In Calculus 1 the fundamental theorem of calculus says $\int_a^b F'(x)\,dx = F(b) - F(a)$: integrating a derivative only needs the endpoint values. Line integrals have the same shortcut when the field is a gradient field. Then no parametrization is needed at all.
Key idea
Fundamental theorem for line integrals. If $C$ is a piecewise smooth curve from $A$ to $B$ and $f$ is continuously differentiable on an open region containing $C$, then
$$\int_C \nabla f \cdot d\mathbf{r} = f(B) - f(A).$$
In particular, the integral is the same along every path from $A$ to $B$, and it is ${}0$ around every closed curve.
Why it works
Let $\mathbf{r}(t)$, $a \le t \le b$, trace $C$. By the chain rule,
$$\frac{d}{dt} f(\mathbf{r}(t)) = f_x x'(t) + f_y y'(t) + f_z z'(t) = \nabla f(\mathbf{r}(t)) \cdot \mathbf{r}'(t).$$
So $$\int_a^b \nabla f(\mathbf{r}(t)) \cdot \mathbf{r}'(t)\,dt = \int_a^b \frac{d}{dt} f(\mathbf{r}(t))\,dt = f(\mathbf{r}(b)) - f(\mathbf{r}(a))$$ by the one-variable theorem. For a piecewise smooth curve, apply this to each piece; the inside endpoints cancel.
Worked examples
Worked example: Endpoints only
Find $\int_C \mathbf{F} \cdot d\mathbf{r}$ for $\mathbf{F} = \langle 2xy, x^2 + 3y^2 \rangle$ along any curve from $(1, 0)$ to $(2, 1)$.
- $\mathbf{F} = \nabla f$ with $f = x^2y + y^3$: $f_x = 2xy$ and $f_y = x^2 + 3y^2$.
- $f(2, 1) - f(1, 0) = (4 + 1) - 0 = 5$.
Answer: ${}5$
Worked example: Along a helix
Find $\int_C \mathbf{F} \cdot d\mathbf{r}$ for $\mathbf{F} = \langle yz, xz, xy + 2z \rangle$ along $\mathbf{r}(t) = \langle \cos t, \sin t, t \rangle$, ${}0 \le t \le 2\pi$.
- $\mathbf{F} = \nabla f$ with $f = xyz + z^2$ (check each partial derivative).
- The curve runs from $\mathbf{r}(0) = (1, 0, 0)$ to $\mathbf{r}(2\pi) = (1, 0, 2\pi)$.
- $$f(1, 0, 2\pi) - f(1, 0, 0) = 4\pi^2 - 0 = 4\pi^2.$$
Answer: ${}4\pi^2$
Worked example: When the theorem does not apply
Find $\oint_C \langle -y, x \rangle \cdot d\mathbf{r}$ around the unit circle, counterclockwise.
- Parametrize: $\mathbf{F} \cdot \mathbf{r}' = \sin^2 t + \cos^2 t = 1$, so the integral is ${}2\pi$.
- A gradient field would give ${}0$ around a closed curve. So $\langle -y, x \rangle$ is not a gradient field, which matches $P_y = -1 \ne 1 = Q_x$.
Answer: ${}2\pi$
Common mistake
Using $f(B) - f(A)$ for a field that is not a gradient. The shortcut needs a potential function. For $\langle -y, x \rangle$ around a closed curve it would give ${}0$, but the true value is ${}2\pi$. Before using endpoints, check that $\mathbf{F} = \nabla f$: find $f$, or at least check that $P_y = Q_x$ on a region without holes.