Question

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

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

Answer 1

The correct option is D $\frac{3}{2}$

$I={\int }_3^6\frac{\sqrt{x}}{\sqrt{9-x}+\sqrt{x}}dx={\int }_3^6\frac{\sqrt{3+6-x}}{\sqrt{9-3-6+x}+\sqrt{3+6-x}}dx={\int }_3^6\frac{\sqrt{9-x}}{\sqrt{x}+\sqrt{9-x}}dx$
$\therefore 2I={\int }_3^6\frac{\sqrt{x}}{\sqrt{9-x}+\sqrt{x}}dx+{\int }_3^6\frac{\sqrt{9-x}}{\sqrt{x}+\sqrt{9-x}}dx={\int }_3^{6}dx=6-3=3.\therefore I=\frac{3}{2}$