CUET Preparation Today
| 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 |
\(\pi\) \(\frac{\pi}{2}\) \(\frac{\pi}{3}\) \(\frac{\pi}{4}\) |
| \(\pi\) |
| Required area \(=\left|\int_{0}^{2}\sqrt{4-x^2}\right|\) |
