Let the area of the region (x,y):x-2y+4≥ 0, x+2y^2≥ 0,x+4y^2≤ 8,y≥ 0 be m n, where m and n are coprime numbers. Then m+n is equal to .

Given,

$$x-2y+4\geq 0,x+2{y}^{2}\geq 0,x+4{y}^{2}\leq 8,y\geq 0$$x−2y+4≥0,x+2y2≥0,x+4y2≤8,y≥0

Now, finding the point of intersection of x-2y+4=0&x+2{y}^{2}=0 we get,

$$(x,y)\equiv (-2,1)$$(x,y)≡(−2,1)

And point of intersection of x-2y+4=0&x+4{y}^{2}=8 will be, $(x,y)\equiv (-1,\frac{3}{2})$(x,y)≡(−1,23​)

Now, plotting the diagramw we get,

Now, from the above diagram, the required area will be,

$$A={\int }_{-2}^{-1}(\frac{x+4}{2}-\sqrt{\frac{-x}{2}})+{\int }_{-1}^{0}(\frac{\sqrt{8-x}}{2}-\sqrt{\frac{-x}{2}})dx+{\int }_{0}^{8}\frac{\sqrt{8-x}}{2}dx$$A=∫−2−1​(2x+4​−2−x​​)+∫−10​(28−x​​−2−x​​)dx+∫08​28−x​​dx

$$\Rightarrow A=\frac{1}{2}{[\frac{{(x+4)}^{2}}{2}]}_{-2}^{-1}-\frac{1}{\sqrt{2}}{[{(-x)}^{\frac{3}{2}}(\frac{-2}{3})]}_{-2}^{-1}+\frac{1}{2}{[{(8-x)}^{\frac{3}{2}}(\frac{-2}{3})]}_{-1}^{0}-\frac{1}{\sqrt{2}}{[{(-x)}^{\frac{3}{2}}(\frac{-2}{3})]}_{-1}^{0}+\frac{1}{2}{[{(8-x)}^{\frac{3}{2}}(\frac{-2}{3})]}_{0}^{8}$$⇒A=21​[2(x+4)2​]−2−1​−2​1​[(−x)23​(3−2​)]−2−1​+21​[(8−x)23​(3−2​)]−10​−2​1​[(−x)23​(3−2​)]−10​+21​[(8−x)23​(3−2​)]08​

$$\Rightarrow A=9+\frac{5}{4}-\frac{4}{3}$$⇒A=9+45​−34​

$$\Rightarrow A=\frac{107}{12}$$⇒A=12107​

$$\Rightarrow m+n=119$$⇒m+n=119