RECALL: $\displaystyle\Int_a^b{df\over dx}\,dx = f \Big|_a^b$

This is the Fundamental Theorem of Calculus, which just says that the integral of a derivative is the function you started with. We could also write this simply as $$\int df = f$$ But recall the master formula (1) of Section 3 which says $$df = \grad{f} \cdot d\rr$$ Putting this all together, we get the fundamental theorem for line integrals, which says that $$\Lint \grad{f} \cdot d\rr = f \Big|_A^B$$ for any curve $C$ starting at some point $A$ and ending at some point $B$.

Notice that the right-hand side does not depend on the curve $C$! 1) This behavior leads us directly to the notion of path independence. A line integral of the form $\DS\Lint\FF\cdot d\rr$ is said to be independent of path if its value depends only on the endpoints $A$ and $B$ of the curve $C$, not on the particular curve connecting them. If a line integral is independent of path, we no longer need to specify the path $C$, so we instead write $$\Int_A^B \FF \cdot d\rr$$ If you know that a line integral is independent of path, you may {\it choose} a different path (with the same endpoints) which makes evaluating the integral as simple as possible!

The fundamental theorem implies that vector fields of the form $\FF=\grad{f}$ are special; the corresponding line integrals are always independent of path. One way to think of this is to imagine the level curves of $f$; the change in $f$ depends only on where you start and end, not on how you get there. These special vector fields have a name: A vector field $\FF$ is said to be conservative if there exists a potential function $f$ such that $\FF=\grad{f}$.

If $\FF$ is conservative, then $\DS\Lint\FF\cdot d\rr$ is independent of path; the converse is also true. But how do you know if a given vector field $\FF$ is conservative? That's the next lesson.

GOALS