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

1. 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?

2. anonymous

I have never seen a indefinite integral represented like this.

3. anonymous

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

4. 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

5. anonymous

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

6. anonymous

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

7. anonymous

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

8. anonymous

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

9. anonymous

What's wrong piece wise representation?

10. anonymous

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

11. anonymous

so, either we are both wrong or right

12. anonymous

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

13. anonymous

and your integral does seems better

14. anonymous

ohh yeah

15. anonymous

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

16. anonymous

so, there are multiple ways of representing same integral?

17. anonymous

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

18. anonymous

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

19. anonymous

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

20. anonymous

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

21. anonymous

Yes and why is it negative??

22. anonymous

I have confused myself lolol

23. anonymous

i am confused myself and blabbering :/

24. anonymous

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

25. anonymous

Ok I got it. Taking example always helps

26. anonymous

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

27. anonymous

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

28. anonymous

Your first integral was right as well

29. anonymous

My piece wise representation is pelletty :/

30. anonymous

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

31. anonymous

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

32. anonymous

33. anonymous

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

34. anonymous

lol

35. anonymous

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

36. anonymous

@God

37. anonymous

0 fans lol

38. anonymous

I beat God! by 455 fans

39. anonymous

LOLOL

40. anonymous

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

41. anonymous

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