$\int\limits^{2}_{0}\frac{\sqrt{x}}{\sqrt{x}+\sqrt{2-x}}dx$ is equal to :
Answer & explanation
Correct answer: option 4
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)
$⇒2I=\int\limits^{2}_{0}\frac{\sqrt{x}+\sqrt{2-x}}{\sqrt{x}+\sqrt{2-x}}dx$
$2I=\int\limits^{2}_{0}1dx⇒2I=2$
$⇒I=1$