Taking the square of a formula

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Homework Help Overview

The discussion revolves around the algebraic manipulation of the expression involving squares, specifically whether the equation \(a^2 + u^2 - 2au = (a-u)^2 = (u-a)^2\) holds true. Participants also explore the implications of this expression in the context of integrating the function \(x^{-\frac{3}{2}}\) over specific bounds.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants examine the validity of the algebraic expression and its implications for integration. There is a focus on the potential misunderstanding regarding the square root of \((a - u)^2\) and its absolute value interpretation.

Discussion Status

Some participants have provided insights into the algebraic expressions, while others have raised questions about the assumptions made regarding the square root. The discussion appears to be progressing with participants sharing their work and clarifying concepts.

Contextual Notes

There is mention of the original poster's difficulty in using LaTeX due to mobile constraints, which may affect the clarity of the shared work. Additionally, the integration bounds are specified as \((a-u)^2\) to \((a+u)^2\), which may be relevant to the discussion.

Pual Black
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Homework Statement


Hi sorry if the titel is wrong

I want to know if i can write this

##a^2 + u^2 -2au= (a-u)^2 = (u-a)^2##
I get different results when integrating ##x^{-\frac{3}{2}}## in the range ##(a-u)^2## to ##(a+u)^2##
 
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Pual Black said:

Homework Statement


Hi sorry if the titel is wrong

I want to know if i can write this

##a^2 + u^2 -2au= (a-u)^2 = (u-a)^2##
Yes, these expressions are all equal.
Pual Black said:
I get different results when integrating ##x^{-\frac{3}{2}}## in the range ##(a-u)^2## to ##(a+u)^2##
It would be helpful to see your work.
 
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I've uploaded my work as images.
Sorry for not using LaTex but I am on mobile phone.
 

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Pual Black said:
I've uploaded my work as images.
Sorry for not using LaTex but I am on mobile phone.
I think the problem might be from taking the square root of (a - u)2. It's not necessarily equal to a - u. What is true is that ##\sqrt{(a - u)^2} = |a - u|## which in turn is equal to ##|u - a|##.
 
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Ok i got it.
Thank you for your help
 

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