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anonymous

  • one year ago

How do you integrate from 0 to x: (t^2)*sqrt(2+5t^4)? I've tried integration by parts where t^2 is dv/dt and sqrt(2+5t^4) is u but it ends up funny and possibly more complex. Integration by substitution also leaves me with a funny number. I'm really stuck, so any ideas would help. Thanks

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  1. Astrophysics
    • one year ago
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    \[\large \int\limits_{0}^{x}t^2 \sqrt{2+5t^4})dt\] correct?

  2. anonymous
    • one year ago
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    Yup, that's the integral we need to find

  3. Astrophysics
    • one year ago
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    I was thinking fundamental theorem of calculus part 1 but that's more for derivative, hmm.

  4. anonymous
    • one year ago
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    try to do it by multiplying and dividing by t and then apply by parts by that way you can easily integrate the radical part

  5. anonymous
    • one year ago
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    @divu.mkr I'm not quite sure what you mean by multiplying and dividing by t. Do you mean I should multiply the outside t into the square root?

  6. phi
    • one year ago
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    according to wolfram, this is not a simple integral

  7. anonymous
    • one year ago
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    What's a simple integral? Is it still a simple integral without the boundaries (0 and x)?

  8. phi
    • one year ago
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    it looks ugly either way

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  9. phi
    • one year ago
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    without limits http://www.wolframalpha.com/input/?i=integral+%28t^2%29*sqrt%282%2B5t^4%29+dt

  10. anonymous
    • one year ago
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    Wholey, the integral got symbols I've never seen before! On the other hand, it does make a very nice looking curve.

  11. phi
    • one year ago
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    That is what I meant by not simple. I have no idea how to get that answer.

  12. Empty
    • one year ago
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    Yeah, I get the feeling that the real question was "Differentiate this with respect to x" haha

  13. anonymous
    • one year ago
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    Hmm in that case then, maybe Astrophysics was on the right track... The full question is: calculate lim (x-->infinite) (x^5) \[\lim_{x \rightarrow \infty}\frac{ x^5 }{ \int\limits_{0}^{x}(t^2)*\sqrt(2+5t^4) }\]

  14. anonymous
    • one year ago
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    Opps there's a line between the x^5 and the bottom integral

  15. anonymous
    • one year ago
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    @Empty yea, I should've posted the whole question

  16. Empty
    • one year ago
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    Ok, yeah you are still asked to do a limit, so almost the same thing. Especially if you use L'Hopital's rule ;) \[\lim_{x \rightarrow \infty}\frac{ x^5 }{ \int\limits_{0}^{x}(t^2)*\sqrt{2+5t^4}dt }\] Also, to make the square root work use {} instead of ().

  17. anonymous
    • one year ago
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    @Empty Haha and that would have been a much less stressful question.

  18. phi
    • one year ago
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    you have and infty/infty so take the derivative of top and bottom

  19. anonymous
    • one year ago
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    Oh good old L'Hopital's rule! Thank you so much!

  20. Empty
    • one year ago
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    To show that we can do L'Hopital's rule in a more convincing way here, notice: \[\large \int\limits_{0}^{x}t^2 \sqrt{2+5t^4})dt \ge \int\limits_{0}^{x}t^2 \sqrt{t^4}dt\] Which is a lot easier to solve and compare to, so since this integral diverges the other one will too in case you weren't sure.

  21. anonymous
    • one year ago
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    Do you mean that we can use the right hand function as the denominator in place of the left one since they both diverge to similar degrees? Oh and, was the condition for being able to use L'Hopital's rule that your function is infinity/infinity?

  22. phi
    • one year ago
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    no, you keep the original Empty is just showing we have an infinity / infinity (specifically the denominator) so it is valid to use L'Hopital

  23. Empty
    • one year ago
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    Yeah that ^ hah

  24. phi
    • one year ago
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    I am getting \( \sqrt{5} \)

  25. anonymous
    • one year ago
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    Ah right I see (since t^2*sqrt(t^4) diverges)! Yup, the answer is sqrt5 I'm just a bit stuck on trying to find the limit... I've differentiated twice and got to \[\frac{ 10x }{ \sqrt{(2+5x^4)^3}*20t^2 }\] But I can't divide by x^4 inside the sqrt as I don't know how to take it out.

  26. anonymous
    • one year ago
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    Oh wait hang on, I found my silly mistake (as always). Got it! Thank you guys and everyone else so much, having had to be baffled by trying to integrate the question. Sorry for not posting the whole question earlier!

  27. phi
    • one year ago
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    you should only differentiate once you get \[ \frac{5x^4}{x^2 \sqrt{2+5x^4} }\] divide top and bottom by x^4 \[ \frac{5}{x^{-2} \sqrt{2+5x^4} }\] bring the x^-2 inside the root \[ \frac{5}{\sqrt{2x^{-4} +5} }\] now take the limit. the 2/x^4 tends to zero and we get \[ \frac{5}{\sqrt{5}}= \sqrt{5}\]

  28. anonymous
    • one year ago
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    Thanks, I eventually got onto the right track. I made the mistake of multiplying the denominator by 1/x^2 after dividing the inside by x^4 instead of multiplying by x^2 outside

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