Practicing Success
The value of the integral $\int\limits_3^6 \frac{\sqrt{x}}{\sqrt{9-x}+\sqrt{x}} d x$, is |
$\frac{1}{2}$ $\frac{3}{2}$ 2 1 |
$\frac{3}{2}$ |
$I=\int\limits_3^6 \frac{\sqrt{x}}{\sqrt{9-x}+\sqrt{x}} d x$ ...(1) so $I=\int\limits_3^6\frac{\sqrt{9-x}}{\sqrt{9-x}+\sqrt{x}}dx$ ...(2) Eq. (1) + Eq. (2) $⇒2I=\int\limits_3^61dx=3$ $I=\frac{3}{2}$ |