Q. The value of the integral $\int_{3}^{6}\frac{\sqrt{x}}{\sqrt{9 - x} + \sqrt{x}}$ $dx$ is
A) $\frac{3}{2}$
B) 2
C) 1
D) $\frac{1}{2}$
Explanation
(a)
Let $I = \int_{3}^{6}{\frac{\sqrt{x}}{\sqrt{9 - x} + \sqrt{x}}dx}$ …(i)
$= \int_{3}^{6}\frac{\sqrt{9 - x}}{\sqrt{9 - 9 + x} + \sqrt{9 - x}}dx$
$\Longrightarrow I = \int_{3}^{6}\frac{\sqrt{9 - x}}{\sqrt{x} + \sqrt{9 - x}}dx\$ …(ii)
On adding Eqs. (i) and (ii), we get
$2I = \int_{3}^{6}{1\ dx = \lbrack x\rbrack_{3}^{6} = 6 - 3 \Longrightarrow I = \frac{3}{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.