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The area lying in the first quadrant and bounded by the circle $x^2+y^2=4$ and lines $x=0 \text{ and } x=1$ is given by

  1. $\frac{\pi}{3}+\frac{\sqrt{3}}{2}$
  2. $\frac{\pi}{6}+\frac{\sqrt{3}}{4}$
  3. $\frac{\pi}{3}-\frac{\sqrt{3}}{2}$
  4. $\frac{\pi}{6}+\frac{\sqrt{3}}{2}$
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1 Answer

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$Area =\int_{0}^{1} \sqrt(4-x^2)$

which on solving will give option A

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