Question

Evaluate: ${\int }_3^6\frac{\sqrt{x}dx}{\sqrt{9-x}+\sqrt{x}}$

Answer 1

Let $I={\int }_3^6\frac{\sqrt{x}dx}{\sqrt{9-x}+\sqrt{x}}$ ........$\left(1\right)$

Replace $x\to 3+6-x=9-x$

$\Rightarrow I={\int }_3^6\frac{\sqrt{9-x}dx}{\sqrt{9-9+x}+\sqrt{9-x}}$ $\left(\because \underset{0}{\overset{a}{\int }}f\left(x\right)dx=\underset{0}{\overset{a}{\int }}f(a-x)dx\right)$

$\Rightarrow I={\int }_3^6\frac{\sqrt{9-x}dx}{\sqrt{x}+\sqrt{9-x}}$ .........$\left(2\right)$

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

$2I={\int }_3^6\frac{\sqrt{x}dx}{\sqrt{9-x}+\sqrt{x}}+{\int }_3^6\frac{\sqrt{9-x}dx}{\sqrt{x}+\sqrt{9-x}}$

$2I={\int }_3^6\frac{(\sqrt{x}+\sqrt{9-x})dx}{\sqrt{9-x}+\sqrt{x}}$

$2I={\int }_3^{6}dx$

$2I={\left[x\right]}_3^6$

$\Rightarrow 2I=6-3=3$

$\therefore I=\frac{3}{2}$