- anonymous

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

- jamiebookeater

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- anonymous

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?

- anonymous

I have never seen a indefinite integral represented like this.

- anonymous

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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## More answers

- anonymous

\[\int\limits_{}^{}|x|^3 dx = \frac{x^3*|x|}{4}+c\]
I hope this is right @FoolForMath and congrats on being a new moderator

- anonymous

This can't be right, |x| stays positive but the x^3 can get negative.

- anonymous

I am not sure though. and I can be wrong.

- anonymous

For negative x ,
|x| = -x and therfore integral value would be positive. Though I can be wrong @ this

- anonymous

For negative x, x^3 * |x| = (-x)^3 (-(-x)) = -x^4

- anonymous

What's wrong piece wise representation?

- anonymous

wait... my piece wise is gonna give same result as yours

- anonymous

so, either we are both wrong or right

- anonymous

\[\huge \int\limits\limits_{}^{}|x|^3 dx = {sgn(x)}* \frac{x^4}{4}+c\]

- anonymous

and your integral does seems better

- anonymous

ohh yeah

- anonymous

sgn(x) how could i not think about it!?!?! :(

- anonymous

so, there are multiple ways of representing same integral?

- anonymous

because every integral on this page is gonna get you the same result

- anonymous

so, how do you rule out others? and make one right.

- anonymous

@Ishaan94 , Is my integral right because substitute -x and check please

- anonymous

-x^4/4 is what we will get.
sgn(-x)x^4/4 = -1* x^4/4

- anonymous

Yes and why is it negative??

- anonymous

I have confused myself lolol

- anonymous

i am confused myself and blabbering :/

- anonymous

Same here LOL. However I wolframed it and it is giving my equation. The sgn(x) one. Ditto . ??

- anonymous

Ok I got it. Taking example always helps

- anonymous

In that case how is my first equation wrong , the very first one I mentioned ??

- anonymous

\[\int_{-1}^1 |x|^3 = \int_{-1}^0 -x^3 + \int_{0}^1 x^3 = -(-1/4) + \frac14 = \frac12 \]
\[\left[sgn(x) \cdot \frac{x^4}{4}\right]_{-1}^1 = \frac{1}4 + \frac14\]

- anonymous

Your first integral was right as well

- anonymous

My piece wise representation is pelletty :/

- anonymous

Haha. Maths always confuses us even when we have the right answers . :P

- anonymous

yeah hahaha
i never had to integrate such indefinite integrals before.

- anonymous

Why isn't FoolForMath replying?

- anonymous

Only 2 people know the answer to your question @Ishaan94 , @FoolForMath and god himself :P

- anonymous

lol

- anonymous

Lol, I had the answer but I only figured it out why is it working. Thanks to M.se again ;)

- anonymous

- anonymous

0 fans lol

- anonymous

I beat God! by 455 fans

- anonymous

LOLOL

- anonymous

So, the answer is \[ \frac 1 4 |x|^3 \cdot x +C \]

- anonymous

Does it matter, @FoolForMath the position of modulus on either x^3 or x :P

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