Chapter 6 - Applications of the Integral - 6.2 Setting Up Integrals: Volume, Density, Average Value - Exercises - Page 298: 45

$-4$

The average is given as follows $$ \frac{1}{3-(-1)} \int_{-1}^{3}\left(2 x^{3}-6 x^{2}\right) d x=\frac{1}{4} \int_{-1}^{3}\left(2 x^{3}-6 x^{2}\right) d x=\left.\frac{1}{4}\left(\frac{1}{2} x^{4}-2 x^{3}\right)\right|_{-1} ^{3}=\frac{1}{4}\left(-\frac{27}{2}-\frac{5}{2}\right)=-4 $$