Answer
$\int_{0}^{1}(x^3-3x^2)~dx = -\frac{3}{4}$
Work Step by Step
We can use the definition of the integral in Theorem 4 to evaluate the integral:
$\int_{a}^{b}f(x)~dx = \lim\limits_{n \to \infty}\sum_{i=1}^{n}f(x_i)\Delta x$
$\Delta x = \frac{b-a}{n} = \frac{1-0}{n} = \frac{1}{n}$
$x_i = \frac{i}{n}$
$\int_{0}^{1}(x^3-3x^2)~dx = \lim\limits_{n \to \infty}\sum_{i=1}^{n}f(x_i)\Delta x$
$= \lim\limits_{n \to \infty}\sum_{i=1}^{n}[(\frac{i}{n})^3-3(\frac{i}{n})^2]~(\frac{1}{n})$
$= \lim\limits_{n \to \infty}~\frac{1}{n}~\sum_{i=1}^{n}(\frac{i^3}{n^3}-\frac{3i^2}{n^2})$
$= \lim\limits_{n \to \infty}~(\frac{1}{n})[\frac{n^2(n+1)^2}{4n^3}-\frac{3n(n+1)(2n+1)}{6n^2}~]$
$= \lim\limits_{n \to \infty}~(\frac{n^2+2n+1}{4n^2}-\frac{6n^2+9n+3}{6n^2}~)$
$= \lim\limits_{n \to \infty}~(\frac{1}{4}+\frac{1}{2n}+\frac{1}{4n^2}-1-\frac{3}{2n}-\frac{1}{2n^2}~)$
$= (\frac{1}{4}+0+0-1-0-0)$
$ = -\frac{3}{4}$
