Question 440073
Looking at the graph:
{{{graph( 300, 300, -7, 7, -7, 7, 5x - x^2 )}}}

The parabolic segment of which we want the area has width w = 5, and height h = 25/4.
Hence by the formula of a parabolic segment, 

{{{A = (2/3)wh = (2/3)*5*(25/4) = 125/6}}} square units.

If we used integral, we get the same area value:
{{{A = int((5x - x^2), dx, 0, 5) = ((5x^2)/2 - x^3/3)[0]^5}}}

={{{125/2 - 125/3 = 125/6}}}