The Midpoint Rule

The midpoint rule approximates a definite integral using the function value at the middle of each subinterval, not the endpoints. Its error bound is |E_M| ≤ (K(b−a)³)/(24n²), where K is the maximum of |f″(x)| on the interval. The formula, a worked example, the error bound with a source, and a comparison with the trapezoidal rule are given.

For a function f continuous on [a,b] and a uniform partition into n subintervals of width Δx = (b−a)/n, the midpoint rule is

Mₙ = Σ_{i=1}^{n} f(𝑥̄ᵢ) Δx

where 𝑥̄ᵢ = (x_{i−1} + xᵢ)/2 is the midpoint of the i-th subinterval. The notation matches Stewart, Calculus, section 5.2 and OpenStax Calculus Volume 2, section 3.6. The midpoint Riemann sum uses the sample point at the centre of each rectangle. This formula is the finite approximation before the limit; the definite integral is the limit as n→∞ of Mₙ.

Finding the Midpoints: Worked Step

Given [a,b] = [0,2] and n = 4, Δx = (2−0)/4 = 0.5. The partition points are x₀ = 0, x₁ = 0.5, x₂ = 1.0, x₃ = 1.5, x₄ = 2.0. The midpoints are the average of consecutive endpoints: 𝑥̄₁ = (0+0.5)/2 = 0.25, 𝑥̄₂ = (0.5+1.0)/2 = 0.75, 𝑥̄₃ = (1.0+1.5)/2 = 1.25, 𝑥̄₄ = (1.5+2.0)/2 = 1.75. This is the most common error point: students use an endpoint or the wrong index. The midpoint rule requires the centre, not the left or right edge.

Let f(x) = x², a = 0, b = 2, n = 4. Δx = 0.5. The midpoints are 0.25, 0.75, 1.25, 1.75. Evaluate f at each: f(0.25) = 0.0625, f(0.75) = 0.5625, f(1.25) = 1.5625, f(1.75) = 3.0625. Multiply each by Δx = 0.5: 0.03125 + 0.28125 + 0.78125 + 1.53125 = 2.625. The exact integral ∫₀² x² dx = (1/3)(2³ − 0³) = 8/3 ≈ 2.6667. The midpoint rule gives 2.625, an absolute error of about 0.0417. For comparison, a right Riemann sum with the same n gives 3.75, a left sum gives 1.75. The midpoint rule is closer here, typical for smooth functions.

This example shows the midpoint rule formula in action. The midpoint Riemann sum matches the integral more closely than endpoint rules for this function because the function is increasing and concave up. The error bound formula below explains why.

The midpoint rule error bound is |E_M| ≤ (K(b−a)³)/(24n²). K is the maximum of |f″(x)| on the closed interval [a,b]. This bound comes from Stewart, Calculus, 8th edition, section 7.7, and the same form appears in OpenStax Calculus Volume 2, 2016 edition, section 3.6, with M in place of K. The bound guarantees the absolute error is no larger than that value, not that it equals it.

What K Means in Practice

For f(x) = x² on [0,2], f″(x) = 2, so |f″(x)| = 2 on the whole interval. Then K = 2. Plug in: (b−a) = 2, n = 4, so (2·2³)/(24·16) = (2·8)/(384) = 16/384 = 0.04167. The actual error was 0.0417. The bound matches the actual error exactly here because the second derivative is constant. For most functions, the bound is larger than the real error. The bound is a worst-case guarantee, not a prediction. If K is unknown because you lack a formula (e.g., from a table), you cannot compute the bound; this is a failure mode for table-based problems. In those cases, use a finer approximation to estimate the error instead.

The midpoint rule and the trapezoidal rule both approximate the integral with error proportional to 1/n² for smooth functions. The midpoint rule’s error bound is roughly half the trapezoidal rule’s bound: the constant factor 24 in the denominator vs. 12 for the trapezoid. Specifically, the trapezoidal rule error bound is |E_T| ≤ (K(b−a)³)/(12n²) from Stewart section 7.7. For the same function and n, the midpoint rule is typically about twice as accurate. However, this depends on concavity. For a concave-up function, the trapezoidal rule overestimates and the midpoint rule underestimates; for concave-down, the reverse. The actual error sign is opposite, so the sum of the two errors times 2 gives the Simpson's rule error, which is even smaller. On a table of values, the trapezoidal rule reuses the same endpoint values as left and right sums, so it costs fewer function evaluations for the same n. But the midpoint rule requires evaluating f at the midpoints, which are new points. For functions you cannot sample further (like a one-time measurement), the trapezoidal rule is the only option. Neither rule is universally better; the midpoint rule wins on smooth functions with known second derivatives, but the trapezoidal rule works with the same data as left and right sums. The College Board tests both on AP Calculus AB and BC, and scoring guidelines require monotonicity justification for over/underestimate, not concavity, for the left and right rules. The midpoint rule’s over/underestimate depends on concavity, not monotonicity. The trapezoidal rule, Simpson's rule, and Riemann sum overestimate/underestimate criteria are separate topics.

The honest caveat: the midpoint rule error bound formula contains K, the maximum of |f″(x)|. If you cannot bound the second derivative, because the function is not twice differentiable, or you only have a table, the bound is unusable. The midpoint rule itself still works; you just lose the guarantee. In those cases, compute Mₙ for increasing n and watch the value converge. That is the practical approach when the error bound formula fails.