Question

Evaluate: ${\int }_0^2\sqrt{4-x^2}dx.$

Answer 1

Let $I={\int }_0^2\sqrt{4-x^2}dx$

$I={\int }_0^2\sqrt{4-x^2}dx={\int }_0^2\sqrt{{\left(2\right)}^2-x^2}dx$

$={[\frac{x}{2}\sqrt{4-x^2}+\frac{4}{2}{\sin }^{-1}\frac{x}{2}]}_0^2$

$=(\frac{2}{2}\times 0+2{\sin }^{-1}1)-(0-2{\sin }^{-1}0)$

$=2{\sin }^{-1}\left(\sin \frac{\pi }{2}\right)$

$=\pi$

Hence ${\int }_0^2\sqrt{4-x^2}dx=\pi$.