How can derivatives and integrals describe opposite parts of an accumulation story?

Imagine a tank whose amount of water changes over time. Let A(t) be the amount of water in liters at time t, measured in minutes. Its derivative A′(t) is the rate of change at that moment, measured in liters per minute. The derivative describes how quickly the amount is changing, not how much water is in the tank.

An integral of that rate over a time interval adds up the changes during that interval. In symbols, if v(t) = A′(t), then the net change from time a to time b is ∫ₐᵇ v(t) dt = A(b) − A(a). In the other direction, the derivative of an accumulated rate gives the rate at that moment, under the continuity condition in the Fundamental Theorem of Calculus. [S1]

So the two operations answer opposite questions: differentiation asks “How fast is the amount changing now?” and integration asks “How much did it change over this interval?”

That is the key distinction: integrating a rate gives the change, not automatically the quantity itself. To find the amount at time t, you need its starting amount as well:

A(t) = A(a) + ∫ₐᵗ v(s) ds.

Here is a small worked example. Assume a tank starts with 5 liters at t = 0, and its rate of change is v(t) = 2t + 1 liters per minute for 0 ≤ t ≤ 3. These numbers and this rate are just an illustrative example.

The accumulated change over the three minutes is ∫₀³ (2t + 1) dt = [t² + t]₀³ = 12 liters. Therefore, the amount after three minutes is 5 liters + 12 liters = 17 liters.

Check the units: integrating liters per minute over minutes gives liters. The starting value is also in liters, so adding it to the change makes sense. The starting-value formula is the standard way to recover the quantity from its rate. [S1]

Keep the rate, net change, and amount separate. A positive rate means the quantity is increasing at that moment; a negative rate means it is decreasing at that moment. When you integrate a rate that changes sign, positive and negative contributions count in opposite directions, so the result is the net change over the interval—not necessarily the total amount of activity in both directions.

For a check, differentiate your proposed amount formula. If A(t) = 5 + t² + t in the example, then A′(t) = 2t + 1, which matches the given rate. Also check the initial condition: A(0) = 5. Together, those checks confirm that the formula matches both the rate and the starting amount. The general relationship between integrating a derivative and net change is stated by the Fundamental Theorem. [S1]

Use this sequence: name the quantity, identify its rate and units, write down the starting value, integrate the rate over the requested interval, and add the resulting net change to the starting amount. Finally, differentiate your answer to check the rate and substitute the start time to check the initial value.

A useful fill-in template is: “The quantity is ___, measured in ___. Its rate is ___ per ___. At time ___, its value is ___. The change from time ___ to ___ is ∫ ___, so the ending value is starting value + change = ___.”

The example establishes the amount only because both the rate and starting amount were specified. A rate by itself determines accumulated change, but without a starting value it does not identify one unique amount function. [S1]