Why an Accumulation Can Pause
An accumulation function records the net amount added or removed by a rate as the input changes. If the rate is zero throughout an interval, no net amount is added anywhere on that interval. The accumulated value therefore stays fixed while the input moves across it.
This is a direct consequence of the interval-increment identity: the change in accumulation between two inputs is the integral of the rate between them. The rate graph makes this relationship visible. A portion of the graph lying along the horizontal axis contributes zero signed area, so the corresponding accumulation graph has a plateau.
Definition and Interval-Increment Identity
Let \(r\) be an integrable rate on an interval containing a basepoint \(c\). An accumulation function with base value \(C\) is defined by integrating the rate from \(c\) to the input \(x\), then adding \(C\):
The integral is signed: reversing its limits reverses its sign. For any two inputs \(x\) and \(y\) in the domain, splitting the integral at \(x\) gives the interval-increment identity:
This identity works regardless of the location of the basepoint. It says that the change in accumulated value from \(x\) to \(y\) depends only on the rate between those inputs.
Why Zero Rate Produces a Plateau
Suppose \(r(t)=0\) for every \(t\) in an interval \(I\). Choose any \(x,y\in I\). Every point between \(x\) and \(y\) is also in \(I\), so the rate is zero throughout the interval of integration. Therefore:
Thus \(A(y)=A(x)\) for every pair of inputs in \(I\), which is precisely the definition of being constant on \(I\). This argument also covers \(y<x\), because the integral is still zero when the limits are reversed.
The last condition follows in both directions from the interval-increment identity. If \(A\) is constant, then \(A(y)-A(x)=0\), so \(\int_x^y r(t)\,dt=0\) for every \(x,y\in I\). Conversely, if all those integrals vanish, the identity gives \(A(y)-A(x)=0\) for every pair of inputs.
Worked Example: Finding the Level of a Plateau
Worked Example: Finding the Level of a Plateau
Suppose \(A(0)=3\), and the rate is \(r(t)=2\) for \(0\le t<2\), \(r(t)=0\) for \(2\le t\le5\), and \(r(t)=-1\) for \(5<t\le7\). Find the value of \(A(x)\) on the zero-rate interval.
For any \(x\) with \(2\le x\le5\), split the accumulated change from \(0\) to \(x\) at \(2\):
The first integral is \(2(2-0)=4\), and the second is \(0(x-2)=0\). Hence:
So the plateau has level \(7\). Check the result using the interval-increment identity: for any \(x,y\in[2,5]\), the rate is zero between them, so \(A(y)-A(x)=\int_x^y0\,dt=0\). In particular, \(A(2)=7\) and \(A(5)=7\), as required.
A plateau’s height is determined by accumulation before it begins, together with the chosen base value. The zero-rate interval explains why the value does not change there; it does not by itself tell us what that value is.
Reading a Plateau From a Rate Graph
In a rate graph, a zero-rate interval appears as a segment lying on the horizontal axis. To locate the matching plateau in the accumulation, use the same input interval: the accumulation has no net change between any two inputs in that segment. To find the plateau’s height, first determine the accumulated value at one point in the interval. That value may come from given initial data or from the signed area accumulated earlier.
For example, suppose a rate graph is described by these horizontal segments: \(r(t)=2\) from \(t=0\) to \(t=1\), \(r(t)=0\) from \(t=1\) to \(t=3\), and \(r(t)=-1\) from \(t=3\) to \(t=5\). If \(A(0)=1\), the rate graph’s middle segment identifies the plateau interval. Its level can be found from the signed area before the plateau:
For any \(x\in[1,3]\), the rate between \(1\) and \(x\) is zero. Thus:
The accumulation graph therefore stays at height \(3\) from \(t=1\) to \(t=3\). The segment where the rate is \(-1\), which comes after the plateau, does not change that conclusion about the plateau’s level.
Worked Example: Reading and Checking a Graph-Based Plateau
Use the rate segments just described, with \(A(0)=1\), to find \(A(4)\) and check how the later negative rate changes the accumulated value.
From \(0\) to \(1\), the rate is \(2\), giving signed area \(2(1-0)=2\). From \(1\) to \(3\), the rate is zero, giving area \(0(3-1)=0\). From \(3\) to \(4\), the rate is \(-1\), giving signed area \((-1)(4-3)=-1\). Therefore:
Check by starting at the plateau level: \(A(3)=3\), and then the rate contributes \(\int_3^4(-1)\,dt=-1\), so \(A(4)=3-1=2\). The check agrees. The accumulation is constant on \([1,3]\), but it is not constant on the entire interval \([1,4]\).
What Constancy Implies About the Rate
The zero-rate condition has an important qualification. A rate that is zero at every point of an interval certainly gives a plateau, but a plateau does not force an integrable rate to be zero at every single point.
If the rate is continuous, constancy does force pointwise zero rate. To see why, let \(A\) be constant on an interval and let \(x\) be an interior point where \(r\) is continuous. For sufficiently small nonzero \(h\), both \(x\) and \(x+h\) lie in that interval. Constancy and the interval-increment identity give:
As \(h\) approaches zero, the average value of a function continuous at \(x\) over the shrinking interval from \(x\) to \(x+h\) approaches \(r(x)\). Since the displayed average is zero for every such \(h\), its limit is zero. Hence \(r(x)=0\). If \(r\) is continuous on the whole interval, this reasoning gives zero at interior points, and continuity gives zero at the endpoints as well.
For a Riemann-integrable rate that is not necessarily continuous, constancy implies that the rate is zero almost everywhere on the interval. Here, “almost everywhere” means that any points where the rate is nonzero form a set of measure zero. Indeed, the same average argument shows \(r(x)=0\) at every continuity point. A Riemann-integrable function is continuous except on a set of measure zero, so any remaining nonzero values occur only on a set of measure zero. Conversely, a rate that is zero almost everywhere has zero integral on every subinterval and therefore produces a constant accumulation there.
Worked Example: A Nonzero Point That Does Not Change the Accumulation
Let \(r(t)=1\) at \(t=\tfrac12\) and \(r(t)=0\) at every other point of \([0,1]\). Define \(A(x)=7+\int_0^x r(t)\,dt\). This rate is nonzero at one point, but the integral over any subinterval is zero: changing an integrable function at a single point does not change its integral.
Thus, for every \(x\in[0,1]\), including \(x=\tfrac12\), the accumulated value is:
Check any pair \(x,y\in[0,1]\): the integral \(\int_x^y r(t)\,dt\) is still zero, whether or not the interval contains \(\tfrac12\). Therefore \(A(y)-A(x)=0\). The accumulation is constant even though the rate is not zero at every point.
A Common Mistake: Treating One Zero as a Whole Zero Interval
A rate being zero at one input does not establish a plateau. For a concrete example, take \(r(t)=t-1\) on \([0,2]\) and \(A(x)=\int_0^x(t-1)\,dt\). The rate is zero at \(t=1\), but it is not zero throughout any interval around \(1\).
Direct calculation gives:
These accumulated values are different. In fact, the change from \(1\) to \(\tfrac32\) is:
The calculation checks the change directly: it is nonzero, so the accumulation is not constant between those inputs. A single zero of the rate is not enough. What matters for constancy is that the integral vanish over every subinterval; a zero-rate interval guarantees that condition, while one point alone does not.
How to Identify and Verify a Plateau
When working from a rate formula or graph, use the interval-increment identity rather than relying on the appearance of one point. A reliable check is:
Find a full interval on which the rate graph lies on the horizontal axis, or on which the given rate is zero.
Confirm that the rate is zero throughout that interval if using the pointwise sufficient condition. If the rate is merely integrable, remember that pointwise zero is sufficient but not necessary.
Compute the accumulated value at one input in the interval, using the base value and the signed accumulation up to that input.
For arbitrary inputs \(x,y\) in the candidate interval, check that \(\int_x^y r(t)\,dt=0\). The interval-increment identity then gives \(A(y)=A(x)\).
Why Plateaus Matter
A plateau identifies a period during which the accumulated quantity does not change. In an application, the rate may represent flow, production, or another changing quantity; a zero rate over an interval means the total accumulated amount remains at its current level there. The plateau’s position reflects when the rate stops contributing, while its height records how much has accumulated before that pause.
Rate graphs are therefore useful not only for showing increases and decreases. They also reveal intervals of no net accumulation. Interpret that evidence carefully: a whole segment of zero rate guarantees a plateau, but a single zero point does not. For integrable rates, the precise condition for a plateau is zero integral over every subinterval, which allows isolated nonzero values that contribute no area.
Check Your Understanding
Use the interval-increment identity and the conditions stated above to answer these questions.
- If \(r(t)=0\) throughout \([a,b]\), what is \(A(y)-A(x)\) for \(x,y\in[a,b]\)?
- If a rate graph lies on the horizontal axis from \(t=2\) to \(t=5\), what can you conclude about the accumulation on that interval?
- Why does a single input where \(r(t)=0\) fail to establish a plateau?
- Can a Riemann-integrable rate be nonzero at an isolated point while its accumulation is constant? Explain.
- What extra condition lets you conclude that a constant accumulation has a rate that is zero at every point of the interval?
