Integral Calculus> Anyone can help me fast in doing this sum...

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

Mayeen Avez , 8 Years ago

Grade 12

1 Answers

jagdish singh singh

Last Activity: 8 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.$$