Mathematics - Area Under Curves Question with Solution | TestHub

The equation of the circle is $x^2+y^2=4$, which represents a circle with centre at $(0,0)$ and radius $2$.

The required area lies in the first quadrant and is bounded by $x=0$ and $x=2$. This region is exactly one-fourth of the total area of the circle.

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

Area $={\left[\frac{x}{2}\sqrt{4-x^2}+\frac{4}{2}{\sin }^{-1}\left(\frac{x}{2}\right)\right]}_0^2$

Area $=\left(0+2{\sin }^{-1}(1)\right)-\left(0+2{\sin }^{-1}(0)\right)$

Area $=2\times \frac{\pi }{2}=\pi$ sq. units.

Answer: $\pi$