Question

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

• A. $\frac{3}{2}$
• B. $2$
• C. $1$

Answer 1

The correct option is A

$\frac{3}{2}$

Explanation for the correct option

Given: ${\int }_3^6\frac{\sqrt{x}}{\sqrt{9-x}+\sqrt{x}}dx$

Let $I={\int }_3^6\frac{\sqrt{x}}{\sqrt{9-x}+\sqrt{x}}dx-----\left(1\right)$

Putting $x\to 6+3-x$

$I={\int }_3^6\frac{\sqrt{6+3-x}}{\sqrt{9-(9-x)}+\sqrt{6+3-x}}dx$

using property ${\int }_b^{a}f\left(x\right)dx={\int }_b^{a}f(a+b-x)dx$

$I={\int }_3^6\frac{\sqrt{9-x}}{\sqrt{x}+\sqrt{9-x}}dx----\left(2\right)$

adding equations $\left(1\right)$ and $\left(2\right)$

$$\begin{gathered} I+I={\int }_3^6\frac{\sqrt{x}}{\sqrt{9-x}+\sqrt{x}}dx+{\int }_3^6\frac{\sqrt{9-x}}{\sqrt{9-x}+\sqrt{x}}dx \\ 2I={\int }_3^6\frac{\sqrt{9-x}+\sqrt{x}}{\sqrt{9-x}+\sqrt{x}}dx \\ 2I={\int }_3^6\left(1\right)dx \\ 2I={\left|x\right|}_3^6 \\ 2I=6-3 \\ I=\frac{3}{2} \end{gathered}$$

Hence, option (A) is correct.