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 \sqrt{X}

I plug it into the formula and get

S = \int\sqrt{1+\sqrt{X}} 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 x^{1/4}. Let \theta <br /> be the angle opposite x^{1/4}.

Use this to put \sqrt{1 + \sqrt{x} } and dx in terms of \theta 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.