Question

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

Answer 1

Let $I{\int }_0^a\frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}}dx.I={\int }_0^a\frac{\sqrt{a-x}}{\sqrt{a-x}+\sqrt{a-(a-x)}}dx[\because {\int }_0^{a}f(x)dx={\int }_0^{a}f(a-x\left)dx\right]={\int }_0^a\frac{\sqrt{a-x}}{\sqrt{a-x+\sqrt{x}}}dx$
On adding Eqs. (i) and (ii), we get
$\Rightarrow 2I={\int }_0^a\frac{\sqrt{x}+\sqrt{a-x}}{\sqrt{a-x}+\sqrt{x}}dx,2I={\int }_0^{a}1dx=[x{]}_0^a=a-0=a\Rightarrow I=\frac{a}{2}$