Question

The value of $I={\int }_0^a\sqrt{\frac{a-x}{x}}dx$ is

• A. $\frac{a}{2}$
• B. $\frac{a}{4}$
• C. $\frac{\pi a}{2}$
• D. $\frac{\pi a}{4}$

Answer 1

The correct option is C

$\frac{\pi a}{2}$

The explanation for the correct answer.

Solve for the value of integral.

$I={\int }_0^a\sqrt{\frac{a-x}{x}}dx$

Let, $x=a{\cos }^2\left(\theta \right)$

$\therefore dx=-2a\cos \left(\theta \right)\sin \left(\theta \right)d\theta$

$$\begin{gathered} I={\int }_{\frac{\mathrm{\pi }}{2}}^0\sqrt{\frac{a-a{\cos }^2\left(\theta \right)}{a{\cos }^2\left(\theta \right)}}\left(-2a\cos \left(\theta \right)\sin \left(\theta \right)d\theta \right) \\ =-2a{\int }_{\frac{\mathrm{\pi }}{2}}^0\sqrt{\frac{a{\sin }^2\left(\theta \right)}{a{\cos }^2\left(\theta \right)}}\left(\cos \left(\theta \right)\sin \left(\theta \right)d\theta \right) \\ =-2a{\int }_{\frac{\mathrm{\pi }}{2}}^0\tan \left(\theta \right)\cos \left(\theta \right)\sin \left(\theta \right)d\theta \\ =-2a{\int }_{\frac{\mathrm{\pi }}{2}}^0{\sin }^2\left(\theta \right)d\theta \\ =-a{\int }_{\frac{\mathrm{\pi }}{2}}^0\left(1-\cos \left(2\theta \right)\right)d\theta \\ =a{\left[\theta -\frac{\sin \left(2\theta \right)}{1}\right]}_0^{\frac{\mathrm{\pi }}{2}} \\ =a\left\{\left[\frac{\mathrm{\pi }}{2}-\frac{\sin \left(\mathrm{\pi }\right)}{2}\right]-0\right\} \\ =\frac{a\mathrm{\pi }}{2} \end{gathered}$$

Hence, option(C) is the correct answer.