Question

By using the properties of definite integrals, evaluate the integral ${\int }_0^a\frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}}dx$

Answer 1

Let $I={\int }_0^a\frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}}dx$ ............... (1)
It is known that, $\left({\int }_0^{a}f\right(x)dx={\int }_0^{a}f(a-x\left)dx\right)$
$I={\int }_0^a\frac{\sqrt{a-x}}{\sqrt{a-x}+\sqrt{x}}dx$ ............ (2)
Adding (1) and (2), we obtain
$2I={\int }_0^a\frac{\sqrt{x}+\sqrt{a-x}}{\sqrt{x}+\sqrt{a-x}}dx$
$\Rightarrow 2I={\int }_0^{a}1\cdot dx$
$\Rightarrow 2I=[x{]}_0^a$
$\Rightarrow 2I=a\Rightarrow I=\frac{a}{2}$