$t_{A/C} = \dfrac{1}{EI}(Area_{AC}) \, \bar{X}_A$

$t_{A/C} = \frac{1}{2}a \left( \dfrac{Pa}{2EI} \right)(\frac{2}{3}a) + a \left( \dfrac{Pa}{3EI} \right)(\frac{3}{2}a) + \frac{1}{2}a \left( \dfrac{2Pa}{3EI} - \dfrac{Pa}{3EI} \right)(\frac{5}{3}a)$

$t_{A/C} = \dfrac{Pa^3}{6EI} + \dfrac{Pa^3}{2EI} + \dfrac{5Pa^3}{18EI}$

$t_{A/C} = \dfrac{17Pa^3}{18EI}$

Therefore,

$\delta_{midspan} = \dfrac{17Pa^3}{18EI}$ *answer*