Reduction formula/Integration by parts problem

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Homework Help Overview

The problem involves determining a reduction formula for a specific integral related to integration by parts. The original poster expresses confusion regarding the role of certain terms in the integral, particularly the absence of an extra r in the denominator and the significance of r^(n-1).

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to understand the integration by parts setup, questioning the absence of an additional r in the denominator and the purpose of r^(n-1). Some participants discuss the choice of u and dv in the integration by parts process.

Discussion Status

Participants are engaging in clarifying the integration by parts method, with some providing insights into the differentiation of terms involved. There is an exchange of ideas regarding the formulation of v and its relationship to dv.

Contextual Notes

The original poster references a picture for the integral in question, indicating that visual context may be necessary for full understanding. The discussion is focused on the mathematical reasoning behind the integration process without reaching a definitive conclusion.

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Homework Statement


The problem asks me to: Determine a reduction formula for the stated integral (See Picture and problem at:

http://garciarussellchem.angelfire.com/Photo/

Integration_by_parts.jpg

Please help me out with this problem. I don't understand why they do not include an extra r in the denominator when integrating dv. I also don't understand what purpose the r^(n-1) serves.

This is a bit confusing.

Thank you.
 
Last edited:
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They are using [tex]u = r^{n-1}[/tex] and [tex]dv = re^{-ar^2}dr[/tex] then doing integration by parts with those...
 
This means that v must equal"

v= (re^-(ar^2))/-2ar in order for us to get the dv back right?

THANK YOU FOR THE QUICK REPLY :) !
 
No prob.

If you differentiate [tex]\frac{-1}{2a}e^{-ar^2}[/tex] you get [tex]re^{-ar^2}[/tex], so that's why [tex]v = \frac{-1}{2a}e^{-ar^2}[/tex]
 
Thanks a bunch.
This site rocks.

(thank you)^(10)
 
Last edited:

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