Chapter 6.4, Problem 39E

To determine

To evaluate the given integral.

Answer to Problem 39E

  ${\displaystyle \underset{-1}{\overset{1}{\int }}{\left(r+1\right)}^{2}dr}=\frac{8}{3}$

Explanation of Solution

Given:

  ${\displaystyle \underset{-1}{\overset{1}{\int }}{\left(r+1\right)}^{2}dr}$

Calculation:

  $\begin{array}{l}{\displaystyle \underset{-1}{\overset{1}{\int }}{\left(r+1\right)}^{2}dr}={\displaystyle \underset{-1}{\overset{1}{\int }}\left(r^2+2r+1\right)dr}\\ ={\left[\frac{r^3}{3}+2\cdot \frac{r^2}{2}+r\right]}_{-1}^1\\ ={\left[\frac{r^3}{3}+r^2+r\right]}_{-1}^1\\ =\left[\frac{{\left(1\right)}^3}{3}+{\left(1\right)}^2+\left(1\right)\right]-\left[\frac{{\left(-1\right)}^3}{3}+{\left(-1\right)}^2+\left(-1\right)\right]\\ =\frac{1}{3}+2+\frac{1}{3}\\ =\frac{8}{3}\end{array}$