# Can't figure out integral question

1. Mar 19, 2006

### Flatland

what is the proper method for finding the average of an integral? For example, the question I'm trying to answer is this:

"Find the average value of the function f(x) = 4 - x² over the interval [-2, 2]."

Now the answer is suppost to be 8/3. But I keep getting an answer of -11/3 no matter what I do. WTF!?!?

2. Mar 19, 2006

### Galileo

f is nonnegative on the interval [-2,2], so how can the average possibly be negative?

3. Mar 19, 2006

### cliowa

Well, what did you do to get -11/3?

The function you are given here is $$\geq 0$$ in [-2,2]. This makes things quite a lot easier.
By integration you can find out the area under the curve. When you say you'd like to know the average value, it means this: replace the function by a constant one that has the same area under its curve (this time it must be a rectangle).
I hope it's clear now.
Best regards

Cliowa

4. Mar 19, 2006

### d_leet

How did you get -11/3? Are you sure you integrated the function correctly?

5. Mar 19, 2006

### Rich B.

Greetings Flatland:

For any continuous function, f(x), on the interval [a,b], the average value of f on that interval is given by:

f_ave = 1/(b-a) * Integral f(x) dx ; [a,b]. So, in the present case,

f_ave = 1/(2+2) * Integral (4 - x^2) dx; [-2,2]
= (1/4)(4x - (1/3) x^3) = (x/12)(12 - x^2); [-2,2]
= (1/6)(12-4) - (-1/6)(12-4)
= (1/6)(8+8)
= 8/3

I hope this helps.

Regards,

Rich B.

6. Mar 19, 2006

### Flatland

nvm I figured out what I did wrong thx