Reduction formula/Integration by parts problem

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The discussion centers on determining a reduction formula for a specific integral involving integration by parts. The user expresses confusion about the absence of an extra r in the denominator when integrating dv and the role of r^(n-1). Participants clarify that using u = r^(n-1) and dv = re^(-ar^2) leads to v = (-1/(2a))e^(-ar^2), which is necessary for correct integration. The differentiation of this expression confirms that it yields the required dv. Overall, the thread emphasizes the importance of understanding integration by parts in this context.
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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.
 
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They are using u = r^{n-1} and dv = re^{-ar^2}dr 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 \frac{-1}{2a}e^{-ar^2} you get re^{-ar^2}, so that's why v = \frac{-1}{2a}e^{-ar^2}
 
Thanks a bunch.
This site rocks.

(thank you)^(10)
 
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Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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