The area (in sq. units) of the region lying in the first quadrant and enclosed by the X -axis, the straight…

Required area$=Ar\left(∆OAB\right)+{\int }_{\sqrt{3}}^2\sqrt{4-x^2}dx$

Now,

$Ar\left(∆OAB\right)=\frac{1}{2}\times \sqrt{3}\times 1=\frac{\sqrt{3}}{2} ...\left(i\right)$

And, let

$I={\int }_{\sqrt{3}}^2\sqrt{4-x^2}dx$

  $={\left[\frac{1}{2}x\sqrt{4-x^2}+\frac{4}{2}{\sin }^{-1}\left(\frac{x}{2}\right)\right]}_{\sqrt{3}}^2$

$=\left[\left\{\frac{1}{2}·2·\sqrt{4-4}+2{\sin }^{-1}\left(\frac{2}{2}\right)\right\}-\left\{\frac{\sqrt{3}}{2}+2{\sin }^{-1}\left(\frac{\sqrt{3}}{2}\right)\right\}\right]$

$=\left[\left\{2\times \frac{\pi }{2}\right\}-\left\{\frac{\sqrt{3}}{2}+2\left(\frac{\pi }{3}\right)\right\}\right]$

$=\frac{\pi }{3}-\frac{\sqrt{3}}{2} ...\left(ii\right)$

By $\left(i\right) \& \left(ii\right)$, we get

Required area$=\frac{\pi }{3} \mathrm{sq}.\mathrm{units}$.