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.
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.
$I=\lim_{n\to\infty}\int_0^1(x^n+(1-x)^n)^{1/n}\ dx=\lim_{n\to\infty}\int_0^1x(1+((1-x)/x)^n)^{1/n}dx$. 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 $\int_0^1xdx=1/2.$