Tutorials › Calculus I › Signed Error in a Rectangle Estimate

Accumulation from Sums · Tutorial 930 of 1495

Signed Error in a Rectangle Estimate

Compare each rectangle with the graph on its subinterval to find the direction and size of its error, then combine the local errors correctly.

Beginner 9 min read

What You'll Learn

  • Define local and total rectangle-estimate error using estimate minus actual accumulation
  • Decide the sign of a local error by comparing a rectangle with the graph
  • Add signed local errors across a partition, including when they partly cancel
  • Recognize when a graph crosses a rectangle height and simple visual comparison is insufficient
  • Avoid confusing overestimates with underestimates or assuming a sample rule always determines the error sign

Why the Sign of an Error Matters

A rectangle sum is an estimate of an accumulated quantity. Knowing only how far the estimate is from the actual amount may not be enough: it can also matter whether the estimate is too large or too small. For example, an estimate of accumulated distance that is too high may lead to a different decision from an estimate that is too low.

The sign records this direction. To use it consistently, we first choose an order for subtracting the quantities. In this tutorial, the error is defined as estimate minus actual accumulation. With that convention, a positive error means an overestimate, and a negative error means an underestimate.

Definition: On one subinterval, the signed local error is the rectangle's estimated contribution minus the exact accumulated contribution on that subinterval. The signed total error is the sum of the local errors. Positive error means the estimate is too high; negative error means it is too low.

This convention applies whether the rate is positive or negative. For a rate below zero, the accumulated contribution is negative. The comparison still concerns whether the rectangle's signed contribution is greater or less than the graph's signed contribution.

Comparing One Rectangle with the Graph

Suppose a subinterval has width \(\Delta x\), and the rectangle uses the sample value \(f(c)\) as its height. The rectangle contributes \(f(c)\Delta x\). Let \(A\) denote the exact signed accumulation represented by the graph over that same subinterval. The local error is therefore

$$ e=f(c)\Delta x-A. $$

Geometrically, the difference is the signed area between the horizontal top of the rectangle and the graph. If the rectangle is above the graph throughout the subinterval, its contribution is greater, so \(e\) is positive. If the graph is above the rectangle throughout, \(e\) is negative. If the graph and rectangle cross, parts of the difference have opposite signs; their effects may cancel, so a visual comparison at just one point does not determine the local error.

Sign test: If the rectangle's height is at least the graph's height everywhere on a subinterval, its local error is nonnegative. If the rectangle's height is at most the graph's height everywhere, its local error is nonpositive. A strict inequality over a region of positive width gives a strictly positive or negative error, respectively.

These comparisons concern the signed vertical values, not the distance from the graph to the horizontal axis. For example, a rectangle at height \(-2\) is above a graph at height \(-3\), even though both are below the axis. The rectangle's signed contribution is then greater, because \(-2\Delta x>-3\Delta x\) when \(\Delta x>0\).

From Local Errors to Total Error

Divide the interval into adjacent subintervals. On each one, the rectangle contribution and actual accumulation can be compared separately. The estimated total is the sum of the rectangle contributions, and the actual total is the sum of the actual contributions. Subtracting the actual total from the estimate gives the sum of the local differences.

$$ \begin{aligned} E &=\text{estimated total}-\text{actual total}\\ &=\sum_{i=1}^{n}R_i-\sum_{i=1}^{n}A_i\\ &=\sum_{i=1}^{n}(R_i-A_i)\\ &=\sum_{i=1}^{n}e_i. \end{aligned} $$

Here \(R_i\) is the rectangle contribution, \(A_i\) is the exact accumulation, and \(e_i=R_i-A_i\) is the signed local error on the \(i\)th subinterval. This identity explains why it is important to retain signs. Positive and negative local errors can cancel; adding only their magnitudes would not give the error of the total estimate.

Result: The signed total error equals the sum of the signed local errors. Therefore, if every rectangle lies above the graph on its subinterval, the total estimate is an overestimate. If every rectangle lies below the graph, it is an underestimate. If local signs differ, add the signed errors before deciding the direction of the total error.

Worked Example: A Rectangle Below an Increasing Graph

Worked Example: A Rectangle Below an Increasing Graph

Let the rate be \(f(x)=x+1\) on \([0,2]\). Use one left-endpoint rectangle. Determine its signed error.

The sample point is \(x=0\), so the rectangle height is \(f(0)=0+1=1\). Its width is \(2-0=2\), giving the estimated contribution

$$ R=1\cdot 2=2. $$

The graph is a line, rising from height \(f(0)=1\) to height \(f(2)=3\). Its exact accumulated amount is the area of a trapezoid with parallel sides \(1\) and \(3\), and width \(2\):

$$ A=\frac{1+3}{2}\cdot 2=2\cdot 2=4. $$

Thus the signed error is

$$ e=R-A=2-4=-2. $$

The sign agrees with the graph: the left-endpoint rectangle has height \(1\), while the line is at least \(1\) throughout the interval. The rectangle is below the graph, so the estimate is an underestimate. Checking by substitution, \(R-A=2-4=-2\), as calculated.

Worked Example: Local Errors Can Partly Cancel

Worked Example: Local Errors Can Partly Cancel

Suppose a rate is linear from height \(3\) at \(x=0\) to height \(1\) at \(x=1\), then linear from height \(1\) at \(x=1\) to height \(2\) at \(x=2\). Use left-endpoint rectangles on \([0,1]\) and \([1,2]\). Find the signed local errors and the total error.

Each subinterval has width \(1\). On \([0,1]\), the sample height is \(3\), so the rectangle contribution is

$$ R_1=3\cdot 1=3. $$

The exact contribution is the area of a trapezoid:

$$ A_1=\frac{3+1}{2}\cdot 1=2. $$

Therefore, the first local error is \(e_1=R_1-A_1=3-2=1\), an overestimate. On \([1,2]\), the left-endpoint sample height is \(1\), so

$$ R_2=1\cdot 1=1, \qquad A_2=\frac{1+2}{2}\cdot 1=\frac{3}{2}. $$

The second local error is \(e_2=R_2-A_2=1-\frac{3}{2}=-\frac{1}{2}\), an underestimate. Add the signed errors:

$$ E=e_1+e_2=1-\frac{1}{2}=\frac{1}{2}. $$

As a check, the estimated total is \(R_1+R_2=3+1=4\), while the actual total is \(A_1+A_2=2+\frac{3}{2}=\frac{7}{2}\). Their difference is \(4-\frac{7}{2}=\frac{1}{2}\), matching the sum of the local errors. The total estimate is an overestimate, although one of its two rectangles underestimates.

Worked Example: A Rectangle Above a Graph Below the Axis

Worked Example: A Rectangle Above a Graph Below the Axis

On \([0,2]\), let \(f(x)=-x-1\), and use one rectangle sampled at the left endpoint \(x=0\). Find the signed error.

The rectangle height is \(f(0)=-1\), and its width is \(2\). Its signed contribution is

$$ R=(-1)\cdot 2=-2. $$

The line goes from height \(-1\) to height \(f(2)=-3\). Its exact signed accumulation is the trapezoid area below the axis, so it is negative:

$$ A=-\frac{1+3}{2}\cdot 2=-4. $$

Therefore,

$$ e=R-A=(-2)-(-4)=2. $$

The error is positive: the rectangle at height \(-1\) lies above the graph, which falls from \(-1\) to \(-3\). The estimated contribution \(-2\) is greater than the actual contribution \(-4\). Checking the subtraction, \(-2-(-4)=-2+4=2\), so the positive sign is correct.

A Practical Method

For a rectangle estimate over several subintervals, organize the comparison one interval at a time. This prevents an error on one subinterval from hiding an error of the opposite sign elsewhere.

1
Find each rectangle contribution.
Multiply its sample height by its subinterval width, keeping negative heights negative.
2
Compare the rectangle with the graph.
Check whether the rectangle is above or below the graph over the whole subinterval. Record a positive sign for an overestimate and a negative sign for an underestimate.
3
Find or determine each local error.
When exact accumulation is available, subtract it from the rectangle contribution. When only a graph is given, use the signed area between the graph and rectangle if it can be determined.
4
Add the signed errors.
Combine positive and negative local errors. The sign of the result determines whether the total estimate is too high or too low.

Common Mistakes

A frequent mistake is to reverse the subtraction without changing the interpretation of the sign. In the first worked example, the estimate is \(2\) and the actual amount is \(4\). The convention here gives

$$ \text{estimate}-\text{actual}=2-4=-2. $$

Writing \(4-2=2\) calculates actual minus estimate instead. That number is not the signed error defined here. It describes the same difference in the opposite order, so it has the opposite sign. Always state the subtraction order before interpreting the result.

Another mistake is to assume that a particular sampling rule always produces an overestimate or always produces an underestimate. The direction depends on the graph's position relative to each rectangle, not just on whether a sample is at the left, right, or midpoint. A left-endpoint rectangle underestimates on the increasing line in the first example, but overestimates on the decreasing segment in the second example.

Finally, if the graph crosses the rectangle height within a subinterval, do not decide the sign by checking only the sample point or one endpoint. The part of the graph above the rectangle contributes error in one direction, and the part below contributes in the other. Their signed effects must be compared or calculated before the local sign is known.

Why Signed Errors Are Useful

Signed errors retain information that absolute differences discard. In the second example, the first rectangle is \(1\) unit too high and the second is \(\frac{1}{2}\) unit too low. The total estimate is not wrong by \(1+\frac{1}{2}=\frac{3}{2}\); the two local errors partly cancel, leaving a total error of \(\frac{1}{2}\). This distinction matters whenever an estimate combines contributions from multiple time periods, intervals, or pieces of data.

When only the direction of the total error is needed, a useful first check is whether every rectangle lies on the same side of the graph over its subinterval. If so, the local errors have the same sign, and their sum has that sign too. When the rectangles lie on different sides, local comparisons still help, but the signed amounts must be combined before the total estimate can be classified.

Check Your Understanding

Use the convention error = estimate minus actual accumulation.

  1. A rectangle lies above the graph throughout a subinterval. What can you conclude about its local error?
  2. One local error is \(3\), and another is \(-5\). What is their combined signed error, and is the combined estimate high or low?
  3. Why can local errors of opposite signs not be combined by adding their absolute values?
  4. A rate and its rectangle height are both negative. How can the rectangle still lie above the graph?
  5. If the graph crosses the rectangle height, what additional comparison is needed before determining the local error's sign?

Test yourself: Calculus I final exam practice

Ready to check what you know? Work a full-length practice final, then open the complete worked solutions.

Want all 1,114 practice problems?

This tutorial covers one topic. Calc Pass covers the full Calculus AB & BC curriculum (the core of Calculus I and II): full review, every practice question, and step-by-step solutions to each one.

QR code to download the Calc Pass app

Get the Calc Pass App

Scan to download and practice on the go, with instant scoring.

Calc Pass Volume 1: 1,114 Practice Questions book cover

Vol. 1: 1,114 Questions

The complete practice question bank, organized by AB & BC topic.

View on Amazon
Calc Pass Volume 2: Solutions Manual book cover

Vol. 2: Solutions Manual

Fully worked solutions to every question in Volume 1.

View on Amazon