Area lying in the 6rst quadrant and bounded by circle x² + y² = 4 and the lines x = 0 and x = 2 is

The area bounded by the circle and the lines in the first quadrant is represented as :

$\mathrm{A}={\int }_0^2 \mathrm{ydx}$

  $\begin{array}{l}={\int }_0^2 \sqrt{4-{\mathrm{x}}^2}\mathrm{dx}\\ ={\left[\frac{\mathrm{x}}{2}\sqrt{4-{\mathrm{x}}^2}+\frac{4}{2}{\sin }^{-1}\frac{\mathrm{x}}{2}\right]}_0^2\\ =\mathrm{\pi }\text{ sq. units }\end{array}$