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Luigi0210

  • 3 years ago

solve the integral

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

  2. Azteck
    • 3 years ago
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    Lolz...didn't even neded to substitute.

  3. Azteck
    • 3 years ago
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    need*

  4. Azteck
    • 3 years ago
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    waste of your time.

  5. zepdrix
    • 3 years ago
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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.

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

  7. Azteck
    • 3 years ago
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    So there's no solution

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

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

  10. Azteck
    • 3 years ago
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    What are you integrating with respect to? You integrating with respect to zero?

  11. Azteck
    • 3 years ago
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    Okay. Now you can integrate it. Just separate the numerator.

  12. Azteck
    • 3 years ago
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    And you can continue integrating per usual.

  13. Lynncake
    • 3 years ago
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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

  14. Luigi0210
    • 3 years ago
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    My only real problem is finding the anti-derivative

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

  16. zepdrix
    • 3 years ago
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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

  17. Luigi0210
    • 3 years ago
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    Thank you very much

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