Question

Evaluate${\int }_a^b\sqrt{(x-a)(b-x)}dx$

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

Answer 1

The correct option is A

$\frac{\pi (b-a{)}^2}{8}$

Explanation for correct option:

Integrating the given integration:

Let $I={\int }_a^b\sqrt{(x-a)(b-x)}dx$

Substitute $x=a{\cos }^2\theta +b{\sin }^2\theta$

$\begin{array}{l}⟹dx=(-2a\cos \theta \sin \theta +2b\sin \theta \cos \theta )d\theta \left[\because \frac{d}{d\theta }\sin \theta =\cos \theta ,\frac{d}{d\theta }\cos \theta =-\sin \theta \right]\\ ⟹dx=(-a\sin 2\theta +b\sin 2\theta )d\theta [\because \sin 2\theta =2\sin \theta \cos \theta ]\\ ⟹dx=(b-a)\sin 2\theta d\theta \end{array}$

$\begin{array}{l}I={\int }_0^{\frac{\pi }{2}}\sqrt{(b-a){\sin }^2\theta \times (b-a){\cos }^2\theta }(b-a)\sin 2\theta d\theta \\ ={\int }_0^{\frac{\pi }{2}}(b-a)\sin \theta \cos \theta (b-a)\sin 2\theta d\theta \\ ={\int }_0^{\frac{\pi }{2}}(b-a{)}^2\times \frac{2}{2}\sin \theta \cos \theta \times \sin 2\theta d\theta \\ ={\int }_0^{\frac{\pi }{2}}(b-a{)}^2\times \frac{1}{2}\sin 2\theta \times \sin 2\theta d\theta \\ ={\int }_0^{\frac{\pi }{2}}(b-a{)}^2\times \frac{1}{2}{\sin }^{2}2\theta d\theta \\ =\frac{(b-a{)}^2}{2}{\int }_0^{\frac{\pi }{2}}{\sin }^{2}2\theta d\theta \\ =\frac{(b-a{)}^2}{2}{\int }_0^{\frac{\pi }{2}}\frac{1-\cos 4\theta }{2}d\theta \left[\because \cos 2\theta =1-2{\sin }^2\theta ⟹2{\sin }^{2}2\theta =1-\cos 4\theta \right]\\ =\frac{(b-a{)}^2}{4}{\left[\theta +\frac{\sin 4\theta }{4}\right]}_0^{\frac{\pi }{2}}d\theta \\ =\frac{(b-a{)}^2}{4}\left[\frac{\pi }{2}+\frac{\sin 4\frac{\pi }{2}}{4}-0-0\right]\\ =\frac{(b-a{)}^2}{4}\left[\frac{\pi }{2}+0-0-0\right]\\ =\frac{\pi (b-a{)}^2}{8}\end{array}$

Thus, ${\int }_a^b\sqrt{(x-a)(b-x)}dx=$$\frac{\pi (b-a{)}^2}{8}$

Hence, option (A) is the correct answer