What is the arclength of f(x)=x/e^(3x) on x in [1,2]?

Feb 19, 2016

Approximately 1.

Explanation:

The arc length is calculated from the following integral:

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

$L = {\int}_{1}^{2} \sqrt{1 + {\left[\left(1 - 3 x\right) {e}^{- 3 x}\right]}^{2}} \mathrm{dx}$

I haven't found an analytic expression for the integral. However, when we plot the function $f \left(x\right)$ it becomes clear that the function has decayed to very small values.

What is deceiving about this graph is that arc-length is calculated in normal Cartesian space, which implies that the scale of $x$ and $y$ should be equal as in the following version:

This implies that the actual arc length between 1 and 2 will be very close to the distance on the axis. Doing the integration numerically, one gets

$L \cong 1.0013 \ldots$

which is very close to 1 as expected. A possible refinement would be to approximate the arc length with a straight line between $f \left(1\right)$ and $f \left(2\right)$ giving the two points

$\left(1 , 0.050\right) \mathmr{and} \left(2 , 0.005\right)$

$L \cong \sqrt{1 + {\left(0.050 - 0.005\right)}^{2}} = 1.0010 \ldots$