Evaluate $\lim_{n \to \infty} \int_{0}^1 [x^n + (1-x)^n ]^{1/n} \ \mathrm{d}x$ 
Evaluate
$$\lim_{n \to \infty} \int_{0}^1 [x^n + (1-x)^n ]^{1/n} \ \mathrm{d}x$$

I simplified the limit to
$$\dfrac{1}{2}\lim_{n \to \infty} \int_{0}^{1/2} \left[\left(\frac{1}{2}+x\right)^n + \left(\frac{1}{2}-x\right)^n \right]^{1/n} \ \mathrm{d}x$$
using properties of definite integrals, but I can't solve further.
Any help will be appreciated.
Thanks.

Note that for $1\ge x>1/2$, 
$$xe^{\frac1n \left(\left(\frac{1-x}{x}\right)^n-\frac12\left(\frac{1-x}{x}\right)^{2n}\right)}<\left(x^n+(1-x)^n\right)^{1/n}=x\left(1+\left(\frac{1-x}{x}\right)^n\right)^{1/n}<xe^{\frac1n \left(\frac{1-x}{x}\right)^n}$$
where we used the estimates $x-\frac12 x^2<\log (1+x)<x$.  Thus, by the Squeeze Theorem 
$$\lim_{n\to \infty}\left(x^n+(1-x)^n\right)^{1/n}=x \tag 1$$
for $1\ge x>1/2$.  Exploiting symmetry, we have for $0\le x<1/2$ 
$$\lim_{n\to \infty}\left(x^n+(1-x)^n\right)^{1/n}=1-x \tag 2$$
Therefore, putting $(1)$ and $(2)$ together reveals
$$\lim_{n\to \infty}\left(x^n+(1-x)^n\right)^{1/n}=\frac12 +\left|x-\frac12\right|$$
And since $1/2<\left(x^n+(1-x)^n\right)^{1/n}\le 1$, we can use the Dominated Convergence Theorem to obtain
$$\begin{align}
\lim_{n\to \infty}\int_0^1 \left(x^n+(1-x)^n\right)^{1/n}\,dx&=\int_0^1 \lim_{n\to \infty}\left(x^n+(1-x)^n\right)^{1/n}\,dx\\\\
&=\int_0^1 \left( \frac12 +\left| x-\frac12 \right| \right) \,dx\\\\
&=\frac34
\end{align}$$
as expected!
