Step-by-Step Solution

The region is defined by $x^2 + 1 \le y \le 3-x$. First, we find the intersection points of the curves $y = x^2 + 1$ and $y = 3-x$. $x^2 + 1 = 3 - x \implies x^2 + x - 2 = 0 \implies (x+2)(x-1) = 0$. So, $x = -2$ and $x = 1$. The intersection points are $(-2, 5)$ and $(1, 2)$. The area of the region is given by $A = \int_{-2}^{1} (3-x - (x^2+1)) dx = \int_{-2}^{1} (2 - x - x^2) dx = [2x - \frac{x^2}{2} - \frac{x^3}{3}]_{-2}^{1} = (2 - \frac{1}{2} - \frac{1}{3}) - (-4 - 2 + \frac{8}{3}) = 2 - \frac{1}{2} - \frac{1}{3} + 6 - \frac{8}{3} = 8 - \frac{1}{2} - \frac{9}{3} = 8 - \frac{1}{2} - 3 = 5 - \frac{1}{2} = \frac{9}{2}$. Now, we divide the region by the line $x = -1$. We need to find the area to the left of $x = -1$ and to the right of $x = -1$. Area to the left of $x = -1$: $A_1 = \int_{-2}^{-1} (3-x - (x^2+1)) dx = \int_{-2}^{-1} (2 - x - x^2) dx = [2x - \frac{x^2}{2} - \frac{x^3}{3}]_{-2}^{-1} = (-2 - \frac{1}{2} + \frac{1}{3}) - (-4 - 2 + \frac{8}{3}) = -2 - \frac{1}{2} + \frac{1}{3} + 6 - \frac{8}{3} = 4 - \frac{1}{2} - \frac{7}{3} = 4 - \frac{3+14}{6} = 4 - \frac{17}{6} = \frac{24-17}{6} = \frac{7}{6}$. Area to the right of $x = -1$: $A_2 = \int_{-1}^{1} (3-x - (x^2+1)) dx = \int_{-1}^{1} (2 - x - x^2) dx = [2x - \frac{x^2}{2} - \frac{x^3}{3}]_{-1}^{1} = (2 - \frac{1}{2} - \frac{1}{3}) - (-2 - \frac{1}{2} + \frac{1}{3}) = 2 - \frac{1}{2} - \frac{1}{3} + 2 + \frac{1}{2} - \frac{1}{3} = 4 - \frac{2}{3} = \frac{12-2}{3} = \frac{10}{3}$. The ratio is $A_1 : A_2 = \frac{7}{6} : \frac{10}{3} = \frac{7}{6} : \frac{20}{6} = 7 : 20$. So, $m = 7$ and $n = 20$. $m+n = 7+20 = 27$.