$\int\limits_1^2\frac{\sqrt{x}}{\sqrt{3-x}+\sqrt{x}}dx$ is equal to
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $\frac{1}{2}$
Let $I=\int_{1}^{2}\frac{\sqrt{x}}{\sqrt{3-x}+\sqrt{x}}\,dx$
Use property $\int_{a}^{b}f(x)\,dx=\int_{a}^{b}f(a+b-x)\,dx$ Here $a=1,\ b=2 \Rightarrow a+b=3$
$I=\int_{1}^{2}\frac{\sqrt{3-x}}{\sqrt{x}+\sqrt{3-x}}\,dx$
Add both:
$2I=\int_{1}^{2}\frac{\sqrt{x}+\sqrt{3-x}}{\sqrt{x}+\sqrt{3-x}}\,dx =\int_{1}^{2}1\,dx=1$
∴ $I=\frac{1}{2}$
Final Answer: $\frac{1}{2}$