$EI ~ \theta_{AB} = 0$
$\frac{1}{2}(\frac{1}{2}L)(\frac{1}{4}w_oL^2) - \frac{1}{2}L(M_{wall}) - \frac{1}{3}(\frac{1}{2}L)(\frac{1}{8}w_oL^2) = 0$
$\frac{1}{16}w_oL^3 - \frac{1}{2}M_{wall}L - \frac{1}{48}w_oL^3 = 0$
$\frac{1}{2}M_{wall}L = \frac{1}{24}w_oL^3$
$M_{wall} = \frac{1}{12}w_oL^2$ answer
Note that the actual bending moment is a negative moment (bending the beam downward) as shown in the figure. The answer above is positive which indicates that our assumption of downward arrows is correct. Had we assumed positive M_{wall} (upward arrows), the answer would be negative pointing out that the actual moment is negative (or downward arrows).
$\delta_{max} = t_{A/B}$
$EI ~ \delta_{max} = EI ~ t_{A/B}$
$EI ~ \delta_{max} = (Area_{AB}) \cdot \bar X_A$
$EI ~ \delta_{max} = \frac{1}{2}(\frac{1}{2}L)(\frac{1}{4}w_oL^2)[ \, \frac{2}{3}(\frac{1}{2}L) \, ] - \frac{1}{2}L(M_{wall})[ \, \frac{1}{2}(\frac{1}{2}L) \, ]$
$~ ~ ~ ~ ~ ~ ~~~~ ~~~ ~ ~ ~ - \frac{1}{3}(\frac{1}{2}L)(\frac{1}{8}w_oL^2)[ \, \frac{3}{4}(\frac{1}{2}L) \, ]$
$EI ~ \delta_{max} = \frac{1}{16}w_oL^3(\frac{1}{3}L) - \frac{1}{2}M_{wall}L(\frac{1}{4}L) - \frac{1}{48}w_oL^3(\frac{3}{8}L)$
$EI ~ \delta_{max} = \frac{1}{48}w_oL^4 - \frac{1}{8}M_{wall}L^2 - \frac{1}{128}w_oL^4$
$EI ~ \delta_{max} = \frac{1}{48}w_oL^4 - \frac{1}{8}(\frac{1}{12}w_oL^2)L^2 - \frac{1}{128}w_oL^4$
$EI ~ \delta_{max} = \frac{1}{48}w_oL^4 - \frac{1}{96}w_oL^4 - \frac{1}{128}w_oL^4$
$EI ~ \delta_{max} = \frac{1}{384}w_oL^4$ answer