Evaluating a definite integral as signed area (triangle below the axis)

The line $f(x) = 3x-9$ crosses the $x$-axis where $3x - 9 = 0$, i.e. at $x=3$. At $x=0$, $f(0) = -9$. So on $[0,3]$ the graph forms a right triangle with base $3$ (from $x=0$ to $x=3$) and height $9$ (the vertical distance $|f(0)|$), lying entirely below the $x$-axis.

The area of this triangle is $\frac{1}{2}(3)(9) = 13.5$. Since the region lies below the $x$-axis, the definite integral is the negative of this area, so $\int_0^3 (3x-9)\,dx = -13.5$.