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