Simpson 1/3 (3 pts)
- Created by
- Renato Passos, Eng. de Software
- Reviewed by
- Renato Passos, Eng. de Software
Last updated: Apr 18, 2026
About this calculator
The Simpson 1/3 calculator approximates definite integrals using Simpson's rule with three points. This numerical method divides the interval into two equal parts and applies the formula (h/3)[f(x₀) + 4f(x₁) + f(x₂)], where h is the step between x₀ and x₂. It's effective when the integrand is complex or lacks an analytical solution.
The calculator works by sampling the function at endpoints (x₀ and x₂) and the midpoint (x₁). The formula weights the midpoint with coefficient 4 and endpoints with 1, creating a parabolic approximation through the three points. This yields higher accuracy than the trapezoidal rule for smooth functions.
Use this calculator in scenarios like experimental data analysis, engineering, and physics where analytical integration is impractical. Common precautions include ensuring function smoothness over the interval and choosing a sufficiently small h to minimize errors. Avoid applying it to discontinuous functions.
For reliable results, the interval [a, b] must be split into an even number of subintervals with equal spacing. Accuracy improves with more points but increases computation time. Simpson's 1/3 rule works best for functions approximated by parabolas, though advanced methods exist for complex functions.
Frequently asked questions
Why use Simpson's 1/3 rule?
It provides higher accuracy than the trapezoidal rule by using parabolas to approximate the function over small intervals.
How is the formula (h/3)[f0 +4f1 +f2] derived?
The formula is obtained by integrating a parabola that passes through the three points (x₀, x₁, x₂), weighting function values proportionally to the area under the curve.
What happens if the function isn't smooth?
Accuracy decreases because Simpson's rule assumes the function is approximately parabolic. Irregular curves require adaptive methods.
Can this calculator handle discrete functions?
Yes, as long as you have equally spaced function values at x₀, x₁, and x₂. Use interpolation for discontinuous data.
What's the relationship between step h and precision?
Smaller h increases accuracy but also computational effort. It's a trade-off between precision and efficiency.