Explanation
\begin{aligned}
\mathrm{A} & =\int_{0}^{4}\sqrt{4+y}dy-\int_{0}^{2}\sqrt{4-y}dy-\int_{0}^{4}\sqrt{y}dy \\
& =\left(\frac{(4+y)^{\frac{3}{2}}}{\frac{3}{2}}\right)_{0}^{4}+\left(\frac{(4-y)^{\frac{3}{2}}}{\frac{3}{2}}\right)_{0}^{2}-\left(\frac{y^{\frac{3}{2}}}{\frac{3}{2}}\right)_{0}^{4} \\
& =\frac{40\sqrt{2}}{3}-16=\frac{80\sqrt{2}}{6}-16
\end{aligned}
\begin{aligned}
& \mathrm{A}=\frac{80\sqrt{2}}{\alpha}-\beta \\
& \alpha=6,\beta=16 \\
& \alpha+\beta=22
\end{aligned}
