Exact Accumulation at a Constant Rate

When a Rectangle Estimate Is Exact

A rectangle sum is usually an estimate of accumulated change: on each subinterval, it uses one sampled rate to represent the rate throughout that subinterval. If the rate is constant, however, every sample gives the same rate. The rectangles then combine to give the exact total, no matter how the interval is partitioned or whether the samples are taken at left endpoints, right endpoints, midpoints, or elsewhere.

This special case is useful because it connects a rate to an accumulated change with no approximation error. The total change depends only on the constant rate and the length of the time interval.

Partitions and Rectangle Sums

Consider an interval from \(a\) to \(b\), where \(a<b\). A partition divides it into \(n\) subintervals using points \(x_0,x_1,\ldots,x_n\), with

The width of subinterval \(i\) is \(\Delta x_i=x_i-x_{i-1}\), so every width is positive. Choose a sample point \(\xi_i\) in the \(i\)th subinterval, meaning \(x_{i-1}\leq \xi_i\leq x_i\). If \(r(x)\) is the rate, the corresponding rectangle sum is

Each term is a rectangle's height, the sampled rate, multiplied by its width. In an accumulation setting, rate times interval width has units of accumulated quantity. The sum adds the contributions from all the subintervals.

Why Every Sampling Rule Gives the Same Total

Suppose the rate is constant, so \(r(x)=c\) throughout \([a,b]\). At every possible sample point, the sampled rate is \(c\). This is true for left-endpoint, right-endpoint, midpoint, and arbitrary-point sampling. The rectangle sum therefore becomes \(c\) multiplied by the sum of all subinterval widths.

The widths add to the full interval length: each interior partition point cancels when the differences are added. This gives a complete reason that the rectangle sum is exact.

In the middle line, every interior point \(x_1,\ldots,x_{n-1}\) appears once with a positive sign and once with a negative sign, so those terms cancel. The remaining difference is \(x_n-x_0=b-a\). The result does not depend on how many subintervals there are, how wide they are, or where the sample points lie.

For equal-width partitions, all the widths have the same value \(\Delta x=(b-a)/n\). The sum has \(n\) identical terms, each equal to \(c\Delta x\), so it is \(nc\Delta x=c(b-a)\). Unequal widths give the same conclusion because their sum is also \(b-a\).

Worked Example: Equal-Width Intervals

Unequal Widths Still Add to the Same Total

Equal subdivisions are not required. In an unequal partition, a wider subinterval contributes more than a narrower one because its rectangle has greater width. But the rate height remains the same on every rectangle. Adding the different widths recovers the total interval length, so the total accumulated change is unchanged.

Accumulated Change Can Be Negative

A rate can be negative when the measured quantity is decreasing in the chosen positive direction. The same rectangle-sum reasoning applies: a negative rate gives rectangles with negative signed contributions. A negative accumulated change means the quantity has decreased; it does not mean that a physical amount such as volume is itself negative.

From Accumulated Change to a Final Amount

The rectangle sum gives the change in a quantity, not automatically its final value. If a quantity starts at \(Q_0\), and its rate of change is the constant \(c\) over an interval of length \(b-a\), then the final quantity is the initial quantity plus the accumulated change:

For example, if a reservoir contains \(50\) liters and gains water at \(4\) liters per minute for \(7\) minutes, the accumulated gain is \(4(7)=28\) liters. Its final amount is \(50+28=78\) liters. The starting amount and the change are different quantities, even though they use the same units.

A Common Mistake: Forgetting the Widths

A rectangle contribution is rate multiplied by width. Adding sampled rates alone leaves out the time or other input interval over which each rate acts. For example, on the partition \(0,2,5\), a constant rate of \(5\) units per minute does not produce a total of \(5+5=10\) units just because there are two rectangles. That calculation incorrectly treats both widths as one minute.

The actual widths are \(2-0=2\) minutes and \(5-2=3\) minutes. The correct calculation is

As a check, the full interval is \(5-0=5\) minutes, and the constant rate gives \(5\times5=25\) units. The rectangle sum and the full-interval calculation agree.