\[\int\limits (dx)^2\] = ?
integral of (dx)^2?? o.O
maybe the result will be a first derivative =_=
xdx maybe?
Well, if dx is the infinitesimal change in x, then dx^2 is an even more infinitesimal change in x, so dx^2=dx, so the answer is x :P
Not a real question.
so infinitesimal infinitesimal is just the same infinitesimal?
If you take something infinitely small and then make it smaller, you still have something infinitely small. Just like how \(\infty+1=\infty\).
This notation isn't defined. In any sense.
The one thing I didn't account for is that dx could be negative, which would make dx^2 -dx, which would make the primitive \(\pm x\)
i wasn't geting what it means.
Is this even possible? I dont think it makes any sense
indeed =_=
No, this isn't possible, and it doesn't make sense, I'm just making jokes. Bad math jokes that nobody finds funny.
is this the problem itself @binary3i ? or is it a result of your solution?
That's like saying \(3(+)^{2}3\)
That's not exactly defined.
Guys, guys,... We can only beat a deadhouse an infinitesimal amount of time.
math jokes were never meant to be funny
deadhorse*
An abelian grape!
thanks guys. lol
Somewhat related: No neutrinos are welcome to this bar. A neutrino entered the bar.
neutrinos?
@FoolForGirls, it's a reference to the false discovery that neutrinos travel faster than light.
ommigosh :O
lol @inkyvoyd you have been typing for tooooooo long lad.
It's actually glitched. I haven't even focused on the textbox -_-
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