Let the area of the region \ ( x, y ) : x - 2y + 4 ≥ 0, x + 2 y ^ 2...

Given,

$x - 2y + 4 \ge 0, x + 2{y}^{2} \ge 0, x + 4{y}^{2} \le 8, y \ge 0$

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

\left(x, y\right) ≡ \left(- 2, 1\right)$</p><p>And point of intersection of$x - 2y + 4 = 0 \text{ and } x + 4{y}^{2} = 8$will be, \left(x, y\right) ≡ \left(- 1, \dfrac{3}{2}\right)$

Now, plotting the diagramw we get,

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

$A = {∫}_{- 2}^{- 1}\left(\dfrac{x + 4}{2} - \sqrt{\dfrac{- x}{2}}\right) + {∫}_{- 1}^{0}\left(\dfrac{\sqrt{8 - x}}{2} - \sqrt{\dfrac{- x}{2}}\right) d x + {∫}_{0}^{8}\dfrac{\sqrt{8 - x}}{2} d x$

$⇒ A = \dfrac{1}{2}{\left[\dfrac{{\left(x + 4\right)}^{2}}{2}\right]}_{- 2}^{- 1} - \dfrac{1}{\sqrt{2}}{\left[\left(- x\right)\dfrac{3}{2}\left(\dfrac{- 2}{3}\right)\right]}_{- 2}^{- 1} + \dfrac{1}{2}{\left[\left(8 - x\right)\dfrac{3}{2}\left(\dfrac{- 2}{3}\right)\right]}_{- 1}^{0} - \dfrac{1}{\sqrt{2}}{\left[\left(- x\right)\dfrac{3}{2}\left(\dfrac{- 2}{3}\right)\right]}_{- 1}^{0} + \dfrac{1}{2}{\left[\left(8 - x\right)\dfrac{3}{2}\left(\dfrac{- 2}{3}\right)\right]}_{0}^{8}$

$⇒ A = 9 + \dfrac{5}{4} - \dfrac{4}{3}$

$⇒ A = \dfrac{107}{12}$

$⇒ m + n = 119$