- 2069 Bhadra · 6 marks
Gauss-Legendre 2-point: e^(-x²)
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.
Evaluate ∫ from 2 to 4 of e^(-x²) dx using 2-point Gauss Legendre method.
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 |
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).
✓ Answer checked by Claude Opus 5.5 (high effort)
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