jmc

If $x<1,$ then $$|x-1|+|x-3|=1-x+3-x=4-2x.$$If $1\le x<3,$ then $$|x-1|+|x-3|=x-1+3-x=4.$$And if $3\le x,$ then $$|x-1|+|x-3|=x-1+x-3=2x-4.$$We can then graph $y=|x-1|+|x-3|$ and $y=8.$



Solving $4x-2x=8,$ we find $x=-2.$ Solving $2x-4=8,$ we find $x=6.$ Thus, the two graphs intersect at $(-2,8)$ and $(6,8).$

Thus, the region we are interested in is a trapezoid with bases 2 and 8, and height 6, so its area is $$\frac{2+8}{2}\cdot 6=\boxed{30}.$$