$\dfrac{y}{x} = \dfrac{w_o}{L}$
$y = \dfrac{w_o}{L}x$
$M = \frac{1}{2}L(w_o)(\frac{1}{3}L) = \frac{1}{6}w_o L^2$
$R = \frac{1}{2}w_o L$
Moments about B:
Triangular force to the left of B:
$M_1 = -\frac{1}{2}(L - x)(w_o - y)(\frac{1}{3})(L - x)$
$M_1 = -\frac{1}{6}(L - x)^2 \left( w_o - \dfrac{w_ox}{L} \right)$
$M_1 = -\dfrac{w_o(L - x)^3}{6L}$
Triangular upward force:
$M_2 = \frac{1}{2}(xy)(\frac{1}{3}x) = \frac{1}{6}x^2\dfrac{w_ox}{L}$
$M_2 = \dfrac{w_o x^3}{6L}$
Rectangle (wo by x):
$M_3 = -w_ox(\frac{1}{2}x) = -\frac{1}{2}w_o x^2$
Reactions R and M:
$M_4 = Rx = \frac{1}{2}w_oLx$
$M_5 = -M = -\frac{1}{6}w_oL^2$
Deviation at B with the tangent line through C
$EI \, t_{B/C} = (Area_{BC}) \, \bar{X}_B$
$EI \, t_{B/C} = \frac{1}{4}x\left( \dfrac{w_ox^3}{6L} \right)(\frac{1}{5}x) + \frac{1}{2}x(\frac{1}{2}w_oLx)(\frac{1}{3}x) - (\frac{1}{6}w_oL^2)\,x(\frac{1}{2} x) - \frac{1}{3}x(\frac{1}{2}w_o x^2)(\frac{1}{4}x)$
$EI \, t_{B/C} = \dfrac{w_o}{120L}x^5 + \dfrac{w_oL}{12}x^3 - \dfrac{w_oL^2}{12}x^2 - \dfrac{w_o}{24}x^4$
$EI \, t_{B/C} = \dfrac{w_ox^2}{120L} (x^3 + 10L^2x - 10L^3 - 5Lx^2)$
Therefore,
$EI \, \delta = -\dfrac{w_ox^2}{120L} (x^3 + 10L^2x - 10L^3 - 5Lx^2)$
$EI \, \delta = \dfrac{w_ox^2}{120L} (10L^3 - 10L^2x + 5Lx^2 - x^3)$ answer
