Question

Evaluate ${\int }_2^3\frac{\sqrt{x}dx}{\sqrt{5-x}+\sqrt{x}}$.

Answer 1

Let, $I={\int }_2^3\frac{\sqrt{x}dx}{\sqrt{5-x}+\sqrt{x}}$ .......$\left(1\right)$

We have ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$

$\Rightarrow I={\int }_2^3\frac{\sqrt{x}dx}{\sqrt{5-x}+\sqrt{x}}$

Replace $x\to 3+2-x=5-x$

$I={\int }_2^3\frac{\sqrt{5-x}dx}{\sqrt{5-(5-x)}+\sqrt{5-x}}$

$I={\int }_2^3\frac{\sqrt{5-x}dx}{\sqrt{x}+\sqrt{5-x}}$ .......$\left(2\right)$

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

$2I={\int }_2^3\frac{\sqrt{x}dx}{\sqrt{5-x}+\sqrt{x}}+{\int }_2^3\frac{\sqrt{5-x}dx}{\sqrt{x}+\sqrt{5-x}}$

$2I={\int }_2^3\frac{(\sqrt{x}+\sqrt{5-x})dx}{\sqrt{5-x}+\sqrt{x}}$

$\Rightarrow 2I={\int }_2^{3}dx$

$\Rightarrow 2I={\left[x\right]}_2^3$

$\Rightarrow 2I=3-2=1$

$\therefore I=\frac{1}{2}$