Step-Function Rates And Piecewise Accumulation

From a Step-Function Rate to Accumulation

A rate describes how quickly a quantity changes. When the rate is constant over an interval, the change in the quantity on that interval is the rate multiplied by the elapsed time. On a graph of the rate, this change is the signed area of a rectangle.

If the rate changes from one constant value to another at several times, the accumulation can be built one interval at a time. Each rectangular area is added to the total already accumulated. The resulting function is linear on each interval, and it remains continuous where the rate jumps.

This method is useful when a rate is given in stages, as in a table of constant rates or a graph made of horizontal segments. The essential bookkeeping is to keep track of interval lengths, signs, and the accumulated value at each change point.

Step-Function Rates and Accumulation

The rate can jump at the partition points. Its values at those individual points do not change the accumulated area: changing a function at finitely many points does not alter its definite integral. For calculations, we may assign each partition point to either adjoining interval, provided the constant values on the open intervals are unchanged.

Fix an initial accumulation value \(A\) at \(x_0=a\). The accumulation function is the initial value plus the signed area under the rate from \(a\) to the input \(x\). On a full interval, the area is a rectangle. If \(x\) lies partway through an interval, the partial area is also a rectangle, with width \(x-x_{i-1}\).

A positive rate adds to the accumulation, while a negative rate subtracts from it. The width of each rectangle is the length of its interval, not the endpoint at which the rate changes.

Why the Accumulation Is Continuous and Piecewise Linear

On any one partition interval, the completed areas do not change as \(x\) moves within that interval. Only the partial rectangle changes. Its area is the constant rate multiplied by the distance from the interval's left endpoint, so the formula has the form “constant plus constant times \(x\).” That is a linear function on the interval.

At a change point, the area from the interval just completed has already been included. The next interval contributes zero area at its left endpoint, because its width there is zero. Consequently, the formulas on the two sides give the same accumulated value at the change point.

This conclusion follows directly from the rectangular-area construction; it does not require the rate itself to be continuous. A jump in rate changes the slope of the accumulation, but does not create a jump in its value.

Worked Example: A Rate That Changes Sign

A General Derivation

The formula can be justified by splitting the area at each partition point. Suppose \(x\in[x_{k-1},x_k]\). From \(a=x_0\) to \(x\), the first \(k-1\) intervals are complete. Their signed areas are \(c_i(x_i-x_{i-1})\). The last part runs only from \(x_{k-1}\) to \(x\), where the rate is \(c_k\), so its area is \(c_k(x-x_{k-1})\). Add these areas to the initial value \(A\).

Here the integral is the accumulated signed area; for a step function, each complete or partial piece is a rectangle. The formula also proves continuity at the partition points. At \(x=x_k\), the formula from interval \(k\) gives

The formula on the next interval starts with exactly that same sum, and its partial rectangle has width zero at \(x_k\). Both formulas therefore give the same value there. On each interval the expression is linear in \(x\), so the accumulation is piecewise linear as claimed.

Worked Example: Unequal Intervals and a Nonzero Initial Value

Keeping the Accumulated Total Straight

A frequent error is to treat each new interval as if accumulation starts over at zero. For instance, in the preceding example, one might write \(F(x)=-3(x-1)\) on \([1,4]\). This is the area contributed by the second interval alone, not the full accumulation. It would give \(F(1)=0\) but incorrectly give \(F(4)=-9\) only because the first interval's ending value happened to be zero. On the last interval, starting over would give \(\frac12(x-4)\), which misses the previous accumulated value \(-9\).

The correction is to carry the endpoint value forward. On an interval with left endpoint \(x_{k-1}\), rate \(c_k\), and known accumulated value \(F(x_{k-1})\), use

This formula makes the bookkeeping explicit: the first term is everything accumulated so far, and the second is the new interval's partial signed area.

Another common mistake is to use the next rate for the entire earlier time span. A rate applies only on its own interval. Calculate each rectangle with its own rate and width, then add the rectangles in order. A negative rectangle reduces the running total; it does not erase the earlier areas unless its signed area exactly cancels them.

Worked Example: Building the Function from a Table

Why This Representation Is Useful

A piecewise-linear accumulation formula records both the running total and the effect of each rate stage. Its value at a partition point is obtained by adding all completed signed rectangles. Within an interval, a change in the input changes the accumulation by the interval's rate times that change in input.

This gives a reliable way to translate tables or horizontal-step rate graphs into formulas. It also makes important features visible: positive rates produce rising segments, negative rates produce falling segments, and a zero rate produces a constant segment. At a jump in rate, the accumulation remains joined while the direction or steepness of its linear segment can change.

When checking your work, verify the accumulated value at every partition point from both adjoining formulas, and compare the final value with the initial value plus the signed rectangle areas. These checks catch most errors in widths, signs, and carried totals.