---
title: "Line integral independent of path"
subject: "Mathematics III"
chapter: 2
key: "semester_3/mathematics_iii:line-integral-independent-of-path"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/line-integral-independent-of-path
site: BCT Engineering
---

# Line integral independent of path

Mathematics III · Chapter 2: Line, Surface and Volume Integrals

- Asked in: 2080 Bhadra, 5 marks; 2078 Kartik, 5 marks; 2076 Asoj, 5 marks; 2075 Asoj, 5 marks; 2072 Kartik, 5 marks; 2070 Chaitra, 5 marks (asked 6 times)
- Video: [Path independence for line integrals | Multivariable Calculus | Khan Academy](https://www.youtube.com/watch?v=K_fgnCJOI8I) — Khan Academy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/mathematics_iii/line-integral-independent-of-path
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-2

## Question

Prove that the integral ∫_A^B F·dr is independent of the path joining any two points A and B in a region if and only if ∮_C F·dr = 0 for any simple closed curve C in the region formed by arcs joining the points A and B.

## Short answer

The line integral from A to B is path-independent if and only if the circulation around every closed curve C is zero. This follows by splitting C into two paths between two points on C and comparing their integrals.

## Answer

**Statement.** Let F be a vector field continuous in a region. Then ∫_A^B F·dr has the same value along every path from A to B in the region **if and only if** ∮_C F·dr = 0 for every closed curve C in the region.

#### Part 1: Independent of path ⇒ ∮_C F·dr = 0
Let C be any simple closed curve in the region. Choose two points A and B on C. They split C into two arcs: C₁ = APB (from A to B) and C₂ = BQA (from B back to A).

```text
        P
     .-----.
    A       B
     '-----'
        Q

C₁ = A→B along P ;  C₂ = B→A along Q
```

Then

```text
∮_C F·dr = ∫_{APB} F·dr + ∫_{BQA} F·dr
         = ∫_{APB} F·dr − ∫_{AQB} F·dr
```

(reversing the direction of a path changes the sign of the integral). APB and AQB are two paths from A to B, and by hypothesis their integrals are equal. Hence

```text
∮_C F·dr = 0
```

#### Part 2: ∮_C F·dr = 0 for every closed curve ⇒ independent of path
Let C₁ and C₂ be any two paths from A to B. Let C be the closed curve formed by going along C₁ from A to B and returning along C₂ reversed (from B to A). By hypothesis

```text
0 = ∮_C F·dr = ∫_{C₁} F·dr + ∫_{−C₂} F·dr = ∫_{C₁} F·dr − ∫_{C₂} F·dr
```

so ∫_{C₁} F·dr = ∫_{C₂} F·dr. The value does not depend on which path from A to B is taken. (If the two paths cross each other, the closed curve is split at the crossing points into simple closed loops, each having integral zero, so the same conclusion holds.) ∎

#### Remark
For a simply connected region this property is equivalent to F being conservative: F = ∇φ and curl F = 0, in which case ∫_A^B F·dr = φ(B) − φ(A).

## Check yourself

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

1. Independence of path is equivalent to:
   A. F constant
   B. ∮F·dr=0 for every closed curve
   C. curl F ≠ 0
   D. F·dr=1
   Answer: B. The theorem's equivalence. (Key point: Statement)

2. Closed curve C is split into arcs APB and BQA; the integral becomes:
   A. ∫APB − ∫AQB
   B. ∫APB + ∫AQB
   C. 0 automatically
   D. ∫APB·∫AQB
   Answer: A. Reversing BQA gives −∫AQB. (Key point: Part 1 split)

3. Reversing path direction:
   A. Leaves it unchanged
   B. Doubles it
   C. Makes it zero
   D. Changes the integral's sign
   Answer: D. ∫−C = −∫C. (Key point: Reversal)

4. Given ∮=0, path C₁ out and C₂ back gives:
   A. ∫C₁ = −∫C₂
   B. ∫C₁ = 0 always
   C. ∫C₁ = ∫C₂
   D. ∫C₁ + ∫C₂ = 2
   Answer: C. 0 = ∫C₁ − ∫C₂. (Key point: Part 2)

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