anonymous
  • anonymous
* The average value of an even function is (-ve ,+ve, infinite, none)
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Well that depends really... The average value of its entire domain?
anonymous
  • anonymous
I can think of an even function which has a finite average value, and I can think of one that has an infinite one...
anonymous
  • anonymous
If you were talking about an odd function then it'd be easy...

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anonymous
  • anonymous
This was a question in my exam and i dnt knw how to figure it out..
anonymous
  • anonymous
What about an odd functon??
anonymous
  • anonymous
For an odd function it would be 0, since on the other side of the y axis you have negative values.
anonymous
  • anonymous
so the best ans is none...??
anonymous
  • anonymous
yes, I guess.
anonymous
  • anonymous
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 \(u=-x\)\[=\lim_{a\rightarrow\infty}\frac{1}{2a}(\int_0^af(x)dx-\int_{0}^af(u)du)=0\]

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