Diagnosing Overestimates and Underestimates

Why the Sign of a Rate Can Be Misleading

A rectangle sum estimates how much a rate changes an accumulated quantity over time. When the rate is positive, a larger rate means more positive accumulation. When the rate is negative, it means negative accumulation: the quantity is decreasing over that interval. This can make it tempting to reverse an estimate rule just because the graph is below the horizontal axis.

The correct diagnosis depends on two things: whether the rate is increasing or decreasing, and the numerical order of the estimated and actual signed totals. A negative number can still be larger than another negative number. For example, \(-9>-12\).

Left and Right Endpoint Estimates

Let a rate \(f\) be considered on an interval \([a,b]\), and divide the interval at points \(a=x_0<x_1<\cdots<x_n=b\). The width of subinterval \(i\) is \(\Delta x_i=x_i-x_{i-1}\), which is positive. A left-endpoint estimate uses the rate at the start of each subinterval; a right-endpoint estimate uses the rate at its end.

Write \(A\) for the exact signed accumulation over the whole interval. A positive contribution increases the accumulated quantity; a negative contribution decreases it. The monotone-rate bounds compare endpoint rectangles with the exact accumulation on each subinterval. Adding those subinterval bounds gives the following diagnosis.

The rule identifies the endpoint estimate on either side of the exact signed total. It does not depend on whether the graph lies above or below the axis.

Why the Bounds Hold

For an increasing rate, the rate at the left endpoint of any subinterval is no greater than the rate at its right endpoint. The exact accumulation on that subinterval lies between the two endpoint-rectangle contributions, by the monotone-rate bound. Since the total accumulation is the sum of the subinterval accumulations, add the lower bounds together and add the upper bounds together.

Here \(A_i\) is the exact signed accumulation on subinterval \(i\). Adding the inequalities preserves their order, so

The left sum is \(L_n\), the middle sum is the total accumulation \(A\), and the right sum is \(R_n\). Therefore \(L_n\le A\le R_n\). For a decreasing rate, the rate at the left endpoint is no smaller than the rate at the right endpoint, so the same subinterval bounds occur in the opposite order. Adding them gives \(R_n\le A\le L_n\).

The widths being positive is important: they scale the rate values without reversing their order. The rate values themselves may be negative. Their signs do not change which endpoint value is numerically larger.

Worked Example: Increasing Rate Above Zero

Worked Example: Increasing Rate Below Zero

Worked Example: Decreasing Rate Below Zero

Worked Example: Decreasing Rate That Crosses Zero

Common Pitfall: Comparing Negative Estimates by Magnitude

A common error is to decide that a more negative estimate is numerically larger because its magnitude is larger. In the decreasing-rate example, one might say, “The right sum has magnitude \(12\), which is greater than the exact magnitude \(10.5\), so the right sum is an overestimate.” That conclusion confuses magnitude with signed numerical order.

Thus the right estimate, \(-12\), is an underestimate of the signed accumulation, not an overestimate. The estimate’s magnitude, \(12\), is larger than the magnitude of the actual signed accumulation, \(10.5\); those are different comparisons. If the question concerns net change, compare signed values. If it concerns unsigned geometric area, compare nonnegative areas instead. When a rate crosses zero, monotonicity alone does not make the signed accumulation the same as the total unsigned area.

A Reliable Diagnosis

First determine whether the rate is increasing or decreasing over the interval being estimated. Then identify whether the estimate uses left or right endpoints. Finally, state the comparison using ordinary numerical order. The same procedure works above the axis, below it, and when the graph crosses the axis.

These conclusions require monotonicity on the interval used. If the rate changes from increasing to decreasing, or vice versa, one global endpoint rule need not apply. The graph’s behavior must be assessed on the relevant interval or intervals. Also, “increasing” and “decreasing” here allow a rate to be constant on part or all of an interval; that is why the bounds use “or equal to.”

Why the Diagnosis Matters

Rectangle estimates are useful when a rate is observed at a finite set of times but the accumulated change between observations is not known exactly. Monotonicity gives a way to judge the direction of the likely error without computing the exact total. For example, if a positive rate is increasing, left-endpoint rectangles miss some accumulation on each subinterval, while right-endpoint rectangles include extra. If the rate is negative, the same signed comparison remains valid, but “extra” negative contribution makes the estimate numerically smaller.

Keeping signed accumulation distinct from unsigned area prevents a frequent interpretation error. A negative rate represents a decrease in the accumulated quantity, not a positive increase. When the question is about net change, preserve the signs of the rectangle contributions and compare the resulting real numbers. When the graph crosses zero, positive and negative contributions can cancel in the net change, as in the final example.