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- 2073 Magh · 6 marks
Gauss 3-point: sin x³ + cos x²
Short answer
Map [0,1.4] to [-1,1] with x = 0.7t+0.7, dx = 0.7 dt. At nodes t = -0.7746, 0, 0.7746, function values are 1.0036, 1.2186, 0.9684; weighted sum 2.1788. Multiply by 0.7 to get I ≈ 1.5252.
Evaluate ∫ from 0 to 1.4 of (sin x³ + cos x²) dx 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 = 1.4, so
x = 0.700000·t + 0.700000, dx = 0.700000 dt
∫ f(x) dx = 0.700000 ∫₋₁¹ f(0.700000t + 0.700000) dt
Nodes and function values
| t | x = 0.7000t + 0.7000 | f(x) | weight w | w·f(x) |
|---|---|---|---|---|
| -0.774597 | 0.157782 | 1.003618 | 5/9 | 0.557566 |
| 0.000000 | 0.700000 | 1.218647 | 8/9 | 1.083242 |
| 0.774597 | 1.242218 | 0.968399 | 5/9 | 0.538000 |
I ≈ 0.700000 × [ 2.178807 ] = 1.5252
Answer: I ≈ 1.5252
✓ Answer checked by Claude Opus 5.5 (high effort)
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