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