# How do you find the average value of the function for f(x)=2-1/2x, 0<=x<=4?

Mar 22, 2018

The average value of the function on the given interval is $1$.

#### Explanation:

Average value of a function is given by

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

$A = \frac{1}{4} {\int}_{0}^{4} 2 - \frac{1}{2} x \mathrm{dx}$

$A = \frac{1}{4} {\left[2 x - \frac{1}{4} {x}^{2}\right]}_{0}^{4}$

$A = \frac{1}{4} \left(2 \left(4\right) - \frac{1}{4} {\left(4\right)}^{2}\right)$

$A = \frac{1}{4} \left(8 - 4\right)$

$A = \frac{1}{4} \left(4\right)$

$A = 1$

Hopefully this helps!

Mar 22, 2018

The average value of $f$ over $\left[0 , 4\right]$ is 1.

#### Explanation:

The average value of a function over an interval is its (definite) integral over that interval divided by the length of the interval.

${\int}_{0}^{4} \left(2 - \frac{x}{2}\right) \mathrm{dx} = {\left[2 x - {x}^{2} / 4\right]}_{0}^{4}$

$= 4 - 0$

$= 4$

$\frac{{\int}_{0}^{4} \left(2 - \frac{x}{2}\right) \mathrm{dx}}{4 - 0} = \frac{4}{4} = 1$

The average value of $f$ over $\left[0 , 4\right]$ is 1.