$\Sigma M_{R2} = 0$

$4R_1 = 400(3)(2.5) + 500(2)$

$R_1 = 1000 \, \text{N}$

$\Sigma M_{R1} = 0$

$4R_2 = 400(3)(1.5) + 500(2)$

$R_2 = 700 \, \text{N}$

$(Area_{AB})\,\bar{X}_A = \frac{1}{2}(4)(2800)(\frac{4}{3}) - \frac{1}{2}(2)(1000)(\frac{2}{3}) - \frac{1}{3}(3)(1800)(\frac{3}{4})$

$(Area_{AB})\,\bar{X}_A = 5\,450 \, \text{N}\cdot\text{m}^3$ *answer*

$(Area_{AB})\,\bar{X}_B = \frac{1}{2}(4)(2800)(\frac{8}{3}) - \frac{1}{2}(2)(1000)(\frac{4}{3}) - \frac{1}{3}(3)(1800)(\frac{9}{4} + 1)$

$(Area_{AB})\,\bar{X}_B = 7\,750 \, \text{N}\cdot\text{m}^3$ *answer*

The distance between the centroid of the second triangle i.e. the centroid of the area of the moment due to the load 500 N about $R_1$ and the support $R_2$ seems to be wrong.

Actually, it must be $4-\frac{2}{3}=\frac{10}{3}$.

So the answer would be 5750 $N\cdot m^3$ not $7750\, N\cdot m^3$.

Please correct it.

Yes, it is obviously an error. Thank you for pointing it out.