# Integration of constant: $\int dp = \Delta p$ in impulse formula

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 docs.google.com/file/d/0Bx0Sv7WqdghabS10TWpFcmpLdm8/edit –  raindrop Nov 22 '12 at 0:53

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