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.
 A: 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!
A: It's a little easier to let $x=(1+u)/2$ with $-1\le u\le1$. We have
$$\int_0^1(x^n+(1-x)^n)^{1/n}dx={1\over2}\int_{-1}^1\left(\left(1+u\over2\right)^n+\left(1-u\over2\right)^n \right)^{1/n}du=\int_0^1\left(1+u\over2\right)\left(1+\left(1-u\over1+u\right)^n \right)^{1/n}du$$
where we did two things in the second step: we used the symmetry between $1+u$ and $1-u$ to replace ${1\over2}\int_{-1}^1$ with $\int_0^1$, and we factored out the $\left(1+u\over2\right)^n$.  What you do from here on out depends in part on how finicky you need to be about moving the limit across the integral sign. But since for $0\le u\le1$ we have
$$1\le\left(1+\left(1-u\over1+u\right)^n \right)^{1/n}\le\left(1+\left(1-0\over1+0\right)^n \right)^{1/n}=2^{1/n}$$
we see that
$$\int_0^1\left(1+u\over2\right)du\le\int_0^1\left(1+u\over2\right)\left(1+\left(1-u\over1+u\right)^n \right)^{1/n}du\le2^{1/n}\int_0^1\left(1+u\over2\right)du\to\int_0^1\left(1+u\over2\right)du$$
so the Squeeze Theorem tell us
$$\lim_{n\to\infty}\int_0^1(x^n+(1-x)^n)^{1/n}dx=\int_0^1\left(1+u\over2\right)du={(1+u)^2\over4}\big|_0^1=1-{1\over4}={3\over4}$$
A: Hint: It looks like there may be a mistake in your proposed simplification step, but at any rate, you can break the limit up into $\lim_{n \to \infty} \int_0^{1/2} f_n dx + \int_{1/2}^1 f_n dx$ where $f_n = (x^n + (1 - x)^n)^{1/n}$. You should be able to show that in each piece of the broken up integral, either $x^n$ is hugely dominating or $(1-x)^n$ is hugely dominating (before you take the root). Since then you are taking the $n$th root before integrating, this will hopefully allow you to see what the limit should be.
A: $I=\lim_{n\to\infty}\int_0^1(x^n+(1-x)^n)^{1/n}\ dx=\lim_{n\to\infty}\int_0^1x(1+((1-x)/x)^n)^{1/n}dx$. Now $\lim_{n\to\infty}(1+(1-x)/x)ⁿ)^{1/n}=1$. This can be verified by taking log. Assuming uniform convergence we get the result as $\int_0^1xdx=1/2.$
