\in [x_{i-1},x_i]$
so our question is what is 
$I_n=\lim_{n\to \infty}\sqrt [n]{\lim _{k\to \infty}\sum_{i=1}^k f^n(x_i^*) dx}$

Elementary tools (high school ones) reveal that
$$\lim_
{n\to\infty} \sqrt [n]{\int _a^b f^n(x) dx}=\operatorname{max}(f(x))$$
