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dfresenius

  • 2 years ago

Partial fraction decompostion with integrals..

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

  2. dfresenius
    • 2 years ago
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    |dw:1360273420317:dw|

  3. dfresenius
    • 2 years ago
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    stuck here

  4. dfresenius
    • 2 years ago
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    i need to find A B and C

  5. Jemurray3
    • 2 years ago
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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?

  6. dfresenius
    • 2 years ago
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    |dw:1360273719414:dw|

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

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

  9. dfresenius
    • 2 years ago
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    |dw:1360274047172:dw|

  10. Jemurray3
    • 2 years ago
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    Yep

  11. dfresenius
    • 2 years ago
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    |dw:1360274149320:dw|

  12. dfresenius
    • 2 years ago
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    so far so good?

  13. Jemurray3
    • 2 years ago
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    mhmm

  14. dfresenius
    • 2 years ago
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    |dw:1360274464556:dw|

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

  16. dfresenius
    • 2 years ago
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    oh ya!

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

  18. dfresenius
    • 2 years ago
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    my online homework still says im wrong

  19. dfresenius
    • 2 years ago
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    maybe i need to combine the ln's?

  20. cherio12
    • 2 years ago
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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?

  21. dfresenius
    • 2 years ago
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    the x^2

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

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

  24. SithsAndGiggles
    • 2 years ago
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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.

  25. dfresenius
    • 2 years ago
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    @SithsAndGiggles can you see where I went wrong?

  26. SithsAndGiggles
    • 2 years ago
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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.

  27. SithsAndGiggles
    • 2 years ago
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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 \[-9=Ax(x-3)+B(x-3)+Cx^2\\ -9=(A+C)x^2+(-3A+B)x-3B\\ \text{Matching up coefficients, you get}\\ \begin{cases}A+C=0\\ -3A+B=0\\ -3B=-9\end{cases}\\ \text{Solving, you get}\\ B=3, A=1, C=-1\]

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

  29. SithsAndGiggles
    • 2 years ago
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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.

  30. dfresenius
    • 2 years ago
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    thanks a ton!

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

  32. dfresenius
    • 2 years ago
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    NOOOOOOOOOOOO

  33. dfresenius
    • 2 years ago
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    thank you so much for the help!!!

  34. dfresenius
    • 2 years ago
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    cant believe i missed that!!!

  35. dfresenius
    • 2 years ago
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    |dw:1360279683216:dw|

  36. dfresenius
    • 2 years ago
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    err ln2 instead of ln3

  37. dfresenius
    • 2 years ago
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    and no + C

  38. dfresenius
    • 2 years ago
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    its right!!!!!!!!!

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

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

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