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

This Question is Closed

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Hmm \[ \int x^2 = \left\{\begin{array}{kcc}\frac{1}{4}\cdot x^4, \quad x>0 \\ 0, \quad x = 0 \\ \frac{1}{4}x^4, \quad x<0 \end{array}\right.\]Maybe?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I have never seen a indefinite integral represented like this.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Me neither but I never had to integrate modulus functions without limits before. It has to be like this, I am quite sure.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{}^{}x^3 dx = \frac{x^3*x}{4}+c\] I hope this is right @FoolForMath and congrats on being a new moderator

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0This can't be right, x stays positive but the x^3 can get negative.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I am not sure though. and I can be wrong.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0For negative x , x = x and therfore integral value would be positive. Though I can be wrong @ this

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0For negative x, x^3 * x = (x)^3 ((x)) = x^4

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0What's wrong piece wise representation?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0wait... my piece wise is gonna give same result as yours

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so, either we are both wrong or right

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\huge \int\limits\limits_{}^{}x^3 dx = {sgn(x)}* \frac{x^4}{4}+c\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and your integral does seems better

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0sgn(x) how could i not think about it!?!?! :(

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so, there are multiple ways of representing same integral?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0because every integral on this page is gonna get you the same result

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so, how do you rule out others? and make one right.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@Ishaan94 , Is my integral right because substitute x and check please

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0x^4/4 is what we will get. sgn(x)x^4/4 = 1* x^4/4

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yes and why is it negative??

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I have confused myself lolol

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i am confused myself and blabbering :/

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Same here LOL. However I wolframed it and it is giving my equation. The sgn(x) one. Ditto . ??

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ok I got it. Taking example always helps

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0In that case how is my first equation wrong , the very first one I mentioned ??

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\int_{1}^1 x^3 = \int_{1}^0 x^3 + \int_{0}^1 x^3 = (1/4) + \frac14 = \frac12 \] \[\left[sgn(x) \cdot \frac{x^4}{4}\right]_{1}^1 = \frac{1}4 + \frac14\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Your first integral was right as well

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0My piece wise representation is pelletty :/

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Haha. Maths always confuses us even when we have the right answers . :P

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yeah hahaha i never had to integrate such indefinite integrals before.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Why isn't FoolForMath replying?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Only 2 people know the answer to your question @Ishaan94 , @FoolForMath and god himself :P

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Lol, I had the answer but I only figured it out why is it working. Thanks to M.se again ;)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I beat God! by 455 fans

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So, the answer is \[ \frac 1 4 x^3 \cdot x +C \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Does it matter, @FoolForMath the position of modulus on either x^3 or x :P
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