Let the line $$x = - 1$$ divide the area of the region $$ \left\{(x,y): 1+x^{2}\leq y \leq 3 -x\right\} $$ in the ratio m : n, gcd (m, n) = 1. Then m + n is equal to

We need to find the area of the region $$\{(x,y): 1+x^2 \le y \le 3-x\}$$ and determine how the line $$x = -1$$ divides it.

We find the intersection of $$y = 1+x^2$$ and $$y = 3-x$$ by solving $$1+x^2 = 3-x \implies x^2 + x - 2 = 0 \implies (x+2)(x-1) = 0$$, which gives $$x = -2$$ or $$x = 1$$. Thus the region extends from $$x = -2$$ to $$x = 1$$.

The total area is $$A = \int_{-2}^{1} [(3-x) - (1+x^2)]\,dx = \int_{-2}^{1} (2 - x - x^2)\,dx = \left[2x - \frac{x^2}{2} - \frac{x^3}{3}\right]_{-2}^{1}$$. At $$x = 1$$ we have $$2 - \frac{1}{2} - \frac{1}{3} = \frac{12-3-2}{6} = \frac{7}{6}$$, and at $$x = -2$$ we get $$-4 - 2 + \frac{8}{3} = -6 + \frac{8}{3} = -\frac{10}{3}$$, so $$A = \frac{7}{6} - \left(-\frac{10}{3}\right) = \frac{7}{6} + \frac{10}{3} = \frac{27}{6} = \frac{9}{2}$$.

The area to the left of $$x = -1$$ is $$A_1 = \int_{-2}^{-1}(2-x-x^2)\,dx = \left[2x - \frac{x^2}{2} - \frac{x^3}{3}\right]_{-2}^{-1}$$. At $$x = -1$$ this equals $$-2 - \frac{1}{2} + \frac{1}{3} = \frac{-12-3+2}{6} = -\frac{13}{6}$$, and at $$x = -2$$ we reuse $$-\frac{10}{3}$$ from above, giving $$A_1 = -\frac{13}{6} + \frac{10}{3} = -\frac{13}{6} + \frac{20}{6} = \frac{7}{6}$$.

The area to the right of $$x = -1$$ is $$A_2 = A - A_1 = \frac{9}{2} - \frac{7}{6} = \frac{27-7}{6} = \frac{20}{6} = \frac{10}{3}$$.

Hence $$\frac{A_1}{A_2} = \frac{7/6}{10/3} = \frac{7}{6} \cdot \frac{3}{10} = \frac{7}{20}$$, so the ratio is $$m:n = 7:20$$. Since $$\gcd(7,20)=1$$ we have $$m+n=7+20=27$$.

Option 2: 27