$\dfrac{y}{x} = \dfrac{w_o}{L}$

$y = \dfrac{w_ox}{L}$

Moment at x:

$M = R_Ax - \frac{1}{2}xy(\frac{1}{3}x)$

$M = R_Ax - \frac{1}{6}x^2y$

$M = R_Ax - \dfrac{x^2}{6}\left( \dfrac{w_ox}{L} \right)$

$M = R_Ax - \dfrac{w_ox^3}{6L}$

Thus,

$EI \, y'' = R_Ax - \dfrac{w_ox^3}{6L}$

$EI \, y' = \dfrac{R_Ax^2}{2} - \dfrac{w_ox^4}{24L} + C_1$

$EI \, y = \dfrac{R_Ax^3}{6} - \dfrac{w_ox^5}{120L} + C_1x + C_2$

At x = 0, y = 0, thus C_{2} = 0

At x = L, y’ = 0

$0 = \dfrac{R_AL^2}{2} - \dfrac{w_oL^4}{24L} + C_1$

$C_1 = \dfrac{w_oL^3}{24} - \dfrac{R_AL^2}{2}$

Thus, the deflection equation is

$EI \, y = \dfrac{R_Ax^3}{6} - \dfrac{w_ox^5}{120L} + \left( \dfrac{w_oL^3}{24} - \dfrac{R_AL^2}{2} \right)x$

At x = L, y = 0

$0 = \dfrac{R_AL^3}{6} - \dfrac{w_oL^5}{120L} + \left( \dfrac{w_oL^3}{24} - \dfrac{R_AL^2}{2} \right)L$

$0 = \dfrac{R_AL^3}{6} - \dfrac{w_oL^4}{120} + \dfrac{w_oL^4}{24} - \dfrac{R_AL^3}{2}$

$0 = -\dfrac{R_AL^3}{3} + \dfrac{w_oL^4}{30}$

$\dfrac{R_AL^3}{3} = \dfrac{w_oL^4}{30}$

$R_A = \dfrac{w_oL}{10}$ *answer*