Question

Evaluate ${\int }_1^2\frac{\sqrt{x}}{\sqrt{3-x}+\sqrt{x}}dx$.

Answer 1

Let, $I={\int }_1^2\frac{\sqrt{x}}{\sqrt{3-x}+\sqrt{x}}dx$ .......... (i)
Using properties, ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$
$I={\int }_1^2\frac{\sqrt{3-x}}{\sqrt{3-(3-x)}+\sqrt{3-x}}dx$
$I={\int }_1^2\frac{\sqrt{3-x}}{\sqrt{x}+\sqrt{3-x}}dx$ .......... (ii)
Adding (i) and (ii), $2I={\int }_1^2\frac{\sqrt{x}+\sqrt{3-x}}{\sqrt{x}+\sqrt{3-x}}dx={\int }_1^{2}dx$
$=[x{]}_1^2=2-1=1$
$\therefore I=\frac{1}{2}$