Fundamental theorem for line integrals

WHAT is it?

The key words to cloze:

$\mathbf{F}=\nabla f$ means $\mathbf{F}$ is conservative.
The integral depends only on the endpoints, not the path.
This is called path independence.

WHY is it true? (Derivation from scratch)

We derive it using only the chain rule and the ordinary FTC. No memorising.

Step 1 — Write the line integral as a 1D integral. By definition of a line integral over a parametrised curve, $\int_C \nabla f \cdot d\mathbf{r} = \int_a^b \nabla f(\mathbf{r}(t)) \cdot \mathbf{r}'(t)\, dt.$ Why this step? A line integral is "field dotted with the tiny displacement $d\mathbf{r}=\mathbf{r}'(t)\,dt$ ", summed along the curve. This just substitutes the parametrisation.

Step 2 — Recognise the multivariable chain rule. Let $g(t) = f(\mathbf{r}(t))$ , a plain function of one variable. The chain rule says $\frac{dg}{dt} = \nabla f(\mathbf{r}(t)) \cdot \mathbf{r}'(t).$ Why this step? As $t$ changes, $f$ changes through each coordinate: $\frac{d}{dt}f(x(t),y(t),z(t)) = f_x x' + f_y y' + f_z z'$ , which is exactly the dot product $\nabla f \cdot \mathbf{r}'$ . This is the heart of the proof — the integrand is secretly a perfect derivative.

Step 3 — Apply the ordinary FTC. $\int_a^b \frac{dg}{dt}\, dt = g(b) - g(a).$ Why this step? Once we've shown the integrand is $g'(t)$ , the single-variable FTC finishes the job.

Step 4 — Substitute back. $g(b)-g(a) = f(\mathbf{r}(b)) - f(\mathbf{r}(a)) = f(B)-f(A). \qquad\blacksquare$

Figure — Fundamental theorem for line integrals

HOW to use it (worked examples)

Worked example Example 2 — Find the potential first

Evaluate $\int_C \mathbf{F}\cdot d\mathbf{r}$ where $\mathbf{F}=(2xy,\,x^2+2y)$ , $C$ from $(0,0)$ to $(1,2)$ .

Step 1 — Test conservativeness. In 2D, $\mathbf{F}=(P,Q)$ is conservative (on a simply connected region) iff $\partial P/\partial y=\partial Q/\partial x$ . Here $P_y=2x$ , $Q_x=2x$ . ✓ Equal, so a potential exists. Why? If a potential $f$ exists, $P=f_x,Q=f_y$ , and mixed partials commute: $f_{xy}=f_{yx}$ . Step 2 — Build $f$ . From $f_x=2xy$ , integrate in $x$ : $f=x^2y+h(y)$ . Why? "Antidifferentiate" the $x$ -component; the constant of integration may depend on $y$ . Step 3 — Match the $y$ -component. $f_y=x^2+h'(y)$ must equal $Q=x^2+2y$ , so $h'(y)=2y\Rightarrow h(y)=y^2$ . Thus $f=x^2y+y^2$ . Step 4 — Apply FTLI. $f(1,2)=1\cdot2+4=6$ , $f(0,0)=0$ . Answer $=6-0=\boxed{6}$ .

Common mistakes

Recall Feynman: explain to a 12-year-old

Imagine hiking on a hill. Your "height" is the function $f$ . Walking around adds and subtracts elevation as you go up and down. The vector field $\nabla f$ is like the "steepest slope arrow" at every spot. The Fundamental Theorem says: no matter which winding trail you take from the bottom cabin to the top lodge, the total elevation gained is just (height of lodge) − (height of cabin). The squiggly path doesn't matter — only where you start and stop. And if you hike a loop back to your cabin, your net elevation change is exactly zero.

Active-recall flashcards

What does the Fundamental Theorem for Line Integrals state?

\mathbf{F}=\nabla f

with

\nabla f

continuous on a curve

C

from

A

B

, then

\int_C\nabla f\cdot d\mathbf{r}=f(B)-f(A)

What two ingredients does the proof of FTLI use?

The multivariable chain rule (

\frac{d}{dt}f(\mathbf{r}(t))=\nabla f\cdot\mathbf{r}'

) and the ordinary single-variable FTC.

A field whose line integral depends only on endpoints is called?

Conservative (it equals

\nabla f

for some potential

f

); the property is called path independence.

What is the line integral of a conservative field around any closed loop?

Zero, because the start and end points coincide so

f(B)-f(A)=0

2D test for whether

\mathbf{F}=(P,Q)

is conservative?

On a simply connected region,

\partial P/\partial y=\partial Q/\partial x

Why is

P_y=Q_x

alone not always enough?

The domain must be simply connected; on a region with a hole (e.g. the vortex field around the origin) a closed-loop integral can be nonzero.

How do you reconstruct the potential

f

from

\mathbf{F}=(P,Q)

Integrate

P

x

to get

f=\int P\,dx + h(y)

, then differentiate in

y

and match

f_y=Q

to find

h(y)

Connections

Line integrals of vector fields — the general case FTLI specialises.
Conservative vector fields — fields where FTLI applies.
Potential functions — the $f$ in $\nabla f$ .
Green's theorem — explains the simply-connected subtlety via circulation.
Fundamental Theorem of Calculus — the 1D ancestor.
Gradient — the operator $\nabla$ at the centre of it all.
Curl — vanishing curl is the 3D conservativeness test.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, normal line integral mein tumhe pura path follow karke integrate karna padta hai — bahut mehnat. Lekin agar field $\mathbf{F}$ kisi scalar function $f$ ka gradient hai, yani $\mathbf{F}=\nabla f$ (isko "conservative field" kehte hain), to jaadu ho jaata hai: integral sirf start aur end point pe depend karta hai, beech ka rasta bilkul matter nahi karta. Formula seedha: $\int_C \nabla f\cdot d\mathbf{r}=f(B)-f(A)$ . Bilkul wahi shape jaise single-variable wala FTC $\int_a^b g'(x)dx=g(b)-g(a)$ .

Yeh kyun hota hai? Proof mein bas do cheezein lagti hain. Pehle chain rule: $\frac{d}{dt}f(\mathbf{r}(t)) = \nabla f\cdot\mathbf{r}'(t)$ — yani integrand actually ek perfect derivative hai. Phir normal FTC laga do, aur bach jaate hain sirf endpoints. Imagine karo ek pahaadi pe trek — $f$ tumhari height hai. Chahe jitna tedha rasta lo, total height gain sirf (upar wali jagah ki height) minus (neeche wali jagah ki height) hoti hai. Aur agar loop maar ke wapas wahi aa gaye, net change = zero.

Practical tip: jab bhi $\mathbf{F}=(P,Q)$ diya ho, pehle check karo $P_y=Q_x$ . Agar equal hai (aur region mein koi hole nahi hai), to potential $f$ exist karta hai — $P$ ko $x$ mein integrate karke $f$ banao, phir $y$ -component match karo. Phir bas endpoints daal do. Ek bada saavdhani: agar $P_y\neq Q_x$ , to field conservative nahi hai aur tumhe sach mein pura path integrate karna padega — FTLI shortcut wahan kaam nahi karega. Iss ek check se 20 minute ka kaam 30 second mein ho jaata hai. Mantra: "Gradient In, Endpoints Out."

WHAT is it?

WHY is it true? (Derivation from scratch)

HOW to use it (worked examples)

Common mistakes

Active-recall flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Multivariable Calculus

Connections