Question

Evaluate the following integrals:

${\int }_2^8\frac{\sqrt{10-x}}{\sqrt{x}+\sqrt{10-x}}dx$

Answer 1

Let I = ${\int }_2^8\frac{\sqrt{10-x}}{\sqrt{x}+\sqrt{10-x}}dx$ .....(1)
Then,
$$\begin{gathered} I={\int }_2^8\frac{\sqrt{10-\left(2+8-x\right)}}{\sqrt{2+8-x}+\sqrt{10-\left(2+8-x\right)}}dx\left[{\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f\left(a+b-x\right)dx\right] \\ ={\int }_2^8\frac{\sqrt{x}}{\sqrt{10-x}+\sqrt{x}}dx.....\left(2\right) \end{gathered}$$

Adding (1) and (2), we have

$$\begin{gathered} 2I={\int }_2^8\frac{\sqrt{10-x}+\sqrt{x}}{\sqrt{x}+\sqrt{10-x}}dx \\ \Rightarrow 2I={\int }_2^{8}dx \\ \Rightarrow 2I={\overline{)x}}_2^8 \\ \Rightarrow 2I=8-2=6 \\ \Rightarrow I=3 \end{gathered}$$