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anonymous

  • one year ago

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

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  1. anonymous
    • one year ago
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    \[ F(x) = \int\limits_{0}^{t} (210000 E^x) / (4.1 + 1.2E^x) dx \] F[0] = 0

  2. dan815
    • one year ago
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    |dw:1438639191274:dw|

  3. anonymous
    • one year ago
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    I can see how I might convert it to this.. is there a product rule for integration? \[ F(x) = \int\limits_{0}^{t} (210000 E^x) (4.1 + 1.2E^x)^-1 \]

  4. dan815
    • one year ago
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    use that method i showed u up there

  5. anonymous
    • one year ago
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    lol, dan, that makes no sense to me at all.

  6. dan815
    • one year ago
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    u will have to do integration by parts for the equation in the form 1/(1+e^x)

  7. Astrophysics
    • one year ago
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    Yeah, what exactly did you do danni boi

  8. dan815
    • one year ago
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    well u can transform your equation into that form if u pull out ur constants

  9. Astrophysics
    • one year ago
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    Show

  10. dan815
    • one year ago
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    |dw:1438639475428:dw|

  11. Astrophysics
    • one year ago
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    I mean I know you can pull the constants out

  12. Astrophysics
    • one year ago
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    but the other part

  13. dan815
    • one year ago
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    |dw:1438639533510:dw|

  14. freckles
    • one year ago
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    he added in a 0

  15. dan815
    • one year ago
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    do parts on the 2nd term

  16. Astrophysics
    • one year ago
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    Hey, why can't we just say u = 1+e^x, du = e^x dx,

  17. dan815
    • one year ago
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    yea thats fine too

  18. Astrophysics
    • one year ago
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    Then we would just have \[\int\limits \frac{ 1 }{ u } du\]

  19. Astrophysics
    • one year ago
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    I think that's much more straight forward

  20. anonymous
    • one year ago
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    so we get ln[u] ?

  21. Astrophysics
    • one year ago
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    \[\int\limits \frac{ e^x }{ 1+e^x } dx\] \[u = 1+e^x \implies du = e^x dx \] \[\int\limits \frac{ 1 }{ u } du\]

  22. dan815
    • one year ago
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    hehe this was silly

  23. Astrophysics
    • one year ago
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    :P

  24. dan815
    • one year ago
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    the derivative of the bottom was on the top of the fraction you shuld notice that

  25. Astrophysics
    • one year ago
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    When in doubt try u sub @hughfuve hehe

  26. dan815
    • one year ago
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    |dw:1438639816201:dw|

  27. dan815
    • one year ago
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    |dw:1438640062643:dw|

  28. freckles
    • one year ago
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    I'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

  29. anonymous
    • one year ago
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    Hi 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

  30. Astrophysics
    • one year ago
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    Is it F(x) or F(t)

  31. freckles
    • one year ago
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    should be F(t) if the upper limit is t

  32. Astrophysics
    • one year ago
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    Yeah that's what I thought at first as well, but it says F(x) at the top just making sure

  33. anonymous
    • one year ago
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    I think it would be F[t] given that integral, how much time t before we reach 625k

  34. anonymous
    • one year ago
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    Unfortunately 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.

  35. anonymous
    • one year ago
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    If you can throw me a google link that makes this any easier.. I'd appreciate it.

  36. freckles
    • one year ago
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    so you haven't actually talk about how to integrate like substitution (this was the method used above)

  37. anonymous
    • one year ago
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    have 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.

  38. anonymous
    • one year ago
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    so 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..

  39. anonymous
    • one year ago
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    They 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.

  40. Astrophysics
    • one year ago
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    Youtube 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=qclrs-1rpKI

  41. Astrophysics
    • one year ago
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    should*

  42. anonymous
    • one year ago
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    I 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.

  43. anonymous
    • one year ago
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    thanks astro

  44. Astrophysics
    • one year ago
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    Go 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!!

  45. anonymous
    • one year ago
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    lol, I can't believe this bloody course.. lmao.. its so bad,

  46. Astrophysics
    • one year ago
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    Your welcome :)

  47. Astrophysics
    • one year ago
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    Haha, yeah it can be tedious but it's beautiful.

  48. freckles
    • one year ago
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    \[\int\limits_{}^{} \frac{ a e^x}{b+c e^{x}} dx \\ \text{ \let } u=b+c e^{x} \\ \frac{du}{dx}=0+ce^x \\ \frac{du}{dx}=ce^{x} \\ du=c e^{x} dx \\ \text{ divide } c \text{ on both sides } \\ \frac{1}{c} du=e^x dx \\ \int\limits \frac{ae^x}{b+ce^{x}} dx=a \int\limits \frac{ 1}{b+ce^{x}} e^{x} dx =a \int\limits \frac{1}{u} \frac{1}{c} du \\ \frac{a}{c} \int\limits \frac{1}{u} du=\frac{a}{c} \ln|u|+K=\frac{a}{c}\ln|b+ce^{x}|+K\] 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

  49. anonymous
    • one year ago
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    nice one freckles.... that makes sense.

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

  51. anonymous
    • one year ago
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    sweet thanks.. okay I have to head out for a bit.. but this is very helpful.

  52. anonymous
    • one year ago
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    I wish I could give you all medals

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