 My Math Forum I don't understand this derivation with a definite integral
 User Name Remember Me? Password

 Calculus Calculus Math Forum

 August 14th, 2014, 07:37 AM #1 Member   Joined: Oct 2009 Posts: 50 Thanks: 0 I don't understand this derivation with a definite integral Hello, I'm taking a 101 physics course, and at some point the book makes a calculus derivation that I don't quite understand It starts with the equation: $\displaystyle dP/dy = pg$ and using this it assumes that, P = P1 at y = y1, and P = P2 at y = y2 And derives: $\displaystyle \int^{P2}_{P1}\,dx = - \int^{y2}_{y1} pg\,dy$ I assume that the first step is to move dy to the right: $\displaystyle dP = pg dy$ But I simply don't understand how the step from this to the derived equation could have possibly been made. Why can we integrate one side with P and the other with y? What does it even mean to integrate dP? I understand that we can integrate 1 and get: $\displaystyle \int^{P2}_{P1}\,dx$ But we don't start out with 1, we start out with dP. What am I missing here? Thanks alot in advance, because I've been trying to work this out all afternoon now. August 14th, 2014, 08:26 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,675 Thanks: 2654 Math Focus: Mainly analysis and algebra It's badly written. The variable in a definite integral is a dummy integral - it's defined only within the integral. So the derivation you are looking at actually has two distinct variables called $y$. Better would be to write $$\int_{P_1}^{P_2}\mathbb{d}u = \int_{y_1}^{y_2}pg \,\mathbb{d}v$$ where $p$ and $g$ are written as functions of $v$ (assuming they aren't constants). So the $\mathbb{d}P$ can become anything within the definite integal. One could do it as a definite integral, in which case we would keep the variable names, writing $$\int \mathbb{d}P = \int pg \,\mathbb{d}y$$ We then evaluate the integral at $(P_1,y_1)$ and/or $(P_2,y_2)$ to determine the constant of integration. Thanks from topsquark Last edited by v8archie; August 14th, 2014 at 08:32 AM. Tags definite, derivation, integral, understand Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post cheyb93 Calculus 1 January 13th, 2013 06:37 PM BinaryReader Calculus 3 March 27th, 2012 09:49 PM binny Calculus 11 May 28th, 2011 10:12 AM rowdy3 Calculus 4 April 23rd, 2011 04:51 AM jkmartinez Calculus 2 October 27th, 2009 10:53 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top

Copyright © 2019 My Math Forum. All rights reserved.       