Evaluating a V-shaped integral using triangle areas

The graph of $f(x) = |x-5|$ lies entirely above the $x$-axis on $[2,8]$, since $f(x) \geq 0$ for all $x$. It forms two right triangles: one on $[2,5]$ with base $5-2=3$ and height $f(2) = 3$, and another on $[5,8]$ with base $8-5=3$ and height $f(8) = 3$.

Each triangle has area $\frac{1}{2}(3)(3) = \frac{9}{2}$. Since both regions lie above the $x$-axis, both areas contribute positively to the integral, so $\int_2^8 |x-5|\,dx = \frac{9}{2} + \frac{9}{2} = 9.$