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- 2073 Bhadra · 6 marks
Gauss 3-point: e^(-x²)
Short answer
Change variable x = 0.65t + 0.85, dx = 0.65 dt. Evaluate Gauss nodes ±0.774597 and 0 with weights 5/9, 8/9, 5/9. Function values: 0.8869, 0.4855, 0.1601; weighted sum 1.0132. Multiply by 0.65 gives I ≈ 0.6586.
Evaluate ∫ from 0.2 to 1.5 of e^(-(x²)) dx using the 3 point Gaussian quadrature formula.
The 3-point Gauss–Legendre formula works on [−1, 1]:
∫₋₁¹ g(t) dt ≈ (5/9)·g(−√0.6) + (8/9)·g(0) + (5/9)·g(√0.6), √0.6 = 0.774597
Change of limits
For ∫ₐᵇ f(x) dx put x = (b−a)/2·t + (b+a)/2. Here a = 0.2, b = 1.5, so
x = 0.650000·t + 0.850000, dx = 0.650000 dt
∫ f(x) dx = 0.650000 ∫₋₁¹ f(0.650000t + 0.850000) dt
Nodes and function values
| t | x = 0.6500t + 0.8500 | f(x) | weight w | w·f(x) |
|---|---|---|---|---|
| -0.774597 | 0.346512 | 0.886858 | 5/9 | 0.492699 |
| 0.000000 | 0.850000 | 0.485537 | 8/9 | 0.431588 |
| 0.774597 | 1.353488 | 0.160104 | 5/9 | 0.088947 |
I ≈ 0.650000 × [ 1.013234 ] = 0.6586
Answer: I ≈ 0.6586
✓ Answer checked by Claude Opus 5.5 (high effort)
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