Answer
$\int_{1}^{4}(x^2-4x+2)~dx = -3$
Work Step by Step
We can use the definition of the integral in theorem 4 to evaluate the integral:
$\int_{a}^{b}f(x)~dx = \lim\limits_{n \to \infty}\sum_{i=1}^{n}f(x_i)\Delta x$
$\Delta x = \frac{b-a}{n} = \frac{4-1}{n} = \frac{3}{n}$
$x_i = 1+\frac{3i}{n}$
$\int_{1}^{4}(x^2-4x+2)~dx = \lim\limits_{n \to \infty}\sum_{i=1}^{n}f(x_i)\Delta x$
$= \lim\limits_{n \to \infty}\sum_{i=1}^{n}[(1+\frac{3i}{n})^2-4(1+\frac{3i}{n})+2]~(\frac{3}{n})$
$= \lim\limits_{n \to \infty}(\frac{3}{n})\sum_{i=1}^{n}(1+\frac{6i}{n}+\frac{9i^2}{n^2})+(-4-\frac{12i}{n})+(2)$
$= \lim\limits_{n \to \infty}(\frac{3}{n})\sum_{i=1}^{n}(-1-\frac{6i}{n}+\frac{9i^2}{n^2})$
$= \lim\limits_{n \to \infty}(\frac{3}{n})[~\sum_{i=1}^{n}-1-\sum_{i=1}^{n}\frac{6i}{n}+\sum_{i=1}^{n}\frac{9i^2}{n^2}~]$
$= \lim\limits_{n \to \infty}(\frac{3}{n})[-n-\frac{6}{n}\cdot \frac{n(n+1)}{2}+\frac{9}{n^2}\cdot \frac{n(n+1)(2n+1)}{6}]$
$= \lim\limits_{n \to \infty}(\frac{3}{n})[-n-3n-3+3n+\frac{9}{2}+\frac{3}{2n}]$
$= \lim\limits_{n \to \infty}(\frac{3}{n})[-n+\frac{3}{2}+\frac{3}{2n}]$
$= \lim\limits_{n \to \infty}(-3+\frac{9}{2n}+\frac{9}{2n^2})$
$= -3$
