Question

$\underset{0}{\overset{2}{\int }}\sqrt{4-x^2}dx$

Answer 1

$$\begin{gathered} {\int }_0^2\sqrt{4-x^2}dx \\ ={\int }_0^2\sqrt{2^2-x^2}dx \\ ={\left[\frac{x}{2}\sqrt{4-x^2}+\frac{1}{2}\times 2^2{\sin }^{-1}\frac{x}{2}\right]}_0^2 \\ ={\left[\frac{x}{2}\sqrt{4-x^2}\right]}_0^2+2{\left[{\sin }^{-1}\frac{x}{2}\right]}_0^2 \\ =0+2\left(\frac{\mathrm{\pi }}{2}-0\right) \\ =\mathrm{\pi } \end{gathered}$$