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yrelhan4

  • 3 years ago

(1 - x^3)^n integration. limits 0 to 1 ??

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  1. experimentX
    • 3 years ago
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    try trig sub x^3 = sin^2u

  2. yrelhan4
    • 3 years ago
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    hmm.. i'll try it once and get back to you..

  3. hartnn
    • 3 years ago
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    if that doesn't work, do u= 1-x^3

  4. yrelhan4
    • 3 years ago
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    how ould that help? @hartnn

  5. hartnn
    • 3 years ago
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    du = -3x^2 dx du = - 3(1-u)^(2/3)dx du/[- 3(1-u)^(2/3)] = dx just trying.......

  6. timo86m
    • 3 years ago
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    intigrate with respect to x or n?

  7. experimentX
    • 3 years ago
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    |dw:1356588801737:dw|

  8. yrelhan4
    • 3 years ago
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    @timo86m x

  9. timo86m
    • 3 years ago
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    -x^3/ln(1-x^3) With respect to n

  10. yrelhan4
    • 3 years ago
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    @experimentX yes i did the same. i think i got it now.. thank you..

  11. experimentX
    • 3 years ago
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    this should end up around beta function http://en.wikipedia.org/wiki/Beta_function#Properties

  12. hartnn
    • 3 years ago
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    is this from Gamma Function ?

  13. timo86m
    • 3 years ago
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    \[2/9\,{\frac {\Gamma \left( n+1 \right) \pi \,\sqrt {3}}{\Gamma \left( 2/3 \right) \Gamma \left( n+4/3 \right) }} \] gamma

  14. hartnn
    • 3 years ago
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    because it gets the form u^n (1-u)^n

  15. hartnn
    • 3 years ago
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    u^n (1-u)^m

  16. timo86m
    • 3 years ago
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    oh and i used maple got no idea how to do it :P

  17. hartnn
    • 3 years ago
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    http://www.wolframalpha.com/input/?i=integrate+%281-x%5E3%29%5En+dx+from+0+to+1

  18. experimentX
    • 3 years ago
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    \[ \frac{1}{3} \beta(2(n+1), 4/3)) = \]

  19. wio
    • 3 years ago
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    Umm, can binomial theorem help with this?

  20. wio
    • 3 years ago
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    \[ \Large (a+b)^n = \sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^{k} \]We let \(a = 1\) and \(b = -x^3\)?

  21. yrelhan4
    • 3 years ago
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    can anybody integrate 2/3 [(1-t^2) / (t)^1/3] dt.. from 0 to 1.. i reached till here..

  22. wio
    • 3 years ago
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    Like what about:\[ \Large \int_0^1\sum_{k=0}^{n}\binom{n}{k}(-x^3)^{k}dx \]

  23. hartnn
    • 3 years ago
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    how u got that ?

  24. yrelhan4
    • 3 years ago
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    @wio no idea about it.. its a grade 12 level question.. lol..

  25. yrelhan4
    • 3 years ago
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    @experimentX 's method..

  26. experimentX
    • 3 years ago
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    that's a good strategy!!

  27. timo86m
    • 3 years ago
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    infinity to 2/3 [(1-t^2) / (t)^1/3]

  28. hartnn
    • 3 years ago
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    i don't think this can be solved without beta functions, did u go to wiki link of beta function ?

  29. yrelhan4
    • 3 years ago
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    http://www.wolframalpha.com/input/?i=integrate+%5B%281-t%5E2%29+%2F+%28t%29%5E1%2F3%5D+dt+from+0+to+1 i am talking about this..

  30. yrelhan4
    • 3 years ago
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    @hartnn hmm.. i gotta read beta function once..

  31. experimentX
    • 3 years ago
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    http://www.wolframalpha.com/input/?i=integrate+%5B%281-t%5E2%29+%2F+%28t%29%5E1%2F3%5D+dt

  32. hartnn
    • 3 years ago
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    thats easy to integrate just seperate the denominator.

  33. hartnn
    • 3 years ago
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    t^(-1/3) - t^(-5/3)

  34. yrelhan4
    • 3 years ago
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    i missed the n.. i think i gota read the beta/gamma function once.. thank you everyone..

  35. hartnn
    • 3 years ago
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    yup, whether u put, u=1-x^3 or x^3 = sin^2 u you'll end up with one of the form of beta function.

  36. experimentX
    • 3 years ago
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    for the last one \[ \frac 1 3 \times 2 \int_0^{\pi \over 2} \cos^{2(n+1) - 1}(x) \sin ^{\frac 4 3 - 1 }(x) dx = \frac 13 \beta (n+1, \frac 23 ) = \frac 13 \frac{\Gamma(n+1) \Gamma(2/3)}{\Gamma(n+4/3) } \]

  37. experimentX
    • 3 years ago
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    something went wrong

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