The value of the integral, $\int_3^6 \frac{\sqrt{x}}{\sqrt{9-x}+\sqrt{x}} d x$ is
The value of the integral, $\int_3^6 \frac{\sqrt{x}}{\sqrt{9-x}+\sqrt{x}} d x$ is
1/2
3/2
2
1
Solution
$I=\int_3^6 \frac{\sqrt{x}}{\sqrt{9-x}+\sqrt{x}} d x$
$I=\int_3^6 \frac{\sqrt{9-x}}{\sqrt{9-x}+\sqrt{x}} d x$
$2 I=\int_3^6 d x=3 \Rightarrow I=\frac{3}{2}$