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**SlipEternal** First of all, when is $\displaystyle 3x \ge x^2$ and when is $\displaystyle x^2 \ge 3x$? To figure this out, find out when they are equal.

$\displaystyle x^2 = 3x$ when $\displaystyle x=0, x=3$.

So, at $\displaystyle x=-1, 1 = (-1)^2 > 3(-1) = -3$. Hence $\displaystyle x^2 \ge 3x$ on the interval $\displaystyle (-\infty,0]$.

How about on $\displaystyle [0,3]$? Well, $\displaystyle 1 = (1)^2 < 3(1) = 3$, so $\displaystyle 3x \ge x^2$ on the interval $\displaystyle [0,3]$.

So, the correct set-up for this would be:

$\displaystyle \int_{-1}^0 (x^2-3x)dx + \int_0^2 (3x-x^2)dx$.