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elica85
 2 years ago
integrate...
elica85
 2 years ago
integrate...

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elica85
 2 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{0}^{t} \frac{ Vm }{ Mmt } dt\]

elica85
 2 years ago
Best ResponseYou've already chosen the best response.0take V,m, and M as constants..

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1It might be a little easier to read if pull the constants (in the numerator) outside of the integral, and then apply a `U substitution`.\[\large \int\limits\limits_{0}^{t} \frac{Vm}{Mmt} dt \qquad \rightarrow \qquad Vm\int\limits\limits_{0}^{t} \frac{1}{Mmt}dt\] Are you familiar with U substitution? :)

elica85
 2 years ago
Best ResponseYou've already chosen the best response.0thx, yes, what i get is V{ln[(Mmt)/M]}...

elica85
 2 years ago
Best ResponseYou've already chosen the best response.0now if M1=Mmt, M=Mb+Mf...and you substitue and simplify, i'm suppose to get Vln[1+(Mf/Mb)]

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1I'm so confused, there are too many M's floating around :( lol

elica85
 2 years ago
Best ResponseYou've already chosen the best response.0sorry i will redo with different letters...

elica85
 2 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{0}^{t} \frac{ ab }{ cbt } dt\] all constants except t

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1So integrating, before doing any fancy simplification should give you,\[\large ab\left(\frac{1}{b}\right)\lncbt_0^t\] Moar letters? :U oh boy lol

elica85
 2 years ago
Best ResponseYou've already chosen the best response.0ok so integrate over limits and it should be a[ln(cbt)ln(c)] correct so far?

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1Yah looks good so far c:

elica85
 2 years ago
Best ResponseYou've already chosen the best response.0which is a{ln[(cbt)/c]}

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1Yes, you can bring the negative into the log as an exponent. Which will give us this,\[\large a \ln\left[\left(\frac{cbt}{c}\right)^{1}\right] \qquad = \qquad a \ln\left[\left(\frac{c}{cbt}\right)\right]\] Whichever way makes sense to you though :3

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1Now what's going on with this d and e and f nonsense? XD lol

elica85
 2 years ago
Best ResponseYou've already chosen the best response.0ok, 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

elica85
 2 years ago
Best ResponseYou've already chosen the best response.0ok, i think one of the variables is messed up on the solution. i got it, thx for your help
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