# Thread: Average Value of an abs. value function

1. ## Average Value of an abs. value function

Q: Find the average value of $\displaystyle f(x) = | 7 - x |$ $\displaystyle [4,10]$

My solution: I used $\displaystyle \frac {1}{b-a} \int_{a}^{b} f(x)dx$

So I have $\displaystyle \frac {1}{10-4} \int_{4}^{10} (7 - x) dx$, but the result is 0...

KK

2. You need to find, first
$\displaystyle \int_4^{10}|7-x|dx$
The function is continous therefore you can subdivide this in any way you wish,
Do this,
$\displaystyle \int_4^7 |7-x|dx+\int_7^{10}|7-x|dx$
Now on the interval,
$\displaystyle f(x)=|7-x|=7-x$ on $\displaystyle [4,7]$
And,
$\displaystyle f(x)=|7-x|=x-7$ on $\displaystyle [7,10]$
Thus,
$\displaystyle \int_4^7 7-x dx+\int_7^{10} x-7dx$
--------
the result is 0...
Impossible, since the function is continous and,
$\displaystyle |7-x|>0$
Then the dominance rule says,
$\displaystyle \int_4^{10}|7-x|dx>0$
And when you divide a non-zero number by the length of this interval you do not end with a negative number