Question

Prove that ${\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f(a-x)dx$ and hence evaluate ${\int }_0^a\frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}}dx$.

Answer 1

Let $I={\int }_0^{a}f\left(x\right)dx$ Put $x=a-t\Rightarrow dx=-dt$
when $x=0\Rightarrow t=a$; when $x=a\Rightarrow t=0$

$I={\int }_a^{0}f(a-t)(-dt)$

$={\int }_0^{a}f(a-t)dt$ (using property of definite integrals)

$={\int }_0^{a}f(a-x)dx$ (changing variable $t$ to $x$)

Consider $I={\int }_0^a\frac{\sqrt{x}dx}{\sqrt{x}+\sqrt{a-x}}....\left(1\right)$

$={\int }_0^a\frac{\sqrt{a-x}dx}{\sqrt{a-x}+\sqrt{a-(a-x)}}={\int }_0^a\frac{\sqrt{a-x}dx}{\sqrt{a-x}+\sqrt{a-a+x}}$

$I={\int }_0^a\frac{\sqrt{a-x}dx}{\sqrt{a-x}+\sqrt{x}}.....\left(2\right)$

Adding $\left(1\right)$ and $\left(2\right)$ we get

$2I={\int }_0^a\frac{\sqrt{x}+\sqrt{a-x}}{\sqrt{a-x}+\sqrt{x}}dx={\int }_0^{a}1dx=x{|}_0^a$

$\therefore I=\frac{a}{2}$.