$EI \, t_{A/B} = (Area_{AB}) \, \bar{X}_A$
$EI \, t_{A/B} = \frac{1}{2}(a + b)[\, w_ob \,(a + b) \, ][ \, \frac{2}{3}(a+b) \, ] - \frac{1}{3}b(\frac{1}{2}w_ob^2)(a + \frac{3}{4}b)$
$EI \, t_{A/B} = \frac{1}{3}w_ob(a + b)^3 - \frac{1}{6}w_ob^3(a + \frac{3}{4}b)$
$EI \, t_{A/B} = \frac{1}{3}w_ob(a + b)^3 - \frac{1}{24}w_ob^3 (4a + 3b)$
$EI \, t_{A/B} = \frac{1}{3}w_ob(a + b)^3 - \frac{1}{24}w_ob^3 [ \, (2a + 2b) + (a + b) + a \, ]$
$EI \, t_{A/B} = \frac{1}{3}w_ob(\frac{1}{2}L)^3 - \frac{1}{24}w_ob^3 [ \, L + \frac{1}{2}L + a \, ]$
$EI \, t_{A/B} = \frac{1}{24}w_oL^3b - \frac{1}{24}w_ob^3 (\frac{3}{2}L + a)$
$EI \, t_{A/B} = \frac{1}{24}w_oL^3b - \frac{1}{48}w_ob^3 (3L + 2a)$
$EI \, t_{A/B} = \frac{1}{24}w_oL^3b - \frac{1}{48}w_ob^3 [ \,3L + (L - 2b) \, ]$
$EI \, t_{A/B} = \frac{1}{24}w_oL^3b - \frac{1}{48}w_ob^3 (4L - 2b)$
$EI \, t_{A/B} = \frac{1}{24}w_oL^3b - \frac{1}{24}w_ob^3 (2L - b)$
$EI \, t_{A/B} = \frac{1}{24}w_ob [ \, L^3 - b^2(2L - b) \, ]$
$EI \, t_{A/B} = \frac{1}{24}w_ob (L^3 - 2Lb^2 + b^3)$ answer
When 2b = L; b = ½L
$EI \, t_{A/B} = \frac{1}{24}w_o(\frac{1}{2}L) [ \, L^3 - 2L(\frac{1}{2}L)^2 + (\frac{1}{2}L)^3 \, ]$
$EI \, t_{A/B} = \frac{1}{48}w_oL \, [ \, L^3 - \frac{1}{2}L^3 + \frac{1}{8}L^3 \, ]$
$EI \, t_{A/B} = \frac{1}{48}w_oL \, [ \, \frac{5}{8}L^3 \, ]$
$EI \, t_{A/B} = \frac{5}{384}w_oL^4$ (okay!)
