Solution:

Concept: Distance is the definite integral of velocity. Integrate \( v = At + Bt^2 \) from \( t=1 \) to \( t=2 \). \( \int_{1}^{2} (At + Bt^2) dt = \left[ A\frac{t^2}{2} + B\frac{t^3}{3} \right]_{1}^{2} \). Evaluating this gives \( \left( 2A + \frac{8B}{3} \right) - \left( \frac{A}{2} + \frac{B}{3} \right) = \frac{3A}{2} + \frac{7B}{3} \).