Substitution integral

Exercise:

Compute, $\int_{0}^{2}{\frac{\sqrt{x}}{\sqrt{x}+\sqrt{2-x}}dx}$

Solution: Let $u=2-x\Rightarrow x=2-u\Leftrightarrow du=-dx$

So $I=\int_{0}^{2}{\frac{\sqrt{x}}{\sqrt{x}+\sqrt{2-x}}dx}=-\int_{u\left( 0 \right)}^{u\left( 2 \right)}{\frac{\sqrt{2-u}}{\sqrt{2-u}+\sqrt{u}}du=-\int_{2}^{0}{\frac{\sqrt{2-u}}{\sqrt{2-u}+\sqrt{u}}du}}$

$=\int_{0}^{2}{\frac{\sqrt{2-u}}{\sqrt{2-u}+\sqrt{u}}du}=\int_{0}^{2}{\frac{\sqrt{2-x}}{\sqrt{2-x}+\sqrt{x}}dx}$ Thus $2I=\int_{0}^{2}{dx}=2\Rightarrow I=1$