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

Jun 5, 2018

Recall the distance formula:

$D \left(x\right) = \sqrt{{\left(\Delta x\right)}^{2} + {\left(\Delta y\right)}^{2}}$

Now extend that to arc length:

$s = D \left(x\right) = \sum \sqrt{{\left(\Delta x\right)}^{2} + {\left(\Delta y\right)}^{2} / {\left(\Delta x\right)}^{2} \cdot {\left(\Delta x\right)}^{2}}$

$= \sum \sqrt{1 + {\left(\frac{\Delta y}{\Delta x}\right)}^{2}} \left(\Delta x\right)$

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

This is just a "dynamic", infinitesimally-short-distance formula that accumulates over an interval of constantly increasing $x$. The general strategy is to get common denominators, perhaps complete the square, and get the square root to go away.

Thus, we need the first derivative of $f \left(x\right)$ first:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left[{e}^{3 x} / x + {x}^{2} {e}^{x}\right]$

$= {e}^{3 x} \cdot - \frac{1}{x} ^ 2 + \frac{1}{x} \cdot 3 {e}^{3 x} + {x}^{2} \cdot {e}^{x} + {e}^{x} \cdot 2 x$

$= - {e}^{3 x} / {x}^{2} + \frac{3 {e}^{3 x}}{x} + {x}^{2} {e}^{x} + 2 x {e}^{x}$

$= \left(3 x - 1\right) {e}^{3 x} / {x}^{2} + x \left(x + 2\right) {e}^{x}$

Now, in the expression, we would next square it:

${\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2} = {\left[\left(3 x - 1\right) {e}^{3 x} / {x}^{2} + x \left(x + 2\right) {e}^{x}\right]}^{2}$

Using Wolfram Alpha, this was simplified to:

$= {e}^{2 x} / {x}^{4} {\left[\left(x + 2\right) {x}^{3} + {e}^{2 x} \left(3 x - 1\right)\right]}^{2}$

Inserting it into the equation for arc length, we obtain the integral that we would have attempted:

$\textcolor{g r e e n}{s = {\int}_{1}^{2} \sqrt{1 + {e}^{2 x} / {x}^{4} {\left[\left(x + 2\right) {x}^{3} + {e}^{2 x} \left(3 x - 1\right)\right]}^{2}} \mathrm{dx}}$

There is no solution for this in terms of standard mathematical functions.

The numerical solution is:

$\textcolor{b l u e}{s = 208.471}$