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.
 A: 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 $2\int_0^1\frac{du}{(1+u)^3}=\frac{3}{4}$
A: Use the better estimate $$\int_{0}^{\frac{1}{2}}(1-x)\,dx+ \int_{\frac{1}{2}}^{1}x \,dx< \int_{0}^{1}(x^n+(1-x)^n)^{\frac{1}{n}}\,dx< \int_{0}^{\frac{1}{2}}2^{1/n}(1-x)\,dx+ \int_{\frac{1}{2}}^{1}2^{1/n}x\,dx$$
Because $2^{1/n} \to 1,$ the squeeze theorem tells us the limit is $\int_{0}^{\frac{1}{2}}(1-x)\,dx+ \int_{\frac{1}{2}}^{1}x \,dx = 3/4.$
A: From norm equivalences one remembers
$$
\max(a,b)\le\sqrt[n]{a^n+b^n}\le\max(a,b)\sqrt[n]2
$$
which can be used to get upper and lower bounds for the integral,
$$
\frac34\le\int_0^1\sqrt[n]{x^n+(1-x)^n}\,dx\le \frac34\;\sqrt[n]2
$$
which nicely gives the limit as $\frac34$.
A: Note that $$\int_{0}^{1}\left(x^{n}+\left(1-x\right)^{n}\right)^{1/n}dx\leq\int_{0}^{1}1dx=1
 $$ so from the Dominated Convergence Theorem we get $$\lim_{n\rightarrow\infty}\int_{0}^{1}\left(x^{n}+\left(1-x\right)^{n}\right)^{1/n}dx=\int_{0}^{1}\lim_{n\rightarrow\infty}\left(x^{n}+\left(1-x\right)^{n}\right)^{1/n}dx
 $$ and $$\lim_{n\rightarrow\infty}\left(x^{n}+\left(1-x\right)^{n}\right)^{1/n}=\exp\left(\lim_{n\rightarrow\infty}\frac{\log\left(x^{n}+\left(1-x\right)^{n}\right)}{n}\right)
 $$ $$=\exp\left(\lim_{n\rightarrow\infty}\frac{x^{n}\log\left(x\right)+\left(1-x\right)^{n}\log\left(1-x\right)}{x^{n}+\left(1-x\right)^{n}}\right)=\max\left\{ x,1-x\right\} 
 $$ hence $$\lim_{n\rightarrow\infty}\int_{0}^{1}\left(x^{n}+\left(1-x\right)^{n}\right)^{1/n}dx=\int_{0}^{1}\max\left\{ x,1-x\right\} dx
 $$ $$=\int_{0}^{1/2}\left(1-x\right)dx+\int_{1/2}^{1}xdx=\color{red}{\frac{3}{4}}.$$
A: Substitute 1/2+y for x, y going from -1/2 to 1/2. This a symmetric integral so the positve and negative intervals contribute equally and it suffices to consider only the positve interval. As n goes to infinity (1/2-y)^n is neglible compared to (1/2+y)^n. Hence the root approaches 1/2+ y and the integral approaches 2*Int((1/2+y)*dy) = 1/2+1/4 = 3/4
