Question

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

Answer 1

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

Answer 2

what is w. and r.?

Answer 3

ok lets do this way..

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

OH

Answer 8

did it help?

Answer 9

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

Answer 10

That was another method of doing it

Answer 11

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

Answer 12

do with the first method then

Answer 13

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

Answer 14

Using substitution doesn't help

Answer 15

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

Answer 16

dhatraditya, do you have any input?

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

you have yo change your limits appropriately

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

limits are 0 to 2 according to the problem

Answer 26

ah...my bad

Answer 27

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

Answer 28

わかて、 すみません

Answer 29

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

Answer 30

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

Answer 31

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

Answer 32

oh i see.

Answer 33

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

Answer 34

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,\[t-\log (1+e^t)\]

Answer 35

Sub. in your limits and multiply by 10.

Answer 36

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

Answer 37

period :D

Answer 38

:D

Answer 39

coincidentally
i just figured that out too

Answer 40

which one?

Answer 41

making the integral into two parts

Answer 42

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

Answer 43

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

Answer 44

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

Answer 45

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

Answer 46

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

Answer 47

or yeah

in the second part of the intergral where you have to integrate e^t/e^t=1, substitution works definitely

Answer 48

:)

Answer 49

ahhh thanks so muchh

Answer 50

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