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The 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
A$\pi$
B$\frac{\pi}{2}$
C$\frac{\pi}{3}$
D$\frac{\pi}{4}$
Answer & Solution
Correct answer: A. $\pi$
1. The circle has radius $r = 2$, so its full area is $\pi r^2 = 4\pi$.
2. The bounds $x = 0$ to $x = 2$ in the first quadrant capture exactly one quarter of the disc.
3. Hence the required area $= \frac{1}{4}\cdot 4\pi = \pi$.
4. The fraction is a quarter, not a half, which rules out $\frac{\pi}{2}$.
_Source: NCERT Class 12 Mathematics Ch 8 "Application of Integrals", p.4_
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