
Calculus Calculus Math Forum 
 LinkBack  Thread Tools  Display Modes 
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. Last edited by v8archie; August 14th, 2014 at 08:32 AM. 

Tags 
definite, derivation, integral, understand 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
definite integral  cheyb93  Calculus  1  January 13th, 2013 06:37 PM 
Definite integral  BinaryReader  Calculus  3  March 27th, 2012 09:49 PM 
definite integral  binny  Calculus  11  May 28th, 2011 10:12 AM 
Definite integral.  rowdy3  Calculus  4  April 23rd, 2011 04:51 AM 
definite integral  jkmartinez  Calculus  2  October 27th, 2009 10:53 PM 