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

Circle is $x^{2}+y^{2}=4$ with centre $(0,0)$ and radius $2$.

se required area is shaded area of circle from

$(x=0\ to\ x=2)$

$=\displaystyle\int_{0}^{2}\sqrt{4-x^{2}}dx=\displaystyle\int_{0}^{2}\sqrt{2^{2}-x^{2}}dx$

$=\left.\left(\dfrac{x}{2}\sqrt{2^{2}-x^{2}}+\dfrac{4}{2}\sin^{-1}\dfrac{x}{2}\right)\right]_{0}^{2}$ $\displaystyle\int \sqrt{a^{2}-x^{2}}dx=\dfrac{x}{a}\sqrt{a^{2}-x^{2}}+\dfrac{a^{2}}{2}\sin^{-1}\dfrac{x}{a}$

$=\dfrac{2}{2}\sqrt{2^{2}-2^{2}}+\dfrac{4}{2}\sin^{-1}\dfrac{2}{2}-\dfrac{0}{2}\sqrt{2^{2}-0^{2}}-\dfrac{4}{2}\sin^{-1}\dfrac{0}{2}$

$=1\sqrt{0}+2\sin^{-1}1-0-2\sin^{-1}0$

$=2\times \dfrac{\pi}{2}-2\times 0=\pi\ sq.units$

so answer is $(A)\pi$ unit .....

OR

As circle of radius $2$. so area of circle $=\pi r^{2}$

$\pi \times 2^{2}=4\pi$

As we need area of only $1$ quadrant

$\Rightarrow \dfrac{1}{4}\times$ area of circle $=\dfrac{1}{4}\times 4\pi =\pi\ sq.units$.

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