Riemann Sum Calculator
Approximate the area under a curve f(x) = c·xⁿ using left, right, or midpoint rectangle sums. Ideal for calculus students learning numerical integration or checking definite integral estimates.
About this calculator
A Riemann sum approximates a definite integral by dividing the interval [a, b] into n equal subintervals of width Δx = (b − a) / n, then summing the areas of rectangles. For a function f(x) = coefficient · x^power, the sum is: S = Σ f(xᵢ) · Δx. The choice of xᵢ determines the method: left endpoint uses xᵢ = a + i·Δx, right endpoint uses xᵢ = a + (i+1)·Δx, and midpoint uses xᵢ = a + (i + 0.5)·Δx. As n increases, all three methods converge to the true integral. The midpoint rule typically produces the least error for smooth functions.
How to use
Suppose f(x) = 2x² on [0, 3] with n = 3 rectangles using the left method. Step 1: Δx = (3 − 0) / 3 = 1. Step 2: Left endpoints are x = 0, 1, 2. Step 3: f(0) = 0, f(1) = 2, f(2) = 8. Step 4: S = (0 + 2 + 8) · 1 = 10. Enter coefficient = 2, power = 2, lower bound = 0, upper bound = 3, rectangles = 3, method = left. The exact integral is 18, so more rectangles improve accuracy.
Frequently asked questions
What is the difference between left, right, and midpoint Riemann sums?
Left Riemann sums sample the function at the left edge of each subinterval, right sums use the right edge, and midpoint sums use the center. For increasing functions, left sums underestimate and right sums overestimate the true integral. The midpoint rule is generally the most accurate of the three for smooth functions because its error is proportional to Δx² rather than Δx.
How many rectangles do I need for an accurate Riemann sum approximation?
There is no single answer — accuracy depends on how curved the function is and which method you use. As a rule of thumb, doubling n roughly halves the error for left and right sums, and quarters it for the midpoint rule. For most introductory problems, n = 100 or more gives a result very close to the true integral. If precision matters, consider using the midpoint method with a large n, or switch to Simpson's rule.
Why does the Riemann sum converge to the definite integral as n approaches infinity?
As the number of rectangles increases, each subinterval width Δx shrinks toward zero, so the rectangles hug the curve more and more tightly. The limit of the Riemann sum as n → ∞ is the formal definition of the definite integral in calculus. This convergence holds for any continuous function on a closed interval, guaranteeing that the approximation becomes exact in the limit regardless of which endpoint rule you choose.