The value of $\int\limits^{2}_{1}\frac{\sqrt{x}}{\sqrt{3-x}+\sqrt{x}}dx= $ is : |
1 $\frac{1}{2}$ $\frac{1}{3}$ $\frac{1}{4}$ |
$\frac{1}{2}$ |
The correct answer is Option (2) → $\frac{1}{2}$ $I=\int\limits^{2}_{1}\frac{\sqrt{x}}{\sqrt{3-x}+\sqrt{x}}dx$ ...(1) $I=\int\limits^{2}_{1}\frac{\sqrt{1+2-x}}{\sqrt{3-(1+2-x)}+\sqrt{1+2-x}}dx$ $I=\int\limits^{2}_{1}\frac{\sqrt{3-x}}{\sqrt{3-x}+\sqrt{x}}dx$ ...(2) eq. (1) + eq. (2) $2I=\int\limits^{2}_{1}1dx=1$ so $I=\frac{1}{2}$ |