# Anyone can help me fast in doing this sum from mathematics-integralcaluclus

jagdish singh singh
173 Points
7 years ago
$\hspace{-0.7 cm.}Let I = \int^{1}_{0}x^{1004}(1-x)^{1004}dx and J = \int^{1}_{0}x^{1004}\bigg(1-x^{2010}\bigg)^{1004}dx\\\\\\ Now we will calculate J =J = \int^{1}_{0}x^{1004}\bigg(1-x^{2010}\bigg)^{1004}dx \\\\ Put x^{1005} = t\;, Then x^{1004}dx = \frac{1}{1005}dt and changing limits, We get\\\\\\ J = \frac{1}{1005}\int^{1}_{0}(1-t^2)^{1004}dt = \frac{1}{1005}\int^{1}_{0}\bigg[1-(1-t)^2\bigg]^{1004}dt\\\\\\$

$\hspace{-0.7 cm.}J = \frac{1}{1005}\int^{1}_{0}(2t-t^2)^{1004}dt = \frac{1}{1005}\int^{1}_{0}t^{1004}(2-t)^{1004}dt\;,\\\\\\ Put t=2u and dt=2du and changing limits, We get\\\\ J = \frac{2^{2008}}{1005}\cdot 2\int ^{\frac{1}{2}}_{0}u^{1004}(1-u)^{1004}du = \frac{2^{2008}}{1005}\int^{1}_{0}u^{1004}(1-u)^{1004}du\\\\\\ We have used \int^{2a}_{0}f(x)dx = 2\int^{a}_{0}f(x)dx\;, When f(2a-x) = f(x)$
$\hspace{-0.7 cm.}So J = \frac{2^{2008}}{1005}\int^{1}_{0}u^{1004}(1-u)^{1004}du = \frac{2^{2008}}{105}\int^{1}_{0}x^{1004}(1-x)^{1004}dx\\\\\\ So we get J = \frac{2^{2008}}{1005}I\Rightarrow I = \frac{1005}{2^{2008}}.J Now given 2^{2010}\cdot \frac{I}{J} = \lambda\\\\\\ So \lambda =2^{2010} \cdot \frac{1005}{2^{2008}}\cdot \frac{J}{J} = 2^2\cdot 1005 = 2^2\cdot 3\cdot 5 \cdot 67\\\\\\ So Highest prime factor of \lambda = 67.$