Chapter 5 - Section 5.4 - The Fundamental Theorem of Calculus - Exercises - Page 320: 1

$-\frac{10}{3}$

$\int_{0}^{2} x(x-3)dx=\int_{0}^{2} (x^2-3x)dx=[\frac{x^3}{3}-\frac{3}{2}x^2]_{0}^{2}=[\frac{2^3}{3}-\frac{3}{2}\cdot 2^2]-[\frac{0^3}{3}-\frac{3}{2}\cdot 0^2]=[\frac{8}{3}-6]-0=[\frac{8}{3}-\frac{18}{3}]=-\frac{10}{3}$