---
title: "Gauss 3-point: e^(sin x)"
subject: "Numerical Method"
chapter: 5
key: "semester_4/numerical_method:gauss-3pt-e-sinx"
source: https://www.bctengineer.com/q/semester_4/numerical_method/gauss-3pt-e-sinx
site: BCT Engineering
---

# Gauss 3-point: e^(sin x)

Numerical Method · Chapter 5: Numerical Differentiation and Integration

- Asked in: 2075 Baisakh, 6 marks
- Video: [Problems on Gauss-Quadrature 2-point and 3-point formula by Keshav Jadhav](https://www.youtube.com/watch?v=El0o_vCHL7Y) — Keshav Jadhav
- 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-e-sinx
- Whole chapter: https://www.bctengineer.com/md/semester_4/numerical_method/chapter-5

## Question

Evaluate ∫ from 0 to π/2 of e^(sin x) dx using Gaussian 3-point formula.

## Short answer

Transform to [-1,1] with x = (π/4)t + π/4, dx = π/4 dt. Apply 3-point Gauss-Legendre with nodes ±√0.6,0 and weights 5/9,8/9,5/9 to get I ≈ 3.1039.

## 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 = π/2 = 1.570796, so

```text
x = 0.785398·t + 0.785398,     dx = 0.785398 dt
∫ f(x) dx = 0.785398 ∫₋₁¹ f(0.785398t + 0.785398) dt
```

#### Nodes and function values

| t | x = 0.7854t + 0.7854 | f(x) | weight w | w·f(x) |
|---|---|---|---|---|
| -0.774597 | 0.177031 | 1.192567 | 5/9 | 0.662537 |
| 0.000000 | 0.785398 | 2.028115 | 8/9 | 1.802769 |
| 0.774597 | 1.393765 | 2.676128 | 5/9 | 1.486738 |

```text
I ≈ 0.785398 × [ 3.952044 ] = 3.1039
```

(The accurate value is 3.1044, so the 3-point result is good to about three significant figures.)

**Answer: I ≈ 3.1039**

## 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 -0.577350, 0.577350 with weights 1, 1
   C. Nodes -0.774597, 0, 0.774597 with weights 5/9, 8/9, 5/9
   D. Nodes -1, 0, 1 with weights 1/3, 4/3, 1/3
   Answer: C. 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, π/2] converted to [-1, 1] here?
   A. x = 1.570796t, dx = 1.570796 dt
   B. x = 0.785398t - 0.785398, dx = dt
   C. x = t + 0.785398, dx = dt
   D. x = 0.785398t + 0.785398, dx = 0.785398 dt
   Answer: D. With a = 0, b = π/2 both (b-a)/2 and (b+a)/2 equal 0.785398. (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 multiply by the factor (b-a)/2 from dx
   B. Add the three products and divide by 3
   C. Average the three f(x) values and multiply by b-a
   D. Add the products and multiply by 2/(b-a)
   Answer: A. Since dx = ((b-a)/2) dt, the weighted sum is multiplied by (b-a)/2. (Key point: Final assembly)

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