Solution:

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

$\begin{aligned}
\therefore \text { Area } \mathrm{OAB} &=\int_{0}^{2} y d x \\
&=\int_{0}^{2} \sqrt{4-x^{2}} d x \\
&=\left[\frac{x}{2} \sqrt{4-x^{2}}+\frac{4}{2} \sin ^{-1} \frac{x}{2}\right]_{0}^{2} \\
&=2\left(\frac{\pi}{2}\right) \\
&=\pi \text { units }
\end{aligned}$
As a result, the correct answer is $A$.