A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 3 years ago
integrate...
anonymous
 3 years ago
integrate...

This Question is Closed

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

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

zepdrix
 3 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? :)

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

anonymous
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.1I'm so confused, there are too many M's floating around :( lol

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

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

zepdrix
 3 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

anonymous
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.1Yah looks good so far c:

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

zepdrix
 3 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
 3 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

anonymous
 3 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

anonymous
 3 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
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.