Answer
$-\frac{9}{2}$
Work Step by Step
On the interval $0 \leq x \leq 9$, the graph of $(\frac{1}{3}x-2)$ forms two triangles. One triangle is below the x-axis and one triangle is above the x-axis. We can find the area of these two triangles:
below the x-axis:
$A = \frac{1}{2}(6)(2) = 6$
above the x-axis:
$A = \frac{1}{2}(3)(1) = \frac{3}{2}$
Therefore: $\int_{0}^{9} (\frac{1}{3}x-2)~dx = \frac{3}{2} - 6 = -\frac{9}{2}$
