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tjutta

  • 3 years ago

lim x>infinite 1/x^2 S(integral sign) tan^-1 dt

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  1. ash2326
    • 3 years ago
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    \[\lim_{x\to \infty} \frac 1 {x^2} \int \tan^{-1} t\ dt\]Is this your question?

  2. tjutta
    • 3 years ago
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    yes sorry Im new to this :-)

  3. ash2326
    • 3 years ago
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    No need to be sorry :)

  4. ash2326
    • 3 years ago
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    The integral doesn't involve x, so it can be taken out. \[\int \tan^{-1} t\ dt\times \lim_{x\to \infty} \frac 1 {x^2} \] tell me what's \[\frac 1 \infty=???\]

  5. ash2326
    • 3 years ago
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    @tjutta ???

  6. tjutta
    • 3 years ago
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    \[\infty\]

  7. ash2326
    • 3 years ago
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    nope \[\frac 1 0= \infty\] so \[\frac 1 \infty=???\]

  8. tjutta
    • 3 years ago
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    0

  9. ash2326
    • 3 years ago
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    good :D \[\int \tan^{-1} t\times 0=???\]

  10. tjutta
    • 3 years ago
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    SIN/COS ? :-)

  11. ash2326
    • 3 years ago
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    if you multiply 0 by tan t then what you'd get?

  12. tjutta
    • 3 years ago
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    0

  13. ash2326
    • 3 years ago
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    so it's 0

  14. tjutta
    • 3 years ago
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    what ? the answer is 0 :-) Ok it makes since if you put it that way but for me whos math retarded its difficult:-)

  15. ash2326
    • 3 years ago
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    You know the integral is a variable of t, if you integrate it and multiply by zero. You'd get zero. So it's not necessary to integrate. It's 0

  16. ash2326
    • 3 years ago
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    I doubt that this is so easy. I think you have missed some portion of the question. Maybe limits on the integral. Could you check the question again?

  17. tjutta
    • 3 years ago
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    its just as you put it so I think this is right but I have one more :-) its \[\int\limits_{}^{} \frac{ (x^3)\ }{(x-1)^2(x+2) }\] Can you help me with that?

  18. ash2326
    • 3 years ago
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    okay, did you try this?

  19. tjutta
    • 3 years ago
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    I dont know what to do :-) should I start by divide ?

  20. ash2326
    • 3 years ago
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    What's the expansion of (x-1)^2?

  21. tjutta
    • 3 years ago
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    (x-1)(x+1) ?

  22. ash2326
    • 3 years ago
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    What would you get if you multiply these?

  23. tjutta
    • 3 years ago
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    ok I only have 2 ?

  24. ash2326
    • 3 years ago
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    \[(x-1)\times (x-1)=???\]

  25. tjutta
    • 3 years ago
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    x^2 - 2x + 1

  26. ash2326
    • 3 years ago
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    Good:) Numerator we have x^2, so add and subtract (-2x+1) Could you rewrite the integral using this?

  27. tjutta
    • 3 years ago
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    \[\int\limits_{}^{} \frac{ x }{-2x+1 } \] Im I on the right way? I know Im not very good at this at all :-)

  28. ash2326
    • 3 years ago
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    I'm sorry, i thought it's x^2 in numerator Sorry. Let me think about this

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