Find $\lim\limits_{n \to \infty} \int_0^1 \sqrt[n]{x^n+(1-x)^n} \,dx$. I have the following limit to find:
$$\lim_{n \to \infty} \int_0^1 \sqrt[n]{x^n+(1-x)^n} \, dx$$
And I have to choose between the following options:
A. $0$
B. $1$
C. $\dfrac{3}{4}$
D. $\dfrac{1}{2}$
E. $\dfrac{1}{4}$
This is what I did:
$$\int_0^1 \sqrt[n]{x^n+(1-x)^n}  \, dx =
\int_0^{\frac{1}{2}} \sqrt[n]{x^n+(1-x)^n} \, dx +
\int_{\frac{1}{2}}^1 \sqrt[n]{x^n+(1-x)^n} \, dx$$


*

*If we consider the first term of this sum:


$$\int_0^{\frac{1}{2}} \sqrt[n]{x^n+(1-x)^n} \, dx \ge
\int_0^\frac{1}{2} \sqrt[n]{x^n+x^n} \, dx =
\int_0^\frac{1}{2}2^{\frac{1}{n}}x \, dx$$
And that means:
$$\lim_{n \to \infty} \int_0^{\frac{1}{2}} \sqrt[n]{x^n+(1-x)^n} \, dx \ge
\lim_{n \to \infty} \int_0^\frac{1}{2}2^{\frac{1}{n}}x \, dx =
\int_0^\frac{1}{2}x \, dx = 
\frac{1}{8} \tag 1$$


*

*If we consider the second term of that sum we have:


$$\int_{\frac{1}{2}}^1 \sqrt[n]{x^n+(1-x)^n} \, dx \ge
\int_\frac{1}{2}^1 \sqrt[n]{(1-x)^n+(1-x)^n} \, dx =
\int_\frac{1}{2}^1 2^{\frac{1}{n}}(1-x) \, dx$$
And that means:
$$\lim_{n \to \infty} \int_{\frac{1}{2}}^1 \sqrt[n]{x^n+(1-x)^n} \, dx \ge \lim_{n \to \infty} \int_\frac{1}{2}^1 2^{\frac{1}{n}}(1-x) \, dx = \lim_{n \to \infty} \int_\frac{1}{2}^1 (1-x) \, dx = \frac{1}{8} \tag 2$$
Now we can sum $(1)$ and $(2)$ and we have:
$$\lim_{n \to \infty} \int_0^1 \sqrt[n]{x^n+(1-x)^n} \, dx \ge \frac{1}{4}$$
So now we have a lower bound. We can do something similar for the upper bound.


*

*Again, let's consider the first term of that sum:


$$\int_0^{\frac{1}{2}} \sqrt[n]{x^n+(1-x)^n} \, dx \le
\int_0^{\frac{1}{2}} \sqrt[n]{(1-x)^n+(1-x)^n} \, dx = 
\int_0^{\frac{1}{2}} 2^\frac{1}{n}(1-x) \, dx$$
That means:
$$\lim_{n \to \infty} \int_0^{\frac{1}{2}} \sqrt[n]{x^n+(1-x)^n} \, dx \le
\lim_{n \to \infty} \int_0^{\frac{1}{2}} 2^\frac{1}{n}(1-x) \, dx = 
\int_0^{\frac{1}{2}} (1-x) \, dx = 
\frac{3}{8} \tag 3$$


*

*And if we consider the second part of that sum:


$$\int_{\frac{1}{2}}^1 \sqrt[n]{x^n+(1-x)^n} \, dx \le 
\int_{\frac{1}{2}}^1 \sqrt[n]{x^n+x^n} \, dx = 
\int_{\frac{1}{2}}^1 2^\frac{1}{n} x \, dx$$
And that means:
$$\lim_{n \to \infty} \int_{\frac{1}{2}}^1 \sqrt[n]{x^n+(1-x)^n} \, dx \le
\lim\_{n \to \infty} \int_{\frac{1}{2}}^1 2^\frac{1}{n} x \, dx = 
\lim_{n \to \infty} \int_{\frac{1}{2}}^1 x \, dx = 
\frac{3}{8} \tag 4$$
And now if we sum $(3)$ and $(4)$ we get:
$$\lim_{n \to \infty} \int_0^1 \sqrt[n]{x^n+(1-x)^n} \, dx \le
\frac{3}{4}$$
So after all of that, we have:
$$\frac{1}{4} \le
\lim_{n \to \infty} \int_0^1 \sqrt[n]{x^n+(1-x)^n} \, dx \le \frac{3}{4}$$
But this didn't help me all that much. Choices C, D and E are still consistent with this inequality that I got. So what should I do to find the exact answer? Or did I do something wrong in my calculations?
 A: 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. 
A: 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 $1$.
