Area lying in the first quadrant and bounded by the circle $x^2+y^2=4$ and the lines $x=0$ and $x=2$ is

The area bounded by the circle and the lines, $x=0$ and $x=2$, in the first quadrant is represented as shaded region in the plot.
$\therefore$ Area $OAB={\int }_0^{2}ydx$
$={\int }_0^2\sqrt{4-x^2}dx$
$={\left[\frac{x}{2}\sqrt{4-x^2}+\frac{4}{2}{\sin }^{-1}\frac{x}{2}\right]}_0^2$
$=2\left(\frac{\pi }{2}\right)=\pi$sq. units
Thus, the correct answer is A.