Mumbai University > First Year Engineering > sem 2 > Applied Maths 2

Marks : 3

Year : 2014

Put $\sqrt x=t$

When $x=0,t=0 \space \space x=t^2$

$x=1,t=1\space \space dx=2tdt$

$=\int\limits_0^1\sqrt{t-t^2}2t dt$

$=\int\limits_0^1 t^{1/2}(1-t)^{1/2} 2t dt $

$=2\int\limits_0^1 t^{3/2} (1-t)^{1/2} dt$

$=2B\Bigg(\dfrac 52,\dfrac 32\Bigg)$

$=\dfrac {2\sqrt{\dfrac 52}\sqrt{\dfrac 32}}{\sqrt4} $

$=\dfrac {2\dfrac 32.\dfrac 12\sqrt{\dfrac 12}\dfrac 12\sqrt{\dfrac 12}}{3\times 2\times 1}$

$=\dfrac 18\pi$