Calculus

Integral Calculator

Compute definite integrals numerically using Simpson's rule

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How it works

An integral accumulates a quantity across an interval. The definite form gives the signed area between a curve and the x-axis; the indefinite form gives the whole family of antiderivatives. Definite results here come from Simpson rule, which fits a parabola through each pair of sub-intervals rather than a straight line.

Formula
∫ₐᵇ f(x) dx = F(b) − F(a),  where F′ = f
Power rule: ∫ xⁿ dx = x^(n+1)/(n+1) + C,  for n ≠ −1
Simpson: ∫ₐᵇ f(x) dx ≈ (h/3)[f(x₀) + 4f(x₁) + 2f(x₂) + … + 4f(xₙ₋₁) + f(xₙ)],  h = (b − a)/n
Variables
  • a, blower and upper limits of integration
  • Fany antiderivative of f
  • Cconstant of integration, present only in the indefinite form
  • nnumber of Simpson sub-intervals, which must be even
  • hwidth of each sub-interval
Worked example
Inputs: ∫₀² x³ dx
  1. Antiderivative: F(x) = x⁴/4
  2. F(2) − F(0) = 16/4 − 0 = 4
  3. Simpson with n = 2 and h = 1: (1/3)[f(0) + 4f(1) + f(2)]
  4. (1/3)[0 + 4(1) + 8] = 12/3 = 4
Result: 4, and Simpson rule lands on it exactly
Notes
  • Simpson rule integrates any polynomial up to degree three without error, which is why two sub-intervals already give the exact answer above. Its error term carries a fourth derivative, so it only begins to drift on functions that curve more sharply than a cubic.
  • What you get is signed area. ∫₀^2π sin x dx is zero because the lobes cancel; for physical area, integrate |f(x)| or split the interval at every root first.
  • Quadrature rules break on integrands that blow up inside the interval. ∫₀¹ x^(−1/2) dx converges to 2 analytically, but sampling at x = 0 returns infinity, so improper integrals need a substitution or a limit rather than more sub-intervals.

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