\(\int^{3} _{2}(x^2 - 2x)\)dx =\(\left[\frac{x^3}{3}-\frac{2x^2}{2}\right ]^{3}_{2}\\\left[\frac{x^3}{3}-x^2 + C\right ]^{3}_{2}\\\left[\frac{3^3}{3}-3^2 + C \right ]-\left[\frac{2^3}{3}-2^2 + C \right ]\\9-9-\left[\frac{8}{3}-4 \right ]\) =\(\frac{-8}{3}\) + 4 = \(\frac{4}{3}\)