Estimating Accumulation From Tables

From a Rate Table to an Accumulation Estimate

A rate table records values at selected inputs, but an accumulation depends on the rate throughout an interval. When the rate is known only at sample points, rectangles provide a way to estimate the accumulated change.

Let \(r(t)\) be a rate and fix a starting input \(a\). The accumulation at an endpoint \(x\), with \(x\ge a\), is

The value \(A(x)\) is a signed accumulation: intervals where \(r(t)\) is positive contribute positively, and intervals where it is negative contribute negatively. To estimate it from a table, divide \([a,x]\) into subintervals and approximate the rate on each subinterval by one sampled value.

Left, Right, and Midpoint Sums

Suppose the interval is partitioned at \(a=t_0<t_1<\cdots<t_n=x\). The \(i\)-th subinterval is \([t_{i-1},t_i]\), and its width is \(\Delta t_i=t_i-t_{i-1}\). The three common choices are:

• Left sum: use \(r(t_{i-1})\), the rate at the left endpoint.
• Right sum: use \(r(t_i)\), the rate at the right endpoint.
• Midpoint sum: use \(r\bigl((t_{i-1}+t_i)/2\bigr)\), the rate at the midpoint.

Each rectangle’s signed contribution is its width multiplied by its sampled rate. The width is positive, while the rate may be positive, zero, or negative. Thus a negative rate produces a negative rectangle contribution; it is not replaced by its absolute value.

The midpoint sum can be computed only when the midpoint rate values are available in the table or supplied by other information. Endpoint values alone do not determine the midpoint values.

Why the Rectangle Sum Approximates Accumulation

On a subinterval of width \(\Delta t_i\), imagine replacing the varying rate by the constant sampled value \(c_i\). A constant rate \(c_i\) acting over a width \(\Delta t_i\) contributes \(c_i\Delta t_i\). Adding these contributions over the partition gives the rectangle sum. The approximation comes from treating the rate as nearly constant on each small interval.

This also explains why the signs must be retained. If the rate is negative on part of the interval, that part represents a decrease in the accumulated quantity. Replacing its contribution by a positive area would estimate total amount rather than net accumulation.

Worked Example: Signed Contributions from Endpoint Data

When Left and Right Sums Bound the Integral

A left or right sum is not automatically an upper or lower bound. A direction can be guaranteed when the rate is known to be monotone over the interval. Here is the reason for an increasing rate.

For an increasing rate, on each subinterval \([t_{i-1},t_i]\), its values satisfy \(r(t_{i-1})\le r(t)\le r(t_i)\). Integrating these inequalities over that subinterval and adding the results proves the bounds.

and hence, after summing over all subintervals,

For a decreasing rate, \(r(t_{i-1})\ge r(t)\ge r(t_i)\), so the same argument gives \(R\le\int_a^x r(t)\,dt\le L\). These comparisons remain valid even when some or all of the rates are negative.

Midpoint Estimates and Approximation Direction

A midpoint sum uses a sample from the middle of each subinterval. It can be useful even when the rate is not monotone, but the midpoint value alone does not usually show whether the rectangle lies above or below the graph throughout the interval.

There is a direction guarantee when the rate is known to be convex or concave on each subinterval. A function \(r\) is convex if, for inputs \(u,v\) in its domain and \(0\le\lambda\le1\),

For a convex rate on an interval with midpoint \(m\), symmetry around \(m\) gives \(r(m-s)+r(m+s)\ge2r(m)\). Integrating for \(s\) from \(0\) to half the interval width shows that the integral over the interval is at least the midpoint rectangle. Thus, for a convex rate, the midpoint sum is a lower bound; for a concave rate, the inequality reverses and it is an upper bound. Without such shape information, a midpoint estimate has no guaranteed direction.

What a Table Can—and Cannot—Tell You

A table determines the rectangle sums once the partition and the required sampled values are specified. It does not, by itself, determine the exact accumulation: between sample points, the rate could vary in different ways while keeping the same tabulated values. Nor do increasing or decreasing entries in a table alone prove that the rate is monotone between those entries.

Use a direction claim only when the needed behavior is known on the full intervals. Monotonicity gives left- and right-sum bounds. Convexity or concavity can give a midpoint bound. If none of these conditions is known, report the rectangle result as an estimate without calling it an upper or lower bound.

Common Mistake: Ignoring the Sign or Width

A frequent error is to turn every contribution positive, as if the goal were total accumulated amount rather than net accumulation. For example, the middle midpoint contribution in the last example is

not \(+6\). Replacing it by \(+6\) would give \(8+6+3=17\) kilograms, a different quantity. The correction is to keep the rate’s sign and multiply by the interval width.

Another error is to use the number of intervals as the width. In the midpoint example, the second interval has width \(5-2=3\), not \(1\). Its contribution is \((-2)(3)=-6\), not \(-2\). For unequal partitions, calculate each width separately before summing.

A Reliable Procedure

These estimates connect sampled rate data to variable-limit accumulation: each rectangle represents an approximate signed change over one short interval, and the sum estimates the net change over the full interval.