What Does the Integral of Momentum with Respect to Time Mean?

The expression $\int_{t_1}^{t_2}\mathbf{p}(t)\,dt$ means “add up the momentum vector over a time interval.” It is not one of the usual energy or impulse formulas, so its meaning is easiest to identify from its units and from the relation between momentum and velocity.

For an object with constant mass, the integral has a simple interpretation: it is the object’s mass multiplied by its displacement. If you meant a related quantity such as impulse or work, those use different integrals, explained below.

Meaning and units

For momentum $\mathbf{p}(t)$ between times $t_1$ and $t_2$, write

$$\int_{t_1}^{t_2} \mathbf{p}(t)\,dt.$$

Momentum has units $\mathrm{kg\,m/s}$. Multiplying by time gives

$$\mathrm{kg\,m/s}\times\mathrm{s}=\mathrm{kg\,m}.$$

So the result is not energy, whose units are joules ($\mathrm{kg\,m^2/s^2}$), and not impulse, whose units are $\mathrm{kg\,m/s}$.

Constant-mass interpretation

For a body of constant mass $m$, momentum is $\mathbf{p}=m\mathbf{v}$, where $\mathbf{v}$ is velocity. Therefore,

$$\int_{t_1}^{t_2}\mathbf{p}(t)\,dt =m\int_{t_1}^{t_2}\mathbf{v}(t)\,dt =m\bigl(\mathbf{x}(t_2)-\mathbf{x}(t_1)\bigr).$$

The integral of velocity over time is displacement, not distance. The result points in the direction of the net displacement and can be zero even if the object traveled along a nonzero path and returned to its starting point.

For an indefinite integral, the result is a vector-valued function plus a constant vector:

$$\int \mathbf{p}(t)\,dt=m\mathbf{x}(t)+\mathbf{C},$$

where the choice of the position origin is absorbed into $\mathbf{C}$.

Worked example

Suppose a $2\,\mathrm{kg}$ object moves in one dimension with momentum

$$p(t)=(2t+1)\,\mathrm{kg\,m/s},$$

where $t$ is measured in seconds. From $t=0$ to $t=3\,\mathrm{s}$,

$$\begin{aligned} \int_0^3 p(t)\,dt &=\int_0^3(2t+1)\,dt\;\mathrm{kg\,m/s}\cdot\mathrm{s}\\ &=[t^2+t]_0^3\,\mathrm{kg\,m}\\ &=12\,\mathrm{kg\,m}. \end{aligned}$$

Since the mass is $2\,\mathrm{kg}$, the displacement is $12/2=6\,\mathrm{m}$. You can check this from $v(t)=p(t)/m=t+0.5\,\mathrm{m/s}$: integrating the velocity from 0 to 3 seconds also gives $6\,\mathrm{m}$.

Do not confuse it with impulse or work

These related expressions answer different questions:

Expression	Meaning	Units
$\int \mathbf{p}\,dt$	Momentum accumulated over time; for constant mass, mass times displacement	$\mathrm{kg\,m}$
$\int \mathbf{F}\,dt=\Delta\mathbf{p}$	Impulse, equal to the change in momentum	$\mathrm{kg\,m/s}$
$\int \mathbf{F}\cdot d\mathbf{x}=\Delta K$	Work, equal to the change in kinetic energy under the work–energy theorem	joules

The dot product matters in the work integral: work is a scalar, while momentum and its time integral are vectors. Integrating momentum with respect to time does not, by itself, calculate energy.

A different expression, $\int \mathbf{p}\cdot\mathbf{v}\,dt$, is also not the same as $\int\mathbf{p}\,dt$. For constant mass, $\mathbf{p}\cdot\mathbf{v}=mv^2=2K$, so this latter integral is $2\int K\,dt$.

Common mistakes

• Calling the integral impulse: Impulse is the integral of force over time, not momentum over time.
• Calling it energy: Its units are $\mathrm{kg\,m}$, which are not joules.
• Using distance instead of displacement: The integral of velocity gives net displacement. Direction and any return along the path matter.
• Dropping vector direction: In more than one dimension, integrate each component of momentum. Opposing contributions can cancel.
• Assuming the constant-mass result always applies: If mass changes, $\mathbf{p}=m(t)\mathbf{v}$ and the integral is $\int m(t)\mathbf{v}(t)\,dt$; it does not generally simplify to one fixed mass times displacement. The variable-mass model and the system being tracked must be specified.