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Mohsin007 Group Title

* The average value of an even function is (-ve ,+ve, infinite, none)

  • one year ago
  • one year ago

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  1. vf321 Group Title
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    Well that depends really... The average value of its entire domain?

    • one year ago
  2. vf321 Group Title
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    I can think of an even function which has a finite average value, and I can think of one that has an infinite one...

    • one year ago
  3. vf321 Group Title
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    If you were talking about an odd function then it'd be easy...

    • one year ago
  4. Mohsin007 Group Title
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    This was a question in my exam and i dnt knw how to figure it out..

    • one year ago
  5. Mohsin007 Group Title
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    What about an odd functon??

    • one year ago
  6. vf321 Group Title
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    For an odd function it would be 0, since on the other side of the y axis you have negative values.

    • one year ago
  7. Mohsin007 Group Title
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    so the best ans is none...??

    • one year ago
  8. vf321 Group Title
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    yes, I guess.

    • one year ago
  9. vf321 Group Title
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    But to your previous question: For an odd function \(f\), we know that \(f(x)=-f(-x)\). As such, the average value (assuming \(f\) exists over all \(\mathbb R\)) is\[\lim_{a\rightarrow\infty}\frac{1}{2a}\int_{-a}^af(x)dx=\lim_{a\rightarrow\infty}\frac{1}{2a}(\int_0^af(x)dx+\int_{-a}^0f(-x)dx))\]\[=\lim_{a\rightarrow\infty}\frac{1}{2a}(\int_0^af(x)dx+\int_{-a}^0-f(-x)dx)\]Let \(u=-x\)\[=\lim_{a\rightarrow\infty}\frac{1}{2a}(\int_0^af(x)dx-\int_{0}^af(u)du)=0\]

    • one year ago
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