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

# Fourier series: x + x²

Mathematics III · Chapter 4: Fourier Series

- Asked in: 2078 Bhadra, 5 marks
- Video: [Fourier series of the function f(x)=x+x^2 , Engineering Mathematics ,Diploma Mathematics](https://www.youtube.com/watch?v=z8kpAlULwxc) — GBK Academy 
- 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-x-x2-three
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-4

## Question

Obtain the Fourier series of f(x) = x + x² in −π ≤ x ≤ π.

## Short answer

The Fourier coefficients are a0 = 2π²/3, a_n = 4(-1)^n/n², b_n = 2(-1)^(n+1)/n. The series is x + x² = π²/3 + ∑_{n=1}^∞ [4(-1)^n cos(nx)/n² + 2(-1)^(n+1) sin(nx)/n].

## Answer

f(x) = x + x² on (−π, π). Split into the odd part x and the even part x².

#### Coefficients
```text
a₀ = (1/π) ∫_{−π}^{π} (x + x²) dx = (1/π)(2π³/3) = 2π²/3
aₙ = (1/π) ∫ (x + x²) cos nx dx = (1/π) ∫ x² cos nx dx = 4(−1)ⁿ/n²
bₙ = (1/π) ∫ (x + x²) sin nx dx = (1/π) ∫ x sin nx dx = 2(−1)ⁿ⁺¹/n
```

(x cos nx and x² sin nx are odd, so they integrate to zero.)

#### Series
```text
x + x² = π²/3 + Σ_{n=1}^∞ [ 4(−1)ⁿ cos nx / n² + 2(−1)ⁿ⁺¹ sin nx / n ]
       = π²/3 − 4cos x + cos 2x − (4/9)cos 3x + ...
         + 2 sin x − sin 2x + (2/3) sin 3x − ...
```

## Check yourself

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

1. In x + x^2 on (−π, π), which part gives the sine terms?
   A. The odd part x
   B. The even part x^2
   C. Both parts
   D. Neither
   Answer: A. Sine terms come from the odd part; cosine terms and a0 from the even part. (Key point: Symmetry)

2. aₙ for x + x^2 is:
   A. 4(−1)^(n+1)/n^2
   B. 4(−1)^n/n^2
   C. 2(−1)^n/n
   D. 2π^2/3
   Answer: B. an = (1/π)∫ x^2 cos nx dx = 4(−1)^n/n^2. (Key point: an)

3. bₙ for x + x^2 is:
   A. 4(−1)^n/n^2
   B. 2(−1)^n/n
   C. 2(−1)^(n+1)/n
   D. 0
   Answer: C. bn = (1/π)∫ x sin nx dx = 2(−1)^(n+1)/n. (Key point: bn)

4. The constant term of the series is:
   A. 2π^2/3
   B. π^2/6
   C. 0
   D. π^2/3
   Answer: D. a0/2 = (2π^2/3)/2 = π^2/3. (Key point: Constant term)

## Related in Mathematics III

- [Fourier series: x − x²](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Afourier-series-x-x2) — Mathematics III, chapter 4, asked 6×
- [Fourier series: x³](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Afourier-series-x3) — Mathematics III, chapter 4, asked 6×
- [Fourier series of a function](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Afourier-series-of-a-function) — Mathematics III, chapter 4, asked 6×
- [Fourier series: eˣ](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Afourier-series-e) — Mathematics III, chapter 4, asked 6×
- [Fourier series: x²](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Afourier-series-x2-two) — Mathematics III, chapter 4, asked 2×
- [Fourier series: x − x²](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Afourier-series-x-x2-five) — Mathematics III, chapter 4

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