# Thread: f(avg) why are 2s appearing?

1. ## f(avg) why are 2s appearing?

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?

2. Originally Posted by azurephantom

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:
$\displaystyle 2 \times \frac{1}{6} \left[ \frac{x^3}{3} + 7x \right] ^{3} _{-3}$

3. Oh now I understand. This is really help me speed through some problems.

4. Small correction in red:

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:
$\displaystyle 2 \times \frac{1}{6} \left[ \frac{x^3}{3} + 7x \right] ^{3} _{{\color{red}0}}$
This follows because the function being integrated is even:

$\displaystyle \int_{-a}^{a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx$ when $\displaystyle f(x)$ is an even function.

5. Actually, the way it's done, it would rather be :

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