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

Jun 4, 2018

${\int}_{- 1}^{0} \sqrt{1 + {\left(\frac{{e}^{x} \left(3 x - 4\right) - 3}{1 + {e}^{x}} ^ 2\right)}^{2}} \mathrm{dx} \approx 2.62618234447323 \ldots$

Explanation:

Arc length formula of $f \left(x\right)$ from $x \in \left[a , b\right]$:

${\int}_{a}^{b} \sqrt{1 + {\left(\frac{\mathrm{df}}{\mathrm{dx}}\right)}^{2}} \mathrm{dx}$

Hence, substituting into the formula, we have

Arc Length = ${\int}_{- 1}^{0} \sqrt{1 + {\left(\frac{{e}^{x} \left(3 x - 4\right) - 3}{1 + {e}^{x}} ^ 2\right)}^{2}} \mathrm{dx}$

Since $\frac{\mathrm{df}}{\mathrm{dx}} = \frac{{e}^{x} \left(3 x - 4\right) - 3}{1 + {e}^{x}} ^ 2$

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

${\int}_{- 1}^{0} \sqrt{\frac{{e}^{2 x} \left(9 {x}^{2} + 22\right) + {e}^{x} \left(28 - 6 \left(4 {e}^{x} + 3\right) x\right) + 4 {e}^{3 x} + {e}^{4 x} + 10}{{e}^{x} + 1} ^ 4} \mathrm{dx}$

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

Derivation of Arc Length Formula

Given a function $y = f \left(x\right)$

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

$\Delta s \approx \sqrt{\Delta {x}^{2} + \Delta {y}^{2}}$

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

$\mathrm{ds} = \sqrt{{\mathrm{dx}}^{2} + {\mathrm{dy}}^{2}}$

To manipulate it to be in terms of the x interval $\mathrm{dx}$, we can adjust the equation such that $\mathrm{ds}$ is in terms of $\mathrm{dx}$.

$\mathrm{ds} = \sqrt{{\mathrm{dx}}^{2} + {\left(\mathrm{dx} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}}$
=sqrt(dx^2(1+(dy/dx)^2)
$= \sqrt{1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}} \mathrm{dx}$

Integrating both sides across the interval $x \in \left[a , b\right]$, we have

$s = {\int}_{a}^{b} \sqrt{1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}} \mathrm{dx}$