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

$\begin{gathered} m={\int }_{-1}^{-2}(3-x)-(1+{x}^{2})dx \ m=2+\frac{3}{2}-\frac{7}{3}=\frac{7}{6}\, M={\int }_{1}^{-1}(3-x)-(1+{x}^{2})dx=4-\frac{1}{2}(0)-\frac{1}{3}(1+1)=4-\frac{2}{3}=\frac{10}{3} \ m:M=\frac{7}{6}:\frac{10}{3}=\frac{7}{20} \ m+n=27 \end{gathered}$

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