$\begin{aligned} & \int_0^2[2 x] \mathrm{d} x=\int_0^{1 / 2} 0 \mathrm{~d} x+\int_{\frac{1}{2}}^1 1 \cdot \mathrm{d} x+\int_1^{3 / 2} 2 \mathrm{~d} x+\int_{\frac{3}{2}}^2 3 \mathrm{~d} x \\ & =0+\left(1-\frac{1}{2}\right)+2\left(\frac{3}{2}-1\right)+3\left(2-\frac{3}{2}\right) \\ & =0+\frac{1}{2}+2 \times \frac{1}{2}+3 \times \frac{1}{2}=3\end{aligned}$