What is the arclength of f(x)=e^(1/x)/xf(x)=e1xx on x in [1,2]x∈[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)