Question

If $b>a,$ and $I={\int }_a^b\sqrt{\frac{x-a}{b-x}}dx,$ then $I$ equals

• A. $\frac{\pi }{2}(b-a)$
• B. $\pi (b-a)$
• C. $\pi /2$
• D. $2\pi (b-a)$

Answer 1

The correct option is A $\frac{\pi }{2}(b-a)$

Substitute $b-x=t^2$ so that
$I={\int }_{\sqrt{b-a}}^0\sqrt{\frac{b-t^2-a}{t^2}}(-2t)dt$
$=2{\int }_0^c\sqrt{c^2-t^2}dt$, where $c=\sqrt{b-a}$
$=2{[\frac{1}{2}t\sqrt{c^2-t^2}+\frac{c^2}{2}{\sin }^{-1}\left(\frac{t}{c}\right)]}_0^c$
$=0+c^2{\sin }^{-1}\left(1\right)-0$ $=\frac{\pi }{2}(b-a)$