# Average value for integrals

1. Sep 19, 2013

### Jbreezy

1. The problem statement, all variables and given/known data
I have a question can the average value for an integral be negative. I don't see why not just checking.

You know this evalutation f_ave = (1/b-a) ∫ f(x) dx

2. Relevant equations

thx

3. The attempt at a solution

2. Sep 19, 2013

### pasmith

That is the average value of the function f on [a,b]. It can of course be negative, and will be if f(x) < 0 for all x in [a,b].

3. Sep 19, 2013

### HallsofIvy

Staff Emeritus
If the average value of the function is negative, of course!

More correctly f_ave = (1/(b-a)) ∫ f(x) dx. What you wrote would normally be interpreted
f_ave = ((1/b)-a) ∫ f(x) dx

Of course. Take the simplest example: f(x)= -1 for all x.
Then $$\int_0^1 f(x)dx= -\int_0^1 dx= -(1- 0)= -1$$
Slightly more complicated, if f(x)= -x,
$$\int_0^1 f(x)dx= -\int_0^1 xdx= -\frac{1}{2}$$.