Let the line x = -1 divide the area of the region (x,y) : 1 + x^2 y 3-x in the ratio m:n, …

Question

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

Accepted Answer

The curves $y=1+x^2$ and $y=3-x$ intersect where $x^2+x-2=0$, i.e. $x=-2, 1$. So the region spans $x \in [-2,1]$.

Left part ($x=-2$ to $-1$):
$$m = \int_{-2}^{-1}\left[(3-x)-(1+x^2)\right]dx = \frac{7}{6}$$

Right part ($x=-1$ to $1$):
$$M = \int_{-1}^{1}\left[(3-x)-(1+x^2)\right]dx = \frac{10}{3}$$

Ratio $m : M = \dfrac{7}{6} : \dfrac{10}{3} = 7:20$, so $m+n = 27$.

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