# f(avg) why are 2s appearing?

• Jul 20th 2009, 09:48 PM
azurephantom
f(avg) why are 2s appearing?
http://img24.imageshack.us/img24/1061/ahhr.jpg

Hello. I need help learning how to do this. I actually undersrtand everything except why these 2 are appearing.

Can anyone explain to me why they appear on these (f)avg equations?
• Jul 20th 2009, 09:56 PM
janvdl
Quote:

Originally Posted by azurephantom
http://img24.imageshack.us/img24/1061/ahhr.jpg

Hello. I need help learning how to do this. I actually undersrtand everything except why these 2 are appearing.

Can anyone explain to me why they appear on these (f)avg equations?

I only took a quick look at it, but it would appear that when you substitute (-3) and subtract it, you actually get the same term as when you substitute (+3). (Symmetric function?)

So you actually have this:
$2 \times \frac{1}{6} \left[ \frac{x^3}{3} + 7x \right] ^{3} _{-3}$
• Jul 20th 2009, 10:30 PM
azurephantom
Oh now I understand. This is really help me speed through some problems.
• Jul 21st 2009, 02:08 AM
mr fantastic
Small correction in red:

Quote:

Originally Posted by janvdl
I only took a quick look at it, but it would appear that when you substitute (-3) and subtract it, you actually get the same term as when you substitute (+3). (Symmetric function?)

So you actually have this:
$2 \times \frac{1}{6} \left[ \frac{x^3}{3} + 7x \right] ^{3} _{{\color{red}0}}$

This follows because the function being integrated is even:

$\int_{-a}^{a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx$ when $f(x)$ is an even function.
• Jul 21st 2009, 02:14 AM
Moo
Actually, the way it's done, it would rather be :

Since $g(x)=\frac{x^3}{3}+7x$ is an odd function, then $\left[\frac{x^3}{3}+7x\right]_{-3}^3=g(3)-g(-3)=2g(3)$