$\int\limits^{2}_{0}\frac{\sqrt{x}}{\sqrt{x}+\sqrt{2-x}}dx$ is equal to :

1

The correct answer is Option (1) → 1

$I=\int\limits^{2}_{0}\frac{\sqrt{x}}{\sqrt{x}+\sqrt{2-x}}dx$  ...(1)

$⇒I=\int\limits^{2}_{0}\frac{\sqrt{2-x}}{\sqrt{x}+\sqrt{2-x}}dx$   ...(2)

eq. (1) + eq. (2)

$⇒2I=\int\limits^{2}_{0}\frac{\sqrt{x}+\sqrt{2-x}}{\sqrt{x}+\sqrt{2-x}}dx$

$2I=\int\limits^{2}_{0}1dx⇒2I=2$

$⇒I=1$