Integral Calculator
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Formula: Simpson's Rule: ∫f(x)dx ≈ (h/3)[f(x₀) + 4f(x₁) + 2f(x₂) + 4f(x₃) + ... + f(xₙ)]
Worked example — 0.33333333
Formula
Simpson's Rule: ∫f(x)dx ≈ (h/3)[f(x₀) + 4f(x₁) + 2f(x₂) + 4f(x₃) + ... + f(xₙ)]
Simpson's 1/3 rule uses quadratic interpolation for numerical integration. It requires an even number of subintervals. Error is O(h⁴), making it much more accurate than the trapezoidal rule for smooth functions.
Worked Examples
Example 1: ∫x² from 0 to 1
Problem:f(x) = x², bounds [0, 1]
Solution:∫x²dx = x³/3 |₀¹ = 1/3 = 0.33333333
Result:0.33333333
Example 2: ∫sin(x) from 0 to π
Problem:f(x) = Math.sin(x), bounds [0, 3.14159265]
Solution:∫sin(x)dx = [-cos(x)] from 0 to π = (-cos(π)) - (-cos(0)) = 1 - (-1) = 2
Result:≈ 2.00000000
Example 3: Natural log from a reciprocal integral
Problem:Evaluate the definite integral of 1/x from x = 1 to x = e (about 2.718281828).
Solution:The antiderivative of 1/x is ln|x|. Integral = ln(e) - ln(1) = 1 - 0 = 1 This is the defining property of e: it is exactly the upper limit that makes the area under 1/x, starting at 1, equal to one square unit. Simpson's rule reproduces it to about 10 digits because 1/x is smooth and has no singularity on [1, e].
Result:1.00000000
Frequently Asked Questions
How accurate is Simpson's rule?
Simpson's rule has error proportional to h⁴ (where h = step size). With 1000 intervals, it gives ~10-12 digits of accuracy for smooth functions. It's exact for polynomials up to degree 3.
What functions can I enter?
Use JavaScript math syntax: x*x, Math.sin(x), Math.exp(x), Math.log(x), Math.sqrt(x), Math.pow(x,3), etc. Use * for multiplication.
What is the difference between a definite and indefinite integral?
A definite integral has upper and lower bounds and produces a numerical result representing the net area under a curve. An indefinite integral produces an antiderivative function plus a constant C, representing the general family of functions whose derivative is the integrand. Integral Calculator evaluates definite integrals numerically.
Why might a numerical integral give an incorrect answer?
Singularities (where the function is undefined or infinite, like 1/x at x=0) cause numerical methods to fail or give wildly inaccurate results. Functions with very steep gradients or rapid oscillations may also require many more intervals for accuracy. Always verify results against known antiderivatives when possible.
How is Simpson's rule different from the trapezoidal rule?
The trapezoidal rule approximates the area under each subinterval using a straight line between two points (a trapezoid), with error proportional to h². Simpson's rule instead fits a parabola through each group of three points, capturing curvature that a straight line misses, which is why its error shrinks proportional to h⁴ — dramatically more accurate for the same number of subintervals on smooth functions.
What real-world quantities are computed using definite integrals?
Definite integrals compute the total distance traveled from a velocity function, the total work done by a variable force, the total charge delivered by a varying current over time, the probability of a continuous random variable falling in a range (area under a probability density function), and the total revenue from a continuously varying price/demand function.
Can Integral Calculator evaluate improper integrals (with infinite bounds)?
No — Integral Calculator requires finite numeric upper and lower bounds, since Simpson's rule divides a finite interval into subintervals. Improper integrals with infinite bounds, or with a singularity at an endpoint, require specialized techniques (often approximated using a substitution to transform the interval, or a suitably large finite bound as an approximation) beyond straightforward Simpson's rule evaluation.
Why does Simpson's rule require an even number of subintervals?
Simpson's 1/3 rule fits a parabola through each consecutive triple of points, so the interval must be consumed in pairs of subintervals — one parabola per pair. That is why the weights alternate 1, 4, 2, 4, ..., 4, 1: interior odd-indexed nodes sit at parabola midpoints (weight 4) and even-indexed nodes are shared between adjacent parabolas (weight 2). An odd count leaves one orphan subinterval with no partner.
What does a negative definite integral result mean?
The definite integral returns signed net area, not total area. Where f(x) is below the x-axis the contribution is negative, so regions above and below can cancel. Integrating sin(x) from 0 to 2*pi gives 0 even though the curve encloses real area. If you want total unsigned area, integrate the absolute value of the function, splitting the interval at each root where f changes sign.
How should I choose the number of intervals for a given function?
Because Simpson's error scales as h^4, doubling the interval count cuts the error by roughly sixteen. A practical test is to compute the integral at n and at 2n subintervals: if the two agree to the digits you care about, the result has converged. Smooth, gently curving functions converge with a few hundred subintervals; sharply peaked or oscillatory integrands need far more, and floating-point round-off eventually limits further gains.
References
Reviewed for accuracy by Manoj Kumar, Mathematics Educator · Editorial policy
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