---
title: "Gauss 3-point: sin x³ + cos x²"
subject: "Numerical Method"
chapter: 5
key: "semester_4/numerical_method:gauss-3pt-sin-x3-cos-x2"
source: https://www.bctengineer.com/q/semester_4/numerical_method/gauss-3pt-sin-x3-cos-x2
site: BCT Engineering
---

# Gauss 3-point: sin x³ + cos x²

Numerical Method · Chapter 5: Numerical Differentiation and Integration

- Asked in: 2073 Magh, 6 marks
- Video: [GAUSSIAN INTEGRATION // Gauss Legendre 1, 2, 3 Point Formula // Gauss Quadrature 2,3,4 point Formula](https://www.youtube.com/watch?v=YzSdt1_kIJg) — Arya Anjum
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/numerical_method/gauss-3pt-sin-x3-cos-x2
- Whole chapter: https://www.bctengineer.com/md/semester_4/numerical_method/chapter-5

## Question

Evaluate ∫ from 0 to 1.4 of (sin x³ + cos x²) dx using Gaussian 3-point formula.

## 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.

## Answer

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

```text
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 |

```text
I ≈ 0.700000 × [ 2.178807 ] = 1.5252
```

**Answer: I ≈ 1.5252**

## Check yourself

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

1. What are the nodes and weights of the 3-point Gauss-Legendre formula on [-1, 1]?
   A. Nodes -0.774597, 0, 0.774597 with weights 8/9, 5/9, 8/9
   B. Nodes -0.774597, 0, 0.774597 with weights 5/9, 8/9, 5/9
   C. Nodes -0.577350, 0.577350 with weights 1, 1
   D. Nodes -1, 0, 1 with weights 1/3, 4/3, 1/3
   Answer: B. The 3-point rule uses t = 0, ±sqrt(0.6) with weight 8/9 at the centre and 5/9 at each outer node. (Key point: Gauss 3-point weights)

2. How is the integral over [0, 1.4] converted to [-1, 1] here?
   A. x = 1.4t + 0.7, dx = 1.4 dt
   B. x = 0.7t + 1.4, dx = 0.7 dt
   C. x = 0.7t + 0.7, dx = 0.7 dt
   D. x = 0.7t - 0.7, dx = dt
   Answer: C. With a = 0, b = 1.4, both the scale and the shift equal 0.7. (Key point: Change of limits)

3. After computing w·f(x) at the three nodes, what is the last step?
   A. Add the three products and divide by 3
   B. Average the three f(x) values and multiply by b-a
   C. Add the products and multiply by 2/(b-a)
   D. Add the three products and multiply by the factor (b-a)/2 from dx
   Answer: D. Since dx = ((b-a)/2) dt, the weighted sum is multiplied by (b-a)/2. (Key point: Final assembly)

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