Find the arc length of a curve over an interval

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SUMMARY

The discussion focuses on calculating the arc length of a curve defined by the integral \(\int (y^4 + 2y^2 + 1)^{1/2} dy\). The user initially attempted a u-substitution with \(u = y^2\), leading to a transformation of the integral. However, a key insight was provided that simplifies the expression to \((y^2 + 1)\), which when raised to the power of \(1/2\) simplifies the integration process significantly. This simplification is crucial for correctly determining the arc length.

PREREQUISITES
  • Understanding of integral calculus, specifically arc length calculations.
  • Familiarity with u-substitution techniques in integration.
  • Knowledge of algebraic simplification of polynomial expressions.
  • Ability to manipulate and differentiate functions involving powers and roots.
NEXT STEPS
  • Study the process of simplifying polynomial expressions under square roots.
  • Learn more about u-substitution in integral calculus.
  • Explore the application of arc length formulas in different contexts.
  • Review techniques for integrating rational functions and their simplifications.
USEFUL FOR

Students studying calculus, particularly those focusing on integration techniques and arc length calculations, as well as educators seeking to clarify these concepts.

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Homework Statement


I'm trying to find the arc length of a curve over an interval and I've arrived at
[tex]\int[/tex] (y4 +2y2 +1)1/2 dy
and now I'm pretty sure i should use a u substitution in order to integrate.
I tried using u=y2 so du=2y dy so dy=du/2y
Then you have [tex]\int[/tex] (u2+2u+1)1/2 and eventually it boils down to [tex]\int[/tex] u+1 du which gives you u2/2 +u

But I think I forgot to incorporate the du/2y and I don't know if I can just change the answer to y2/2 + y.

Any thoughts?
 
Last edited:
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You don't have to 'u sub' anything yet. First try and simplify (y^4+2*y^2+1)^(1/2). Isn't y^4+2*y^2+1=(y^2+1)^2? What's the (1/2) power of that?
 

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