anonymous
  • anonymous
integrate...
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[\int\limits_{0}^{t} \frac{ Vm }{ M-mt } dt\]
anonymous
  • anonymous
take V,m, and M as constants..
zepdrix
  • zepdrix
It might be a little easier to read if pull the constants (in the numerator) outside of the integral, and then apply a `U substitution`.\[\large \int\limits\limits_{0}^{t} \frac{Vm}{M-mt} dt \qquad \rightarrow \qquad Vm\int\limits\limits_{0}^{t} \frac{1}{M-mt}dt\] Are you familiar with U substitution? :)

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anonymous
  • anonymous
thx, yes, what i get is -V{ln[(M-mt)/M]}...
anonymous
  • anonymous
now if M1=M-mt, M=Mb+Mf...and you substitue and simplify, i'm suppose to get Vln[1+(Mf/Mb)]
anonymous
  • anonymous
but i don't
zepdrix
  • zepdrix
I'm so confused, there are too many M's floating around :( lol
anonymous
  • anonymous
sorry i will redo with different letters...
anonymous
  • anonymous
\[\int\limits_{0}^{t} \frac{ ab }{ c-bt } dt\] all constants except t
anonymous
  • anonymous
d=c-bt c=e+f
zepdrix
  • zepdrix
So integrating, before doing any fancy simplification should give you,\[\large ab\left(-\frac{1}{b}\right)\ln|c-bt|_0^t\] Moar letters? :U oh boy lol
anonymous
  • anonymous
ok so integrate over limits and it should be -a[ln(c-bt)-ln(c)] correct so far?
zepdrix
  • zepdrix
Yah looks good so far c:
anonymous
  • anonymous
which is -a{ln[(c-bt)/c]}
zepdrix
  • zepdrix
Yes, you can bring the negative into the log as an exponent. Which will give us this,\[\large a \ln\left[\left(\frac{c-bt}{c}\right)^{-1}\right] \qquad = \qquad a \ln\left[\left(\frac{c}{c-bt}\right)\right]\] Whichever way makes sense to you though :3
zepdrix
  • zepdrix
Now what's going on with this d and e and f nonsense? XD lol
anonymous
  • anonymous
ok, so f has to be in there cuz that's what i'm trying to solve. so plugging in the other letters and simplifying, according to the solution, it's suppose to be a[ln(1+(f/e))] but i can't get it. however, i haven't tried with your second expression up there so i will try it now
anonymous
  • anonymous
ok, i think one of the variables is messed up on the solution. i got it, thx for your help
zepdrix
  • zepdrix
yay c:

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