Average Value of an abs. value function

**tttcomrader** · October 25th 2006, 04:35 PM

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

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

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

KK

**ThePerfectHacker** · October 25th 2006, 05:01 PM

You need to find, first
$\int_4^{10}|7-x|dx$
The function is continous therefore you can subdivide this in any way you wish,
Do this,
$\int_4^7 |7-x|dx+\int_7^{10}|7-x|dx$
Now on the interval,
$f(x)=|7-x|=7-x$ on $[4,7]$
And,
$f(x)=|7-x|=x-7$ on $[7,10]$
Thus,
$\int_4^7 7-x dx+\int_7^{10} x-7dx$
--------

the result is 0...

Impossible, since the function is continous and,
$|7-x|>0$
Then the dominance rule says,
$\int_4^{10}|7-x|dx>0$
And when you divide a non-zero number by the length of this interval you do not end with a negative number

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