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anonymous
 one year ago
How do I Integrate
(210000 E^t) / (4.1 + 1.2E^t)
latex coming...
anonymous
 one year ago
How do I Integrate (210000 E^t) / (4.1 + 1.2E^t) latex coming...

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[ F(x) = \int\limits_{0}^{t} (210000 E^x) / (4.1 + 1.2E^x) dx \] F[0] = 0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I can see how I might convert it to this.. is there a product rule for integration? \[ F(x) = \int\limits_{0}^{t} (210000 E^x) (4.1 + 1.2E^x)^1 \]

dan815
 one year ago
Best ResponseYou've already chosen the best response.3use that method i showed u up there

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0lol, dan, that makes no sense to me at all.

dan815
 one year ago
Best ResponseYou've already chosen the best response.3u will have to do integration by parts for the equation in the form 1/(1+e^x)

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Yeah, what exactly did you do danni boi

dan815
 one year ago
Best ResponseYou've already chosen the best response.3well u can transform your equation into that form if u pull out ur constants

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1I mean I know you can pull the constants out

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1but the other part

dan815
 one year ago
Best ResponseYou've already chosen the best response.3do parts on the 2nd term

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Hey, why can't we just say u = 1+e^x, du = e^x dx,

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Then we would just have \[\int\limits \frac{ 1 }{ u } du\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1I think that's much more straight forward

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1\[\int\limits \frac{ e^x }{ 1+e^x } dx\] \[u = 1+e^x \implies du = e^x dx \] \[\int\limits \frac{ 1 }{ u } du\]

dan815
 one year ago
Best ResponseYou've already chosen the best response.3the derivative of the bottom was on the top of the fraction you shuld notice that

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1When in doubt try u sub @hughfuve hehe

freckles
 one year ago
Best ResponseYou've already chosen the best response.2I'm curious @hughfuve was the question really to integrate I notice you wrote an definite integral and called it F(t) \[F(t)=\int\limits_0^t \frac{210000e^x}{4.1+1.2e^x}dx \\ \text{ I mean we could evaluate the integral on the } \\ \text{ right hand side }\] but depending on the question it is asking we may not have to

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Hi freckles, well the actual problem I was trying to work out is given that integral, how long would it take F[t] to reach the value of 625000 if F[0]=0. I figured if I could get to an actual definition of F[t] that I could solve for F[t]=625000 and find t

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Is it F(x) or F(t)

freckles
 one year ago
Best ResponseYou've already chosen the best response.2should be F(t) if the upper limit is t

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Yeah that's what I thought at first as well, but it says F(x) at the top just making sure

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think it would be F[t] given that integral, how much time t before we reach 625k

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Unfortunately I have not been introduced to integration by parts.. (not sure if that is needed) I have only just been introduced to the Fundamental Theorem, and that's about all I know about integration. \[\int\limits_{a}^{b} f(x) dx = F(b)F(a)\] That's the net extent of my knowledge so far.. so I'm looking at all this and wondering how it all relates. Up until now it has been all differentiation. And about all I know is that this equation results in an antiderivative. My class has no textbook and no lectures and no teacher.. so it's really bloody hard. I have to google my way to answer problems, but I do have mathematica installed, and I am allowed to use mathematica as a solution. But for some reason , it just doesn;t like this equation at all, so I have to prove it by hand.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If you can throw me a google link that makes this any easier.. I'd appreciate it.

freckles
 one year ago
Best ResponseYou've already chosen the best response.2so you haven't actually talk about how to integrate like substitution (this was the method used above)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0have not seen anything about substitution in the course.. but I did use substitution to differentiate with the chain rule.. (which was also not in the course, lol) .. but I do have chain rule down pretty good.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so far the recommendation to integrate.. is just work out what you would do to differentiate your way here. Which is fine for most things that I can just look at a table of differentiation rules.. but this one threw me off..

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0They use Dsolve and integrate a lot in the course to show examples of what you are trying to achieve.. they dont actually use the long hand methods.

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Youtube has many videos as the links I've sent you, in 10 minutes you can learn u  substitution, you really take advantage of these videos: https://www.youtube.com/watch?v=qclrs1rpKI

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I have googled substitution rule.. doing some reading now.. I;ll close this out and come back to it later.. thanks everyone, this will get me there.

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Go with the examples on youtube, as you can pause it try it yourself and then watch the person work it out, and once you get good you can also make it 2x speed and watch it in half the time. Youtube is great, take advantage!!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0lol, I can't believe this bloody course.. lmao.. its so bad,

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Haha, yeah it can be tedious but it's beautiful.

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[\int\limits_{}^{} \frac{ a e^x}{b+c e^{x}} dx \\ \text{ \let } u=b+c e^{x} \\ \frac{du}{dx}=0+ce^x \\ \frac{du}{dx}=ce^{x} \\ du=c e^{x} dx \\ \text{ divide } c \text{ on both sides } \\ \frac{1}{c} du=e^x dx \\ \int\limits \frac{ae^x}{b+ce^{x}} dx=a \int\limits \frac{ 1}{b+ce^{x}} e^{x} dx =a \int\limits \frac{1}{u} \frac{1}{c} du \\ \frac{a}{c} \int\limits \frac{1}{u} du=\frac{a}{c} \lnu+K=\frac{a}{c}\lnb+ce^{x}+K\] by the way a,b,c are constants that are you were given and K is the constant of integration which won't matter for your case since you have a definite integral

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0nice one freckles.... that makes sense.

freckles
 one year ago
Best ResponseYou've already chosen the best response.2as you see differentiating that answer should give us what we started with the derivative of K is 0 so we are looking at: \[(\frac{a}{c} \lnb+ce^{x})' \\ \text{ by constant multiple rule } \\ \frac{a}{c}(\lnb+ce^{x})' \\ \text{ by chain rule } \\ \frac{a}{c} \frac{0+ce^{x}}{b+ce^{x}} \\ \frac{a}{c} \frac{ce^{x}}{b+ce^{x}} \\ \text{ we know } \frac{c}{c}=1 \\ a \frac{e^{x}}{b+ce^x} \\ \frac{ae^{x}}{b+ce^{x}} \\ \text{ just if you wanted to check your integration }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sweet thanks.. okay I have to head out for a bit.. but this is very helpful.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I wish I could give you all medals
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