---
title: "Gauss-Legendre 2-point: e^(-x²)"
subject: "Numerical Method"
chapter: 5
key: "semester_4/numerical_method:gauss-2pt-e-x2"
source: https://www.bctengineer.com/q/semester_4/numerical_method/gauss-2pt-e-x2
site: BCT Engineering
---

# Gauss-Legendre 2-point: e^(-x²)

Numerical Method · Chapter 5: Numerical Differentiation and Integration

- Asked in: 2069 Bhadra, 6 marks
- Video: [Chapter 07.05: Lesson: Two Point Gauss Quadrature Rule: Example](https://www.youtube.com/watch?v=sefcWt9cd9s) — numericalmethodsguy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/numerical_method/gauss-2pt-e-x2
- Whole chapter: https://www.bctengineer.com/md/semester_4/numerical_method/chapter-5

## Question

Evaluate ∫ from 2 to 4 of e^(-x²) dx using 2-point Gauss Legendre method.

## Short answer

Use two-point Gauss-Legendre with x = t + 3 on [2,4]. The approximation gives 0.0028, though the exact integral is about 0.0041, so two points is only rough here.

## Answer

Gauss–Legendre 2-point formula: ∫₋₁¹ F(t) dt ≈ F(−1/√3) + F(1/√3), nodes t = ±0.577350, weights 1, 1.

#### Change of limits
For [2, 4]: x = [(4−2)t + (4+2)]/2 = t + 3, dx = dt. So I = ∫₋₁¹ e^{−(t+3)²} dt.

| t | weight | x = t+3 | f(x) = e^(−x²) | w·f(x) |
|---|---|---|---|---|
| -0.5774 | 1.0000 | 2.4226 | 0.002825 | 0.002825 |
| 0.5774 | 1.0000 | 3.5774 | 0.000003 | 0.000003 |

```text
I ≈ 1×0.002825 + 1×0.000003
  = 0.002828
  ≈ 0.0028
```
**Answer: ∫₂⁴ e^(−x²) dx ≈ 0.0028** (values are very small because e^(−x²) decays fast; the exact value is 0.004146, so 2-point Gauss is only roughly accurate here).

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What are the 2-point Gauss-Legendre nodes and weights?
   A. t = ±1/sqrt3 = ±0.57735, weights 1 and 1
   B. t = ±0.774597, weights 5/9
   C. t = 0, weights 2
   D. t = ±1, weights 1
   Answer: A. The 2-point rule is F(-1/√3) + F(1/√3). (Key point: 2-point nodes)

2. For [2, 4], what is the substitution?
   A. x = 2t + 3, dx = 2 dt
   B. x = t + 3, dx = dt
   C. x = t + 2, dx = dt
   D. x = 0.5t + 3, dx = 0.5 dt
   Answer: B. (b-a)/2 = 1, (b+a)/2 = 3. (Key point: Change of limits)

3. Why is the result 0.0028 only rough versus exact 0.004146?
   A. The weights are wrong
   B. The limits are wrong
   C. The integrand decays fast so 2 points represent it poorly
   D. Gauss rule is exact for exponentials
   Answer: C. 2-point Gauss is exact only to degree 3; e^(-x²) varies sharply. (Key point: Accuracy)

4. What is the sum of the two terms?
   A. 0.5
   B. 0.0014
   C. 2.4226
   D. 0.002825 + 0.000003 ≈ 0.0028
   Answer: D. The factor in front is 1. (Key point: Result)

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