Fundamental theorem for line integrals: Visual walkthrough

We assume you know nothing except: what an arrow (a vector) is, and what it means to add up tiny pieces. Everything else — dot product, gradient, "line integral" — is rebuilt from a picture below.

Step 1 — What is a "field" and a "path"?

WHAT. A vector field $F$ is a rule that plants a little arrow at every point of the plane. A path (curve) $C$ is a walking route from a start point $A$ to an end point $B$ .

WHY. Before we can integrate a field along a path, we need both objects clearly in front of us. Everything later is just "how much does the field push along the direction I walk?"

PICTURE. Below, the pale arrows are the field $F$ (one at each grid point). The dark curve is our route $C$ , threading from $A$ (bottom-left) to $B$ (top-right).

Step 2 — Describe the walk with one clock variable $t$

WHAT. We label our position along the curve by a single number $t$ ("time on a clock"), running from $t = a$ at the start to $t = b$ at the end. The position at clock-reading $t$ is written $r (t)$ — a point that depends on $t$ .

$r (t) = (across x (t), up y (t)), start r (a) = A, end r (b) = B .$

WHY. A curve wanders in two dimensions, which is hard to integrate directly. But if a single clock variable $t$ drives it, then everything becomes a function of one variable — and one-variable calculus is a solved problem. This is the trick that lets the ordinary Fundamental Theorem of Calculus eventually finish the job.

PICTURE. The same curve, now with clock-readings $t = a, t_{1}, t_{2}, \dots, t = b$ ticked along it. Each tick is a point $r (t)$ .

Step 3 — The tiny step $d r$ and the velocity arrow $r^{'} (t)$

WHAT. Over a tiny clock-interval $d t$ , your position moves by a tiny displacement arrow $d r$ . That arrow points along the curve, and its length is your speed times $d t$ :

$d r = velocity arrow r^{'} (t) tiny time d t .$

Here $r^{'} (t)$ is the velocity: the direction you're heading and how fast, exactly tangent to the curve.

WHY. A line integral asks "how much does the field help me along my direction of travel?" To answer that we need the direction of travel written as an arrow — that arrow is $r^{'} (t)$ . Multiplying by $d t$ shrinks it to the actual tiny hop we take.

PICTURE. A zoom-in on one tick: the green tangent arrow $r^{'} (t)$ , and the tiny hop $d r$ lying along the curve.

Step 4 — Assemble the line integral

WHAT. At each point we take the field arrow $F (r (t))$ , dot it with our tiny step, and add up all the pieces:

$\int_{C} F \cdot d r = \int_{a}^{b} field where I am F (r (t)) \cdot my direction r^{'} (t) tiny hop d t .$

The wiggly $\int_{C}$ became a plain $\int_{a}^{b}$ over the clock — a genuine ordinary integral.

WHY. Each term $F \cdot r^{'} d t$ is the little bit of "push along the path" gained on one hop. Summing them (that's what $\int$ does) totals the field's help over the whole trip. This is the general line integral — true for any field, conservative or not.

PICTURE. Two field arrows on the curve: one nearly aligned with the step (big positive contribution, coral), one nearly perpendicular (tiny contribution, faded). The dot product weights each hop by alignment.

Step 5 — The special case: the field is a gradient

WHAT. Now suppose the field is not arbitrary but is the gradient of some scalar "height" function $f$ :

$F = \nabla f = (slope in x f_{x}, slope in y f_{y}) .$

Think of $f (x, y)$ as the height of a hill above each point. The gradient arrow $\nabla f$ at any spot points straight uphill, and is longer where the hill is steeper.

WHY. This is the one extra assumption that makes the whole thing collapse. A field built this way is called conservative, and $f$ is its potential. We need $f$ to exist because the next step secretly rebuilds $f$ 's rate of change.

PICTURE. A hill (shaded contours = height $f$ ). At three points, the lavender $\nabla f$ arrows point uphill, perpendicular to the contour lines, longer on steeper ground.

Step 6 — The heart: the integrand is a perfect derivative

WHAT. Define $g (t) = f (r (t))$ : your height as you walk, a plain one-variable function of the clock. The multivariable chain rule says its rate of change is exactly our integrand:

$\frac{d g}{d t} = height change from moving in x f_{x} x^{'} (t) + height change from moving in y f_{y} y^{'} (t) = the very integrand from Step 4 \nabla f (r (t)) \cdot r^{'} (t) .$

WHY. As the clock ticks, $f$ changes through both coordinates at once. Each coordinate contributes (its slope)(how fast that coordinate moves) — and adding those two contributions is precisely a dot product $\nabla f \cdot r^{'}$ . So the thing we are integrating is not just any function: it is $g^{'} (t)$ , the derivative of your height. That is the entire secret.

PICTURE. Left: the hill with the path drawn on it. Right: the height profile $g (t) = f (r (t))$ plotted against the clock $t$ — a simple 1-D curve whose slope at each $t$ is our integrand.

Step 7 — Finish with the ordinary FTC

WHAT. We are now integrating a plain derivative over a plain interval, so the single-variable Fundamental Theorem of Calculus applies:

$\int_{a}^{b} g^{'} (t) \frac{d g}{d t} d t = height at end g (b) - height at start g (a) .$

Substituting $g (t) = f (r (t))$ and $r (a) = A, r (b) = B$ :

$\int_{C} \nabla f \cdot d r = f (B) - f (A) ■$

WHY. The FTC says "the total accumulated change of a derivative equals the net change of the original." Since our integrand is $g^{'}$ , the total pushed-along-the-path equals just (end height) − (start height). Every intermediate wiggle cancelled.

PICTURE. The height profile $g (t)$ : the shaded area under $g^{'} (t)$ equals the vertical drop from $g (a)$ to $g (b)$ — only the two endpoints' heights survive.

Step 8 — Every case: two paths, and the closed loop

WHAT. Because the answer is $f (B) - f (A)$ and contains no trace of the route:

Two different paths $C_{1}, C_{2}$ from $A$ to $B$ give the same value — path independence.
A closed loop ( $A = B$ ) gives $f (A) - f (A) = 0$ .
Degenerate path (stay put, $A = B$ , zero-length curve) also gives $0$ — consistent, no special case needed.

WHY. These aren't new theorems; they're the single formula read in different situations. Cover them all so no scenario surprises you. (Warning kept from the parent: this needs $f$ to actually exist everywhere on the region — see the vortex counterexample there, tied to Green's theorem and the curl test.)

PICTURE. Two coloured trails (lavender, coral) from $A$ to $B$ over the same hill — same height difference. Below, a closed loop returning to $A$ : net climb $= 0$ .

The one-picture summary

Everything on one canvas: the field arrows, the wiggly path, the hidden height-hill, and the collapse to two endpoint heights.

Recall Feynman: tell the whole walk in plain words

Picture a hill; your height is $f$ . At every spot there's an arrow pointing straight uphill — that's the gradient field $\nabla f$ . Now walk any trail from a low cabin $A$ to a high lodge $B$ . On each tiny step you ask, "how much did I climb?" — that's the field arrow dotted with your step. Add up all those tiny climbs. But here's the magic: each tiny climb is exactly the change in your height on that step. So adding them all is just totalling up the height changes — which telescopes into (final height) − (starting height). The messy middle cancels completely. Take a straight trail, take a loop-the-loop trail: same answer, because the hill doesn't care how you got there. And walk a full loop back home? You end at the same height you left — net climb zero.

Connections

Fundamental theorem for line integrals — the parent result these pictures derive.
Line integrals of vector fields — the general integral of Step 4 before the gradient trick.
Gradient — the uphill-arrow operator of Step 5.
Potential functions — the height function $f$ .
Conservative vector fields — the fields where this collapse happens.
Fundamental Theorem of Calculus — the 1-D engine in Step 7.
Green's theorem · Curl — govern when a valid $f$ exists at all.