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

$t_{A/C} = \frac{1}{2}a \left( \dfrac{w_oa^2}{EI} \right)(\frac{2}{3}a) + a\left( \dfrac{2w_oa^2}{3EI} \right)(\frac{3}{2}a) + \frac{1}{2}a \left( \dfrac{2w_oa^2}{3EI} \right)(\frac{5}{3}a) - \frac{1}{3}a \left( \dfrac{w_oa^2}{3EI} \right)(\frac{7}{4}a)$

$t_{A/C} = \dfrac{w_oa^4}{3EI} + \dfrac{w_oa^4}{EI} + \dfrac{5w_oa^4}{9EI} - \dfrac{7w_oa^4}{36EI}$

$t_{A/C} = \dfrac{61w_oa^4}{36EI}$

Therefore,

$\delta_{midspan} = \dfrac{61w_oa^4}{36EI}$ *answer*