Integral dx/sqrt(x - a)integral dx/sqrt(1/ax)

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Discussion Overview

The discussion revolves around the evaluation of integrals involving square roots, specifically the integrals of the forms ∫dx/√(x - a) and ∫dx/√(1/ax). Participants explore substitution methods for solving these integrals, share their learning experiences, and express feelings of embarrassment regarding their understanding.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Exploratory

Main Points Raised

  • One participant presents the integrals ∫dx/√(x - a) and ∫dx/√(1/ax) without initial context or methods for evaluation.
  • Another participant suggests reading suggestions, indicating a desire for further resources or guidance.
  • A participant expresses embarrassment over their initial confusion but acknowledges significant learning in a short time, indicating a personal journey of understanding.
  • A later reply provides a detailed substitution method for the integral ∫√(x - a)dx, suggesting u = x - a and transforming the integral accordingly.
  • Another participant corrects the first integral's expression, indicating it should be ∫dx/√(x - a) = ∫u^(-1/2) du, acknowledging a missed detail in the original post.
  • A subsequent reply agrees with the correction and confirms that the substitution method remains valid despite the initial oversight.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the initial presentation of the integrals, but there is agreement on the validity of the substitution methods discussed. The discussion reflects a mix of confusion, learning, and corrections without a definitive resolution on the integrals themselves.

Contextual Notes

Some participants express uncertainty regarding the initial setup of the integrals and the appropriate substitutions. There are unresolved aspects related to the clarity of the integral expressions and the steps involved in the substitution methods.

rebeka
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integral dx/sqrt(x - a)
integral dx/sqrt(1/ax)
 
Last edited:
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reading suggestions
 
sorry for wasting time I think I just understood substitution, out of embarassment I am never coming back to this forum(bright side of life I learned more in three days than I have in two years)
 
We all embarrass ourselves at one time or another! (Some of us more than others. Believe me, I know about embarrassing myself!) PLEASE come back to PhysicsForums!

(I need someone to make me look good!)

Anyway, for those who are still wondering how to do these:

[tex]\int \sqrt{x-a}dx[/tex]

Let u= x-1 so du= dx and [tex]\sqrt{x-a}= \sqrt{u}= u^{\frac{1}{2}}[/tex]. The integral becomes [tex]\int u^{\frac{1}{2}}du[/tex] which can be done with the power law.

[tex]\int \frac{dx}{\sqrt{\frac{1}{ax}}}[/tex]

[tex]\frac{1}{\sqrt{\frac{1}{ax}}}[/tex] is just [tex]\sqrt{ax}[/tex].

Let u= ax so du= adx, dx= (1/a)du and [tex]\sqrt{ax}= \sqrt{u}= u^{\frac{1}{2}}[/tex]

The integral becomes [tex]\frac{1}{a}\int u^{\frac{1}{2}}du[/tex]
 
Last edited by a moderator:
HallsofIvy said:
[tex]\int \sqrt{x-a}dx[/tex]

HallsofIvy, should not this equation be:
[tex]\int \frac{dx}{\sqrt{x - a}} = \int u^{-\frac{1}{2}} du[/tex]
 
Oh, you're right! I missed the "1/" in the first post. Of course, the substitution would be exactly the same and the u-integral what you show. Thanks.
 

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