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dan815

  • 3 years ago

integrate x/lnx

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  1. Xavier
    • 3 years ago
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    With ln(x) in the denominator like that that's not one of our preatty integrals.

  2. anonymous
    • 3 years ago
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    you will not find a nice closed form for this one

  3. dan815
    • 3 years ago
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    ah

  4. dan815
    • 3 years ago
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    is it one of those repeating integrating ones

  5. anonymous
    • 3 years ago
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    you mean like if you integrate by parts will you go around in a circle? i don't think so

  6. anonymous
    • 3 years ago
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    i think you will not be able to solve in in the sense that you will not be able to find some nice function \(F\) whose derivative is \(\frac{x}{\ln(x)}\)

  7. Xavier
    • 3 years ago
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    It's the type where our general methods of integrating get us nowhere. We have to leave it in the form of an integral and work with it that way.

  8. dan815
    • 3 years ago
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    this is what im trying to evaultate |dw:1363059332628:dw|

  9. dan815
    • 3 years ago
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    is there another method to equate integral like i know this isnt conservative so

  10. Xavier
    • 3 years ago
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    That's completly different. If you are integrating with repect to y first then you don't need to worry about the expressions in x -- they are essentially constants.

  11. inkyvoyd
    • 3 years ago
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    Wolfram says it's not expressible in terms of elementary functions - http://www.wolframalpha.com/input/?i=integrate+x%2Fln+x

  12. inkyvoyd
    • 3 years ago
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    imo you'd best be off solving it numerically

  13. dan815
    • 3 years ago
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    oh hold on i think i got it xD i forgot the lnx cancels if i do dy first

  14. Xavier
    • 3 years ago
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    Yup

  15. dan815
    • 3 years ago
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    xavier u got some time to help ?? im working on one of my assigments due in about 6 hours, gonna need help when i get stuck

  16. Xavier
    • 3 years ago
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    I'm here on and off. I'll try to help. Sure other people will jump onboard if I take too long

  17. dan815
    • 3 years ago
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    just looking thru the questions so far i think im gonna just have some trouble in the volume integral questions ill tell ya when i get to them

  18. dan815
    • 3 years ago
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    look at this one its about setting the right bounds |dw:1363060518856:dw|

  19. dan815
    • 3 years ago
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    i get why for dx its y^2 to 4 but how much for dy just doing 0 to 2 instead is enough how come it cant be be rootX to 4

  20. dan815
    • 3 years ago
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    but how come*

  21. Xavier
    • 3 years ago
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    What is correct there? The diagram or the bounds on your integrals?

  22. dan815
    • 3 years ago
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    the whole thing is correct i just need to covert to an another double integration where im doing it in reverse order dy dx

  23. dan815
    • 3 years ago
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    but before i do that i wanna know why they set up bounds like that

  24. Xavier
    • 3 years ago
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    Either the bounds are wrong or the shaded area is incorrect. Having \[y^2 \le x \le 4\] is the larger of the two areas there.

  25. dan815
    • 3 years ago
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    the shaded region is what im integrating over its in the bounds the function is just like a force equation

  26. dan815
    • 3 years ago
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    where z = ysinx^2

  27. Xavier
    • 3 years ago
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    Then your bounds on x should be \[0 \le x \le y^2\]

  28. dan815
    • 3 years ago
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    oh nvm i see what u mean the diagram is wrong

  29. dan815
    • 3 years ago
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    so the bottom part shud be the shaded region right

  30. dan815
    • 3 years ago
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    |dw:1363061109806:dw|

  31. Xavier
    • 3 years ago
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    Ah ok so you know that the curve your have is y^2=x or y=sqrt(x). So y is bounded above by sqrt(x) and below by 0. And x ranges from 0 to 4

  32. dan815
    • 3 years ago
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    yeah

  33. dan815
    • 3 years ago
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    so if i wanted to switch order of integration to dy dx then

  34. Xavier
    • 3 years ago
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    Just use the bounds on y on the inner integral. x bounds go on outer integral \[0 \le y \le \sqrt{x}\ ; 0 \le x \le 4\]

  35. dan815
    • 3 years ago
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    0 to 4 and 0 to root x

  36. dan815
    • 3 years ago
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    ok got it

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