Which of the following is correct concerning the value of $I = \int\limits_0^a {\sqrt {\frac{x}{{a – x}}} } dx $, $a > 0$?

(A) < 1.5 a
(B) $a\pi $
(C) $2a \pi $
(D) > 1.5 a

Solution

For ${\sqrt {\frac{x}{{a – x}}} }$ to be real, ${\frac{x}{{a – x}}}\geq 0$

Since a > 0, $x \geq 0$ which means the numerator of ${\frac{x}{{a – x}}} \geq 0 $

So, a – x > 0 which means x < a

So, the substitution $x = a{\sin ^2}\theta $ can be used which satisfies the limits imposed on x.

The given integral, $I = \int\limits_0^{\pi /2} {\sqrt {\frac{{a{{\sin }^2}\theta }}{{a – a{{\sin }^2}\theta }}} } 2a\sin \theta \cos \theta d\theta $

$ \Rightarrow I = \int\limits_0^{\pi /2} {\frac{{\sin \theta }}{{\cos \theta }}} 2a\sin \theta \cos \theta d\theta $

$ \Rightarrow I = a\int\limits_0^{\pi /2} {2{{\sin }^2}\theta d\theta } = a\int\limits_0^{\pi /2} {(1 – \cos 2\theta )d\theta } $

$ \Rightarrow I = a\left( {\left. {\theta – \frac{1}{2}\sin 2\theta } \right|_0^{\pi /2}} \right) = a\frac{\pi }{2}$

Hence, (D)