Question

${\int }_a^b\sqrt{(x-a)(b-x)}dx(b>a)$ is equal to

• A. $\frac{\pi {(b-a)}^2}{8}$
• B. $\frac{\pi {(b+a)}^2}{8}$
• C. ${(b-a)}^2$
• D. ${(b+a)}^2$

Answer 1

The correct option is A $\frac{\pi {(b-a)}^2}{8}$

$I={\int }_a^b\sqrt{(x-a)(b-x)}dx(b>a)$
Let $x=a{\cos }^2\theta +b{\sin }^2\theta$
$\therefore dx=(-2a\cos \theta \sin \theta +2b\sin \theta \cos \theta )d\theta =2(b-a)\cos \theta \sin \theta d\theta$
where $x=a$
$a(1-{\cos }^2\theta )=b{\sin }^2\theta \Rightarrow a{\sin }^2\theta =b{\sin }^2\theta$
$(a-b){\sin }^2\theta =0\Rightarrow \theta =0$
when $x=b,\theta =\pi /2$
$\therefore I={\int }_0^{\pi /2}\sqrt{b({\sin }^2\theta -a{\sin }^2\theta )(b{\cos }^2\theta -a{\cos }^2\theta )}\times 2\cos \theta \sin \theta (b-a)d\theta$
$={\int }_0^{\pi /2}2.(b-a)\sin \theta \cos \theta (b-a)\sin \theta \cos \theta d\theta$
$={\int }_0^{\pi /2}\frac{{(b-a)}^2}{2}{\sin }^2\theta d\theta =\frac{{(b-a)}^2}{2}{\int }_0^{\pi /2}{\sin }^2\theta d\theta$
$\Rightarrow I=\frac{{(b-a)}^2}{2}{\int }_0^{\pi /2}\left[\frac{1-\cos 4\theta }{2}\right]d\theta$
$\Rightarrow I=\frac{{(b-a)}^2}{2}\times 2\times \frac{1}{2}\times \frac{\pi }{2}=\frac{\pi }{8}{(b-a)}^2$