Area lying in the first quadrant and bounded by the circle

The area bounded by the circle and the lines, x = 0 and x = 2, in the first quadrant is represented as

$\therefore$ Area $\mathrm{OAB}=\int_{0}^{2} y d x$

$=\int_{0}^{2} \sqrt{4-x^{2}} d x$

$=\left[\frac{x}{2} \sqrt{4-x^{2}}+\frac{4}{2} \sin ^{-1} \frac{x}{2}\right]_{0}^{2}$

$=2\left(\frac{\pi}{2}\right)$

$=\pi$ units

Thus, the correct answer is A.