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Jonask

  • 2 years ago

integration- radicals

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  1. Kanwar245
    • 2 years ago
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    ?

  2. Jonask
    • 2 years ago
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    \[\color{blue}{\huge\int_0^1\sqrt[3]{1-x^7}-\sqrt[7]{1-x^3}dx}\]

  3. UnkleRhaukus
    • 2 years ago
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    \[\color{blue}{\int_0^1\sqrt[3]{1-x^7}dx-\int\sqrt[7]{1-x^3}dx}\]

  4. Jonask
    • 2 years ago
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    yes but when i do that it still seems ugly

  5. UnkleRhaukus
    • 2 years ago
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    do one half at a time

  6. Jonask
    • 2 years ago
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    i shud make a u sub \[u=1-x^7\]

  7. Jonask
    • 2 years ago
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    in the 1st one

  8. UnkleRhaukus
    • 2 years ago
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    i'd say so

  9. Jonask
    • 2 years ago
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    \[u=1-x^7\] \[\int\frac{u^{1/3}}{7x^6}du\]

  10. Jonask
    • 2 years ago
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    the x^6 is a problem

  11. Jonask
    • 2 years ago
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    \[\int\frac{u^{1/3}}{7(1-u)^{1/7}}\]

  12. UnkleRhaukus
    • 2 years ago
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    hmm,

  13. UnkleRhaukus
    • 2 years ago
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    maybe that method isn't going to work

  14. Jonask
    • 2 years ago
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    http://www.wolframalpha.com/input/?i=integral+%281-x%5E7%29%5E%7B1%2F3%7D-%281-x%5E3%29%5E%7B1%2F7%7D ugly as well but i think we are supposed to use 'by parts'

  15. UnkleRhaukus
    • 2 years ago
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    where did you get this question?

  16. Jonask
    • 2 years ago
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    steward calculus advanved problem

  17. Jonask
    • 2 years ago
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    *

  18. shubhamsrg
    • 2 years ago
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    what if we use the property integral ( f(x) from x=a to b = f(a+b-x) from x= a to b )?

  19. Jonask
    • 2 years ago
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    in this case a+b=0+1 int f(1-x)=f(x)

  20. shubhamsrg
    • 2 years ago
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    yep, am not too sure if that'll work though, but probably should work ?

  21. Jonask
    • 2 years ago
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    ok i have to go i;ll try it thanks to everyone for the effort

  22. SithsAndGiggles
    • 2 years ago
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    According to WA, the definite integral is 0: http://www.wolframalpha.com/input/?i=integrate+%28%281-x%5E7%29%5E%281%2F3%29+-+%281-x%5E3%29%5E%281%2F7%29%29+over+%5B0%2C1%5D Maybe there's some detail we're not getting about the function itself.

  23. Jonask
    • 2 years ago
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    interesting result maybe this property is because of the function is actually an inverse of itself i mean \[f(x)=\sqrt[3]{1-x^7}\] then \[f^{-1}(x)=\sqrt[7]{1-x^3}\] should this mean anything

  24. Jonask
    • 2 years ago
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    \[\int_0^1 f(x)-f^{-1}(x) \space dx\] \[f(1)=f^{-1}(1)=0\]

  25. experimentX
    • 2 years ago
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    this is a famous question on M.SE

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