Q:oneDcollision Here we use the Riemann sum to clarify what happens when the force is not constant. If the force is not constant, we can still use F  ̄ Δ t {\displaystyle {\bar {F}}\Delta t} as the impulse, with the understanding that F  ̄ {\displaystyle {\bar {F}}} represents a time average. Recall that the average of a large set of numbers is the sum divided by the N {\displaystyle N} : F  ̄ = ∑ n F n N {\displaystyle {\bar {F}}={\frac {\sum _{n}F_{n}}{N}}} With a bit of algebra, we can turn this into a Riemann sum. For a collision that occurs over a finite time interval, Δ t {\displaystyle \Delta t} , we break that collision time into much smaller intervals δ t {\displaystyle \delta t} . The former might be the collision time between a golf ball and the club, while the latter would be the time interval of an ultra high-speed camera. Note that Δ t / δ t = N {\displaystyle \Delta t/\delta t=N} , where N {\displaystyle N} is the number of frames of the camera. Let F n {\displaystyle F_{n}} be the force associated with the n-th frame. The discretely defined average force associated with that camera is: F  ̄ Δ t = ∑ n F n N ⋅ Δ t = ∑ n = 1 N [ F n ⋅ { Δ t / δ t N } ⋅ δ t ] = ∑ n = 1 N F n ⋅ δ t → ∫ 0 Δ t F ( t ) d t {\displaystyle {\bar {F}}\Delta t={\frac {\sum _{n}F_{n}}{N}}\cdot \Delta t=\sum _{n=1}^{N}\left[F_{n}\cdot \left\{{\frac {\Delta t/\delta t}{N}}\right\}\cdot \delta t\right]=\sum _{n=1}^{N}F_{n}\cdot \delta t\rightarrow \int _{0}^{\Delta t}F(t)\,dt} Footnote: This conversion from discrete to continuous math is easy to grasp, although the details are difficult to master: Other examples of this method include: Descrete and continuous expection values. The Discrete Fourier transform, the Fourier series, and the Fourier transform
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