Question

The value of $2^{2010}\frac{\underset{0}{\overset{1}{\int }}{x}^{1004}{(1-x)}^{1004}dx}{\underset{0}{\overset{1}{\int }}{x}^{1004}{(1-x^{2010})}^{1004}dx}$ is equal to

• A. $1005$
• B. $2010$
• C. $4020$
• D. $8040$

Answer 1

The correct option is C $4020$

Given,
$I=2^{2010}\frac{\underset{0}{\overset{1}{\int }}{x}^{1004}{(1-x)}^{1004}dx}{\underset{0}{\overset{1}{\int }}{x}^{1004}{(1-x^{2010})}^{1004}dx},$
Let,
$I_1=\underset{0}{\overset{1}{\int }}{x}^{1004}{(1-x)}^{1004}dx=2\underset{0}{\overset{1/2}{\int }}{x}^{1004}{(1-x)}^{1004}dx\text{and}{I}_2=\underset{0}{\overset{1}{\int }}{x}^{1004}{(1-x^{2010})}^{1004}dx$
Putting $x^{1005}=t$
$\Rightarrow 1005x^{1004}dx=dt$
$\Rightarrow$$\frac{1}{1005}\underset{0}{\overset{1}{\int }}(1-t^2{)}^{1004}dt$
$=\frac{1}{1005}\underset{0}{\overset{1}{\int }}\left(t\right(2-t){)}^{1004}dt\{\because \underset{0}{\overset{a}{\int }}f\left(x\right)dx=\underset{0}{\overset{a}{\int }}f(a-x)dx\}$
$=\frac{1}{1005}\underset{0}{\overset{1}{\int }}\left(t^{1004}\right(2-t){)}^{1004}dt$
Now put $t=2y$
$\Rightarrow dt=2dy$
$\Rightarrow I_2=\frac{2}{1005}\underset{0}{\overset{1/2}{\int }}\left(2y{)}^{1004}\right(2-2y{)}^{1004}dy=\frac{2}{1005}\cdot 2^{1004}\cdot 2^{1004}\underset{0}{\overset{1/2}{\int }}{y}^{1004}(1-y{)}^{1004}dy=\frac{1}{1005}{2}^{2009}\underset{0}{\overset{1/2}{\int }}{y}^{1004}(1-y{)}^{1004}dy=\frac{1}{1005}{2}^{2008}{I}_1\Rightarrow \frac{I_1}{I_2}=\frac{1005}{2^{2008}}\Rightarrow I=2^{2010}\frac{I_1}{I_2}=4\cdot 1005=4020$