Find the area between the graph y=2e^x/(1+e^x) and 2e^x/(1+e^2x)

Expert Answers
steveschoen eNotes educator| Certified Educator

Well, this depends upon what domain of numbers are you taking.  For instance, if you consider the domain from 0 --> +infinity, the area would be infinity.  Similar if you consider all numbers from -infinity --> 0.  Reason being the lines only intersect at one point.  There is no closed area to calculate an area.  Even if you consider both sides, you would still have infinity.

But, then, if you take only part of those numbers.  For instance, from 0 --> 5, then, we can take the integral of the top graph, here the first equation, and subtract the integral of the bottom graph, where the second equation.

For the first equation, the area under the curve from 0 --> 5 is:

8.627 units

For the second curve, the area is:

1.557 units

So, subtracting those, we would have 7.070 units

I did use an online integral calculator on this one,

embizze eNotes educator| Certified Educator

Find the area between the curves `y=(2e^x)/(1+e^x) ` and `y=(2e^x)/(1+e^(2x)) `

These curves only intersect at x=0. Without limits, the area to the right of zero is infinite. (The first curve is bounded above by 2 while the second curve is bounded below by 0 so for large x the area is approximately 2x.)

To the left of zero both curves converge to zero. The area to the left of zero can be computed:

`int_(-oo)^0 [(2e^x)/(1+e^x)-(2e^x)/(1+e^(2x))]dx `

`=lim_(a --> -oo)int_a^0 [(2e^x)/(1+e^x)-(2e^x)/(1+e^(2x))]dx`

`=lim_(a->-oo)[2tan^(-1)(e^x)-ln(e^x+1)]_a^0 `

`=1/2(pi-2ln(4))~~0.184502 `