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

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