Simpson’s Rule Integration Calculator

Simpson’s rule fits parabolas through successive triples of points and adds their areas. For ∫₀¹ x² dx with n = 2 intervals (h = 0.5, nodes 0, 0.5, 1): (0.5/3)[0 + 4(0.25) + 1] = (1/6)(2) = 0.33333 — exactly 1/3, because the integrand is itself a parabola and Simpson reproduces quadratics perfectly. Error behaves like h⁴: halving the interval count cuts error 16×. For ∫₀¹ x⁴ dx the same two intervals give 0.33333... precisely (1/6)[0 + 4(1/16) + 1] = (1/6)(1.25) = 0.20833 versus the true 0.2 — error 0.00833, which drops to 0.00052 with n = 4. The 1-4-2-4-…-4-1 weight pattern is the signature: interior odd nodes get 4, interior even nodes get 2. The calculator enforces even n, tabulates node values with their weights, and compares against closed forms when available.

1. 1

   Enter f(x), bounds a, b

   The integrand and interval

2. 2

   Choose an even n

   Number of subintervals; 2, 4, 8… are valid

3. 3

   Apply the 1-4-2 pattern

   Node values multiply weights 1, 4, 2, 4, …, 1

4. 4

   Sum and scale

   Multiply by h/3 for the integral estimate

• 1

  Exact for polynomials up to degree 3 — a bonus beyond parabolas

• 2

  n must be even; the weights run 1, 4, 2, 4, …, 4, 1

• 3

  Error shrinks 16× each time n doubles

• Using an odd number of intervals

• Assigning weight 4 to even-indexed interior nodes (they take 2)

• Multiplying by h instead of h/3

• Forgetting the endpoints carry weight 1, not 2 or 4