Q. The value of the integral $\int_{a}^{b}{\frac{\sqrt{x}\ dx}{\sqrt{x} + \sqrt{a + b - x.}}\ }is$
A) $\pi$
B) $\frac{1}{2}(b - a)$
C) $\frac{\pi}{2}$
D) $b - a$
Explanation
(b)
Let $I = \int_{a}^{b}{\frac{\sqrt{x\ dx}}{\sqrt{x +}\sqrt{a + b - x}}\ldots(i)}$
$\therefore I = \int_{a}^{b}\frac{\sqrt{a + b - x}}{\sqrt{a + b - x} + \sqrt{x}}\ \ \ \ldots(ii)$
On adding Eqs. (i) and (ii),we get
$2I = \int_{a}^{b}{dx = \lbrack x\rbrack_{a}^{b}} = b - a$
$\Rightarrow \ I = \frac{b - a}{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.