Question

Just another cute integral:

\[\int |x|^3 \; dx \]

and yes this is indefinite ;)

Answer 1

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?

Answer 2

I have never seen a indefinite integral represented like this.

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

What's wrong piece wise representation?

Answer 10

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

Answer 11

so, either we are both wrong or right

Answer 12

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

Answer 13

and your integral does seems better

Answer 14

ohh yeah

Answer 15

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

Answer 16

so, there are multiple ways of representing same integral?

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

-x^4/4 is what we will get.

sgn(-x)x^4/4 = -1* x^4/4

Answer 21

Yes and why is it negative??

Answer 22

I have confused myself lolol

Answer 23

i am confused myself and blabbering :/

Answer 24

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

Answer 25

Ok I got it. Taking example always helps

Answer 26

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

Answer 27

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

Answer 28

Your first integral was right as well

Answer 29

My piece wise representation is pelletty :/

Answer 30

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

Answer 31

yeah hahaha

i never had to integrate such indefinite integrals before.

Answer 32

Why isn't FoolForMath replying?

Answer 33

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

Answer 34

lol

Answer 35

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

Answer 36

@God

Answer 37

0 fans lol

Answer 38

I beat God! by 455 fans

Answer 39

LOLOL

Answer 40

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

Answer 41

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