# Simpson's Rule, quick question!

• Oct 26th 2009, 09:31 PM
ilovecalculus
Simpson's Rule, quick question!
I'm reading through my textbook, and I am not quite following the derivation of Simpson's rule. The one in my textbook is almost identical to the derivation on this website:
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I just can't figure out why the term "(Bx^2)/2" just dissapears! In my textbook it's there on one line and the next line it's just gone and that's all that changed between the two lines. Could someone please explain to me why this happens? Thank you!
• Oct 26th 2009, 09:49 PM
scorpion007
$\int_{-h}^{h} f(x)\,dx = 0$ if f(x) is odd.

$\int_{-h}^{h} \{ax^2+\underbrace{bx}_{\text{odd}}+c\}\,dx =\int_{-h}^{h} \{ax^2+c\}\,dx$

[-h,h] is a symmetric interval so any odd function cancels when integrated.
• Oct 26th 2009, 09:59 PM
ilovecalculus
Ohh, I think I get it, so since we set the bounds at -h and h, we end up with -Bh and +Bh. But what about the fact that if you integrated it anyway, you'd get 1/2Bh^2, because h^2 and (-h)^2 aren't opposite?
• Oct 26th 2009, 10:25 PM
scorpion007
If you integrate Bx from -h to h you will get 0.

$\int_{-h}^{h} Bx\,dx=\frac{Bx^2}{2}\bigg|_{-h}^h=\frac{B(h)^2}{2}-\frac{B(-h)^2}{2}=0$
• Oct 26th 2009, 10:31 PM
ilovecalculus
Oh, nevermind, I was thinking of it the opposite way! Definitely get it now! Thanks a lot! (Clapping)