Question

The value of the integral $I=\int\limits^{6}_{{3}}$ $\frac{\sqrt{x}}{\sqrt{9-x}+\sqrt{x}}dx$ is :

• A. $\frac{3}{2}$
• B. $2$
• C. $1$
• D. $\frac{1}{2}$

Answer 1

Answer: (A)

$\frac{3}{2}$

Answer 2

Let $I=\int\limits^{6}_{{3}}$ $\frac{\sqrt{x}}{\sqrt{9-x}+\sqrt{x}}dx ...\left(i\right)$ $=\int\limits^{6}_{{3}}$ $\frac{\sqrt{9-x}}{\sqrt{9-9+x}+\sqrt{9-x}}dx$ $\Rightarrow$ $I=\int\limits^{6}_{{3}}$ $\frac{\sqrt{9-x}}{\sqrt{x}+\sqrt{9-x}}dx ...\left(ii\right)$ On adding Eqs. (i) and (ii), we get $2I=\int\limits^{6}_{{3}}$ $\frac{\sqrt{x}+\sqrt{9-x}}{\sqrt{x}+\sqrt{9-x}}dx$ $=\int\limits^{6}_{{3}}$ $1\,dx=\left[x\right]_{3}^{6}$ $=6-3=3$ $\Rightarrow I=\frac{3}{2}$