The value of $\left(\frac{2^{2010}}{1005}\right)\frac{{\displaystyle \underset{0}{\overset{1}{\int }}{x}^{1004}{(1-x)}^{1004}dx}}{{\displaystyle \underset{0}{\overset{1}{\int }}{x}^{1004}{(1-x^{2010})}^{1004}dx}}$ is ___________

$I_1=\underset{0}{\overset{1}{\int }}{x}^{1004}{(1-x)}^{1004}dx$

$I_1=2\underset{0}{\overset{1/2}{\int }}{x}^{1004}{(1-x)}^{1004}dx$

$And I_2=\underset{0}{\overset{1}{\int }}{x}^{1004}{(1-x^{2010})}^{1004}dx$

$Put x^{1005}=t\Rightarrow 1005x^{1004}dx=dt$

${\left.I_2=\frac{1}{1005}{\int }_0^1 1-t^2\right)}^{1004}dt {\int }_0^a f\left(x\right)={\int }_0^a f(a-x)$

$=\frac{1}{1005}\underset{0}{\overset{1}{\int }}t{(2-t)}^{1004}dt$

$=\frac{1}{1005}\underset{0}{\overset{1}{\int }}{t}^{1004}{(2-t)}^{1004}dt$

Put t = 2y

$\Rightarrow dt=2dy$

$I_2=\frac{1}{1005}\underset{0}{\overset{1/2}{\int }}{\left(2y\right)}^{1004}{(2-2y)}^{1004}dt$

$=\frac{1}{1005}\cdot 2\cdot 2^{1004}\cdot 2^{1004}{\int }_0^{1/2} y^{1004}(1-y{)}^{1004}dy (dt=2dy)$

$=\frac{2^{2008}}{1005}.2\underset{0}{\overset{1/2}{\int }}{y}^{1004}{(1-y)}^{1004}dy$

$I_2=\frac{2^{2008}}{1005}.I_1\Rightarrow \frac{I_1}{I_2}=\frac{1005}{2^{2008}}$

$\Rightarrow \frac{2^{2010}{I}_1}{1005I_2}=4$