Question

The value of the integral, ${\int }_3^6\frac{\sqrt{x}}{\sqrt{9-x}+\sqrt{x}}dx$ is?

• A. $1/2$
• B. $3/2$
• C. $2$
• D. $1$

Answer 1

The correct option is B $3/2$

Let $I={\int }_3^6\frac{\sqrt{x}}{\sqrt{9-x}+\sqrt{x}}dx$ (i)
Also $I={\int }_3^6\frac{\sqrt{3+6-x}}{\sqrt{9-3-6+x}+\sqrt{9-x}}dx$
$\Rightarrow$ $I={\int }_3^6\frac{\sqrt{9-x}}{\sqrt{x}+\sqrt{9-x}}dx$ (ii)
Adding equation (i) and (ii), we get
$2I={\int }_3^6\frac{\sqrt{9-x}+\sqrt{x}}{\sqrt{9-x}+\sqrt{x}}dx={\int }_3^{6}dx=6-3=3$
$\Rightarrow$ $I=3/2$
Ans: B