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Math Help - Simpson's Rule, quick question!

  1. #1
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    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:
    Math Forum - Ask Dr. Math

    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!
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  2. #2
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    \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.
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  3. #3
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    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?
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  4. #4
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    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
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  5. #5
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    Oh, nevermind, I was thinking of it the opposite way! Definitely get it now! Thanks a lot!
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