If the area bounded by circle ${\mathrm{x}}^2+{\mathrm{y}}^2=4$, the parabola $\mathrm{y}={\mathrm{x}}^2+\mathrm{x}+1$and the curve $\mathrm{y}=\left[{\sin }^2\frac{\mathrm{x}}{4}+\cos \frac{\mathrm{x}}{4}\right],\text{ (where }[ ]$ denotes the greates integer function) and x-axis is $\left(\sqrt{3}+\frac{2\mathrm{\pi }}{3}-\frac{1}{\mathrm{k}}\right)$, then the numerical quantity k should be:

$$\begin{gathered} \begin{array}{l}{\sin }^2\frac{x}{4}+\cos \frac{x}{4}=1-{\cos }^2\frac{x}{4}+\cos \frac{x}{4}\\ =1-\left({\cos }^2\frac{x}{4}-\cos \frac{x}{4}\right)\\ =\frac{5}{4}-{\left(\cos \frac{x}{4}-\frac{1}{2}\right)}^2 \\ \text{ Now, for }-2\le x\le 2 \\ \begin{array}{l}\Rightarrow -\frac{1}{2}\le \frac{x}{4}\le \frac{1}{2}\\ \Rightarrow {\cos }^2\frac{x}{4}\le \cos \frac{x}{4}\text{ for all }x\in [-2,2]\\ \Rightarrow {\cos }^2\frac{x}{4}+{\sin }^2\frac{x}{4}\le {\sin }^2\frac{x}{4}+\cos \frac{x}{4},\mathrm{\forall }x\in [-2,2]\\ \Rightarrow \left[{\sin }^2\frac{x}{4}+\cos \frac{x}{4}\right]=1\mathrm{\forall }x\in [-2,2] \\ \Rightarrow y=1\end{array}\end{array} \end{gathered}$$

$$\begin{gathered} x=-\sqrt{3}\text{ and }x=\sqrt{3} \\ y=1\text{ and }y=x^2+x+1 \\ \Rightarrow x=-1\text{ and }x=0 \\ \begin{array}{l}={\int }_{-2}^{-\sqrt{3}} \sqrt{4-x^2}dx+{\int }_{-\sqrt{3}}^{-1} dx+{\int }_{-1}^0 \left(x^2+x+1\right)dx \\ +{\int }_0^{\sqrt{3}} dx+{\int }_{\sqrt{3}}^2 \sqrt{4-x^2}dx\\ =\frac{2\pi }{3}+\sqrt{3}-\frac{1}{6}\end{array} \end{gathered}$$