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geerky42
 one year ago
Is it me or whoever answered my question makes no sense?
http://www.peeranswer.com/study/mathematics/557e2b603102d10905d35bd5
My question:
geerky42
 one year ago
Is it me or whoever answered my question makes no sense? http://www.peeranswer.com/study/mathematics/557e2b603102d10905d35bd5 My question:

This Question is Closed

geerky42
 one year ago
Best ResponseYou've already chosen the best response.0There is this property that states that \[\int_k^kf(x)~\mathrm dx = 0\quad \forall~ k\in\mathbb R\] From what I understand, integral is area enclosed by \(f(x)\) and xaxis. So this property makes sense since \(0\cdot f(k)=0\), given that width of line is \(0\). So saying we have \(f(x)=\dfrac{1}{x}\), is \(\displaystyle \int_0^0 f(x)~\mathrm dx \) still \(0\)? Because \(f(x)\) doesn't exist at \(x=0\) hence integral doesn't exist since it is basically \(0\cdot\text{indetermine}\) So I believe property is supposed to be \[\int_k^kf(x)~\mathrm dx = \begin{cases}0 & \text{if }~f(k)~\exists\\\not\exists & \text{if }~f(k)~\not\exists \end{cases}\] Right? If not, why not?

geerky42
 one year ago
Best ResponseYou've already chosen the best response.0@freckles @SithsAndGiggles
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