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dfresenius

Partial fraction decompostion with integrals..

  • one year ago
  • one year ago

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  1. dfresenius
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    \[\int\limits_{4}^{5}\frac{ x^3-3x^2 -9}{ x^3-3x^2 }dx\]

    • one year ago
  2. dfresenius
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    |dw:1360273420317:dw|

    • one year ago
  3. dfresenius
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    stuck here

    • one year ago
  4. dfresenius
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    i need to find A B and C

    • one year ago
  5. Jemurray3
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    Ignore the one for now, just try to decompose \[ \frac{9}{x^2(x-3)} =\frac{A}{x} + \frac{B}{x^2} + \frac{C}{x-3} \] Multiply through by x^2(x-3) to get \[ 9 = Ax(x-3)+ B(x-3) + Cx^2 \] There is a tricky way, which is faster, or a straightforward way, which takes longer. Which would you prefer?

    • one year ago
  6. dfresenius
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    |dw:1360273719414:dw|

    • one year ago
  7. dfresenius
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    so, now, I can plug those in to find A?

    • one year ago
  8. Jemurray3
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    Plug those values in for B and C and just pick any arbitrary value of x to get A.

    • one year ago
  9. dfresenius
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    |dw:1360274047172:dw|

    • one year ago
  10. Jemurray3
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    Yep

    • one year ago
  11. dfresenius
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    |dw:1360274149320:dw|

    • one year ago
  12. dfresenius
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    so far so good?

    • one year ago
  13. Jemurray3
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    mhmm

    • one year ago
  14. dfresenius
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    |dw:1360274464556:dw|

    • one year ago
  15. Jemurray3
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    Shouldn't the 3rd and 6th terms be ln(2) and ln(1)?

    • one year ago
  16. dfresenius
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    oh ya!

    • one year ago
  17. Jemurray3
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    Other than that, looks fine. you can write ln(4) as 2 ln(2) if you'd like.

    • one year ago
  18. dfresenius
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    my online homework still says im wrong

    • one year ago
  19. dfresenius
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    maybe i need to combine the ln's?

    • one year ago
  20. cherio12
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    I think you decompose it to\[\frac{ A }{ x-3 }+\frac{ B+Cx }{ x^2 }\]..not a 100% sure..but there is something different done when you are over a power greater than one. Side note, were did you get A/x from?

    • one year ago
  21. dfresenius
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    the x^2

    • one year ago
  22. dfresenius
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    a/x + b/x^2

    • one year ago
  23. dfresenius
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    im on my last attempt to answer it correctly , damn

    • one year ago
  24. SithsAndGiggles
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    @cherio12, when decomposing (something)/x^2 you have \[\frac{A}{x}+\frac{B}{x^2}\] I'm not familiar with the formal explanation as to why this happens, but I know if you have factors with multiplicity greater than 1 they're treated as normal linear factors.

    • one year ago
  25. dfresenius
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    @SithsAndGiggles can you see where I went wrong?

    • one year ago
  26. SithsAndGiggles
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    Your answer doesn't look wrong at first glance, but I'm still working back. It could be that the program your class uses is very picky about the answer it accepts.

    • one year ago
  27. SithsAndGiggles
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    Okay, so using the longer method mentioned by @Jemurray3, I get different values of A, B, and C: \[1 + \int_{4}^{5}-\frac{9}{x^2(x-3)}dx\] Focusing on the integral: Decomposing into partial fractions, you get \[-9=Ax(x-3)+B(x-3)+Cx^2\\ -9=(A+C)x^2+(-3A+B)x-3B\\ \text{Matching up coefficients, you get}\\ \begin{cases}A+C=0\\ -3A+B=0\\ -3B=-9\end{cases}\\ \text{Solving, you get}\\ B=3, A=1, C=-1\]

    • one year ago
  28. dfresenius
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    strange, how will I know when to use each method?

    • one year ago
  29. SithsAndGiggles
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    The method I used works pretty much every time. You're less prone to making mistakes, I think. I'll check your work again; if the answer really is wrong, it'll probably be an algebra mistake.

    • one year ago
  30. dfresenius
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    thanks a ton!

    • one year ago
  31. SithsAndGiggles
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    Yep, that's what it was.|dw:1360279359173:dw|

    • one year ago
  32. dfresenius
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    NOOOOOOOOOOOO

    • one year ago
  33. dfresenius
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    thank you so much for the help!!!

    • one year ago
  34. dfresenius
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    cant believe i missed that!!!

    • one year ago
  35. dfresenius
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    |dw:1360279683216:dw|

    • one year ago
  36. dfresenius
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    err ln2 instead of ln3

    • one year ago
  37. dfresenius
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    and no + C

    • one year ago
  38. dfresenius
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    its right!!!!!!!!!

    • one year ago
  39. SithsAndGiggles
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    I'm not getting the -ln4 that you are. Here's my work, post-integration: \[1+(\ln 5-\ln 4) - 3\left(\frac{1}{5}-\frac{1}{4}\right)-(\ln 2 - \ln 1)\\ 1+\ln 5-2\ln 2 +\frac{3}{20}-\ln 2\\ \ln 5-3\ln 2 +\frac{23}{20}\\ \ln 5-\ln 8 +\frac{23}{20}\\ \ln {\frac{5}{8}}+\frac{23}{20}\\\]

    • one year ago
  40. SithsAndGiggles
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    I take it back, it was there, I just rewrote it. You're welcome!

    • one year ago
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