Q:

Integration with Limits

Find $$\displaystyle \lim_{n \to \infty} \int^{1}_{0}(x^{n}+(1-x)^{n})^{\frac{1}{n}}dx$$
Now, the solution was hinted like this:
using the property $$f(x)=f(2a-x)$$ the limit becomes half
$$2 \displaystyle \int^{\frac{1}{2}}_{0} (x^{n}+(1-x)^{n})^{\frac{1}{n}} dx$$.....$$(1)$$
plz see carefully.
for $$x\in (0,\frac{1}{2})$$
$$x$$ has smaller value than $$(1-x)$$.
taking common $$(1-x)^{n}$$ in $$(1)$$
also $$\displaystyle \lim_{n\to \infty} (\frac{x}{1-x})^{n} \rightarrow 0$$
the integral reduces to
$$2 \displaystyle \int^{\frac{1}{2}}_{0} (1-x) \times (1^{n})^{\frac{1}{n}}.dx$$
$$2 \displaystyle \int^{\frac{1}{2}}_{0} (1-x).dx$$
$$\frac{3}{4}$$

A:

$$y=(x^{n}+(1-x)^{n})^{\frac{1}{n}}=(1-x)\left(1+\left(\frac{x}{1-x}\right)^{n}\right)^{\frac{1}{n}}$$
Now, in $0\leq x \leq \frac{1}{2}$
$$0\leq \frac{x}{1-x} \leq 1$$
$$0\leq \left(\frac{x}{1-x}\right)^{n} \leq 1$$
$$1\leq \left(1+\left(\frac{x}{1-x}\right)^{n}\right)^{\frac{1}{n}} \leq 2^{\frac{1}{n}}$$
$$(1-x)\leq y \leq 2^{\frac{1}{n}}(1-x)$$
$$\int_0^{\frac{1}{2}}(1-x)dx \leq \int_0^{\frac{1}{2}} y\:dx \leq 2^{\frac{1}{n}}\int_0^{\frac{1}{2}}(1-x)dx$$
$$\frac{3}{8} \leq \int_0^{\frac{1}{2}} y\:dx \leq 2^{\frac{1}{n}} \frac{3}{8}$$
for $x$ tending to infinity  $2^{\frac{1}{n}}$ tends to $1$. So, the limit is :
$$\int_0^{\frac{1}{2}} y\:dx =\frac{3}{8}$$
$$ \lim_{n \to \infty} \int^{1}_{0}(x^{n}+(1-x)^{n})^{\frac{1}{n}}dx=2\lim_{n \to \infty} \int^{\frac{1}{2}}_{0}(x^{n}+(1-x)^{n})^{\frac{1}{n}}dx=\frac{3}{4}$$
In addition. A detailed answer to the question raised in several remarks : How to prove the last equation above ?
$\int_{\frac{1}{2}}^{1}(x^{n}+(1-x)^{n})^{\frac{1}{n}}dx$, with change of variable $x=(1-X)$ or $X=(1-x)$ , then $dx=-dX$. The lower bounday $x=\frac{1}{2}$ becomes $X=1-\frac{1}{2}=\frac{1}{2}$, the upper boundary $x=1$ becomes $X=1-1=0$.
$$
\int_{\frac{1}{2}}^{1}(x^{n}+(1-x)^{n})^{\frac{1}{n}}dx= \int_{\frac{1}{2}}^{0}((1-X)^{n}+(1-(1-X))^{n})^{\frac{1}{n}}(-dX)$$
$$
\int_{\frac{1}{2}}^{1}(x^{n}+(1-x)^{n})^{\frac{1}{n}}dx= -\int_{\frac{1}{2}}^{0}((1-X)^{n}+X^{n})^{\frac{1}{n}}dX= \int^{\frac{1}{2}}_{0}((1-X)^{n}+X^{n})^{\frac{1}{n}}dX
$$
A basic property of the defined integrals : one can change the symbol of the dummy variable without changing the value of the integral. So, one is allowed to remplace $X$ by $x$ or any other symbol.
$$
\int_{\frac{1}{2}}^{1}(x^{n}+(1-x)^{n})^{\frac{1}{n}}dx= \int^{\frac{1}{2}}_{0}((1-x)^{n}+x^{n})^{\frac{1}{n}}dx=\frac{3}{8}
$$
$$
\int^{1}_{0}(x^{n}+(1-x)^{n})^{\frac{1}{n}}dx= \int^{\frac{1}{2}}_{0}(x^{n}+(1-x)^{n})^{\frac{1}{n}}dx+\int_{\frac{1}{2}}^{1}(x^{n}+(1-x)^{n})^{\frac{1}{n}}dx=\frac{3}{8}+\frac{3}{8}=\frac{3}{4}
$$

