The value of $\int\limits^{2}_{1}\frac{\sqrt{x}}{\sqrt{3-x}+\sqrt{x}}dx= $ is :
Answer & explanation
Correct answer: option 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 $I=\frac{1}{2}$