Chapter 4 - Integration - 4.6 The Fundamental Theorem Of Calculus - Exercises Set 4.6 - Page 320: 11

$\int_{-2}^1 f(x) dx = 48$

$\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} $