# What is the average value of a function f(x) x^2 - 2x + 5 on the interval [2,6]?

Apr 21, 2016

$k = \frac{172}{12}$

#### Explanation:

$\text{let's recall that the average value of a function for an interval}$
$\text{ of (a,b) is given by formula:}$
$k = \frac{1}{b - a} {\int}_{a}^{b} f \left(x\right) d x$
where;
$k : \text{average value}$
$k = \frac{1}{6 - 2} {\int}_{2}^{6} \left({x}^{2} - 2 x + 5\right) d x$

$k = \frac{1}{4} \left(| {x}^{3} / 3 - 2 {x}^{2} / 2 + 5 x {|}_{2}^{6}\right)$

$k = \frac{1}{4} \left(| {x}^{3} / 3 - {x}^{2} + 5 x {|}_{2}^{6}\right)$

$k = \frac{1}{4} \left[\left({6}^{3} / 3 - {6}^{2} + 5 \cdot 6\right) - \left({2}^{3} / 3 - {2}^{2} + 5 \cdot 2\right)\right]$

$k = \frac{1}{4} \left[\left(\frac{216}{3} - 36 + 30\right) - \left(\frac{8}{3} - 4 + 10\right)\right]$

$k = \frac{1}{4} \left[\left(\frac{216}{3} - 6\right) - \left(\frac{8}{3} + 6\right)\right]$

$k = \frac{1}{4} \left[\frac{216 - 18}{3} - \frac{8 + 18}{3}\right]$

$k = \frac{1}{4} \left[\frac{198}{3} - \frac{26}{3}\right]$

$k = \frac{1}{4} \left[\frac{172}{3}\right]$

$k = \frac{172}{12}$