Answer
$\frac{9}{8}$
Work Step by Step
$f_{avg}=\frac{1}{b-a}\int_{a}^{b}f(x)dx$
The area under the curve from x=0 to x=8 is 9. We get this value by counting the number of unit boxes and subtracting the positive values, which are above the x-axis, from the negative values (below x-axis).
$f_{avg}=\frac{1}{8-0}(9)=\frac{9}{8}$