Part of the area is below the x-axis and part above. For 0≤x≤3 (x between 0 and 3) the curve is below the x-axis. For 3≤x≤5 it's above the axis. Soi if we want the area in real terms (positive quantities) we need to do two separate integrals of ∫ydx:
|0∫3(x2-3x)dx| + 3∫5(x2-3x)dx=|[x3/3-3x2/2]03|+[x3/3-3x2/2]35=
|9-27/2|+(125/3-75/2)=|-9/2|+25/6=9/2+25/6=26/3.
Note that the first integral is negative when evaluated indicating that this area is below the x-axis. The absolute value of this gives us the area we need. The other integral evaluates positively. Total area=26/3.
If we had simply evaluated the integral between 0 and 5 we would have calculated the net area, that is, the area below the axis would have been subtracted from the area above, giving us a misleading answer, because it would have been negative (-⅓).