# What is the average value of the function f(x)=18x+8 on the interval [0,10]?

Oct 27, 2016

$98$

#### Explanation:

The average value of $f$ on $\left[a , b\right]$ is

$\frac{1}{b - a} {\int}_{a}^{b} f \left(x\right) \mathrm{dx}$.

For this problem, that is

$\frac{1}{10 - 0} {\int}_{0}^{10} \left(18 x + 8\right) \mathrm{dx} = \frac{1}{10} {\left[9 {x}^{2} + 8 x\right]}_{0}^{10}$

$= \frac{1}{10} \left[980\right] = 98$.