A Definite Integral as Net Accumulated Quantity

From a Rate to a Net Change

A rate describes how quickly a quantity changes. If water enters a tank at a rate of \(3\) litres per minute, then over a short interval of \(0.1\) minute the amount entering is approximately \(3(0.1)=0.3\) litres. The shorter the interval, the more accurately the rate at one instant describes the change across that interval.

Over a longer period, the rate may vary, and it may sometimes be positive and sometimes negative. A definite integral combines the contributions across the whole interval. Positive rates add to the quantity being tracked; negative rates subtract from it. The result is the net accumulated change, not necessarily the total amount that moved or the final amount present.

For example, if \(r(t)\) is measured in litres per minute and \(t\) in minutes, then \(\int_a^b r(t)\,dt\) is measured in litres. The integral has the units of a change in volume, not a rate.

Why the Integral Represents Accumulation

Divide the time interval \([a,b]\) into small subintervals. On a subinterval of width \(\Delta t_i\), the rate is approximately \(r(\xi_i)\) for some time \(\xi_i\) in that subinterval. The change over that subinterval is therefore approximately \(r(\xi_i)\Delta t_i\). Adding these approximations gives a Riemann sum:

When \(r\) is Riemann integrable, these sums approach \(\int_a^b r(t)\,dt\) as the largest subinterval width approaches zero. This explains why the integral collects rate-times-time contributions. Negative rate values contribute negative terms, so the integral records increases and decreases with their signs.

The link to change can also be justified directly when the accumulated quantity is a differentiable function. Suppose \(Q\) is continuous on \([a,b]\), differentiable on \((a,b)\), and \(Q'(t)=r(t)\), where \(r\) is continuous on \([a,b]\). For any partition \(a=t_0<t_1<\cdots<t_n=b\), the mean value theorem on each subinterval gives a point \(\xi_i\in(t_{i-1},t_i)\) such that

Adding these equalities makes the intermediate values of \(Q\) cancel:

As the partition widths shrink, the right side approaches the Riemann integral of \(r\). Therefore, under these conditions,

This is a precise version of the accumulation interpretation: the integral of the rate gives the change in the quantity, not its starting value. If the initial amount \(Q(a)\) is known, then the final amount is \(Q(a)+\int_a^b r(t)\,dt\).

Positive and Negative Contributions

A positive rate contributes an increase to the tracked quantity. A negative rate contributes a decrease. Thus, an integral can be zero even when substantial changes occurred, if increases and decreases cancel. The sign of the integral tells the direction of the net change:

• If \(\int_a^b r(t)\,dt>0\), the quantity has a net increase.
• If \(\int_a^b r(t)\,dt<0\), the quantity has a net decrease.
• If \(\int_a^b r(t)\,dt=0\), the net change is zero; the quantity need not have stayed constant.

This is closely related to signed area: portions of the rate graph above the horizontal axis contribute positively, and portions below it contribute negatively. Adding the signed contributions gives the integral. It does not give the sum of all contributions with their signs removed.

Worked Example: Net Change in a Tank’s Volume

Motion: Displacement Is Not Distance

For motion along a line, velocity is a signed rate of change of position. Positive velocity moves in one direction and negative velocity moves in the opposite direction. The integral of velocity over time gives displacement, the net change in position. Total distance travelled instead counts movement in both directions as positive.

Worked Example: A Flow Rate That Changes Sign

A Common Mistake: Making Every Contribution Positive

A frequent error is to add the magnitudes of positive and negative rates when the question asks for net change. In the tank example, this incorrect calculation would be

That answer treats the \(6\) litres leaving the tank as though they entered. The correction is to preserve the negative sign for outflow:

The value \(18\) litres can describe the total amount transferred in both directions, if that is what is being asked. It is not the net change. Before calculating, identify whether the question asks for net accumulation, total movement, or the final amount.

Reading an Accumulation Question Carefully

For an applied problem, organize the interpretation before doing arithmetic. Identify the quantity that changes, decide which direction is positive, and check the units of the rate and interval. Then preserve the signs of all contributions, combine them to get net change, and use the initial amount only if the question asks for a final amount.

Remember that a zero integral says only that the net change is zero. An object could travel forward and then return to its starting position; a container could receive water and later lose the same amount. In either case, the net change is zero even though activity occurred throughout the interval.