Question

${\int }_{\alpha }^{\beta }\sqrt{(x-\alpha )(\beta -x)}dx$ equals

• A. $\frac{\pi }{2}(\beta -\alpha )$
• B. $\frac{\pi }{8}(\beta -\alpha )$
• C. $\frac{\pi }{8}{(\beta -\alpha )}^2$
• D. None of these

Answer 1

The correct option is C $\frac{\pi }{8}{(\beta -\alpha )}^2$

Let $I={\int }_{\alpha }^{\beta }\sqrt{(x-\alpha )(\beta -x)}dx$
Put $t=\frac{1}{2}(x-\alpha +x-\beta )=x-\frac{1}{2}(\alpha +\beta )$
So that $(x-\alpha )(\beta -x)=(t+c)(c-t)=c^2-t^2$
Where $c=\frac{1}{2}(\beta -\alpha )$
Thus $I={\int }_{-c}^c\sqrt{c^2-t^2}dt=2{\int }_0^c\sqrt{c^2-t^2}dt$
$=2{[\frac{t}{2}\sqrt{c^2-t^2}+\frac{c^2}{2}{\sin }^{-1}\frac{t}{c}]}_0^c$
$=\frac{\pi }{8}{(\beta -\alpha )}^2$