Exploring Accumulation of Change

Picture a graph of speed versus time. Speed is measured in miles per hour, time in hours. If you slice the time axis into a thin strip of width dt, the car's speed during that strip is roughly constant, so distance covered in that strip is speed times dt — a thin rectangle's area. Add up every strip from start to finish and you get total distance. That sum of infinitely many thin rectangles is exactly what an integral computes, which is why integral of rate with respect to time equals total accumulated amount.

This works for any rate: a flow rate gives accumulated volume, a population growth rate gives change in population, a velocity gives displacement. The units always check out: rate's units times time's units equals the accumulated quantity's units. For a graph made of flat pieces, like a car driving at constant speeds, you don't need calculus at all — just add up rectangle areas directly, since each piece already has constant height.

The catch is that area under a graph is signed, not automatically positive. When the rate function dips below the horizontal axis, that piece contributes negative area to the total. For a velocity graph, negative velocity means moving backward, so that interval subtracts from the running total instead of adding to it. This signed-area idea is exactly what separates displacement from total distance traveled in Unit 8: displacement is the plain signed integral, while total distance requires flipping the negative pieces positive first, since distance traveled never shrinks no matter which direction you're moving.