Given: $f\left(x\right)=\left\{\begin{array}{ll}{x}^2+3x+a,& x\le 1\\ bx+2,& x>1\end{array}\right.$

Now, for function to be continuous,

$\Rightarrow {\left(1\right)}^2+3\times 1+a=b\times 1+2$

$\Rightarrow 4+a=b+2$

$\Rightarrow a-b=-2 ...\left(i\right)$

Now, differentiating and putting the value of $x$,

 $\Rightarrow 2x+3=b$

$\Rightarrow 2\left(1\right)+3=b$

$\Rightarrow b=5$

$\Rightarrow a=3$

$\Rightarrow {\int }_{-2}^{2}f\left(x\right)dx={\int }_{-2}^1\left(x^2+3x+a\right)dx+{\int }_1^2\left(bx+2\right)dx$

$\Rightarrow {\int }_{-2}^{2}f\left(x\right)dx={\int }_{-2}^1\left(x^2+3x+3\right)dx+{\int }_1^2\left(5x+2\right)dx$

$\Rightarrow {\int }_{-2}^{2}f\left(x\right)dx={\left[\frac{x^3}{3}+\frac{3x^2}{2}+3x\right]}_{-2}^1+{\left[\frac{5x^2}{2}+2x\right]}_1^2$

$\Rightarrow {\int }_{-2}^{2}f\left(x\right)dx=\left[\left(\frac{1}{3}+\frac{3}{2}+3\right)-\left(\frac{-8}{3}+6-6\right)\right]+\left[\left(\frac{5\times 4}{2}+2\times 2\right)-\left(\frac{5}{2}+2\right)\right]$

$\Rightarrow {\int }_{-2}^{2}f\left(x\right)dx=\left[\frac{1}{3}+\frac{3}{2}+3+\frac{8}{3}\right]+\left[14-\frac{9}{2}\right]$

$\Rightarrow {\int }_{-2}^{2}f\left(x\right)dx=\frac{15}{2}+14-\frac{9}{2}$

$\Rightarrow {\int }_{-2}^{2}f\left(x\right)dx=17$