Chapter 5 - Section 5.4 - The Fundamental Theorem of Calculus - Exercises - Page 320: 19

$-\frac{8}{3}$

$\int_{1}^{-1} (r+1)^2 dr=\int_{1}^{-1} (r^2+2r+1) dr=[\frac{r^3}{3}+r^2+r]_{1}^{-1}=[\frac{(-1)^3}{3}+(-1)^2+(-1)]-[\frac{1^3}{3}+1^2+1]=[\frac{-1}{3}+1-1]-[\frac{1}{3}+1+1]=-\frac{1}{3}-\frac{7}{3}=-\frac{8}{3}$