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Luigi0210

solve the integral

  • one year ago
  • one year ago

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  1. Luigi0210
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    \[\int\limits_{0}^{4}(1-\sqrt{u})/(\sqrt{u})\]

    • one year ago
  2. Azteck
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    Lolz...didn't even neded to substitute.

    • one year ago
  3. Azteck
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    need*

    • one year ago
  4. Azteck
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    waste of your time.

    • one year ago
  5. zepdrix
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    Yah the substitution was kinda silly :) If you do a substitution, don't forget to change the limits of integration also.

    • one year ago
  6. Azteck
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    And there's no du at the back of the integral

    • one year ago
  7. Azteck
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    So there's no solution

    • one year ago
  8. Azteck
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    You can't continue on integrating that without respect of anything.

    • one year ago
  9. Luigi0210
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    Ops, sorry forgot about the du.. It is in respect to du

    • one year ago
  10. Azteck
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    What are you integrating with respect to? You integrating with respect to zero?

    • one year ago
  11. Azteck
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    Okay. Now you can integrate it. Just separate the numerator.

    • one year ago
  12. Azteck
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    And you can continue integrating per usual.

    • one year ago
  13. Lynncake
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    \[\int\limits_{}^{}\frac{ 1 }{ \sqrt{u} }du-\int\limits_{}^{}1du\]\[\int\limits_{}^{}u^{-\frac{ 1 }{ 2 }}du-u\]\[2u^{\frac{ 1 }{ 2 }}-u\]Then just plug in the limit from 0 to 4

    • one year ago
  14. Luigi0210
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    My only real problem is finding the anti-derivative

    • one year ago
  15. Azteck
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    Anti-differentiating is just differentiating in reverse. Try and use reverse psychology when integrating if you can.

    • one year ago
  16. zepdrix
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    \[\large \frac{1-\sqrt u}{\sqrt u} \qquad = \qquad \frac{1}{\sqrt u}-\frac{\sqrt u}{\sqrt u} \qquad = \qquad u^{-1/2}-1\] Yah you just apply the `Power Rule for Integration`! :D

    • one year ago
  17. Luigi0210
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    Thank you very much

    • one year ago
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