Chapter 5 - The Integral - 5.4 The Fundamental Theorem of Calculus, Part I - Exercises - Page 258: 43

$\frac{707}{12}.$

We have $$ \int_{-2}^3f(x)dx=\int_{-2}^2f(x)dx+\int_{2}^3f(x)dx\\ =\int_{-2}^2(12-x^2)dx+\int_{2}^3x^3dx\\ =(12x-\frac{1}{3}x^3)_{-2}^2+\frac{1}{4}x^4|_{2}^3\\ =48-\frac{16}{3}+\frac{3^4}{4}-\frac{2^4}{4}=\frac{707}{12}. $$