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- 2071 Bhadra · 5 marks
Gauss 3-point: 1/(x²+2x+1)
Short answer
Map [0,2] to [-1,1] with x = t+1, dx = dt. At Gauss nodes t = -0.7746, 0, 0.7746, f(x)=1/(x+1)² gives 0.6660, 0.2500, 0.1299; weighted sum 0.6644. Integral ≈ 0.6644.
Evaluate ∫ from 0 to 2 of dx/(x² + 2x + 1) using Gaussian 3 point 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, b = 2, so
x = 1.000000·t + 1.000000, dx = 1.000000 dt
∫ f(x) dx = 1.000000 ∫₋₁¹ f(1.000000t + 1.000000) dt
Nodes and function values
| t | x = 1.0000t + 1.0000 | f(x) | weight w | w·f(x) |
|---|---|---|---|---|
| -0.774597 | 0.225403 | 0.665950 | 5/9 | 0.369972 |
| 0.000000 | 1.000000 | 0.250000 | 8/9 | 0.222222 |
| 0.774597 | 1.774597 | 0.129897 | 5/9 | 0.072165 |
I ≈ 1.000000 × [ 0.664360 ] = 0.6644
(The exact value is ∫₀² dx/(x+1)² = 1 − 1/3 = 0.6667, so the 3-point result is accurate to about two decimal places.)
Answer: I ≈ 0.6644
✓ Answer checked by Claude Opus 5.5 (high effort)
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