How come the inverse of a derivative is an integral?

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I can’t conceptually grasp this the same why I can grasp addition/subtraction (the opposite of adding stuff is to take it away), but if the derivative of a function is the slope at any given point, then why would the area under the curve be the opposite of that?

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Anonymous 0 Comments

Let’s talk in terms of position and velocity since that is something most folks find intuitive.

When we take the derivative of the position function, we’re basically finding out the rate of change of position over time… but we’re doing that work as a limit process to find the exact rate-of-change at any particular infinitesimal slice of time. In the end we go from units of “distance” in our position function, to units of “distance-over-time” in our velocity graph.

So, how would we go in reverse- from a velocity graph to a position graph? Well, the units go from “distance-over-time” to just “distance” so we must be multiplying by “time” to make that happen. This is the same math as *”I drove 60mi/hr for 2hr… so I must have driven 120mi!”*… but we’re doing that work as a limit process to find the exact distance traversed over any particular infinitesimal slice of time.

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