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anonymous
 4 years ago
* The average value of an even function is (ve ,+ve, infinite, none)
anonymous
 4 years ago
* The average value of an even function is (ve ,+ve, infinite, none)

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Well that depends really... The average value of its entire domain?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I can think of an even function which has a finite average value, and I can think of one that has an infinite one...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0If you were talking about an odd function then it'd be easy...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0This was a question in my exam and i dnt knw how to figure it out..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0What about an odd functon??

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0For an odd function it would be 0, since on the other side of the y axis you have negative values.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so the best ans is none...??

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0But 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\[\lim_{a\rightarrow\infty}\frac{1}{2a}\int_{a}^af(x)dx=\lim_{a\rightarrow\infty}\frac{1}{2a}(\int_0^af(x)dx+\int_{a}^0f(x)dx))\]\[=\lim_{a\rightarrow\infty}\frac{1}{2a}(\int_0^af(x)dx+\int_{a}^0f(x)dx)\]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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