Question

The value of definite integral $\frac{2^{2010}{\int }_0^1{t}^{1004}{(1-t)}^{1004}dt}{{\int }_0^1{t}^{1004}{\left({1-t}^{2010}\right)}^{1004}dt}$ is

Answer 1

wt, the given integral

$I={\int }_0^1{t}^{1004}(1-t{)}^{1004}dt$

$I={\int }_0^1{t}^{1004}(1-t^{2010}{)}^{1004}dt$

For, J

Let $t^{1005}=z$

$\Rightarrow 1005\times t^{1005-1}=\frac{dz}{dt}$

$\Rightarrow t^{1004}dt=\frac{dz}{1005}$

Now,

$I={\int }_0^1\frac{1}{1005}\times (1-z^2{)}^{1004}dz$

$I={\int }_0^1\frac{1}{1005}\times \left(\right(1-z\left)\right(1+z){)}^{1004}dz$

Using

${\int }_a^{b}f\left(x\right)dx={\int }_a^b(a+b-x)dx$

$J={\int }_0^1\frac{1}{1005}\times z^{1004}\times (2-z{)}^{1004}dz$

Substituting,

$z=2y$

$\Rightarrow dz=2dy$

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

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

$=\frac{1}{1005}\times 2^{2009}\times {\int }_0^{1/2}{y}^{1004}\times (1-y{)}^{1004}\times (1-y{)}^{1004}\times dy$

now,
$I={\int }_0^1{t}^{1004}\times (1-t{)}^{1004}dt$

using, propr

${\int }_0^{2a}f\left(x\right)dx=2{\int }_0^{a}f\left(x\right)dx$

$I=2{\int }_0^{1/2}{x}^{1004}(1-x{)}^{1004}dx$

Replacing x with y

$I=2{\int }_0^{1/2}{y}^{1004}(1-y{)}^{1004}dy$

$\therefore$ required integral

$=2^{2010}\times \frac{I}{j}$

$=2^{2010}\times \frac{2\times {\int }_0^{1/2}{y}^{1004}(1-y{)}^{1004}dy}{\frac{2^{2009}}{1005}\times {\int }_0^{1/2}{y}^{1004}\times (1-y{)}^{1004}dy}$

$=\frac{2^{2010}\times 2\times 1005}{2^{2009}}=2^{2010-2009}\times 2\times 1005$

$=2\times 2\times 1005$

$=4\times 1005=4020$