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roselin

need help with :definite integral from 0 to 1 x^3/ Square root of x^4+9

  • one year ago
  • one year ago

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  1. ChmE
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    \[\int\limits_{0}^{1}{x^3 \over \sqrt{x^4+9}}dx\]Is this the correct equation?

    • one year ago
  2. roselin
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    yes, that is the one.

    • one year ago
  3. ChmE
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    u-substitute for x^4 + 9

    • one year ago
  4. roselin
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    let u be x^4+9?

    • one year ago
  5. roselin
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    but how do you know which one to put as" u"?

    • one year ago
  6. ChmE
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    u = x^4 + 9 du = 4x^3 dx du / 4x^3 = dx\[\int\limits\limits_{0}^{1}{x^3 \over \sqrt{u}}{du \over 4x^3}\]\[\int\limits\limits_{0}^{1}{du \over 4\sqrt{u}}\]Can you finish it? I think you are actually supposed to re-evaluate the limits but I don't remember how to do that. It is not necessary if you replace u with what you substituted it for before you apply the limits

    • one year ago
  7. roselin
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    let me try replacing it

    • one year ago
  8. roselin
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    |dw:1355885411954:dw|

    • one year ago
  9. TweT226
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    to re-evaluate the limits plug in 0 and 1 into u=x^4 + 9

    • one year ago
  10. roselin
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    then, what do I need to do after that?

    • one year ago
  11. ChmE
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    I don't know why you did that.. you need to integrate the last equation I gave you that has u in it

    • one year ago
  12. ChmE
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    \[\int\limits\limits\limits_{9}^{10}{du \over 4\sqrt{u}}\]

    • one year ago
  13. ChmE
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    @TweT226 thx

    • one year ago
  14. ChmE
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    After you do u-substitution to re-evaluate the limits see Twe's post

    • one year ago
  15. roselin
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    when i plug in 0 i get 9 and whne i plug in 1 , I get 10

    • one year ago
  16. ChmE
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    \[\int\limits\limits\limits\limits_{9}^{10}{1 \over 4}u^{-1/2}du\]

    • one year ago
  17. roselin
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    how did u get u^-1/2?

    • one year ago
  18. ChmE
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    \[{{1 \over 4}u^{1/2} \over 1/2}|_9^{10}\]\[{1 \over 2}u^{1/2}|_9^{10}\]\[{1 \over 2}10^{1/2}-{1 \over 2}9^{1/2}\]

    • one year ago
  19. ChmE
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    \[\sqrt x=x^{1/2}\]\[{1 \over x^n}=x^{-n}\]

    • one year ago
  20. roselin
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    oh,okay.i see

    • one year ago
  21. ChmE
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    re-evaluating the limits is the short method. If you don't do that you can plug what you substituted u for then use the old limits after integration. You should get the same answer

    • one year ago
  22. roselin
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    okay.

    • one year ago
  23. ChmE
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    \[\int\limits\limits_{0}^{1}{x^3 \over \sqrt{x^4+9}}dx\]u-substitution: u = x^4 + 9 du = 4x^3 dx du / 4x^3 = dx Re-evaluating the limits u = (0)^4 + 9 = 9 u = (1)^4 + 9 = 10\[\int\limits\limits\limits_{9}^{10}{x^3 \over \sqrt{u}}{du \over 4x^3}\]\[\int\limits\limits\limits\limits_{9}^{10}{du \over 4\sqrt{u}}\]\[\int\limits\limits\limits\limits\limits_{9}^{10}{1 \over 4}u^{-1/2}du\]\[{{1 \over 4}u^{1/2} \over 1/2}|_9^{10}\]\[{1 \over 2}u^{1/2}|_9^{10}\]\[{1 \over 2}10^{1/2}-{1 \over 2}9^{1/2}\]

    • one year ago
  24. roselin
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    thanks

    • one year ago
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