area under the curve

Find the area enclosed by the curve y=3x(x-2) AND the x-axis
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3x(x-2) kross x-axis at x=0 & x=2... kerv is BELO the axis from 0 tu 2...me thank thats the area yu want... integral from 0 tu 2 av(3x2 -6x)=x^3-3x^2, evaluated at x=2 - values at x=0... at x=0, value=0... at x=2, value=8-12=-4...formula sae area=-4 kauz it belo axis... reel area=+4
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