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

- anonymous

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

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- anonymous

put 1 +e^t= z
diff. w.r.t. z , we get
e^t dt= dz

- anonymous

what is w. and r.?

- anonymous

ok lets do this way..

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- anonymous

divide the num and den. by e^t/2

- anonymous

lol, w.r.t means with respect to.

- anonymous

Sorry, I know you're trying to use substitution, but probably a different method from what I was taught

- anonymous

OH

- anonymous

did it help?

- anonymous

What's the point of dividing num. and den. by e^t/2 though?

- anonymous

That was another method of doing it

- anonymous

I do not think dividing by E^t/2 is a correct step in integrating this though

- anonymous

do with the first method then

- 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

Using substitution doesn't help

- anonymous

1/(1+e^t) is lnl1+e^tl ???? how did u get that??

- anonymous

dhatraditya, do you have any input?

- anonymous

This is integration....so I need to find anti-derivative

- anonymous

but ur retriceumption is false......1/(1+e^t) is lnl1+e^tl??? no way

- anonymous

So what is the correct integration of 1/1+e^t; could you show me with steps?

- anonymous

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

- anonymous

you have yo change your limits appropriately

- anonymous

it is indefinite integral...therefore no lower n upper limits

- anonymous

I actually don't care about the limits, I don't know how to antidifferentiate this

- anonymous

zuuto: will u please substitue what I said.....this works ....I've just done it on paper

- anonymous

limits are 0 to 2 according to the problem

- anonymous

ah...my bad

- anonymous

ankur is right, use substitution. It is the easiest method for definite integrals.

- anonymous

わかて、 すみません

- anonymous

oooh. Is that chinese? Sorry I dont understand it.

- anonymous

sorry, my intention was to understand how to differentiate this, I'm trying it right now

- anonymous

It was a mistake to type it, it's japanese

- anonymous

oh i see.

- anonymous

dont forget to change the limits like dhatraditya said....u can't ignore them when u have integral in z

- 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

Sub. in your limits and multiply by 10.

- anonymous

lokisan is right. His method is superior. and faster.

- anonymous

period :D

- anonymous

:D

- anonymous

coincidentally
i just figured that out too

- anonymous

which one?

- anonymous

making the integral into two parts

- anonymous

so it becomes t-lnle^t+1l I think

- anonymous

but u need to get used to substitution as well....u can't just rely on making two integrals every time

- anonymous

Most cases substitution is required, but I'm not how substitution works in this

- anonymous

yes, substitution is a general method. Splitting was more elegant in this case and lokisan correctly spotted that.

- anonymous

I'd say now u have the answer ...try with subs. too

- 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

ahhh thanks so muchh

- anonymous

now I just need to apply this to the problem, which was 10/1+e^t

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