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