Comparing Numerical Sum Methods on the Same Data

Why Compare the Methods?

Each numerical sum method estimates accumulated change by dividing an interval into pieces and using selected function values to represent each piece. The methods can give different answers even when they use the same partition, because they sample the function at different locations.

A fair comparison therefore keeps the interval and subintervals fixed. Then differences in the estimates come from the choice of sample locations and, when the function is known, the way its shape behaves between those locations.

Four Methods on a Common Partition

Let \(x_0<x_1<\cdots<x_n\) be the partition points of an interval, and let \(\Delta x_i=x_i-x_{i-1}\) be the width of subinterval \(i\). Each width is positive. A sum is formed by multiplying each width by a chosen height and adding the products.

• The left sum uses the function value at the left endpoint of each subinterval.
• The right sum uses the value at the right endpoint.
• The midpoint sum uses the value at the midpoint of each subinterval.
• The trapezoidal sum uses the average of the two endpoint values on each subinterval.

These formulas also clarify what information is needed. Left, right, and trapezoidal sums can be calculated from endpoint values. A midpoint sum requires the function value at every subinterval midpoint; endpoint data alone do not generally provide those values.

Worked Example: Comparing Four Sums for a Quadratic

A Relationship Between the Left, Right, and Trapezoidal Sums

On each subinterval, the trapezoidal method uses the average of the left and right endpoint heights. Summing that fact gives a useful relationship: for any partition, including one with unequal widths, the trapezoidal sum is the average of the left and right sums.

To see why, write the two endpoint sums using the same subinterval widths. Adding them pairs the left and right heights for each subinterval. Dividing by two gives the trapezoidal contribution on that subinterval.

Because an average of two numbers lies between them, \(T\) is between \(L\) and \(R\), or equal to both if they are equal. This does not mean that the midpoint sum must lie between the left and right sums; midpoint heights are not formed by averaging endpoint heights.

Worked Example: All Four Methods for a Linear Function

How Function Shape Affects the Comparison

If \(f\) is increasing across every subinterval, its value at each left endpoint is no greater than its value at the corresponding right endpoint. Since the widths are positive, multiplying by the widths and adding gives \(L\le R\). If \(f\) is decreasing, the inequalities reverse, so \(R\le L\). These comparisons concern the endpoint sums; they do not by themselves locate the midpoint sum.

Concavity helps interpret midpoint and trapezoidal estimates relative to the true accumulated area. For a function that is concave up throughout the interval, the midpoint method tends to underestimate and the trapezoidal method to overestimate. For a function that is concave down, the direction reverses. These conclusions depend on the stated concavity holding across the interval, not just at a few sample points.

Worked Example: A Concave-Down Function

Comparing Methods When Only Data Are Given

For a table of sampled data, first check which locations have recorded values. Endpoint data are enough for the left, right, and trapezoidal sums on a given partition. A midpoint sum needs observations at the subinterval midpoints. If those observations are missing, the midpoint sum cannot be calculated from the table alone without making an additional assumption about values between recorded points.

A Common Pitfall: Inventing Midpoint Values

A frequent mistake is to calculate a midpoint height by averaging the endpoint heights, then call the result a midpoint-method height. That average is the trapezoidal height for the subinterval. It equals the actual midpoint function value only in special cases, such as when the function is linear on that subinterval.

For example, on \([0,1]\), the function \(f(x)=x^2\) has endpoint values \(f(0)=0\) and \(f(1)=1\). Incorrectly averaging them gives a supposed midpoint height of \((0+1)/2=0.5\). But the actual midpoint value is \(f(0.5)=(0.5)^2=0.25\). For the two-subinterval example on \([0,2]\), replacing both actual midpoint values \(0.25\) and \(2.25\) by endpoint averages \(0.5\) and \(2.5\) produces \(3\), not the correct midpoint sum \(2.5\). The correction is to evaluate the function at each midpoint, or to say explicitly that a value is being estimated under an additional assumption.

What to Keep in Mind

Use the same partition when comparing methods. Check the required sample locations before calculating, and interpret the results in light of monotonicity and concavity when that information is available. The trapezoidal sum is always the average of the left and right sums on that partition; the midpoint sum has no such general averaging identity.