Midpoint Coordinates in the Midpoint Rule

The midpoint of a subinterval is the arithmetic mean of its two endpoints: (x_{i-1} + x_i)/2. For a uniform partition from a to b with n subintervals of width Δx = (b-a)/n, the i-th midpoint coordinates in the midpoint rule take the closed form a + (i, 0.5)Δx. This single coordinate replaces the need to compute each pair of endpoints individually. The calculation directly yields the sample point where the rectangle height is evaluated.

Take f(x) = x² on [0,2] with n = 4. First compute Δx = (2-0)/4 = 0.5. The midpoints come from the formula a + (i, 0.5)Δx. For i = 1: 0 + (1-0.5) × 0.5 = 0.25. For i = 2: 0 + (2-0.5) × 0.5 = 0.75. For i = 3: 0 + (3-0.5) × 0.5 = 1.25. For i = 4: 0 + (4-0.5) × 0.5 = 1.75. These four midpoints are the sample points.

Now evaluate the function at each: f(0.25) = 0.0625, f(0.75) = 0.5625, f(1.25) = 1.5625, f(1.75) = 3.0625. Multiply each height by Δx = 0.5 and sum: 0.5 × (0.0625 + 0.5625 + 1.5625 + 3.0625) = 0.5 × 5.25 = 2.625. That is the midpoint approximation. The exact integral of x² from 0 to 2 is 8/3 ≈ 2.6667, so the error is about 0.0417, or 1.6%.

The rectangle height determines the rectangle area, and the choice of sample point biases the approximation. A left endpoint uses the function value at the start of the subinterval. An increasing function produces an underestimate from left sums and an overestimate from right sums. The midpoint sits halfway between the two endpoints, so for a linear function it yields the exact average of the left and right heights. This balancing reduces the systematic bias that left and right sums introduce.

For smooth functions, the error of the midpoint rule shrinks proportionally to 1/n², while left and right sums shrink proportionally to 1/n. That means doubling n cuts the midpoint error by a factor of about four, but only halves the left or right error. This faster convergence makes the midpoint rule the preferred rectangle method when you can evaluate the function at any point. Use a midpoint calculator when the subinterval count is moderate and you want a better estimate than left or right without switching to a trapezoidal or Simpson's rule.

Given the interval [a,b] and n subintervals, compute Δx = (b, a)/n. The first subinterval is [a, a + Δx]. Its midpoint is a + 0.5Δx, which equals a + (1-0.5)Δx from the general formula. The second subinterval is [a + Δx, a + 2Δx], with midpoint a + 1.5Δx = a + (2-0.5)Δx. The pattern continues: the i-th midpoint is always a + (i, 0.5)Δx.

Common Student Mistake

A frequent error is using the left endpoint plus half of Δx but forgetting to index correctly. For i = 1, the left endpoint is a, so a + 0.5Δx is correct. For i = 2, the left endpoint is a + Δx, so the midpoint is (a + Δx) + 0.5Δx = a + 1.5Δx. The formula a + (i, 0.5)Δx directly gives this without recomputing the left endpoint each time.

Non-Uniform Partitions

If the subinterval widths vary, you cannot use the shortcut formula. For a non-uniform partition, each subinterval [x_{i-1}, x_i] has its own width Δx_i = x_i, x_{i-1}. The midpoint is simply (x_{i-1} + x_i)/2. The sum becomes Σ f(x_i*) Δx_i, where each Δx_i carries its own value. AP exam rate tables often use uniform partitions, but college problems may present non-uniform data. Always check the partition type before applying the formula.

The midpoint rule approximates the definite integral ∫_a^b f(x) dx by the limit of Riemann sums as n approaches infinity. For any finite n, the sum is an approximation, not the exact area. The midpoint sample point reduces the error compared to left or right for most continuous functions because it samples the subinterval at its center, where the function's average height tends to be closer to the subinterval's true average height.

Over- and underestimate behavior depends on concavity, not monotonicity. For a concave-up function, the midpoint rule underestimates the integral. For a concave-down function, it overestimates. This is the opposite of the trapezoidal rule, which overestimates concave-up functions and underestimates concave-down ones. When you have no information about concavity, the midpoint rule still usually outperforms left or right sums for the same n, but it is not guaranteed. The error bound involves the second derivative, f''(ξ), which measures curvature. A larger curvature means a larger possible error for the same n.

Consider f(x) = 1 if x ≠ 1.5, and f(1.5) = 100, on the interval [0,3] with n = 3. Δx = 1. Subintervals: [0,1], [1,2], [2,3]. Midpoints: 0.5, 1.5, 2.5. The midpoint sample point for the second subinterval falls exactly at x = 1.5, where the function has a sharp spike to 100. The midpoint sum becomes 1 × (f(0.5) + 100 + f(2.5)) ≈ 1 × (1 + 100 + 1) = 102. The true integral of this function (ignoring the single point) is the area of three unit rectangles at height 1, approximately 3. The midpoint approximation is wildly wrong because the lone sample point does not represent the subinterval's behavior.

This failure is extreme, but it illustrates the general vulnerability: the midpoint rule trusts a single sample point per subinterval. If that point lies at a discontinuity, a sharp peak, or a deep valley that does not reflect the rest of the subinterval, the approximation degrades. Left or right sums would face the same problem if the discontinuity fell at an endpoint, but they might miss it if it sits inside the subinterval. No rectangle method with a single sample point is immune to such pathologies.

AP Calculus AB and BC students need the midpoint rule for table-based and graph-based problems where they must compute a sum and justify its behavior. First-year college calculus students benefit from understanding the error scaling and the connection to the definite integral as a limit. Physics and engineering students convert velocity tables into Riemann sums for distance problems; the midpoint rule gives better accuracy for the same effort.

Students who have not yet learned to evaluate a function at a point should first master that prerequisite. Anyone needing the exact value of an integral should use the antiderivative or a symbolic integrator, not a Riemann sum. If the problem supplies a table with unevenly spaced data and asks for a left or right sum specifically, the midpoint rule is irrelevant.

The most common error is treating the midpoint sum as the exact integral. The sum is an approximation; the definite integral is the limit as n approaches infinity. For any finite n, an error remains. Students who forget this will apply the midpoint rule when the problem expects a limit argument, or they will stop at a finite sum and claim it is the exact area. Always verify whether the question asks for an approximation or the exact value.