Let $∆$be the area of the region $\left\{(\mathrm{x},\mathrm{y})\in {\mathrm{ℝ}}^2:{\mathrm{x}}^2+{\mathrm{y}}^2\le 21,{\mathrm{y}}^2\le 4\mathrm{x},\mathrm{x}\ge 1\right\}$. Then $\frac{1}{2}\left(\mathrm{\Delta }-21{\sin }^{-1}\frac{2}{\sqrt{7}}\right)$ is equal to 

$\begin{array}{l}{\mathrm{x}}^2+{\mathrm{y}}^2\le 21,{\mathrm{y}}^2\le 4\mathrm{x},\mathrm{x}\ge 1\\ {\mathrm{x}}^2+4\mathrm{x}\le 21\\ {\mathrm{x}}^2+4\mathrm{x}-21=0\\ (\mathrm{x}+7)(\mathrm{x}-3)=0\\ \mathrm{x}=3,-7\\ ∆=2{\int }_1^3 2\sqrt{\mathrm{x}}\mathrm{dx}+2{\int }_3^{\sqrt{21}} \sqrt{21-{\mathrm{x}}^2}\mathrm{dx}\\ =4\cdot \frac{2}{3}{\left({\mathrm{x}}^{\frac{3}{2}}\right)}_1^3+2{\left[\frac{\mathrm{x}}{2}\sqrt{21-{\mathrm{x}}^2}+\frac{21}{2}{\sin }^{-1}\frac{\mathrm{x}}{\sqrt{21}}\right]}_3^{\sqrt{21}}\\ =\frac{8}{3}[3\sqrt{3}-1]+\left[\left(0+21\frac{\mathrm{\pi }}{2}\right)-\left(3\cdot 2\sqrt{3}+21{\sin }^{-1}\frac{3}{\sqrt{21}}\right)\right]\\ =8\left(\sqrt{3}-\frac{1}{3}\right)+21\frac{\mathrm{\pi }}{2}-6\sqrt{3}-21{\sin }^{-1}\left(\frac{\sqrt{3}}{\sqrt{7}}\right)\\ =8\sqrt{3}-\frac{8}{3}+21\frac{\mathrm{\pi }}{2}-6\sqrt{3}-21{\sin }^{-1}\left(\frac{\sqrt{3}}{\sqrt{7}}\right)\\ =2\sqrt{3}-\frac{8}{3}+21\frac{\mathrm{\pi }}{2}-21{\sin }^{-1}\frac{\sqrt{3}}{\sqrt{7}}\\ =2\sqrt{3}-\frac{8}{3}+21{\sin }^{-1}\frac{2}{\sqrt{7}}\\ \frac{1}{2}\left(\mathrm{\Delta }-21{\sin }^{-1}\frac{2}{\sqrt{7}}\right)=\sqrt{3}-\frac{4}{3}\end{array}$