# The total area between the curve y=x³-3x²+2x,0

8 years ago
y=(-2 + x) (-1 + x) x
y=0, imply x=2,1,0
area between 0 to 2 of curve y,
$\int x^3-3x^2+x=x^2/2 - x^3 + x^4/4$
$\left \| \int_{0}^{2}y \right \|=5$