Limit question related to integration Find the limit,
$$L=\lim_{n\to \infty}\int_{0}^{1}(x^n+(1-x)^n)^{\frac{1}{n}}dx$$
My try:
$$ \int_{0}^{\frac{1}{2}}2^{\frac{1}{n}}xdx+ \int_{\frac{1}{2}}^{1}2^{\frac{1}{n}}(1-x)dx< \int_{0}^{1}(x^n+(1-x)^n)^{\frac{1}{n}}dx< \int_{0}^{\frac{1}{2}}2^{\frac{1}{n}}(1-x)dx+ \int_{\frac{1}{2}}^{1}2^{\frac{1}{n}}xdx$$
Now taking the limit I get that,
$$\frac{1}{4}<L<\frac{3}{4}$$
But, how can I get the exact answer!!
This is Problem 11941 from the American Mathematical Monthly.

Hint/Intuition: 
For $x,y\ge0, \lim_{n\to\infty} \left(x^n+y^n\right)^{1/n}=\max\{x,y\}$.
Alternative Approach:
$$I=\int_{0}^{1}(x^n+(1-x)^n)^{\frac{1}{n}}dx=2\int_{0}^{1/2}(x^n+(1-x)^n)^{\frac{1}{n}}dx$$
$$=2\int_0^{1/2}(1-x)\left(1+\left(\frac{x}{1-x}\right)^n\right)^{1/n}$$
Let $u=\frac{x}{1-x}\implies dx=\frac{du}{(1+u)^2}$
$$I=2\int_0^1\frac{1}{1+u}(1+u^n)^{1/n}\frac{du}{(1+u)^2}$$
Note that for $y>0, 1<(1+y)^{1/n}<1+\frac{y}{n}$, so:
$$2\int_0^1\frac{du}{(1+u)^3}\le I \le 2\int_0^1\left(1+\frac{u^n}{n}\right)\frac{du}{(1+u)^3}$$
Now:
$$0\le I - 2\int_0^1\frac{du}{(1+u)^3}\le \frac{2}{n}\int_0^1\frac{u^n\,du}{(1+u)^3}\le \frac{2}{n}\int_0^1\frac{du}{(1+u)^3}=\frac{3}{4n}$$
So, the integral converges to $2\int_0^1\frac{du}{(1+u)^3}=\frac{3}{4}$
