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dfresenius

  • one year ago

Partial fraction decompostion with integrals..

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

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

  3. dfresenius
    • one year ago
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    stuck here

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

  5. Jemurray3
    • one year 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
    • one year ago
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    |dw:1360273719414:dw|

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

  8. Jemurray3
    • one year 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
    • one year ago
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    |dw:1360274047172:dw|

  10. Jemurray3
    • one year ago
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    Yep

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

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

  13. Jemurray3
    • one year ago
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    mhmm

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

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

  16. dfresenius
    • one year ago
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    oh ya!

  17. Jemurray3
    • one year 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
    • one year ago
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    my online homework still says im wrong

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

  20. cherio12
    • one year 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
    • one year ago
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    the x^2

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

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

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

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

  29. SithsAndGiggles
    • one year 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
    • one year ago
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    thanks a ton!

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

  32. dfresenius
    • one year ago
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    NOOOOOOOOOOOO

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

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

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

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

  37. dfresenius
    • one year ago
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    and no + C

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

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

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