My guess is that you need the shaded area under the curve g(x)=8x+2x2-x3. g(x) has part of its curve beneath the x-axis and part of its curve above the x-axis. We're not told which area is the shaded area.
We need to find out where the curve intersects the x-axis, that is, when g(x)=0.
8x+2x2-x3=x(8+2x-x2)=x(4-x)(2+x). So the intersection points are (0,0), (4,0) and (-2,0).
Consider the area above the x-axis. This lies between x=0 and 4 so these are the limits of integration:
0∫4(8x+2x2-x3)dx=[4x2+⅔x3-¼x4]04=64+128/3-64=128/3. Area above the x-axis=128/3.
Consider the area below the x-axis:
-2∫0(8x+2x2-x3)dx=[4x2+⅔x3-¼x4]-20=(0-(16-16/3-4))=-20/3. The negative sign means the area is below the x-axis.
If we need the total area (the sum of the magnitudes of the two areas) we get 128/3+20/3=148/3 sq units.