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Just another cute integral: \[\int |x|^3 \; dx \] and yes this is indefinite ;)

Mathematics
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Hmm \[ \int |x|^2 = \left\{\begin{array}{kcc}\frac{1}{4}\cdot x^4, \quad x>0 \\ 0, \quad x = 0 \\ \frac{-1}{4}x^4, \quad x<0 \end{array}\right.\]Maybe?
I have never seen a indefinite integral represented like this.
Me neither but I never had to integrate modulus functions without limits before. It has to be like this, I am quite sure.

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Other answers:

\[\int\limits_{}^{}|x|^3 dx = \frac{x^3*|x|}{4}+c\] I hope this is right @FoolForMath and congrats on being a new moderator
This can't be right, |x| stays positive but the x^3 can get negative.
I am not sure though. and I can be wrong.
For negative x , |x| = -x and therfore integral value would be positive. Though I can be wrong @ this
For negative x, x^3 * |x| = (-x)^3 (-(-x)) = -x^4
What's wrong piece wise representation?
wait... my piece wise is gonna give same result as yours
so, either we are both wrong or right
\[\huge \int\limits\limits_{}^{}|x|^3 dx = {sgn(x)}* \frac{x^4}{4}+c\]
and your integral does seems better
ohh yeah
sgn(x) how could i not think about it!?!?! :(
so, there are multiple ways of representing same integral?
because every integral on this page is gonna get you the same result
so, how do you rule out others? and make one right.
@Ishaan94 , Is my integral right because substitute -x and check please
-x^4/4 is what we will get. sgn(-x)x^4/4 = -1* x^4/4
Yes and why is it negative??
I have confused myself lolol
i am confused myself and blabbering :/
Same here LOL. However I wolframed it and it is giving my equation. The sgn(x) one. Ditto . ??
Ok I got it. Taking example always helps
In that case how is my first equation wrong , the very first one I mentioned ??
\[\int_{-1}^1 |x|^3 = \int_{-1}^0 -x^3 + \int_{0}^1 x^3 = -(-1/4) + \frac14 = \frac12 \] \[\left[sgn(x) \cdot \frac{x^4}{4}\right]_{-1}^1 = \frac{1}4 + \frac14\]
Your first integral was right as well
My piece wise representation is pelletty :/
Haha. Maths always confuses us even when we have the right answers . :P
yeah hahaha i never had to integrate such indefinite integrals before.
Why isn't FoolForMath replying?
Only 2 people know the answer to your question @Ishaan94 , @FoolForMath and god himself :P
lol
Lol, I had the answer but I only figured it out why is it working. Thanks to M.se again ;)
0 fans lol
I beat God! by 455 fans
LOLOL
So, the answer is \[ \frac 1 4 |x|^3 \cdot x +C \]
Does it matter, @FoolForMath the position of modulus on either x^3 or x :P

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