anonymous
  • anonymous
Partial fraction decompostion with integrals..
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
\[\int\limits_{4}^{5}\frac{ x^3-3x^2 -9}{ x^3-3x^2 }dx\]
anonymous
  • anonymous
|dw:1360273420317:dw|
anonymous
  • anonymous
stuck here

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anonymous
  • anonymous
i need to find A B and C
anonymous
  • anonymous
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?
anonymous
  • anonymous
|dw:1360273719414:dw|
anonymous
  • anonymous
so, now, I can plug those in to find A?
anonymous
  • anonymous
Plug those values in for B and C and just pick any arbitrary value of x to get A.
anonymous
  • anonymous
|dw:1360274047172:dw|
anonymous
  • anonymous
Yep
anonymous
  • anonymous
|dw:1360274149320:dw|
anonymous
  • anonymous
so far so good?
anonymous
  • anonymous
mhmm
anonymous
  • anonymous
|dw:1360274464556:dw|
anonymous
  • anonymous
Shouldn't the 3rd and 6th terms be ln(2) and ln(1)?
anonymous
  • anonymous
oh ya!
anonymous
  • anonymous
Other than that, looks fine. you can write ln(4) as 2 ln(2) if you'd like.
anonymous
  • anonymous
my online homework still says im wrong
anonymous
  • anonymous
maybe i need to combine the ln's?
anonymous
  • anonymous
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?
anonymous
  • anonymous
the x^2
anonymous
  • anonymous
a/x + b/x^2
anonymous
  • anonymous
im on my last attempt to answer it correctly , damn
anonymous
  • anonymous
@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.
anonymous
  • anonymous
@SithsAndGiggles can you see where I went wrong?
anonymous
  • anonymous
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.
anonymous
  • anonymous
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 \[-9=Ax(x-3)+B(x-3)+Cx^2\\ -9=(A+C)x^2+(-3A+B)x-3B\\ \text{Matching up coefficients, you get}\\ \begin{cases}A+C=0\\ -3A+B=0\\ -3B=-9\end{cases}\\ \text{Solving, you get}\\ B=3, A=1, C=-1\]
anonymous
  • anonymous
strange, how will I know when to use each method?
anonymous
  • anonymous
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.
anonymous
  • anonymous
thanks a ton!
anonymous
  • anonymous
Yep, that's what it was.|dw:1360279359173:dw|
anonymous
  • anonymous
NOOOOOOOOOOOO
anonymous
  • anonymous
thank you so much for the help!!!
anonymous
  • anonymous
cant believe i missed that!!!
anonymous
  • anonymous
|dw:1360279683216:dw|
anonymous
  • anonymous
err ln2 instead of ln3
anonymous
  • anonymous
and no + C
anonymous
  • anonymous
its right!!!!!!!!!
anonymous
  • anonymous
I'm not getting the -ln4 that you are. Here's my work, post-integration: \[1+(\ln 5-\ln 4) - 3\left(\frac{1}{5}-\frac{1}{4}\right)-(\ln 2 - \ln 1)\\ 1+\ln 5-2\ln 2 +\frac{3}{20}-\ln 2\\ \ln 5-3\ln 2 +\frac{23}{20}\\ \ln 5-\ln 8 +\frac{23}{20}\\ \ln {\frac{5}{8}}+\frac{23}{20}\\\]
anonymous
  • anonymous
I take it back, it was there, I just rewrote it. You're welcome!

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