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anonymous

  • 5 years ago

How is 10/(1+e^t) supposed to be integrated from limits 0 to 2? Please show steps.

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  1. anonymous
    • 5 years ago
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    put 1 +e^t= z diff. w.r.t. z , we get e^t dt= dz

  2. anonymous
    • 5 years ago
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    what is w. and r.?

  3. anonymous
    • 5 years ago
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    ok lets do this way..

  4. anonymous
    • 5 years ago
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    divide the num and den. by e^t/2

  5. anonymous
    • 5 years ago
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    lol, w.r.t means with respect to.

  6. anonymous
    • 5 years ago
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    Sorry, I know you're trying to use substitution, but probably a different method from what I was taught

  7. anonymous
    • 5 years ago
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    OH

  8. anonymous
    • 5 years ago
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    did it help?

  9. anonymous
    • 5 years ago
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    What's the point of dividing num. and den. by e^t/2 though?

  10. anonymous
    • 5 years ago
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    That was another method of doing it

  11. anonymous
    • 5 years ago
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    I do not think dividing by E^t/2 is a correct step in integrating this though

  12. anonymous
    • 5 years ago
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    do with the first method then

  13. anonymous
    • 5 years ago
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    My problem is actually, I know that 1/(1+e^t) is lnl1+e^tl, but I don't know what to do with the 10

  14. anonymous
    • 5 years ago
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    Using substitution doesn't help

  15. anonymous
    • 5 years ago
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    1/(1+e^t) is lnl1+e^tl ???? how did u get that??

  16. anonymous
    • 5 years ago
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    dhatraditya, do you have any input?

  17. anonymous
    • 5 years ago
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    This is integration....so I need to find anti-derivative

  18. anonymous
    • 5 years ago
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    but ur retriceumption is false......1/(1+e^t) is lnl1+e^tl??? no way

  19. anonymous
    • 5 years ago
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    So what is the correct integration of 1/1+e^t; could you show me with steps?

  20. anonymous
    • 5 years ago
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    the first method works for sure......after putting 1+e^t=z u will get this : intergartion (dz/ (z-1)z

  21. anonymous
    • 5 years ago
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    you have yo change your limits appropriately

  22. anonymous
    • 5 years ago
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    it is indefinite integral...therefore no lower n upper limits

  23. anonymous
    • 5 years ago
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    I actually don't care about the limits, I don't know how to antidifferentiate this

  24. anonymous
    • 5 years ago
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    zuuto: will u please substitue what I said.....this works ....I've just done it on paper

  25. anonymous
    • 5 years ago
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    limits are 0 to 2 according to the problem

  26. anonymous
    • 5 years ago
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    ah...my bad

  27. anonymous
    • 5 years ago
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    ankur is right, use substitution. It is the easiest method for definite integrals.

  28. anonymous
    • 5 years ago
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    わかて、 すみません

  29. anonymous
    • 5 years ago
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    oooh. Is that chinese? Sorry I dont understand it.

  30. anonymous
    • 5 years ago
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    sorry, my intention was to understand how to differentiate this, I'm trying it right now

  31. anonymous
    • 5 years ago
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    It was a mistake to type it, it's japanese

  32. anonymous
    • 5 years ago
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    oh i see.

  33. anonymous
    • 5 years ago
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    dont forget to change the limits like dhatraditya said....u can't ignore them when u have integral in z

  34. anonymous
    • 5 years ago
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    Can I make a suggestion? You might want to recognise\[\frac{1}{1+e^t}=\frac{1+e^t-e^t}{1+e^t}=\frac{1+e^t}{1+e^t}-\frac{e^t}{1+e^t}=1-\frac{e^t}{1+e^t}\]You can integrate that directly,\[t-\log (1+e^t)\]

  35. anonymous
    • 5 years ago
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    Sub. in your limits and multiply by 10.

  36. anonymous
    • 5 years ago
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    lokisan is right. His method is superior. and faster.

  37. anonymous
    • 5 years ago
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    period :D

  38. anonymous
    • 5 years ago
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    :D

  39. anonymous
    • 5 years ago
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    coincidentally i just figured that out too

  40. anonymous
    • 5 years ago
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    which one?

  41. anonymous
    • 5 years ago
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    making the integral into two parts

  42. anonymous
    • 5 years ago
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    so it becomes t-lnle^t+1l I think

  43. anonymous
    • 5 years ago
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    but u need to get used to substitution as well....u can't just rely on making two integrals every time

  44. anonymous
    • 5 years ago
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    Most cases substitution is required, but I'm not how substitution works in this

  45. anonymous
    • 5 years ago
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    yes, substitution is a general method. Splitting was more elegant in this case and lokisan correctly spotted that.

  46. anonymous
    • 5 years ago
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    I'd say now u have the answer ...try with subs. too

  47. anonymous
    • 5 years ago
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    or yeah in the second part of the intergral where you have to integrate e^t/e^t=1, substitution works definitely

  48. anonymous
    • 5 years ago
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    :)

  49. anonymous
    • 5 years ago
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    ahhh thanks so muchh

  50. anonymous
    • 5 years ago
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    now I just need to apply this to the problem, which was 10/1+e^t

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