Question

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

Answer 1

Recall the distance formula:

`D(x) = sqrt((Deltax)^2 + (Deltay)^2)`

Now extend that to arc length:

`s = D(x) = sumsqrt((Deltax)^2 + (Deltay)^2/(Deltax)^2*(Deltax)^2)`

`= sumsqrt(1 + ((Deltay)/(Deltax))^2)(Deltax)`

`=> color(blue)(s = int_a^b sqrt(1 + ((dy)/(dx))^2)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(x)` first:

`(dy)/(dx) = d/(dx)[e^(3x)/x + x^2e^x]`

`= e^(3x) cdot -1/x^2 + 1/x cdot 3e^(3x) + x^2 cdot e^x + e^x cdot 2x`

`= -e^(3x)/x^2 + (3e^(3x))/x + x^2e^x + 2xe^x`

`= (3x - 1)e^(3x)/x^2 + x(x + 2)e^x`

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

`((dy)/(dx))^2 = [(3x - 1)e^(3x)/x^2 + x(x + 2)e^x]^2`

Using Wolfram Alpha, this was simplified to:

`= e^(2x)/x^4 [(x+2)x^3 + e^(2x)(3x - 1)]^2`

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

`color(green)(s = int_(1)^(2) sqrt(1 + e^(2x)/x^4 [(x+2)x^3 + e^(2x)(3x - 1)]^2)dx)`

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

The numerical solution is:

`color(blue)(s = 208.471)`