Question

The value of ${\int }_1^3\frac{\sqrt{x}}{\sqrt{4-x}+\sqrt{x}}dx$ is

Answer 1

$I={\int }_1^3\frac{\sqrt{x}}{\sqrt{4-x}+\sqrt{x}}dx$ ...(1)
Using ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$
$I={\int }_1^3\frac{\sqrt{4-x}}{\sqrt{4-4+x}+\sqrt{4-x}}dx={\int }_1^3\frac{\sqrt{4-x}}{\sqrt{x}+\sqrt{4-x}}dx$ ...(2)
Adding (1) and (2), we get
$\Rightarrow 2I={\int }_1^3\frac{\sqrt{x}+\sqrt{4-x}}{\sqrt{4-x}+\sqrt{x}}dx={\int }_1^{3}dx={\left[x\right]}_1^3=2\Rightarrow I=1$