Find $\lim\limits_{n \to \infty} \int_0^1 \sqrt[n]{x^n+(1-x)^n} \,dx$.
The present problem is proposed by Prof. Ovidiu Furdui and appeared in Teme de analiza matematica. The solution presented is straightforward (using Squeeze theorem)
$$\small\frac{3}{4}=\int_0^{1/2}(1-x)\textrm{d}x+\int_{1/2}^1 x \textrm{d}x\le I_n \le \int_0^{1/2} \sqrt[n]{(1-x)^n+(1-x)^n}\textrm{d}x +\int_{1/2}^1\sqrt[n]{x^n+x^n}\textrm{d}x=\frac{3}{4}\sqrt[n]{2},$$
where the integral under the limit is denoted by $I_n$.
Letting $n\to\infty$ the desired limit follows, that is $3/4$.
More information: If interested in limits with such a structure, you may consult the book Limits, Series, and Fractional Part Integrals Problems in Mathematical Analysis by Ovidiu Furdui.
For example, on page $7$ you may find the trigonometric version
$$\lim_{n\to\infty} \int_0^{\pi/2}\sqrt[n]{\sin^n(x)+\cos^n(x)}\textrm{d}x,$$
which leads to $\sqrt{2}$ and that can be finished by a similar strategy.
Let the integrand be $f(x)$. Obviously $f(0)=f(1)=1$ and the function achieves a single minimum $\left(\frac12,\frac{\sqrt[n]2}2\right)$. Hence we have certainly $\frac12<I_n<1$.
If you want to compute the limit anyway,
$$\sqrt[n]{x^n+(1-x)^n}dx=x\sqrt[n]{1+\left(\frac{1-x}x\right)^n}\to x$$ when $1-x<x$, and symmetrically. Hence the limit function is simply
$$f(x):=\begin{cases}x\le\frac12\to1-x,\\x\ge\frac12\to x\end{cases}$$ and the area under it is $\frac34$.
Plot for $n=15$:
Update:
The initial argument is wrong, as it only proves $$\frac12\le I\le 1.$$
For a better proof, observe that for all $x$, $$\sqrt{x^2+(1-x)^2}\ge f(x)\ge\max(1-x,x),$$ which excludes $\frac12$ and $1$.