Fundamental Theorem for Line Integrals
Textbook Reference
| Primary source | OpenStax Calculus Volume 3, Section 6.3: “Conservative Vector Fields” |
| Direct link | https://openstax.org/books/calculus-volume-3/pages/6-3-conservative-vector-fields |
| Textbook used in class | Stewart, Calculus, Section 16.3: “The Fundamental Theorem for Line Integrals” (Theorem 2, Examples 3, 4) |
Quick Reference
Fundamental Theorem for Line Integrals (FTLI): If $\mathbf{F} = \nabla f$ is a conservative field and $C$ goes from point $A$ to point $B$, then: $$\int_C \mathbf{F}\cdot d\mathbf{r} = f(B) - f(A).$$
Consequence: The integral depends only on the endpoints, not the path.
Motivation
The single-variable FTC Part 2 says $\int_a^b f'(x)\,dx = f(b)-f(a)$: the integral of a derivative depends only on the endpoint values. The FTLI is the exact analogue: $\int_C\nabla f\cdot d\mathbf{r} = f(B)-f(A)$. The vector field is a gradient (the multivariable derivative of $f$), and the line integral depends only on the values of $f$ at the endpoints.
This makes evaluating line integrals of conservative fields trivial: find the potential function, evaluate at the two endpoints, subtract.
Key Concept
Process for using the FTLI:
- Check that $\mathbf{F}$ is conservative ($P_y = Q_x$).
- Find a potential function $f$ such that $\nabla f = \mathbf{F}$.
- Identify the endpoints $A$ and $B$ of $C$.
- Compute $f(B) - f(A)$.
No need to parametrize the curve.
Worked Example
Evaluate $\int_C\mathbf{F}\cdot d\mathbf{r}$ where $\mathbf{F}(x,y) = \langle 2xy+1, x^2+e^y\rangle$ and $C$ goes from $(0,0)$ to $(1,1)$ along any smooth path. (From the previous skill.)
We found that $\mathbf{F}$ is conservative with potential function $f(x,y) = x^2 y + x + e^y$.
By FTLI: $$\int_C\mathbf{F}\cdot d\mathbf{r} = f(1,1) - f(0,0) = (1\cdot 1 + 1 + e^1) - (0 + 0 + e^0) = (2+e) - 1 = 1+e.$$
No parametrization required.
FTLI says the line integral is zero for all conservative fields. FTLI says the integral equals $f(B)-f(A)$, which is zero only when $A = B$ (a closed curve). For an open path from $A \neq B$, the value is generally nonzero. The statement “$\oint_C\mathbf{F}\cdot d\mathbf{r} = 0$ for closed curves” is a consequence of FTLI applied to $A = B$, not the theorem itself.
Common Misconceptions
to apply the Fundamental Theorem for Line Integrals, the curve must be a straight line.
This is the concept-image-conflicts-definition error. The FTLI holds for any piecewise smooth curve from $A$ to $B$, regardless of shape. The entire point of the theorem is that the path does not matter: only the endpoints $A$ and $B$ determine the value $f(B) - f(A)$. A winding spiral from $(0,0)$ to $(1,1)$ and the straight segment between the same two points give identical results when $\mathbf{F}$ is conservative.
Leveled Practice
Problem 1. Evaluate $\int_C(e^y\,dx + xe^y\,dy)$ from $(0,0)$ to $(2,1)$, where $\mathbf{F} = \langle e^y, xe^y\rangle$.
Show answer
$P_y = e^y$, $Q_x = e^y$. Conservative.
$f_x = e^y \implies f = xe^y + g(y)$.
$f_y = xe^y + g'(y) = xe^y \implies g'(y) = 0 \implies g = C$.
$f(x,y) = xe^y$.
$\int_C = f(2,1) - f(0,0) = 2e - 0 = 2e$.