## anonymous 5 years ago How is 10/(1+e^t) supposed to be integrated from limits 0 to 2? Please show steps.

1. anonymous

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

2. anonymous

what is w. and r.?

3. anonymous

ok lets do this way..

4. anonymous

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

5. anonymous

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

6. anonymous

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

7. anonymous

OH

8. anonymous

did it help?

9. anonymous

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

10. anonymous

That was another method of doing it

11. anonymous

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

12. anonymous

do with the first method then

13. 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

14. anonymous

Using substitution doesn't help

15. anonymous

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

16. anonymous

dhatraditya, do you have any input?

17. anonymous

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

18. anonymous

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

19. anonymous

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

20. anonymous

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

21. anonymous

you have yo change your limits appropriately

22. anonymous

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

23. anonymous

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

24. anonymous

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

25. anonymous

limits are 0 to 2 according to the problem

26. anonymous

27. anonymous

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

28. anonymous

わかて、 すみません

29. anonymous

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

30. anonymous

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

31. anonymous

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

32. anonymous

oh i see.

33. anonymous

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

34. 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,$t-\log (1+e^t)$

35. anonymous

Sub. in your limits and multiply by 10.

36. anonymous

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

37. anonymous

period :D

38. anonymous

:D

39. anonymous

coincidentally i just figured that out too

40. anonymous

which one?

41. anonymous

making the integral into two parts

42. anonymous

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

43. anonymous

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

44. anonymous

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

45. anonymous

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

46. anonymous

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

47. anonymous

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

48. anonymous

:)

49. anonymous

ahhh thanks so muchh

50. anonymous

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