---
title: "Fourier series: 2x − x²"
subject: "Mathematics III"
chapter: 4
key: "semester_3/mathematics_iii:fourier-series-2x-x2-three"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/fourier-series-2x-x2-three
site: BCT Engineering
---

# Fourier series: 2x − x²

Mathematics III · Chapter 4: Fourier Series

- Asked in: 2070 Asar, 5 marks
- Video: [Fourier Series of 2x-x^2 over  (0, 3) || 18mat31 || Dr Prashant Patil](https://www.youtube.com/watch?v=5dScBLo99LY) — Dr Prashant Patil
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/mathematics_iii/fourier-series-2x-x2-three
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-4

## Question

Find a Fourier series to represent f(x) = 2x − x² in the range (0, 3).

## Short answer

For period 3, a₀ = (2/3)∫₀³ (2x−x²)dx = 0. Then aₙ = −9/(n²π²), bₙ = 3/(nπ). The Fourier series is 2x−x² = Σ_{n=1}^∞ [ −9/(n²π²) cos(2nπx/3) + (3/(nπ)) sin(2nπx/3) ].

## Answer

For a function on (0, 3) taken with period 3 (so 2l = 3 and l = 3/2): f(x) = a₀/2 + Σ [aₙ cos(2nπx/3) + bₙ sin(2nπx/3)], where a₀ = (2/3)∫₀³ f dx, aₙ = (2/3)∫₀³ f cos(2nπx/3) dx, bₙ = (2/3)∫₀³ f sin(2nπx/3) dx.

For f(x) = 2x − x²:

Let k = 2nπ/3. At x = 3, kx = 2nπ, so sin kx = 0 and cos kx = 1.

```text
a₀ = (2/3)∫₀³ (2x − x²) dx = (2/3)[x² − x³/3]₀³ = (2/3)(9 − 9) = 0

∫₀³ (2x − x²) cos kx dx = [(2x − x²) sin kx/k]₀³ − (1/k)∫₀³ (2 − 2x) sin kx dx
                        = 0 − (1/k)(6/k) = −6/k²
aₙ = (2/3)(−6/k²) = −4/k² = −9/(n²π²)

∫₀³ (2x − x²) sin kx dx = [−(2x − x²) cos kx/k]₀³ + (1/k)∫₀³ (2 − 2x) cos kx dx
                        = 3/k + 0
bₙ = (2/3)(3/k) = 2/k = 3/(nπ)
```

(Here ∫₀³ (2 − 2x) sin kx dx = 6/k and ∫₀³ (2 − 2x) cos kx dx = 0, by one more integration by parts.)

**Fourier series**

2x − x² = Σ (n=1 to ∞) [ −(9/(n²π²)) cos(2nπx/3) + (3/(nπ)) sin(2nπx/3) ],  0 < x < 3

Since a₀ = 0 there is no constant term.

## Check yourself

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

1. For f on (0,3) with period 3, what is l?
   A. 6
   B. 1
   C. 3/2
   D. 3
   Answer: C. Period 2l = 3, so l = 3/2 and the arguments become 2 n pi x/3. (Key point: Period and l)

2. Why is there no constant term?
   A. The series has only sine terms
   B. a0 = (2/3)(9 - 9) = 0
   C. a0 is infinite
   D. Constants are not allowed
   Answer: B. The integral of 2x - x^2 from 0 to 3 is 0. (Key point: Constant term)

3. What is an?
   A. 3/(n pi)
   B. -9/(n^2 pi^2)
   C. -3/(n pi)
   D. 9/(n^2 pi^2)
   Answer: B. an = -4/k^2 with k = 2n pi/3, giving -9/(n^2 pi^2). (Key point: Cosine coefficient)

4. What is bn?
   A. -3/(n pi)
   B. 9/(n^2 pi^2)
   C. 3/(n pi)
   D. 6/(n pi)
   Answer: C. bn = 2/k = 3/(n pi). (Key point: Sine coefficient)

## Close variants set in other exams

- [Fourier series: x − x²](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Afourier-series-x-x2-two) — Mathematics III, chapter 4, asked 2×

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