If p and q are respectively the global maximum and global minimum of the function $f\left(x\right)=x^2{e}^{2x}$  on the interval $[-2,2],$ then $p{e}^{-4}+q{e}^4=\mathrm{.....}$

$f\left(x\right)=x^2{e}^{2x}\Rightarrow f\text{'}\left(x\right)=2x^2{e}^{2x}+2x{e}^{2x},f\text{'}\text{'}\left(x\right)=4x^2{e}^{2x}+8x{e}^{2x}+2e^{2x}.$

$f\text{'}\left(x\right)=0\Rightarrow (2x^2+2x)e^{2x}=0\Rightarrow 2x(x+1)=0\Rightarrow x=0or-1.$

$f\left(0\right)=0,f(-1)=e^{-2},f(-2)=4e^{-4},f\left(2\right)=4e^4.$

$\therefore$ Global maximum, $p=4e^4,$ Globar minimu, q=0. Now $p{e}^{-4}+q{e}^4=4e^4{e}^{-4}+O{e}^4=4.$