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Math Help - f(avg) why are 2s appearing?

  1. #1
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    Thumbs up 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?
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  2. #2
    Bar0n janvdl's Avatar
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    Quote Originally Posted by azurephantom View Post


    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}
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    Oh now I understand. This is really help me speed through some problems.
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  4. #4
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    Small correction in red:

    Quote Originally Posted by janvdl View Post
    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.
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  5. #5
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    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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