Answer
$\int_{-2}^{0}(x^2+x)~dx = \frac{2}{3}$
Work Step by Step
We can use the definition of the integral in theorem 4 to evaluate the integral:
$\int_{a}^{b}f(x)~dx = \lim\limits_{n \to \infty}\sum_{i=1}^{n}f(x_i)\Delta x$
$\Delta x = \frac{b-a}{n} = \frac{0-(-2)}{n} = \frac{2}{n}$
$x_i = -2+\frac{2i}{n}$
$\int_{-2}^{0}(x^2+x)~dx = \lim\limits_{n \to \infty}\sum_{i=1}^{n}f(x_i)\Delta x$
$= \lim\limits_{n \to \infty}\sum_{i=1}^{n}[(-2+\frac{2i}{n})^2+(-2+\frac{2i}{n})]~(\frac{2}{n})$
$= \lim\limits_{n \to \infty}(\frac{2}{n})\sum_{i=1}^{n}(4-\frac{8i}{n}+\frac{4i^2}{n^2})+(-2+\frac{2i}{n})$
$= \lim\limits_{n \to \infty}(\frac{2}{n})\sum_{i=1}^{n}(2-\frac{6i}{n}+\frac{4i^2}{n^2})$
$= \lim\limits_{n \to \infty}(\frac{2}{n})[~\sum_{i=1}^{n}2-\sum_{i=1}^{n}\frac{6i}{n}+\sum_{i=1}^{n}\frac{4i^2}{n^2}~]$
$= \lim\limits_{n \to \infty}(\frac{2}{n})[2n-\frac{6}{n}\cdot \frac{n(n+1)}{2}+\frac{4}{n^2}\cdot \frac{n(n+1)(2n+1)}{6}]$
$= \lim\limits_{n \to \infty}(\frac{2}{n})[2n-3n-3+\frac{4n}{3}+2+\frac{2}{3n}]$
$= \lim\limits_{n \to \infty}(\frac{2}{n})[\frac{n}{3}-1+\frac{2}{3n}]$
$= \lim\limits_{n \to \infty}(\frac{2}{3}-\frac{2}{n}+\frac{4}{3n^2})$
$= \frac{2}{3}$