Computing $\lim_{n \to \infty} \sqrt[n]{ \int_{0}^{1} (1+x^n)^n dx}$ I tried to compute the limit by using Newton's binomial formula, variable change,  studying the function $1+x^n$ to establish an inequality, but failed to solve it.
$$\lim_{n \to \infty} \sqrt[n]{ \int_{0}^{1} (1+x^n)^n dx}$$
 A: Quite trivially $ I_n=\sqrt[n]{\int_{0}^{1}(1+x^n)^n\,dx} $ is less than two.
However, $1+x^n$ is a convex function on $[0,1]$, hence by considering the tangent in $x=1$ we have:
$$ I_n \geq \sqrt[n]{\int_{0}^{\frac{2}{n}}(2-nx)^n\,dx}=\sqrt[n]{\frac{1}{n}\int_{0}^{2}(2-x)^n\,dx}=\sqrt[n]{\frac{2^{n+1}}{n}\int_{0}^{1}(1-x)^n\,dx} $$
hence:
$$ I_n \geq 2\,\left(\frac{2}{n(n+1)}\right)^{1/n} $$
proves that $\lim_{n\to +\infty}I_n = \color{red}{2}$.
A: (Of course I did some numerical experiments before starting on a proof.)
Let
$$J_n:=\left(\int_0^1 (1+x^n)^n\>dx\right)^{1/n}\ .$$
From $1\leq 1+x^n\leq 2$ $\>(0\leq x\leq 1)$ it immediately follows that $1\leq J_n\leq 2$ for all $n\geq1$, in particular $\lim\sup_{n\to\infty} J_n\leq2$.
Let a small $\alpha>0$ be given. When $1-{\alpha\over n}\leq x\leq 1$ then Bernoulli's inequality gives
$$x^n\geq\left(1-{\alpha\over n}\right)^n\geq1-\alpha\ ,$$
so that we obtain
$$1+x^n\geq 2-\alpha\qquad\left(1-{\alpha\over n}\leq x\leq 1\right)\ .$$
From this we conclude
$$\int_0^1(1+x^n)^ndx\geq \int_{1-\frac{\alpha}{n}}^1 (2-\alpha)^ndx= {\alpha\over n}(2-\alpha)^n$$
which then implies
$$J_n\geq{\root n\of{\alpha\over n}}(2-\alpha)\ .$$
Taking the $\lim\inf$ on both sides proves $\lim\inf_{n\to\infty} J_n\geq 2-\alpha$, and since this is true for all $\alpha>0$ we conclude that $\lim\inf_{n\to\infty} J_n\geq2$.
It follows that in fact $\lim_{n\to\infty} J_n=2$.
A: Or a general claim: if $\;f(x): [a,b]\to\Bbb R\;$ is continuous, then
$$\lim_{n\to\infty}\left(\int_a^b f(x)^ndx\right)^{1/n}=\max_{x\in [a,b]} f(x)$$
so in this case we have
$$\max_{x\in[0,1]}(1+x^n)=2$$
