Let $f_{0}(x)=|x|, f_{1}(x)=\left|f_{0}(x)-1\right|, f_{2}(x)=\mid f_{1}(x)-$ $2 \mid$, find the area of the closed part of the figure enclosed by the graph of the function $y=f_{2}(x)$ and the $x$-axis.
Figure 6-3 is the graph of the function $f_{0}(x)=|x|$. Moving this graph down by 1 unit yields the graph of the function $f_{0}(x)=|x|-1$. Creating the symmetric graph of the part below the $x$-axis with respect to the $x$-axis results in Figure 6-4, where the part above the $x$-axis is the graph of $f_{1}(x)=\left|f_{0}(x)-1\right|$. Moving this graph down by 2 units gives the graph of $f_{0}(x)=|x|-2$. Creating the symmetric graph of the part below the $x$-axis with respect to the $x$-axis results in Figure 6-5, where the part above the $x$-axis is the graph of $f_{2}(x)=\left|f_{1}(x)-2\right|$. It is easy to find that the required area is 7.