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



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Ishaan94
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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?

FoolForMath
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I have never seen a indefinite integral represented like this.

Ishaan94
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Me neither but I never had to integrate modulus functions without limits before. It has to be like this, I am quite sure.

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

Ishaan94
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This can't be right, x stays positive but the x^3 can get negative.

Ishaan94
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I am not sure though. and I can be wrong.

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

Ishaan94
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For negative x, x^3 * x = (x)^3 ((x)) = x^4

Ishaan94
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What's wrong piece wise representation?

Ishaan94
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wait... my piece wise is gonna give same result as yours

Ishaan94
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so, either we are both wrong or right

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

Ishaan94
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and your integral does seems better

Ishaan94
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ohh yeah

Ishaan94
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sgn(x) how could i not think about it!?!?! :(

Ishaan94
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so, there are multiple ways of representing same integral?

Ishaan94
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because every integral on this page is gonna get you the same result

Ishaan94
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so, how do you rule out others? and make one right.

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

Ishaan94
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x^4/4 is what we will get.
sgn(x)x^4/4 = 1* x^4/4

shivam_bhalla
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Yes and why is it negative??

shivam_bhalla
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I have confused myself lolol

Ishaan94
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i am confused myself and blabbering :/

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

shivam_bhalla
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Ok I got it. Taking example always helps

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

Ishaan94
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\[\int_{1}^1 x^3 = \int_{1}^0 x^3 + \int_{0}^1 x^3 = (1/4) + \frac14 = \frac12 \]
\[\left[sgn(x) \cdot \frac{x^4}{4}\right]_{1}^1 = \frac{1}4 + \frac14\]

Ishaan94
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Your first integral was right as well

Ishaan94
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My piece wise representation is pelletty :/

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

Ishaan94
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yeah hahaha
i never had to integrate such indefinite integrals before.

Ishaan94
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Why isn't FoolForMath replying?

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

Ishaan94
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lol

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

Ishaan94
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@God

Ishaan94
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0 fans lol

Ishaan94
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I beat God! by 455 fans

shivam_bhalla
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LOLOL

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

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