Area (in sq.units) lying in the first quadrant and bounded by the circle $x^2+y^2=4$ and the lines $x=0$ and $x=2$ is

• Correct answer
  A

  $\pi$
• B

  $\frac{\pi}{2}$
• C

  $\frac{\pi}{3}$.
• D

  $\frac{\pi}{4}$

Solution

$\begin{aligned} & x^2+y^2=4 \\ & \Rightarrow y^2=4-x^2\end{aligned}$
$\begin{aligned} & =\int_0^2 \sqrt{4-x^2} d x \\ & =\left[\frac{x}{2} \sqrt{4-x^2}+\frac{4}{2} \sin ^{-1}\left(\cdot \frac{x}{2}\right)\right]_0^2\end{aligned}$ $=\left[\frac{2}{2} \sqrt{4-(2)^2}+\frac{4}{2} \sin ^{-1}\left(\frac{2}{2}\right)-\frac{0}{2} \sqrt{4-0}-\frac{4}{2} \sin ^{-1}\left(\frac{0}{2}\right)\right]$ $\begin{aligned} & =2 \times \frac{\pi}{2} \\ & =\pi \text { sq. }\end{aligned}$

Hence, the correct answer is option: 'A'.