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In University Physics, it has something like:

$$\int \sum F dt = \int \frac{dp}{dt} dt = \int dp = \underbrace{p_2 - p_1}_{\Delta p?}$$

But I thought $\int dp = p$? Though my maths is really rusty ... $p$ refers to momentum, $F$ is force

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Here's an explanation of $\int_{t_1}^{t_2} \Delta v dt$ to relate Deltas in integrals – raindrop Nov 22 '12 at 0:53
up vote 2 down vote accepted

The indefinite integral is:

$$ \int \textrm{d}p = p $$

but here, you implicitly evaluate a definite integral (sloppy notation):

$$ \int \textrm{d}x \textrm{ }\hat = \int_a^b \textrm{d}x = x(b) - x(a) $$

and with the short-hand notation $p \equiv p(t)$, we have

$$ \int_1^2 \textrm{d}p = p(2) - p(1) = p_2 - p_1 \equiv \Delta p \quad. $$

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Suggestion to the answer(v1): Replace the term explicit integral with definite integral. – Qmechanic Nov 21 '12 at 17:21

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