$\dfrac{y}{3 - x} = \dfrac{4}{3}$; $y = \frac{4}{3}(3 - x)$
$\dfrac{6A_1\bar{a}_1}{L_1} = \dfrac{Pa}{L}(L^2 - a^2) - \dfrac{M}{L}(3a^2 - L^2)$
$\displaystyle \dfrac{6A_1\bar{a}_1}{L_1} = \int_0^3 \dfrac{(y \, dx)(x)}{5}(5^2 - x^2) - \dfrac{4}{5}[ \, 3(4^2) - 5^2 \, ]$
$\displaystyle \dfrac{6A_1\bar{a}_1}{L_1} = \int_0^3 \dfrac{[ \, \frac{4}{3}(3 - x) \, dx \, ](x)}{5}(25 - x^2) - 18.4$
$\displaystyle \dfrac{6A_1\bar{a}_1}{L_1} = \frac{4}{15}\int_0^3 x(3 - x)(25 - x^2) \, dx - 18.4 = 26.76 - 18.4$
$\dfrac{6A_1\bar{a}_1}{L_1} = 8.36 ~ \text{kN}\cdot\text{m}^3$
$M_1L_1 + 2M_2(L_1 + L_2) + M_3L_2 + \dfrac{6A_1\bar{a}_1}{L_1} + \dfrac{6A_2\bar{b}_2}{L_2} = 0$
$-9(5) + 2M_{wall}(5 + 0) + 0 + 8.36 + 0 = 0$
$M_{wall} = 3.664 ~ \text{kN}\cdot\text{m}$ answer
$M_{wall} = \Sigma M_{\text{to the left of wall}}$
$3.664 = 5R - 6(6.5) - \frac{1}{2}(3)(4)[ \, 2 + \frac{2}{3}(3) ] + 4$
$3.664 = 5R - 39 - 24 + 4$
$5R = 62.664$
$R = 12.5328 ~ \text{kN}$ answer
Check by Area-Moment Method
$EI \, t_{B/C} = 0$
$(Area_{BC}) \cdot \bar{X}_B = 0$
$4(1)(5 - \frac{1}{2}) + \frac{1}{4}(5)(\frac{250}{9})[ \, \frac{4}{5}(5) \,] + \frac{1}{2}(5)(5R)[ \, \frac{2}{3}(5) \,] - \frac{1}{2}(5)(9)[ \, \frac{1}{3}(5) \, ]$
$- \frac{1}{2}(5)(39)[ \, \frac{2}{3}(5) \, ] - \frac{1}{3}(5)(50)[ \, \frac{3}{4}(5) \, ] - \frac{1}{4}(2)(\frac{16}{9})[ \, 3 + \frac{4}{5}(2) \, ] = 0$
$18 + \frac{1250}{9} + \frac{125}{3}R - 37.5 - 325 - 312.5 - \frac{184}{45} = 0$
$\frac{125}{3}R = 522.2$
$R = 12.5328 ~ \text{kN}$ okay
