The region lies in the first quadrant ($x\ge 0,y\ge 0$), under the parabola $y=4-x^2$ (which meets the axes at $(0,4)$ and $(2,0)$), and above the line $y=1-2x$ (which crosses the x-axis at $x=1/2$ and the y-axis at $y=1$).

For $x\in [0,1/2]$, the lower bound is the line $y=1-2x$; for $x\in [1/2,2]$, the lower bound is $y=0$. The upper bound throughout is $y=4-x^2$.

Area: ${\int }_0^2(4-x^2)dx-{\int }_0^{1/2}(1-2x)dx={\left[4x-\frac{x^3}{3}\right]}_0^2-{\left[x-x^2\right]}_0^{1/2}=\frac{16}{3}-\frac{1}{4}=\frac{64-3}{12}=\frac{61}{12}$.

So $\alpha =61$, $\beta =12$, $\alpha +\beta =73$.