To evaluate the definite integral \(\int^{3} _{2}(x^2 - 2x)dx\), we first need to find the antiderivative of the integrand, which is \(x^2 - 2x\). The antiderivative of this function is \(\frac{x^3}{3} - x^2 + C\), where C is the constant of integration.

Now, we can use the Fundamental Theorem of Calculus to evaluate the definite integral. This theorem states that if F(x) is an antiderivative of f(x) on an interval [a, b], then \(\int^{b} _{a}f(x)dx = F(b) - F(a)\). In this case, we have:

\(\int^{3} _{2}(x^2 - 2x)dx = (\frac{3^3}{3} - 3^2) - (\frac{2^3}{3} - 2^2)\)

\(= (9 - 9) - (\frac{8}{3} - 4)\)

\(= 0 - (\frac{8}{3} - \frac{12}{3})\)

\(= \frac{4}{3}\)

So, the value of the definite integral \(\int^{3} _{2}(x^2 - 2x)dx\) is 4/3.