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# 508

Tired of adding C? - April 28, 2012
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Warning: You may lose marks depending on the teacher!

I don't care how many marks I lose; I am adding monkey from now on.

If you lose points, lose the teacher.

∫4x3 dx = x4 + <.)#)))≦

Hah. I used to add octopus :-)

I'm quite a fan of the tan(C) option...if you say it like it looks, you're really adding a Tank to the math problem...which seems quite explosively fun!

I would add Desmond. Because he is my constant ;).

i always raised things to the flower power, whenever i needed to use a variable exponent.

In Communist Russia, C adds you.

No man. It isn't so. Believe me!

I prefer to add DC... I guess that's why I'm an electrical engineer.

I generally have (integral of f'(x)dx)=f(x)+C+k, where k is the constant such that C=0.

Doesn't work so nicely for the integral of f'(u)du ;-)

I lol'd

or maybe, we could add the last paragraph of today's headline news...

Actually, you shouldn't add a constant, but a locally constant function. If you are integrating 1/x^2, the answer should be -1/x + C(x), where C(x) may take different values for positive x and negative x.

But integrating 1/x^2 from a0 makes no real sense anyway, because the integral is infinite - so the "varying" constant doesn't matter.
In standard Riemann integration (without an improper integral) such an integration isn't even allowed, and in that case, the constant is most definitely constant.

If the notation more loosely refers to just an anti-derivative instead of an actually existing integral from k to x (for some constant k), then I guess a "varying" constant function could make sense, so long as the constant changes at a point where there is a singularity anyway, which sort of takes us back to the fact that if you're integrating over a singularity (say without some sort of normalization like: http://en.wikipedia.org/wiki/Cauchy_principal_value - in which case adding varying constants completely defeats the purpose of taking a principal value!) then you're probably doing something wrong :)

And anyway, the examples given have no singularities, so introducing a "variable" constant would arguably be less correct (in either the integral or the anti-derivative interpretation) - as an integral of a bounded function is continuous, but having a varying constant introduces a discontinuity at whatever point it varies. Okay, yes, it doesn't affect things locally, if the constant function doesn't change locally. But if you're only looking at the function locally it *doesn't matter* what the function does outside of that locality.

TLDR: (Being a complete hypocrite:) Stop being so picky :P

Note: Sorry for sounding pretentious, only trying to encourage some maths chat :)

sorry the a0 should be "a<0 to b%gt;0"
PS: Apologies again, that message comes across as slightly aggressive and it wasn't supposed to at all, it was supposed to be light-hearted/jokey! [in particular the "picky" comment was supposed to just be a joke :P] But I can't delete it now unfortunately, so I just want to reiterate no offence intended! :)

+ (speed of light)

Very funny!

Lol
Let me do one
Int(10x)=5x^2-tan(monkey)+42, where monkey is any constant

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