$\Sigma M_A = 0$

$4R_2 + M = 100(4)(2)$

$R_2 = 200 - \frac{1}{4}M$

$\Sigma M_B = 0$

$4R_1 = 100(4)(2) + M$

$R_1 = 200 + \frac{1}{4}M$

$(Area_{AB}) \, \bar{X}_A = 0$

$\frac{1}{2}(4)(800 - M)(\frac{4}{3}) - \frac{1}{3}(4)(800)(1) = 0$

$\frac{8}{3}(800 - M) = \frac{3200}{3}$

$M = 400 \, \text{lb}\cdot\text{ft}$ *answer*

The uniform load over span AB will cause segment AB to deflect downward. The moment load equal to 400 lb·ft applied at the free end will cause the slope through B to be horizontal making the deviation of A from the tangent through B equal to zero. The downward deflection therefore due to uniform load will be countered by the moment load.