Can the Average Value of an Integral Be Negative?

[Jbreezy]

Homework Statement

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

Homework Equations

thx

The Attempt at a Solution

 

[pasmith]

Homework Statement

I have a question can the average value for an integral be negative.

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

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].

 

[HallsofIvy]

Homework Statement

I have a question can the average value for an integral be negative. I don't see why not just checking.

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

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

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

Homework Equations

thxx

The Attempt at a Solution

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,
$$\int_0^1 f(x)dx= -\int_0^1 xdx= -\frac{1}{2}$$.