## Calculus, 10th Edition (Anton)

$\int_{-2}^1 f(x) dx = 48$
$\int_{-2}^1 (x^2-6x+12) dx$ $=[\frac{x^3}{3} -\frac{6x^2}{2} + \frac{12x}{1}]_{-2}^1$ $=[\frac{1^3}{3} -(3)1^2 + 12(1)]-[\frac{(-2)^3}{3}-3(-2)^2 + 12(-2)]$ $= [\frac{1}{3} -3+12]-[-\frac{8}{3}-12-24]$ $=[\frac{28}{3}]-[-\frac{116}{3}]$ $=\frac{144}{3}$ $=\boxed{48}$
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