$\begin{aligned} \int_{0}^{4}|x-2| d x &=\int_{0}^{2}(2-x) d x+\int_{2}^{4}(x-2) d x \\ &=2[x]_{0}^{2}-\frac{1}{2}\left[x^{2}\right]_{0}^{2}+\frac{1}{2}\left[x^{2}\right]_{2}^{4}-2[x]_{2}^{4} \\ &=2(2)-\frac{1}{2}(4)+\frac{1}{2}(16-4)-2(4-2) \\ &=4-2+6-4=4 \end{aligned}$