---
title: "Half range sine series: x²"
subject: "Mathematics III"
chapter: 4
key: "semester_3/mathematics_iii:half-range-sine-series-x2-three"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/half-range-sine-series-x2-three
site: BCT Engineering
---

# Half range sine series: x²

Mathematics III · Chapter 4: Fourier Series

- Asked in: 2072 Kartik, 5 marks
- Video: [VTU Half Range Fourier Series | f(x) = x² | Sine Series | EE301, EC301, ME301](https://www.youtube.com/watch?v=9OONqYUzAv4) — Mathematics Tutor
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/mathematics_iii/half-range-sine-series-x2-three
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-4

## Question

Obtain the half range sine series for the function f(x) = x² in the interval (0, 3).

## Short answer

With L=3, bₙ = (2/3)∫₀³ x² sin(nπx/3)dx = 18(−1)ⁿ⁺¹/(nπ) − 36(1−(−1)ⁿ)/(n³π³). The half-range sine series is x² = Σ_{n=1}^∞ bₙ sin(nπx/3).

## Answer

The half range sine series on (0, L) is f(x) = Σ bₙ sin(nπx/L), with bₙ = (2/L)∫₀ᴸ f(x) sin(nπx/L) dx. Here L = 3.

**Coefficient.** Integrate by parts with u = x², dv = sin(nπx/3)dx; repeated integration gives

```text
bₙ = (2/3)∫₀³ x² sin(nπx/3) dx
   = (2/3)[ −x²(3/nπ) cos(nπx/3) + 2x(3/nπ)² sin(nπx/3) + 2(3/nπ)³ cos(nπx/3) ]₀³
   = (2/3)[ −(27/nπ)(−1)ⁿ + (54/n³π³)((−1)ⁿ − 1) ]
   = 18(−1)ⁿ⁺¹/(nπ) − 36(1 − (−1)ⁿ)/(n³π³)
```

- n even: bₙ = −18/(nπ)
- n odd: bₙ = 18/(nπ) − 72/(n³π³)

**Series**

x² = Σ (n=1 to ∞) [ 18(−1)ⁿ⁺¹/(nπ) − 36(1 − (−1)ⁿ)/(n³π³) ] sin(nπx/3),  0 < x < 3

First terms: b₁ = 18/π − 72/π³ ≈ 3.41, b₂ = −9/π ≈ −2.86.

## Check yourself

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

1. For L = 3, what is bn?
   A. (2/3) integral of x^2 sin(n pi x/3) dx from 0 to 3
   B. (1/3) integral of x^2 sin(n pi x/3) dx
   C. (2/3) integral of x^2 sin(2 n pi x/3) dx
   D. (2/3) integral of x^2 cos(n pi x/3) dx
   Answer: A. Half-range sine uses sin(n pi x/L) with factor 2/L. (Key point: Formula for interval (0, L))

2. How is the integral evaluated?
   A. Differentiating under integral
   B. Partial fractions
   C. Substitution x = sin t
   D. Repeated integration by parts
   Answer: D. x^2 times sine needs two integrations by parts. (Key point: Method)

3. What is bn for even n?
   A. 18/(n pi)
   B. 0
   C. -18/(n pi)
   D. 18/(n pi) - 72/(n^3 pi^3)
   Answer: C. For even n, 1 - (-1)^n = 0, so only the first term -18/(n pi) remains. (Key point: Even/odd n)

4. What is b1 approx?
   A. 3.41
   B. 9.0
   C. -2.86
   D. 2.86
   Answer: A. b1 = 18/pi - 72/pi^3 = 5.73 - 2.32 = 3.41. (Key point: First coefficient)

## Close variants set in other exams

- [Half range sine series: x²](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Ahalf-range-sine-series-x2-two) — Mathematics III, chapter 4, asked 9×
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