Calculating Arc Length for f(x) = 4/5*X^5/4 from [0,4]: Step-by-Step Guide

[Chandasouk]

I need to find the arc length of the function f(x) = 4/5*X^5/4 from [0,4].

You have to find f '(x) first and that would be X^1/4

I square f '(x) and obtain X^1/2 or [tex]\sqrt{X}[/tex]

I plug it into the formula and get

S = [tex]\int[/tex][tex]\sqrt{1+\sqrt{X}}[/tex] from [0,4]

I don't know how to evaluate the integral from here though

 

[vela]

Don't forget the dx. Try a u-substitution or two.

 

[Raskolnikov]

Imagine a right triangle with the two legs as 1 and [tex] x^{1/4}. [/tex] Let [tex] \theta
[/tex] be the angle opposite [tex] x^{1/4}. [/tex]

Use this to put [tex] \sqrt{1 + \sqrt{x} } [/tex] and [tex] dx [/tex] in terms of [tex] \theta [/tex] by using some trig operations. Can you get the rest from here?

 

[vela]

There's really no reason to resort to a trig substitution. There are a couple of obvious substitutions to try, and they will result in a integrand that's straightforward to integrate.