$\begin{align}
\displaystyle \left( \dfrac{6A\bar{a}}{L} \right)_{AB} & = \sum \dfrac{Pa}{L}(L^2 - a^2) \\
& = \int_{x_1}^{x_2} \dfrac{(y \, dx) \, x}{2}(2^2 - x^2) \\
& = \frac{1}{2}\int_{x_1}^{x_2} xy(2^2 - x^2) \, dx \\
& = \frac{1}{2}\int_0^2 x(-0.4x + 2)(2^2 - x^2) \, dx \\
& = \frac{236}{75} ~ \text{m}^2
\end{align}$
$\begin{align}
\displaystyle \left( \dfrac{6A\bar{b}}{L} \right)_{CB} & = \sum \dfrac{Pb}{L}(L^2 - b^2) \\
& = \int_{x_1}^{x_2} \dfrac{(y \, dx)(3 - x)}{3} \left[ 3^2 - (3 - x)^2 \right] \\
& = \frac{1}{3}\int_{x_1}^{x_2} (3 - x)y \left[ 3^2 - (3 - x)^2 \right] \, dx \\
& = \frac{1}{3}\int_0^3 (3 - x)(-0.4x + 1.2) \left[ 3^2 - (3 - x)^2 \right] \, dx \\
& = \frac{108}{25} ~ \text{m}^2
\end{align}$