Find the arc length of f(x) (x^(5/4))/5

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Homework Help Overview

The problem involves finding the arc length of the function f(x) = (x^(5/4))/5, with integration limits from 0 to 4. The arc length formula is referenced, which includes the derivative of the function.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to compute the arc length by first finding the derivative and setting up the integral. They express difficulty in simplifying the integral involving square roots. Some participants suggest various substitutions to aid in the simplification process.

Discussion Status

Participants are actively discussing different substitution methods to simplify the integral. There is acknowledgment of a mistake by the original poster, indicating a productive exchange of ideas, though no consensus on a single approach has been reached.

Contextual Notes

The discussion includes attempts to clarify the setup of the integral and the algebraic manipulations involved, with no resolution on the simplification challenge presented by the original poster.

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Homework Statement



find the arc length of f(x) (x^(5/4))/5.
The integration limits are from 0 to 4.

Homework Equations



The arc length formula is integrate sqrt(1 + (f'(x))^2)


The Attempt at a Solution



f'(x) = (5/4)*(1/5)*x^(1/4) = x^(1/4)/4

integral of sqrt(1 + (x^(1/4)/4)^2) = integral of sqrt(1 + sqrt(x)/16)

The part I am stuck on is getting rid of either sqrt roots.

Ive tried this: integral of sqrt(sqrt(x)/sqrt(x)/ + x/(sqrt(x)16))
but that didnt get me any where...Any idea how to simplify this? I am thinking that there is some sort of algebraic simplification that I am missing.
 
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Why not start by trying the substitution u=1+\frac{\sqrt{x}}{16}[/itex]?
 
Try letting x = u2 followed by v = 1 + (1/16)u.
 
just make the integral like this >>> int of (1/4)Sqrt(16+ Sqrt(x)) dx

then you can substitute >> u = 16 + sqrt(x)

it'll be easy ;)
 
Last edited:
Wow...what a novice mistake I made! Thanks!
 

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