f(avg) why are 2s appearing?

**azurephantom** · July 20th 2009, 08:48 PM

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?

**janvdl** · July 20th 2009, 08:56 PM

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

**azurephantom** · July 20th 2009, 09:30 PM

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

**mr fantastic** · July 21st 2009, 01:08 AM

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

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.

**Moo** · July 21st 2009, 01:14 AM

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

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

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