Find intersection points of $y=1+x^2$ and $y=3-x$: $1+x^2=3-x\Rightarrow x^2+x-2=0\Rightarrow x=-2$ or $x=1$.

Total region spans $x\in [-2,1]$. Line $x=-1$ splits it into a left strip $x\in [-2,-1]$ and a right strip $x\in [-1,1]$.

Heights: $(3-x)-(1+x^2)=2-x-x^2$.

Left strip area $={\int }_{-2}^{-1}(2-x-x^2)dx={\left[2x-\frac{x^2}{2}-\frac{x^3}{3}\right]}_{-2}^{-1}=\left(-2-\frac{1}{2}+\frac{1}{3}\right)-\left(-4-2+\frac{8}{3}\right)=\frac{7}{6}$.

Right strip area $={\int }_{-1}^1(2-x-x^2)dx=\frac{20}{6}=\frac{10}{3}$.

Ratio $=\frac{7/6}{20/6}=\frac{7}{20}$, so $m=7$, $n=20$ (or read in the order the question gives — either way $m+n=27$).