Question

Find the area of the region bounded by the x-axis and the graph of the function:
f(x)= (eˣ)/(1+eˣ)²
We are studying improper integrals if that helps

Answer 1

first thing to note is that this is an even function, so you can integrate
\[\int_0^{\infty} \frac{e^x}{(1+e^x)^2}dx\] and double the result

Answer 2

why would you double it?

Answer 3

because you are being asked for
\[\int_{-\infty}^{\infty}\frac{e^x}{(1+e^x)^2}dx\]

Answer 4

and so since it is even we can take the integral from 0 to infinity, and then multiply by 2

Answer 5

okay I guess I don't understand even integrals

Answer 6

"even function" not "even integral" even function means
\[f(x)=f(-x)\]

Answer 7

and how do you know its from -∞ to ∞?

Answer 8

okay and it's even because anything squared will always be positive

Answer 9

so that what you see to the left of the y - axis is what you see to the right. i.e. it is symmetric with respect to the y axis. so if you want the integral from -5 to 5, say, you can compute the integral from 0 to 5 and double

Answer 10

okay gotcha

Answer 11

you know you are integrating from - infinity to + infinity because of the way the question is asked. area above the x - axis and below the curve

Answer 12

you are not given any bounds other than the x - axis below and the function above

Answer 13

how do you know it's not below the x-axis?

Answer 14

because this function is positive

Answer 15

okay. so now I integrate

Answer 16

before we start,
it is not even because of the "square" it is even because if
\[f(x)=\frac{e^x}{(1+e^x)^2}\] then
\[f(-x)=\frac{e^{-x}}{(1+e^{-x})^2}\] and then if you do some algebra we get this is the same as
\[f(x)=\frac{e^x}{(1+e^x)^2}\]

Answer 17

so first we find an anti-derivative of \[f(x)=\frac{e^x}{(1+e^x)^2}\]

Answer 18

are you sure the function is even satellite?

Answer 19

oh wait...

Answer 20

ok you may proceed

Answer 21

its even

Answer 22

by a u substitution. you let
\[u=1+e^x\]and
\[du=e^xdx\] and so you have
\[\int\frac{1}{u^2}du=-\frac{1}{u}=-\frac{1}{(1+e^x)^2}\]

Answer 23

and yes i am sure it is even because
\[\frac{e^{-x}}{(1+e^{-x})^2}=\frac{\frac{1}{e^x}}{1+\frac{1}{2e^x}+\frac{1}{e^{2x}}}\] which, if you multiply top and bottom by
\[e^{2x}\] gives \[\frac{e^x}{(1+e^x)^2}\]

Answer 24

wait... theintegral of 1/u².. isn't that 1/2 u^-1?

Answer 25

yes i know satellite i already did it

Answer 26

never mind!

Answer 27

@shortiesport, no it is just regular old
\[-\frac{1}{u}\]

Answer 28

ok so now we have the anti derivative. now we get to evaluate but remember that
\[\int_0^{\infty} \frac{e^x}{(1+e^x)^2}dx\] means
\[\lim_{r\rightarrow \infty}\int_0^r \frac{e^x}{(1+e^x)^2}dx\]

Answer 29

do I always have to find if it's even or odd? i'm confuseed too because this chapter was kinda about the limit as b approaches inifinity

Answer 30

so we take
\[-\frac{1}{(1+e^x)}\] and replace x by r and 0

Answer 31

no you don't have to do the even odd thing

Answer 32

no, sometimes it is even, sometimes it is odd, and most time neither. now comes the integral part, where we evaluate and then take the limit as r goes to infinity

Answer 33

so do I write this as the limit as b -> ∞ of -1/(1+eˣ)?

Answer 34

at r you get
\[-\frac{1}{1+e^r}\] and at 0 you get
\[-\frac{1}{1+e^0}=-\frac{1}{2}\] so
\[\lim_{r\rightarrow \infty}\int_0^r \frac{e^x}{(1+e^x)^2}dx=\lim_{r\rightarrow \infty}-\frac{1}{1-e^r}+\frac{1}{2}\]\]

Answer 35

which pretty clearly is
\[\frac{1}{2}\]

Answer 36

since
\[\lim_{r\rightarrow \infty}-\frac{1}{1-e^r}=0\]

Answer 37

and therefore your "final answer" is 1

Answer 38

please notice that although you did not need to use the fact that this was even, it was a huge help. you have an integral that goes from - infinity to infinity, so you are going to have to break it up somewhere.

Answer 39

in other words it is an improper integral two times, you have to compute the limit at both ends, so being even cut the work in half

Answer 40

okay.. i think i'm going to have to go back through the steps in slow motion lol. I understand why we found the integral of the original problem = -1/(1+eˣ)². I just don't understand what we did after that

Answer 41

here is a picture of your function. you can see that it is even, that it sits above the x - axis, and that it approaches 0 nice and fast
http://www.wolframalpha.com/input/?i=y%3De^x%2F%281%2Be^x%29^2

Answer 42

good job satellite
you can pass cal 2 again

Answer 43

like what is the limit even telling me

Answer 44

ok after that we computed this:
\[\int_0^r\frac{e^x}{(1+e^x)^2}dx\] which gave
\[-\frac{1}{1+e^r}+\frac{1}{2}\] by plugging in upper limit and lower limit into the anti derivative

Answer 45

in other words just like you would for a "regular " integral, once you compute the anti derivative
\[F(x)\] you find
\[F(b)-F(a)\]

Answer 46

okay so when did we double it?

Answer 47

(btw I plugged in 0 and b and got the same things)

Answer 48

at the very end. you get 1/2 as your integral, double it and get 1

Answer 49

b, r, whatever your book uses and you like

Answer 50

most actually use l

Answer 51

okay so after I find the integral from 0 to b... why do I just have 1/2?

Answer 52

the integral from 0 to b gave you
\[-\frac{1}{1+e^b}+\frac{1}{2}\] right?

Answer 53

yes

Answer 54

now you take the limit as b goes to infinity, because that is the definition of the "improper" integral

Answer 55

\[\int_0^{\infty} \frac{e^x}{(1+e^x)^2}dx=\lim_{b\rightarrow \infty}\int_0^b \frac{e^x}{(1+e^x)^2}dx=\lim_{b\rightarrow \infty}-\frac{1}{1+e^b}+\frac{1}{2}=0+\frac{1}{2}\]