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

# Half range sine series: x − x²

Mathematics III · Chapter 4: Fourier Series

- Asked in: 2071 Chaitra, 5 marks; 2068 Baishakh, 5 marks (asked 2 times)
- Video: [Half Range Sine Series For F(x) = lx - x^2 in ( o, l ) - Fourier Series - Engineering Mathematics 3](https://www.youtube.com/watch?v=BetZOt2cCgc) — Ekeeda
- 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-x-x2
- 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) = x − x² for 0 < x < 1.

## Short answer

For f(x)=x-x^2 on (0,1), the half-range sine series has coefficients b_n = 8/(n^3 pi^3) for odd n (0 for even n), giving x - x^2 = (8/pi^3) [sin(pi x)/1^3 + sin(3pi x)/3^3 + sin(5pi x)/5^3 + ...].

## Answer

Half-range sine series on (0, 1) (l = 1): f(x) = Σ bₙ sin nπx with

```text
bₙ = 2 ∫₀¹ (x − x²) sin nπx dx
```

Integrate by parts, with u = x − x² (u′ = 1 − 2x, u″ = −2) and ∫ sin nπx dx = −cos nπx/(nπ):

```text
∫₀¹ (x − x²) sin nπx dx
 = [ −(x − x²) cos nπx/(nπ) ]₀¹ + (1/(nπ)) ∫₀¹ (1 − 2x) cos nπx dx
 = 0 + (1/(nπ)) { [ (1 − 2x) sin nπx/(nπ) ]₀¹ + (2/(nπ)) ∫₀¹ sin nπx dx }
 = (1/(nπ)) (2/(nπ)) [ (1 − cos nπ)/(nπ) ]
 = 2 (1 − (−1)ⁿ) / (nπ)³
```

```text
bₙ = 4 (1 − (−1)ⁿ) / (n³π³)  = 8/(n³π³) for odd n,  0 for even n
```

**Answer:**

```text
x − x² = (8/π³) [ sin πx/1³ + sin 3πx/3³ + sin 5πx/5³ + ... ],   0 < x < 1
```

## Check yourself

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

1. On (0,1), f = sum bn sin(n pi x) with bn =
   A. (2/pi) integral from 0 to pi
   B. 2 integral from 0 to 1 f sin(n pi x) dx
   C. integral from 0 to 1 f sin(n pi x) dx
   D. 2 integral f cos(n pi x)
   Answer: B. l = 1. (Key point: Interval)

2. integral of (x - x^2) sin n pi x from 0 to 1 =
   A. (1 - (-1)^n)/(n pi)
   B. 2(1 + (-1)^n)/(n pi)^3
   C. 2(1 - (-1)^n)/(n pi)^3
   D. 0
   Answer: C. Twice by parts. (Key point: Integral)

3. bn is
   A. 8/(n^3 pi^3) for even n
   B. 4/(n pi) for odd n
   C. 0 for odd n
   D. 8/(n^3 pi^3) for odd n, 0 for even n
   Answer: D. 4(1-(-1)^n)/(n pi)^3. (Key point: bn)

## 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-three) — Mathematics III, chapter 4, asked 9×
- [Half range sine series: eˣ](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Ahalf-range-sine-series-e) — Mathematics III, chapter 4, asked 12×
- [Half range sine series: πx − x²](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Ahalf-range-sine-series-x-x2-four) — Mathematics III, chapter 4, asked 12×
- [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×
- [Half range sine series: π − x](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Ahalf-range-sine-series-x) — Mathematics III, chapter 4, asked 13×
- [Half range sine series: eˣ](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Ahalf-range-sine-series-e-two) — Mathematics III, chapter 4, asked 6×

## Related in Mathematics III

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