---
title: "Simpson 1/3: 4e^x/(1+x³)"
subject: "Numerical Method"
chapter: 5
key: "semester_4/numerical_method:simpson-13-4e-x"
source: https://www.bctengineer.com/q/semester_4/numerical_method/simpson-13-4e-x
site: BCT Engineering
---

# Simpson 1/3: 4e^x/(1+x³)

Numerical Method · Chapter 5: Numerical Differentiation and Integration

- Asked in: 2070 Magh, 5 marks
- Video: [1/3rd Rule Numerical Problem | Simpson's Technique](https://www.youtube.com/watch?v=hQxGVepFQ0I) — Mathematics Tutor
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/numerical_method/simpson-13-4e-x
- Whole chapter: https://www.bctengineer.com/md/semester_4/numerical_method/chapter-5

## Question

Evaluate the following using Simpson's 1/3 rule (take h = 0.2): ∫ from 0 to 2 of 4e^x/(1 + x³) dx

## 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.

## Answer

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 |

```text
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**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. With a = 0, b = 2, h = 0.2, how many sub-intervals?
   A. n = 11
   B. n = 10, which is even so the rule applies
   C. n = 5, odd
   D. n = 2
   Answer: B. n = (b-a)/h = 10. (Key point: n)

2. Which weights apply to y1, y2, y3, y4?
   A. 2, 4, 2, 4
   B. 3, 3, 3, 3
   C. 4, 2, 4, 2
   D. 4, 4, 2, 2
   Answer: C. Odd indices weight 4, even interior 2. (Key point: Weights)

3. Which expression gives the integral?
   A. (h/3)[7.2840 + 2(24.1516) + 4(20.2520)]
   B. (3h/8)[7.2840 + 4(24.1516) + 2(20.2520)]
   C. h[7.2840 + 24.1516 + 20.2520]
   D. (h/3)[7.2840 + 4(24.1516) + 2(20.2520)]
   Answer: D. Ends once, odds ×4, evens ×2, times h/3. (Key point: Formula)

4. What is the result?
   A. 9.6263
   B. 14.3946
   C. 7.2840
   D. 4.8132
   Answer: A. 0.066667 × 144.3946 = 9.6263. (Key point: Result)

## Related in Numerical Method

- [Derive Simpson's 1/3 rule](https://www.bctengineer.com/question.md?key=semester_4%2Fnumerical_method%3Aderive-simpson-13) — Numerical Method, chapter 5
- [Composite Simpson 1/3 algorithm](https://www.bctengineer.com/question.md?key=semester_4%2Fnumerical_method%3Asimpson-13-algorithm) — Numerical Method, chapter 5
- [Composite Simpson 3/8: e^x + sin 2x](https://www.bctengineer.com/question.md?key=semester_4%2Fnumerical_method%3Asimpson-38-composite) — Numerical Method, chapter 5
- [Simpson 3/8 pseudo-code](https://www.bctengineer.com/question.md?key=semester_4%2Fnumerical_method%3Asimpson-38-pseudocode) — Numerical Method, chapter 5

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