What Is the Midpoint Rule?
The midpoint rule approximates a definite integral by rectangles whose heights are sampled at the center of each subinterval:
Mₙ = h · [f(m₁) + f(m₂) + … + f(mₙ)], h = (b − a)/n, mᵢ = a + (i − ½)h
Among the Riemann-sum family, left, right, midpoint, the center choice is the accuracy champion: on each slice, sampling at the middle lets the rectangle overshoot on one half and undershoot on the other, and the two errors largely cancel. The result converges with error proportional to h², a full order faster than the left and right rules' h, and even edges out the trapezoidal rule by a factor of two. This calculator sums the rectangles with every height displayed, draws them under the curve, and compares the estimate against the trapezoid rule and a high-accuracy reference so the error hierarchy is visible in numbers.
How to Use the Midpoint Rule Calculator
Enter a function, limits, and the number of subintervals n. The steps state the rule and step size, list the sampled midpoint heights, assemble the sum, and then report three numbers side by side: the midpoint estimate M, the trapezoid estimate T at the same n, and a 20 000-interval Simpson reference standing in for the exact value. The final line notes the signature relationship: the midpoint error is about half the trapezoid's and of the opposite sign, which is why the weighted blend (2M + T)/3 reproduces Simpson's rule exactly, a fact the tool verifies with your inputs. The rectangles drawn on the graph make the cancellation intuition concrete.
Worked Example
Approximate ∫₀² x² dx (exact value 8/3 ≈ 2.6667) with n = 4, so h = 0.5 and midpoints 0.25, 0.75, 1.25, 1.75:
M₄ = 0.5 · (0.0625 + 0.5625 + 1.5625 + 3.0625) = 0.5 · 5.25 = 2.625
The error is 0.0417. The trapezoid rule at the same n gives 2.75, error +0.0833: twice as large, opposite direction, exactly the pattern promised. Blending, (2·2.625 + 2.75)/3 = 2.6667, the exact answer, because Simpson's rule is exact on parabolas. Doubling n to 8 cuts the midpoint error by four (to about 0.0104), the h² law in action.
Why the Middle Beats the Ends
On one slice, expand f around the midpoint m: f(x) ≈ f(m) + f′(m)(x − m) + ½f″(m)(x − m)². Integrating over the slice, the linear term contributes exactly zero, positive on one half, negative on the other, symmetric cancellation. The rectangle f(m)·h therefore captures the integral through first order automatically, and the leading error comes only from the curvature term: (h³/24)f″(m) per slice, totaling ((b−a)h²/24)·f″(ξ) overall. The trapezoid rule's error constant is 1/12, double the midpoint's 1/24, with opposite sign because a chord over-arcs a concave-down curve exactly where a midpoint rectangle under-arcs it. A useful reframing: the midpoint rectangle equals the trapezoid built on the tangent line at the midpoint, so the rule is secretly a tangent rule, which is why it feels like it should be worse than trapezoid but is in fact better.
Common Mistakes to Avoid
- Sampling at subinterval edges. The rule's entire advantage lives at the center points a + (i − ½)h. Using edges silently degrades it to a left or right sum with order-h error.
- Confusing n rectangles with n + 1 points. The midpoint rule uses exactly n samples, none of them the endpoints a and b, an advantage when f misbehaves at an endpoint (1/√x at 0), where trapezoid cannot even start.
- Expecting error to halve when n doubles. It quarters: the rule is second order. If your errors only halve, an implementation bug (edge sampling) is the usual culprit.
- Applying it across a discontinuity. The error theory assumes a smooth integrand; a jump inside one slice contributes an error the h² law never repairs. Split at the discontinuity first.
- Ignoring sign conventions below the axis. Rectangles under the x-axis carry negative heights and subtract, as the definite integral requires; the drawn rectangles show this directly.
Real-World Applications
The midpoint rule is the default when data can be sampled anywhere but only sparsely, since it extracts second-order accuracy from n evaluations with no endpoint samples. It is the standard rectangle method of physics course laboratories (impulse from force samples, charge from current), and its open-endpoint character makes it the textbook remedy for integrable endpoint singularities. In economics and operations it appears as the "mid-year convention": summing a year's continuous cash flow by sampling mid-year is precisely one midpoint slice, chosen over end-of-year sampling for exactly the cancellation reason above. Computer graphics integrates shading along rays with midpoint samples per segment; probability discretizes continuous densities onto bin centers, midpoint logic again, to keep first moments unbiased. And within numerical analysis it is a building block: paired with the trapezoid rule it generates Simpson's rule, and iterated on ODEs it becomes the midpoint (RK2) method of the predictor-corrector family. Compare it against its siblings with the Riemann sum calculator and trapezoidal rule calculator.
Frequently Asked Questions
How accurate is the midpoint rule?
Second order: the error is ((b−a)h²/24)·f″(ξ) for some ξ in the interval, so doubling n divides the error by four. For smooth integrands it is the most accurate of the simple Riemann-type rules, beating the trapezoid rule's constant by a factor of two.
Why does sampling at the midpoint cancel so much error?
By symmetry: over a slice, the linear part of f contributes equally and oppositely on the two half-slices around the center, integrating to zero. The rectangle therefore matches the integral of the local tangent line exactly, leaving only curvature effects, one order higher in h.
Is the midpoint rule really better than the trapezoidal rule?
For smooth functions, yes, its error is about half as large and opposite in sign. The trapezoid rule wins in different circumstances: when only tabulated endpoint data exists (you cannot sample midpoints of given data) and for periodic integrands over a full period, where trapezoid becomes spectacularly accurate.
What is the relationship to Simpson's rule?
The weighted average (2M + T)/3 of midpoint and trapezoid estimates at the same n is algebraically identical to composite Simpson on 2n subintervals. The opposite-sign errors annihilate through fourth order, which is why the worked example lands exactly on 8/3.
Why is it called an "open" rule, and when does that matter?
Open rules never evaluate the endpoints a and b. That matters for integrands undefined or singular at an endpoint, like 1/√x on (0, 1] or ln x near 0: the midpoint rule proceeds untroubled where closed rules (trapezoid, Simpson) would divide by zero on their first sample.
How do I choose n for a target accuracy?
From the error bound: |E| ≤ (b−a)³·max|f″|/(24n²). Solve for n. In practice, compute at n and 2n and compare: for a second-order rule, the difference between the two estimates is about three times the finer one's error, a free practical error estimate.
Is the midpoint rule a Riemann sum?
Yes, the one with sample points chosen at the centers, so it inherits the Riemann-sum guarantee: convergence to the exact integral for every continuous integrand as n grows. Left, right, midpoint, and random-point rules all converge; they differ only in speed, where midpoint leads.
What happens with a discontinuous integrand?
Convergence still holds for piecewise-continuous f, but the rate degrades to first order because the slice containing the jump carries an O(h) error regardless of sampling. The fix is standard: split the integral exactly at the discontinuity and apply the rule to each smooth piece.
Why did my midpoint estimate come out below the true value on a concave-up curve?
That is the rule's signature: on concave-up stretches the tangent-line trapezoid (equivalently the midpoint rectangle) sits under the curve, so M underestimates while T overestimates. Concave-down stretches flip both. The sign of the error tells you the prevailing concavity of the integrand.
Does the midpoint idea appear in solving differential equations?
Directly: the midpoint (RK2) method advances an ODE using the slope sampled at the interval's temporal midpoint, the same symmetry trick lifting Euler's method from first to second order. Numerical integration rules and ODE steppers are parallel families, and midpoint sampling upgrades both.