anonymous
  • anonymous
How is 10/(1+e^t) supposed to be integrated from limits 0 to 2? Please show steps.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
put 1 +e^t= z diff. w.r.t. z , we get e^t dt= dz
anonymous
  • anonymous
what is w. and r.?
anonymous
  • anonymous
ok lets do this way..

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anonymous
  • anonymous
divide the num and den. by e^t/2
anonymous
  • anonymous
lol, w.r.t means with respect to.
anonymous
  • anonymous
Sorry, I know you're trying to use substitution, but probably a different method from what I was taught
anonymous
  • anonymous
OH
anonymous
  • anonymous
did it help?
anonymous
  • anonymous
What's the point of dividing num. and den. by e^t/2 though?
anonymous
  • anonymous
That was another method of doing it
anonymous
  • anonymous
I do not think dividing by E^t/2 is a correct step in integrating this though
anonymous
  • anonymous
do with the first method then
anonymous
  • anonymous
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
anonymous
  • anonymous
Using substitution doesn't help
anonymous
  • anonymous
1/(1+e^t) is lnl1+e^tl ???? how did u get that??
anonymous
  • anonymous
dhatraditya, do you have any input?
anonymous
  • anonymous
This is integration....so I need to find anti-derivative
anonymous
  • anonymous
but ur retriceumption is false......1/(1+e^t) is lnl1+e^tl??? no way
anonymous
  • anonymous
So what is the correct integration of 1/1+e^t; could you show me with steps?
anonymous
  • anonymous
the first method works for sure......after putting 1+e^t=z u will get this : intergartion (dz/ (z-1)z
anonymous
  • anonymous
you have yo change your limits appropriately
anonymous
  • anonymous
it is indefinite integral...therefore no lower n upper limits
anonymous
  • anonymous
I actually don't care about the limits, I don't know how to antidifferentiate this
anonymous
  • anonymous
zuuto: will u please substitue what I said.....this works ....I've just done it on paper
anonymous
  • anonymous
limits are 0 to 2 according to the problem
anonymous
  • anonymous
ah...my bad
anonymous
  • anonymous
ankur is right, use substitution. It is the easiest method for definite integrals.
anonymous
  • anonymous
わかて、 すみません
anonymous
  • anonymous
oooh. Is that chinese? Sorry I dont understand it.
anonymous
  • anonymous
sorry, my intention was to understand how to differentiate this, I'm trying it right now
anonymous
  • anonymous
It was a mistake to type it, it's japanese
anonymous
  • anonymous
oh i see.
anonymous
  • anonymous
dont forget to change the limits like dhatraditya said....u can't ignore them when u have integral in z
anonymous
  • anonymous
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,\[t-\log (1+e^t)\]
anonymous
  • anonymous
Sub. in your limits and multiply by 10.
anonymous
  • anonymous
lokisan is right. His method is superior. and faster.
anonymous
  • anonymous
period :D
anonymous
  • anonymous
:D
anonymous
  • anonymous
coincidentally i just figured that out too
anonymous
  • anonymous
which one?
anonymous
  • anonymous
making the integral into two parts
anonymous
  • anonymous
so it becomes t-lnle^t+1l I think
anonymous
  • anonymous
but u need to get used to substitution as well....u can't just rely on making two integrals every time
anonymous
  • anonymous
Most cases substitution is required, but I'm not how substitution works in this
anonymous
  • anonymous
yes, substitution is a general method. Splitting was more elegant in this case and lokisan correctly spotted that.
anonymous
  • anonymous
I'd say now u have the answer ...try with subs. too
anonymous
  • anonymous
or yeah in the second part of the intergral where you have to integrate e^t/e^t=1, substitution works definitely
anonymous
  • anonymous
:)
anonymous
  • anonymous
ahhh thanks so muchh
anonymous
  • anonymous
now I just need to apply this to the problem, which was 10/1+e^t

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