If  $f\left(x\right)=3\left\{\frac{\left[x\right]}{3}\right\}, g\left(x\right)=\left[\frac{1}{1+x^2}\right]$  and   $\varphi \left(x\right)=g\left(f\left(x\right)\right)\cdot {\left\{x\right\}}^2+g(f\left(x\right)-1)+g(f\left(x\right)-2)\cdot (1-\left\{x\right\}),$  where  $\left[x\right]$ and  $\left\{x\right\}$ denote greatest integer function and fractional parts of ‘x’ respectively, then  $\underset{0}{\overset{2023}{\int }}\varphi \left(x\right)dx=$

$f\left(x\right)=\{\begin{array}{cc}0,& if\left[x\right]=3k\\ 1,& if\left[x\right]=3k+1\\ 2,& if\left[x\right]=3k+2\end{array} \Rightarrow f\left(x\right)$  is periodic with period 3

$g\left(x\right)=\{\begin{array}{cc}0,& ifx\ne 0\\ 1,& ifx=0\end{array}$

$\varphi \left(x\right)=\{\begin{array}{cc}{\left\{x\right\}}^2,& if\left[x\right]=3k\\ 1,& if\left[x\right]=3k+1\\ 1-\left\{x\right\},& if\left[x\right]=3k+2\end{array} \Rightarrow \varphi \left(x\right)$  is periodic with period 3

$$\begin{gathered} \underset{0}{\overset{3}{\int }}\varphi \left(x\right)dx=\underset{0}{\overset{1}{\int }}{x}^{2}dx+\underset{1}{\overset{2}{\int }}1dx+\underset{2}{\overset{3}{\int }}(3-x)dx =\frac{1}{3}+4-\frac{5}{2} \\ \underset{0}{\overset{2023}{\int }}\varphi \left(x\right)dx=\underset{0}{\overset{1}{\int }}{x}^{2}dx+\underset{1}{\overset{2023}{\int }}\varphi \left(x\right)dx =\frac{1}{3}+674\underset{0}{\overset{3}{\int }}\varphi \left(x\right)dx =1236 \end{gathered}$$