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

##### 1 Answer

**Wolframalpha** states that the integral you would have to encounter for the determination of the arc length doesn't have a solution in terms of standard mathematical functions.

That just means that you are only expected to evaluate it ** numerically** (or get to the integrand without evaluating it).

A quick derivation on the arc length reveals that it came from a derivatives treatment of the distance formula.

Arc Length:

This is just a "dynamic", infinitesimally-short-distance formula that accumulates over an interval of constantly increasing

So, you can see that the first thing we could do is take the derivative and then square it.

#= e^"1/x"*(-1/x^2) + 1/x * e^"1/x"*(-1/x^2)#

#= -(e^"1/x")/(x^2) - (e^"1/x")/(x^3)#

Okay... yeah. Let's square... this.

#= [-(e^"1/x")/(x^2) - (e^"1/x")/(x^3)]*[-(e^"1/x")/(x^2) - (e^"1/x")/(x^3)]#

#= [(e^"1/x")/(x^2) + (e^"1/x")/(x^3)][(e^"1/x")/(x^2) + (e^"1/x")/(x^3)]#

#= (e^"2/x")/(x^4) + (2e^"2/x")/(x^5) + (e^"2/x")/(x^6)#

Now, let's simplify as much as possible. How about common denominators by multiplying until we get

#= (x^2e^"2/x")/(x^6) + (2xe^"2/x")/(x^6) + (e^"2/x")/(x^6)#

#= e^"2/x"[(x^2)/(x^6) + (2x)/(x^6) + 1/(x^6)]#

Since

#= color(green)((e^"2/x"(x + 1)^2)/x^6)#

*But Wolframalpha states that this integral doesn't have a solution in terms of standard mathematical functions.*

#s = int_(1)^(2) sqrt(1 + (-(e^"1/x")/(x^2) - (e^"1/x")/(x^3))^2)dx#

#= color(blue)(int_(1)^(2) sqrt(1 + (e^"2/x"(x + 1)^2)/x^6)dx)#

That just means that you are only expected to evaluate this ** numerically** (or get to the integrand without evaluating it).

#s = int_(1)^(2) sqrt((x^6 + e^"2/x"(x + 1)^2)/x^6)dx#

#~~ color(blue)(2.2103)#