Arc length of f(x) = 4/5·x^(5/4) from [0,4]

[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 $$\theta$$ 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.