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

# Half range sine series: x²

Mathematics III · Chapter 4: Fourier Series

- Asked in: 2078 Bhadra, 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
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-4

## Question

Express f(x) = x² as a half-range sine series in 0 < x < 3.

## Short answer

For l=3, b_n = (2/3)∫_0^3 x² sin(nπx/3) dx = 18(-1)^(n+1)/(nπ) + 36[(-1)^n - 1]/(n³π³). Thus x² = ∑_{n=1}^∞ [18(-1)^(n+1)/(nπ) + 36((-1)^n - 1)/(n³π³)] sin(nπx/3).

## Answer

Half-range sine series on (0, 3): l = 3, f(x) = Σ bₙ sin(nπx/3), with

```text
bₙ = (2/3) ∫₀³ x² sin(nπx/3) dx
```

Let k = nπ/3. Integrating by parts twice:

```text
∫₀³ x² sin kx dx = [ −x² cos kx/k + 2x sin kx/k² + 2 cos kx/k³ ]₀³
                 = −9(−1)ⁿ/k + 2[(−1)ⁿ − 1]/k³
                 = −27(−1)ⁿ/(nπ) + 54[(−1)ⁿ − 1]/(nπ)³
```

(at x = 3, kx = nπ so sin kx = 0, cos kx = (−1)ⁿ.)

```text
bₙ = (2/3) × above = 18(−1)ⁿ⁺¹/(nπ) + 36[(−1)ⁿ − 1]/(n³π³)
```

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

**Answer:**

```text
x² = Σ_{n=1}^∞ { 18(−1)ⁿ⁺¹/(nπ) + 36[(−1)ⁿ − 1]/(n³π³) } sin(nπx/3),   0 < x < 3
   = 3.407 sin(πx/3) − 2.865 sin(2πx/3) + 1.824 sin(πx) − 1.432 sin(4πx/3) + ...
```

## Check yourself

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

1. For the half-range sine series on (0, 3), bn is:
   A. (2/3)∫0^3 x^2 sin(nπx/3) dx
   B. (2/π)∫0^3 x^2 sin nx dx
   C. (1/3)∫0^3 x^2 sin(nπx/3) dx
   D. (2/3)∫0^3 x^2 cos(nπx/3) dx
   Answer: A. l = 3, so bn = (2/l)∫ f sin(nπx/l) dx. (Key point: Half-period)

2. At x = 3 with k = nπ/3, which values are used?
   A. sin kx = (−1)^n and cos kx = 0
   B. sin kx = 0 and cos kx = (−1)^n
   C. sin kx = 1 and cos kx = 1
   D. both equal (−1)^n
   Answer: B. kx = nπ at x = 3. (Key point: Evaluating at x = 3)

3. For even n, bn equals:
   A. 0
   B. 18/(nπ)
   C. −18/(nπ)
   D. −72/(n^3π^3)
   Answer: C. For even n, (−1)^n − 1 = 0 so only 18(−1)^(n+1)/(nπ) = −18/(nπ) remains. (Key point: Even n)

4. For odd n, bn equals:
   A. 18/(nπ)
   B. −72/(n^3π^3)
   C. 0
   D. 18/(nπ) − 72/(n^3π^3)
   Answer: D. Both terms survive for odd n. (Key point: Odd n)

## Close variants set in other exams

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

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