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dfresenius Group TitleBest ResponseYou've already chosen the best response.0
\[\int\limits_{4}^{5}\frac{ x^33x^2 9}{ x^33x^2 }dx\]
 one year ago

dfresenius Group TitleBest ResponseYou've already chosen the best response.0
dw:1360273420317:dw
 one year ago

dfresenius Group TitleBest ResponseYou've already chosen the best response.0
stuck here
 one year ago

dfresenius Group TitleBest ResponseYou've already chosen the best response.0
i need to find A B and C
 one year ago

Jemurray3 Group TitleBest ResponseYou've already chosen the best response.1
Ignore the one for now, just try to decompose \[ \frac{9}{x^2(x3)} =\frac{A}{x} + \frac{B}{x^2} + \frac{C}{x3} \] Multiply through by x^2(x3) to get \[ 9 = Ax(x3)+ B(x3) + Cx^2 \] There is a tricky way, which is faster, or a straightforward way, which takes longer. Which would you prefer?
 one year ago

dfresenius Group TitleBest ResponseYou've already chosen the best response.0
dw:1360273719414:dw
 one year ago

dfresenius Group TitleBest ResponseYou've already chosen the best response.0
so, now, I can plug those in to find A?
 one year ago

Jemurray3 Group TitleBest ResponseYou've already chosen the best response.1
Plug those values in for B and C and just pick any arbitrary value of x to get A.
 one year ago

dfresenius Group TitleBest ResponseYou've already chosen the best response.0
dw:1360274047172:dw
 one year ago

dfresenius Group TitleBest ResponseYou've already chosen the best response.0
dw:1360274149320:dw
 one year ago

dfresenius Group TitleBest ResponseYou've already chosen the best response.0
so far so good?
 one year ago

dfresenius Group TitleBest ResponseYou've already chosen the best response.0
dw:1360274464556:dw
 one year ago

Jemurray3 Group TitleBest ResponseYou've already chosen the best response.1
Shouldn't the 3rd and 6th terms be ln(2) and ln(1)?
 one year ago

Jemurray3 Group TitleBest ResponseYou've already chosen the best response.1
Other than that, looks fine. you can write ln(4) as 2 ln(2) if you'd like.
 one year ago

dfresenius Group TitleBest ResponseYou've already chosen the best response.0
my online homework still says im wrong
 one year ago

dfresenius Group TitleBest ResponseYou've already chosen the best response.0
maybe i need to combine the ln's?
 one year ago

cherio12 Group TitleBest ResponseYou've already chosen the best response.0
I think you decompose it to\[\frac{ A }{ x3 }+\frac{ B+Cx }{ x^2 }\]..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

dfresenius Group TitleBest ResponseYou've already chosen the best response.0
the x^2
 one year ago

dfresenius Group TitleBest ResponseYou've already chosen the best response.0
a/x + b/x^2
 one year ago

dfresenius Group TitleBest ResponseYou've already chosen the best response.0
im on my last attempt to answer it correctly , damn
 one year ago

SithsAndGiggles Group TitleBest ResponseYou've already chosen the best response.0
@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

dfresenius Group TitleBest ResponseYou've already chosen the best response.0
@SithsAndGiggles can you see where I went wrong?
 one year ago

SithsAndGiggles Group TitleBest ResponseYou've already chosen the best response.0
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

SithsAndGiggles Group TitleBest ResponseYou've already chosen the best response.0
Okay, so using the longer method mentioned by @Jemurray3, I get different values of A, B, and C: \[1 + \int_{4}^{5}\frac{9}{x^2(x3)}dx\] Focusing on the integral: Decomposing into partial fractions, you get \[9=Ax(x3)+B(x3)+Cx^2\\ 9=(A+C)x^2+(3A+B)x3B\\ \text{Matching up coefficients, you get}\\ \begin{cases}A+C=0\\ 3A+B=0\\ 3B=9\end{cases}\\ \text{Solving, you get}\\ B=3, A=1, C=1\]
 one year ago

dfresenius Group TitleBest ResponseYou've already chosen the best response.0
strange, how will I know when to use each method?
 one year ago

SithsAndGiggles Group TitleBest ResponseYou've already chosen the best response.0
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

dfresenius Group TitleBest ResponseYou've already chosen the best response.0
thanks a ton!
 one year ago

SithsAndGiggles Group TitleBest ResponseYou've already chosen the best response.0
Yep, that's what it was.dw:1360279359173:dw
 one year ago

dfresenius Group TitleBest ResponseYou've already chosen the best response.0
NOOOOOOOOOOOO
 one year ago

dfresenius Group TitleBest ResponseYou've already chosen the best response.0
thank you so much for the help!!!
 one year ago

dfresenius Group TitleBest ResponseYou've already chosen the best response.0
cant believe i missed that!!!
 one year ago

dfresenius Group TitleBest ResponseYou've already chosen the best response.0
dw:1360279683216:dw
 one year ago

dfresenius Group TitleBest ResponseYou've already chosen the best response.0
err ln2 instead of ln3
 one year ago

dfresenius Group TitleBest ResponseYou've already chosen the best response.0
and no + C
 one year ago

dfresenius Group TitleBest ResponseYou've already chosen the best response.0
its right!!!!!!!!!
 one year ago

SithsAndGiggles Group TitleBest ResponseYou've already chosen the best response.0
I'm not getting the ln4 that you are. Here's my work, postintegration: \[1+(\ln 5\ln 4)  3\left(\frac{1}{5}\frac{1}{4}\right)(\ln 2  \ln 1)\\ 1+\ln 52\ln 2 +\frac{3}{20}\ln 2\\ \ln 53\ln 2 +\frac{23}{20}\\ \ln 5\ln 8 +\frac{23}{20}\\ \ln {\frac{5}{8}}+\frac{23}{20}\\\]
 one year ago

SithsAndGiggles Group TitleBest ResponseYou've already chosen the best response.0
I take it back, it was there, I just rewrote it. You're welcome!
 one year ago
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