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:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\pi\)
The circle has radius 2, and x = 0 to x = 2 covers the whole quarter circle in the first quadrant:
\(\displaystyle A = \int_{0}^{2}\sqrt{4-x^2}\,dx\)
\(\displaystyle = \left[\frac{x}{2}\sqrt{4-x^2}+\frac{4}{2}\sin^{-1}\frac{x}{2}\right]_{0}^{2}\)
\(\displaystyle = 0+2\cdot\frac{\pi}{2}-0 = \pi\)
Shortcut: A quarter of a circle of radius 2 has area \(\frac{1}{4}\pi(2)^2 = \pi\).
NCERT (2026-27 edition):
Area Under Simple Curves
