Question

The value of ${\int }_2^8\frac{\sqrt{10-x}}{\sqrt{x}+\sqrt{10-x}}\mathrm{dx}$ is

Answer 1

$\mathrm{Let}I={\int }_2^8\frac{\sqrt{10-x}}{\sqrt{x}+\sqrt{10-x}}\mathrm{dx}....\left(1\right]\Rightarrow I={\int }_2^8\frac{\sqrt{10-\left(10-x\right)}}{\sqrt{\left(10-x\right)}+\sqrt{10-\left(10-x\right)}}\mathrm{dx}\Rightarrow I={\int }_2^8\frac{\sqrt{x}}{\sqrt{x}+\sqrt{10-x}}\mathrm{dx}....\left(2\right]\mathrm{Adding}\left(1\right]\mathrm{and}\left(2\right],\mathrm{we}\mathrm{get}:\Rightarrow 2I={\int }_2^{8}1\cdot \mathrm{dx}\Rightarrow I=\frac{1}{2}{\left(x\right]}_2^8=\frac{1}{2}\left(8-2\right)=3$