Fundamental theorem of line integrals

The fundamental theorem of line integrals (FTLI) is the multivariable analog of the fundamental theorem of calculus.

If $\mathbf{F}=\nabla \varphi$ is a gradient field (with $\varphi$ continuously differentiable) and $C$ is a smooth curve from point $A$ to point $B$, then

${\int }_C\mathbf{F}\cdot d\mathbf{r}=\varphi (B)-\varphi (A).$

Integrating the gradient along a curve gives the difference of $\varphi$ at the endpoints. Same theorem as the FTC, one dimension higher.

Proof

Parameterize $C$ by $\mathbf{r}(t)$, $a\le t\le b$, with $A=\mathbf{r}(a)$, $B=\mathbf{r}(b)$. Define the composition $g(t)=\varphi (\mathbf{r}(t))$. By the multivariable chain rule,

$g^{\prime }(t)=\nabla \varphi (\mathbf{r}(t))\cdot {\mathbf{r}}^{\prime }(t)=\mathbf{F}(\mathbf{r}(t))\cdot {\mathbf{r}}^{\prime }(t).$

So the integrand of the line integral is exactly $g^{\prime }(t)$, and by the ordinary FTC

${\int }_C\mathbf{F}\cdot d\mathbf{r}={\int }_a^b{g}^{\prime }(t)dt=g(b)-g(a)=\varphi (B)-\varphi (A).■$

The theorem is just chain rule plus FTC.

Three immediate consequences

1. Path independence. The right side depends only on the endpoints $A,B$, not on the path. Two different curves $C_1,C_2$ from $A$ to $B$ in the same domain give the same value of $\int \mathbf{F}\cdot d\mathbf{r}$.

2. Closed loops give zero. If $C$ is closed ($A=B$), then

${\oint }_C\mathbf{F}\cdot d\mathbf{r}=\varphi (A)-\varphi (A)=0.$

3. Computational shortcut. Once you know $\varphi$, you skip the parameterize-and-integrate procedure entirely: just evaluate $\varphi$ at the endpoints and subtract.

When FTLI applies (and when it doesn’t)

FTLI requires $\mathbf{F}$ to be a gradient field — i.e., conservative. The cross-partial test (zero curl on a simply connected domain) is how you check.

For non-conservative fields, FTLI doesn’t apply. You have to parameterize and integrate. The path matters.

Worked example

$\mathbf{F}=⟨2x,2y⟩$ on ${\mathbb{R}}^2$. Check: $P_y=0=Q_x$. Conservative.

Find potential: ${\varphi }_x=2x\Rightarrow \varphi =x^2+g(y)$. Then ${\varphi }_y=g^{\prime }(y)=2y$, so $g(y)=y^2$. Potential $\varphi =x^2+y^2$.

For any path $C$ from $(0,0)$ to $(2,4)$:

${\int }_C\mathbf{F}\cdot d\mathbf{r}=\varphi (2,4)-\varphi (0,0)=(4+16)-0=20.$

Verify along the parabola $y=x^2$: $\mathbf{r}(t)=⟨t,t^2⟩$, $\mathbf{F}(\mathbf{r})=⟨2t,2t^2⟩$, ${\mathbf{r}}^{\prime }=⟨1,2t⟩$, dot $=2t+4t^3$, integral ${\int }_0^2(2t+4t^3)dt=4+16=20$. ✓

Same answer along the L-shape from $(0,0)$ to $(2,0)$ to $(2,4)$: along $x$-axis $\int 2xdx=4$; along $y$-axis at $x=2$ $\int 2ydy=16$. Total $20$. ✓

In the FTC family

FTLI is the second rung of a ladder of theorems, each generalizing the FTC to one higher dimension:

Theorem	Boundary integral	Interior integral	Geometry
FTC	$F(b)-F(a)$	$\int F^{\prime }dx$	0D bdry, 1D interior
FTLI	$\varphi (B)-\varphi (A)$	${\int }_C\nabla \varphi \cdot d\mathbf{r}$	0D bdry, 1D curve in 3D
Green’s theorem (circ)	$\oint \mathbf{F}\cdot d\mathbf{r}$	$\iint (Q_x-P_y)dA$	1D bdry, 2D plane
Stokes’ theorem	$\oint \mathbf{F}\cdot d\mathbf{r}$	$\iint (\nabla \times \mathbf{F})\cdot d\mathbf{S}$	1D bdry, 2D surface in 3D
Divergence theorem	$\iint \mathbf{F}\cdot d\mathbf{S}$	$\iiint \nabla \cdot \mathbf{F}dV$	2D bdry, 3D volume

All of these are the same statement at successive dimensions: integrating a derivative on the inside equals integrating the original on the boundary. In modern mathematics they all collapse into one theorem on differential forms, the generalized Stokes’ theorem ${\int }_{M}d\omega ={\int }_{\partial M}\omega$.