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

1. Ishaan94

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. FoolForMath

I have never seen a indefinite integral represented like this.

3. Ishaan94

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

4. shivam_bhalla

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

5. Ishaan94

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

6. Ishaan94

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

7. shivam_bhalla

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

8. Ishaan94

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

9. Ishaan94

What's wrong piece wise representation?

10. Ishaan94

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

11. Ishaan94

so, either we are both wrong or right

12. shivam_bhalla

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

13. Ishaan94

and your integral does seems better

14. Ishaan94

ohh yeah

15. Ishaan94

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

16. Ishaan94

so, there are multiple ways of representing same integral?

17. Ishaan94

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

18. Ishaan94

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

19. shivam_bhalla

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

20. Ishaan94

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

21. shivam_bhalla

Yes and why is it negative??

22. shivam_bhalla

I have confused myself lolol

23. Ishaan94

i am confused myself and blabbering :/

24. shivam_bhalla

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

25. shivam_bhalla

Ok I got it. Taking example always helps

26. shivam_bhalla

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

27. Ishaan94

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

28. Ishaan94

Your first integral was right as well

29. Ishaan94

My piece wise representation is pelletty :/

30. shivam_bhalla

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

31. Ishaan94

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

32. Ishaan94

33. shivam_bhalla

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

34. Ishaan94

lol

35. FoolForMath

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

36. Ishaan94

@God

37. Ishaan94

0 fans lol

38. Ishaan94

I beat God! by 455 fans

39. shivam_bhalla

LOLOL

40. FoolForMath

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

41. shivam_bhalla

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