∫_0^3[x]dx= (where [·] denotes the greatest integer function).

\(\int_{0}^{3}[x]dx=\int_{0}^{1}[dx]dx+\int_{1}^{2}[x]dx+\int_{2}^{3}[x]dx\)

$=\int_{0}^{1}0dx+\int_{1}^{2}1dx+\int_{2}^{3}2dx$

$=1×(2-1)+2×(3-2)$

$=3$