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FoolForMath

  • 4 years ago

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

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  1. Ishaan94
    • 4 years ago
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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?

  2. FoolForMath
    • 4 years ago
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    I have never seen a indefinite integral represented like this.

  3. Ishaan94
    • 4 years ago
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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.

  4. shivam_bhalla
    • 4 years ago
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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

  5. Ishaan94
    • 4 years ago
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    This can't be right, |x| stays positive but the x^3 can get negative.

  6. Ishaan94
    • 4 years ago
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    I am not sure though. and I can be wrong.

  7. shivam_bhalla
    • 4 years ago
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    For negative x , |x| = -x and therfore integral value would be positive. Though I can be wrong @ this

  8. Ishaan94
    • 4 years ago
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    For negative x, x^3 * |x| = (-x)^3 (-(-x)) = -x^4

  9. Ishaan94
    • 4 years ago
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    What's wrong piece wise representation?

  10. Ishaan94
    • 4 years ago
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    wait... my piece wise is gonna give same result as yours

  11. Ishaan94
    • 4 years ago
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    so, either we are both wrong or right

  12. shivam_bhalla
    • 4 years ago
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    \[\huge \int\limits\limits_{}^{}|x|^3 dx = {sgn(x)}* \frac{x^4}{4}+c\]

  13. Ishaan94
    • 4 years ago
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    and your integral does seems better

  14. Ishaan94
    • 4 years ago
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    ohh yeah

  15. Ishaan94
    • 4 years ago
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    sgn(x) how could i not think about it!?!?! :(

  16. Ishaan94
    • 4 years ago
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    so, there are multiple ways of representing same integral?

  17. Ishaan94
    • 4 years ago
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    because every integral on this page is gonna get you the same result

  18. Ishaan94
    • 4 years ago
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    so, how do you rule out others? and make one right.

  19. shivam_bhalla
    • 4 years ago
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    @Ishaan94 , Is my integral right because substitute -x and check please

  20. Ishaan94
    • 4 years ago
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    -x^4/4 is what we will get. sgn(-x)x^4/4 = -1* x^4/4

  21. shivam_bhalla
    • 4 years ago
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    Yes and why is it negative??

  22. shivam_bhalla
    • 4 years ago
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    I have confused myself lolol

  23. Ishaan94
    • 4 years ago
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    i am confused myself and blabbering :/

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

  25. shivam_bhalla
    • 4 years ago
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    Ok I got it. Taking example always helps

  26. shivam_bhalla
    • 4 years ago
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    In that case how is my first equation wrong , the very first one I mentioned ??

  27. Ishaan94
    • 4 years ago
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    \[\int_{-1}^1 |x|^3 = \int_{-1}^0 -x^3 + \int_{0}^1 x^3 = -(-1/4) + \frac14 = \frac12 \] \[\left[sgn(x) \cdot \frac{x^4}{4}\right]_{-1}^1 = \frac{1}4 + \frac14\]

  28. Ishaan94
    • 4 years ago
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    Your first integral was right as well

  29. Ishaan94
    • 4 years ago
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    My piece wise representation is pelletty :/

  30. shivam_bhalla
    • 4 years ago
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    Haha. Maths always confuses us even when we have the right answers . :P

  31. Ishaan94
    • 4 years ago
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    yeah hahaha i never had to integrate such indefinite integrals before.

  32. Ishaan94
    • 4 years ago
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    Why isn't FoolForMath replying?

  33. shivam_bhalla
    • 4 years ago
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    Only 2 people know the answer to your question @Ishaan94 , @FoolForMath and god himself :P

  34. Ishaan94
    • 4 years ago
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    lol

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

  36. Ishaan94
    • 4 years ago
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    @God

  37. Ishaan94
    • 4 years ago
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    0 fans lol

  38. Ishaan94
    • 4 years ago
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    I beat God! by 455 fans

  39. shivam_bhalla
    • 4 years ago
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    LOLOL

  40. FoolForMath
    • 4 years ago
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    So, the answer is \[ \frac 1 4 |x|^3 \cdot x +C \]

  41. shivam_bhalla
    • 4 years ago
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    Does it matter, @FoolForMath the position of modulus on either x^3 or x :P

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