U might have seen it before......

$I=\displaystyle \int_{0}^{1} ((1-x^{a})^{\frac{1}{a}}-x)^{2}.dx$

let $x^{a}=t$

$ax^{a-1}dx=dt$

$I=\frac{1}{a} \displaystyle \int_{0}^{1}((1-t)^{\frac{1}{a}}-t^{\frac{1}{a}})^{2}~t^{\frac{1}{a}-1}dt$

let $\frac{1}{a}=c$

$I=c \displaystyle \int_{0}^{1}((1-t)^{c}-t^{c})^{2}t^{c-1}dt$ ......... $(1)$

using $\displaystyle \int_{a}^{b}f(x)dx=\displaystyle \int_{a}^{b}f(a+b-x)dx$

$I=c \displaystyle \int_{0}^{1}((1-t)^{c}-t^{c})^{2}(1-t)^{c-1}dt$ ...... $(2)$

adding $1$ and $2$

$2I=c \displaystyle \int_{0}^{1}((1-t)^{c}-t^{c})^{2}((1-t)^{c-1}+t^{c-1})dt$

let $(1-t)^{c}-t^{c}=z$

$2I=-\displaystyle \int_{1}^{-1} z^{2}dz$

$2I= \displaystyle \int_{-1}^{1}z^{2}dz$

$2I=2 \displaystyle \int_{0}^{1}z^{2}dz$

$I=\frac{1}{3}$

X is in (0 1),so x^2014=0