# How do you find the average value of f(x)=-2x^3+10x^2-7x+5 as x varies between [1/2,3]?

Jan 8, 2017

$\frac{595}{48}$

#### Explanation:

f(x) is continuous on the closed interval $\left[\frac{1}{2} , 3\right]$

The average value of f(x) is found using.

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{\frac{1}{b - a} {\int}_{a}^{b} f \left(x\right) \mathrm{dx}} \textcolor{w h i t e}{\frac{2}{2}} |}}}$
$\text{where } \left[a , b\right] = \left[\frac{1}{2} , 3\right]$

$\Rightarrow \frac{1}{\frac{5}{2}} {\int}_{\frac{1}{2}}^{3} \left(- 2 {x}^{3} + 10 {x}^{2} - 7 x + 5\right) \mathrm{dx}$

$= \frac{2}{5} {\left[- \frac{1}{2} {x}^{4} + \frac{10}{3} {x}^{3} - \frac{7}{2} {x}^{2} + 5 x\right]}_{\frac{1}{2}}^{3}$

$= \frac{2}{5} \left[\left(- \frac{81}{2} + 90 - \frac{63}{2} + 15\right) - \left(- \frac{1}{32} + \frac{5}{12} - \frac{7}{8} + \frac{5}{2}\right)\right]$

$= \frac{2}{5} \left[33 - \frac{193}{96}\right] = \frac{595}{48}$