Chapter 5 - The Integral - 5.2 The Definite Integral - Exercises - Page 246: 54

$-16$

We have $$ \int_{-2}^{2}\left(2 x^{3}-3x^2\right) d x\\ =\int_{-2}^{0}\left(2 x^{3}-3x^2\right) d x +\int_{0}^{2}\left(2 x^{3}-3x^2\right) d x\\ =-\int_{0}^{-2}\left(2 x^{3}-3x^2\right) d x +\int_{0}^{2}\left(2 x^{3}-3x^2\right) d x\\ =-(2\frac{(-2)^4}{4}-3\frac{(-2)^3}{3})+(2\frac{(2)^4}{4}-3\frac{(2)^3}{3})=-16. $$