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August 14th, 2014, 07:37 AM   #1
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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.
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August 14th, 2014, 08:26 AM   #2
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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.
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Last edited by v8archie; August 14th, 2014 at 08:32 AM.
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