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Gauss-Legendre Quadrature is a numerical integration technique designed to yield maximum algebraic precision with a minimal number of function evaluations. Unlike standard Newton-Cotes methods (such as the Trapezoidal or Simpson's rules) that use fixed, equally spaced evaluation points, Gauss-Legendre quadrature treats both the evaluation locations (nodes) and their corresponding multipliers (weights) as free variables.

Mathematical Foundation To evaluate a definite integral over a standardized interval $[-1, 1]$ :

\int_{-1}^{1} f(x) , dx \approx \sum_{i=1}^{n} w_i f(x_i)

The nodes $x_i$ are selected as the roots of the n-th degree Legendre polynomial $P_n(x)$ . The weights $w_i$ are calculated by integrating the Lagrange interpolating polynomials over the interval:

w_i = \frac{2}{(1 - x_i^2) [P_n'(x_i)]^2}

An n-point Gauss-Legendre rule integrates polynomials of degree up to 2n - 1 exactly.

Domain Transformation (Mapping) For integrals over an arbitrary real interval [a, b], a linear change of variables maps the domain to [-1, 1]:

x(t) = \frac{b - a}{2} t + \frac{a + b}{2}

$dx = \frac{b - a}{2} dt$

The transformation yields:

\int_{a}^{b} f(x) , dx = \frac{b - a}{2} \sum_{i=1}^{n} w_i f\left( \frac{b - a}{2} x_i + \frac{a + b}{2} \right)

Applied Examples (Solved via Simple Solver)

.. Admonition:: Example 1 : Exact Polynomial Integration (2-Point Rule)

Evaluate

\int_{0}^{2} (x^3 - 3x + 2) , dx

Example 1C#

var I = Integral(x => Polyval([1.0, 0.0, -3.0, 2.0], x), 0, 2);
Console.WriteLine($"I = {I:F4}");

OutputFrom the book

I = 2.0000

Exact Analytical Result:

\left[ \frac{x^4}{4} - \frac{3x^2}{2} + 2x \right]_0^2 = (4 - 6 + 4) - 0 = 2

.. Admonition:: Example 2 : Transcendental Function

\int_{0}^{\pi} \sin(x) , dx

Example 2C#

var I = Integral(x => Sin(x), 0, pi);
Console.WriteLine($"I = {I:F4}");

OutputFrom the book

I = 2.0000

Exact Analytical Result:

[-\cos(x)]_0^\pi = 1 - (-1) = 2

.. Admonition:: Example 3 : Infinite bound

Evaluate

\int_{0}^{\infty} e^{-x^2}(\ln(x))^2 , dx

Example 3C#

var I = Integral(x => Exp(-x * x) * Pow(Log(x), 2), 0, inf);
Console.WriteLine($"I = {I:F5}");

OutputFrom the book

I = 1.94752

Exact Analytical Result:

\int_{0}^{\infty} e^{-x^2}(\ln(x))^2 , dx = \frac{\sqrt{\pi}}{16}\left(\pi^2 + 2(\gamma + \ln(4))^2\right) = 1.94752

Multiple Integral

SepalSolver can handle multiple integral via IntegralN. IntegralN is built on Integral running over multiply direction recursively. It is abole to compute upto 4 dimensional integrals efficiently.

.. Admonition:: Example 4 : Hypercube domain in 3 dimensions

Integrate the function $f(x, y, z) = xyz$ over the region where $x$ ranges from $0$ to $1$ , $y$ ranges from $1$ to $2$ , and $z$ ranges from $2$ to $3$ , which can be expressed as:

\int_{0}^{1} \int_{1}^{2} \int_{2}^{3} x y z \, dz \, dy \, dx

Example 4C#

// Define the function to integrate
Func<double, double, double, double> f = (x, y, z) => x * y * z;
// Set the lower bound of x
double x_1 = 0;
// Set the upper bound of x
double x_2 = 1;
// Set the lower bound of y
double y_1 = 1;
// Set the upper bound of y
double y_2 = 2;
// Set the lower bound of z
double z1 = 2;
// Set the upper bound of z
double z2 = 3;
// Calculate the integral
double integral = Integral3(f, x_1, x_2, y_1, y_2, z1, z2);
// Print the result
Console.WriteLine($"The triple integral of x*y*z is approximately: {integral:F4}");

OutputFrom the book

The triple integral of x*y*z is approximately: 1.8750

.. Admonition:: Example 5 :

Integrate the function $f(x, y) = xy$ over the region where $x$ ranges from $0$ to $1$ , $y$ ranges from $x^2$ to $2$ , which can be expressed as:

\int_{0}^{1} \int_{x^2}^{2} x y \, dy \, dx

Example 5C#

// Define the function to integrate
Func<double, double, double> f = (x, y) => x * y;
// Set the lower bound of x
double x_1 = 0;
// Set the upper bound of x
double x_2 = 1;
// Define the lower bound of y as a function of x
Func<double, double> y_1 = x => x * x;
// Set the upper bound of y
double y_2 = 2;
// Calculate the integral
double integral = Integral2(f, x_1, x_2, y_1, y_2);
// Print the result
Console.WriteLine($"The triple integral of x*y*z is approximately: {integral:F4}");

OutputFrom the book

The triple integral of x*y*z is approximately: 0.9167

.. Admonition:: Example 6 : Functional boundary

Integrate the function $f(x, y, z) = xyz$ over the region where $x$ ranges from $0$ to $1$ , $y$ ranges from $x^2$ to $\sqrt{x}$ , and $z$ ranges from $2$ to $3$ , which can be expressed as:

\int_{0}^{1} \int_{x^2}^{\sqrt{x}} \int_{2}^{3} x y z \, dz \, dy \, dx

Example 6C#

// Define the function to integrate
Func<double, double, double, double> f = (x, y, z) => x * y * z;
// Set the lower bound of x
double x_1 = 0;
// Set the upper bound of x
double x_2 = 1;
// Set the lower bound of y as a function of x
Func<double, double> y_1 = x => x * x;
// Define the upper bound of y as a function of x
Func<double, double> y_2 = x => Sqrt(x);
// Set the lower bound of z
double z_1 = 2;
// Set the upper bound of z
double z_2 = 3;
// Calculate the integral
double integral = Integral3(f, x_1, x_2, y_1, y_2, z_1, z_2);
// Print the result
Console.WriteLine($"The triple integral of x*y*z is approximately: {integral:F4}");

OutputFrom the book

The triple integral of x*y*z is approximately: 0.2083

.. Admonition:: Example 7 : Functional boundary

Integrate the function $f(x, y, z) = xyz$ over the region where $x$ ranges from $0$ to $1$ , $y$ ranges from $x^2$ to $\sqrt{x}$ , and $z$ ranges from $xy$ to $x+y$ , which can be expressed as:

\int_{x_1}^{x_2} \int_{x^2}^{\sqrt{x}} \int_{xy}^{x+y} x y z \, dz \, dy \, dx

Example 7C#

// Define the function to integrate
Func<double, double, double, double> f = (x, y, z) => x * y * z;
// Set the lower bound of x
double x_1 = 0;
// Set the upper bound of x
double x_2 = 1;
// Define the lower bound of y as a function of x
Func<double, double> y_1 = x => x * x;
// Define the upper bound of y as a function of x
Func<double, double> y_2 = x => Sqrt(x);
// Define the lower bound of z as a function of x and y
Func<double, double, double> z_1 = (x, y) => x * y;
// Define the upper bound of z as a function of x and y
Func<double, double, double> z_2 = (x, y) => x + y;
// Calculate the integral
double integral = Integral3(f, x_1, x_2, y_1, y_2, z_1, z_2);
// Print the result
Console.WriteLine($"The triple integral of x*y*z is approximately: {integral:F4}");

OutputFrom the book

The triple integral of x*y*z is approximately: 0.0641

The hard way, then the Reineur Solver way

How many points do you need?

Integrate e^(−x²) from 0 to 3. It has no closed form, so every answer is an approximation — the only question is how good, and at what cost.

By handThe trapezoidal rule with a fixed width. Add more points, get more digits.

1Func<double, double> f = x => Exp(-x * x);2double a = 0, b = 3;34foreach (int n in new[] { 10, 100, 1000, 10000 })5{6 double h = (b - a) / n, sum = 0;7 for (int i = 1; i < n; i++) sum += f(a + i * h);8 double area = (h / 2) * (f(a) + 2 * sum + f(b));9 Console.WriteLine($"n = {n,-6} area = {area:F12}");10}

Output

n = 10     area = 0.886202033637
n = 100    area = 0.886207292750
n = 1000   area = 0.886207347704
n = 10000  area = 0.886207348254

With Reineur SolverGauss-Legendre picks its own points, and picks them well.

1double area = Integral(x => Exp(-x * x), 0, 3);2Console.WriteLine($"Integral area = {area:F12}");

Output

Integral   area = 0.886207348260

Takes a free account.

Gauss Legendre

Applied Examples (Solved via Simple Solver)

Multiple Integral