- 2070 Magh · 5 marks
Simpson 1/3: 4e^x/(1+x³)
Short answer
Use Simpson's 1/3 rule with h = 0.2 and n = 10. Combining endpoints, odd, and even weighted values gives the integral approximately 9.6263.
Evaluate the following using Simpson's 1/3 rule (take h = 0.2): ∫ from 0 to 2 of 4e^x/(1 + x³) dx
Simpson's 1/3 rule (n even): ∫ f dx ≈ (h/3)[y₀ + yₙ + 4(y₁+y₃+…) + 2(y₂+y₄+…)].
Here a = 0, b = 2, h = 0.2, so n = (2−0)/0.2 = 10 sub-intervals (even, so the rule applies). f(x) = 4eˣ/(1+x³).
| i | xᵢ | yᵢ = f(xᵢ) | weight |
|---|---|---|---|
| 0 | 0.0000 | 4.0000 | 1 |
| 1 | 0.2000 | 4.8468 | 4 |
| 2 | 0.4000 | 5.6084 | 2 |
| 3 | 0.6000 | 5.9938 | 4 |
| 4 | 0.8000 | 5.8877 | 2 |
| 5 | 1.0000 | 5.4366 | 4 |
| 6 | 1.2000 | 4.8682 | 2 |
| 7 | 1.4000 | 4.3325 | 4 |
| 8 | 1.6000 | 3.8878 | 2 |
| 9 | 1.8000 | 3.5419 | 4 |
| 10 | 2.0000 | 3.2840 | 1 |
y₀ + y₁₀ = 4.0000 + 3.2840 = 7.2840
Odd sum (y₁+y₃+y₅+y₇+y₉) = 24.1516
Even sum (y₂+y₄+y₆+y₈) = 20.2520
I ≈ (0.2/3)[ 7.2840 + 4(24.1516) + 2(20.2520) ]
= 0.066667 × 144.3946
= 9.6263
Answer: ∫₀² 4eˣ/(1+x³) dx ≈ 9.6263
✓ Answer checked by Claude Opus 5.5 (high effort)
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