UNIT 3 • NUMERICAL INTEGRATION

Gauss Quadrature
Step-by-Step Friendly Guide

Gauss-Legendre Quadrature is a remarkably accurate numerical integration technique. With just n strategically placed points, it integrates polynomials of degree up to $2n-1$ exactly — making it far more efficient than Simpson's Rule or Trapezoidal for smooth functions.

THE GAUSS-LEGENDRE FORMULAOPTIMAL POINT PLACEMENT

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

where $t_i$ are Gauss nodes on [−1, 1] and $w_i$ are their corresponding weights.

Key Idea: Optimal NodesUnlike Trapezoidal / Simpson, nodes are not equally spaced — they are roots of Legendre polynomials

P_n(t)

Interval TransformationMap

[a,b] \to [-1,1]

via

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

, then apply weights.

Precision GuaranteeAn

n

-point rule is exact for all polynomials of degree

\leq 2n-1

(e.g., 3-point → degree ≤ 5).

Standard Gauss-Legendre Tables

REFERENCE TABLES

2-POINTExact for degree ≤ 3

i	Node tᵢ	Weight wᵢ
1	−0.5773502692	1.0000000000
2	+0.5773502692	1.0000000000

3-POINTExact for degree ≤ 5

i	Node tᵢ	Weight wᵢ
1	−0.7745966692	0.5555555556
2	0.0000000000	0.8888888889
3	+0.7745966692	0.5555555556

Step-by-Step Worked Example

HAND-HOLDING TUTORIAL

OUR PROBLEM TO SOLVE

Evaluate $\int_0^1 (x^2 + 1)\,dx$ using 3-point Gauss-Legendre Quadrature. (Exact answer: $4/3 \approx 1.333...$ )

STEP 01 • MAP INTERVAL [0,1] TO [-1,1]

Apply the transformation: $x = \frac{b-a}{2}t + \frac{a+b}{2} = \frac{1}{2}t + \frac{1}{2}$ and $\frac{b-a}{2} = \frac{1}{2}$ .

\int_0^1 f(x)\,dx = \frac{1}{2}\sum_{i=1}^{3} w_i\,f\!\left(\frac{t_i+1}{2}\right)

STEP 02 • EVALUATE AT EACH GAUSS POINT

x_1 = \frac{-0.7746+1}{2} = 0.1127 \quad f(x_1) = (0.1127)^2+1 = 1.01270

x_2 = \frac{0+1}{2} = 0.5000 \quad f(x_2) = (0.5)^2+1 = 1.25000

x_3 = \frac{0.7746+1}{2} = 0.8873 \quad f(x_3) = (0.8873)^2+1 = 1.78730

STEP 03 • APPLY WEIGHTS AND SUM

I = \frac{1}{2}\bigl[0.5556 \cdot 1.01270 + 0.8889 \cdot 1.25000 + 0.5556 \cdot 1.78730\bigr]

I = \frac{1}{2}\bigl[0.5626 + 1.1111 + 0.9932\bigr] = \frac{1}{2} \times 2.6669 \approx 1.3333

FINAL CONCLUSION

Result: ≈ 1.33333 (Exact!)

The 3-point Gauss rule gave the exact answer because $x^2+1$ is a degree-2 polynomial, which is within the degree-5 precision guarantee of the 3-point rule. For comparison, the Trapezoidal Rule would need many more subdivisions for the same accuracy.

LABORATORY

Interactive Live Calculator

INTEGRATION LABORATORY

Gauss Quadrature Interactive Engine

PreviousWeddle's Rule

Unit 3•17 of 34

NextNumerical Differentiation

∑

f(x)∫dxπ

NETZ Engine

Evaluating numerical algorithm

UNIT 3 • NUMERICAL INTEGRATION

Gauss Quadrature
Step-by-Step Friendly Guide

THE GAUSS-LEGENDRE FORMULAOPTIMAL POINT PLACEMENT

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

where $t_i$ are Gauss nodes on [−1, 1] and $w_i$ are their corresponding weights.

Key Idea: Optimal NodesUnlike Trapezoidal / Simpson, nodes are not equally spaced — they are roots of Legendre polynomials

P_n(t)

Interval TransformationMap

[a,b] \to [-1,1]

via

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

, then apply weights.

Precision GuaranteeAn

n

-point rule is exact for all polynomials of degree

\leq 2n-1

(e.g., 3-point → degree ≤ 5).

Standard Gauss-Legendre Tables

REFERENCE TABLES

2-POINTExact for degree ≤ 3

i	Node tᵢ	Weight wᵢ
1	−0.5773502692	1.0000000000
2	+0.5773502692	1.0000000000

3-POINTExact for degree ≤ 5

i	Node tᵢ	Weight wᵢ
1	−0.7745966692	0.5555555556
2	0.0000000000	0.8888888889
3	+0.7745966692	0.5555555556

Step-by-Step Worked Example

HAND-HOLDING TUTORIAL

OUR PROBLEM TO SOLVE

Evaluate $\int_0^1 (x^2 + 1)\,dx$ using 3-point Gauss-Legendre Quadrature. (Exact answer: $4/3 \approx 1.333...$ )

STEP 01 • MAP INTERVAL [0,1] TO [-1,1]

Apply the transformation: $x = \frac{b-a}{2}t + \frac{a+b}{2} = \frac{1}{2}t + \frac{1}{2}$ and $\frac{b-a}{2} = \frac{1}{2}$ .

\int_0^1 f(x)\,dx = \frac{1}{2}\sum_{i=1}^{3} w_i\,f\!\left(\frac{t_i+1}{2}\right)

STEP 02 • EVALUATE AT EACH GAUSS POINT

x_1 = \frac{-0.7746+1}{2} = 0.1127 \quad f(x_1) = (0.1127)^2+1 = 1.01270

x_2 = \frac{0+1}{2} = 0.5000 \quad f(x_2) = (0.5)^2+1 = 1.25000

x_3 = \frac{0.7746+1}{2} = 0.8873 \quad f(x_3) = (0.8873)^2+1 = 1.78730

STEP 03 • APPLY WEIGHTS AND SUM

I = \frac{1}{2}\bigl[0.5556 \cdot 1.01270 + 0.8889 \cdot 1.25000 + 0.5556 \cdot 1.78730\bigr]

I = \frac{1}{2}\bigl[0.5626 + 1.1111 + 0.9932\bigr] = \frac{1}{2} \times 2.6669 \approx 1.3333

FINAL CONCLUSION

Result: ≈ 1.33333 (Exact!)

LABORATORY

Interactive Live Calculator

INTEGRATION LABORATORY

Gauss Quadrature Interactive Engine

PreviousWeddle's Rule

Unit 3•17 of 34

NextNumerical Differentiation

Gauss Quadrature Step-by-Step Friendly Guide

Standard Gauss-Legendre Tables

Step-by-Step Worked Example

Result: ≈ 1.33333 (Exact!)

Interactive Live Calculator

Gauss Quadrature Interactive Engine

NETZ Engine

Gauss Quadrature Step-by-Step Friendly Guide

Standard Gauss-Legendre Tables

Step-by-Step Worked Example

Result: ≈ 1.33333 (Exact!)

Interactive Live Calculator

Gauss Quadrature Interactive Engine

Gauss Quadrature
Step-by-Step Friendly Guide

Gauss Quadrature
Step-by-Step Friendly Guide