My Math Forum I don't understand this derivation with a definite integral

 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 Display Modes Linear 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