What is the arclength of #f(x)=(1-3x)/(1+e^x)# on #x in [-1,0]#?

Arc length formula of #f(x)# from #x in [a,b]#:

#int_a^b sqrt(1+((df)/dx)^2) dx#

Hence, substituting into the formula, we have

Arc Length = #int_(-1)^0 sqrt(1+((e^x (3 x - 4) - 3)/(1 + e^x)^2)^2)dx#

Since #(df)/dx=(e^x (3 x - 4) - 3)/(1 + e^x)^2#

You could try to simplify this integral, but soon one will realise that this does not have a "nice" closed form

#int_(-1)^0 sqrt((e^(2 x) (9 x^2 + 22) + e^x (28 - 6 (4 e^x + 3) x) + 4 e^(3 x) + e^(4 x) + 10)/(e^x + 1)^4) dx#

At this point, you can only approximate this integral using Riemann sums or infinite series, and the answer you get will be around #2.62618234447323...#

Derivation of Arc Length Formula

Given a function #y=f(x)#

To find the arc length #Deltas# over an interval of time #Deltax#, you can try to approximate it with a straight line, using the hypotenuse of the right-angled triangle with sides #Deltax, Deltay#, which is the change in the x- and y- coordinates over the interval #Deltax#.

#Deltas ~~ sqrt(Deltax^2+Deltay^2)#

Of course, this will often be wildly inaccurate since most curves aren't exactly a straight line.

However, for a continuous function, the smaller x interval you choose, the more accurate the above approximation is (you can imagine zooming in on a function's graph: the more you zoom, the closer the function looks like a straight line)

Thus, at the limit when you shorten the time interval, you get

#ds = sqrt(dx^2+dy^2)#

To manipulate it to be in terms of the x interval #dx#, we can adjust the equation such that #ds# is in terms of #dx#.

#ds = sqrt(dx^2+(dx*dy/dx)^2)#
#=sqrt(dx^2(1+(dy/dx)^2)#
#=sqrt(1+(dy/dx)^2)dx#

Integrating both sides across the interval #x in [a, b]#, we have

#s=int_a^b sqrt(1+(dy/dx)^2)dx#