A three-part examination of the Fundamental Theorem of Line Integrals. Learn how to use this theorem in multivariable calculus. Simplify the process of solving line integrals using the FTLI.

1. 1
Concepts and Applications of the
Fundamental Theorem of Line Integrals
Samson Axelrod, Jacob Braginsky, Pranav Dwarakanath, Param Thakkar
May 7, 2023

2. The Fundamental Theorem of Line Integrals
Section 1.1
Introduction to the Fundamental Theorem of Line Integrals
Background Information: Gradient Fields and Path Independence
Understanding gradient fields is essential in order to apply the Fundamental Theorem of Line Integrals.
Consider , a continuous vector field that satisfies = ∇ . This scalar-valued function is known as the
𝐹
→
𝐹
→
𝑓 𝑓
potential function of . This property classifies as a conservative vector field, or a gradient field. Any
𝐹
→
𝐹
→
gradient field defined on an open, connected domain will produce path-independent line integrals. This means
that
d = d
𝐶1
∫𝐹
→
· 𝑠
→
𝐶2
∫𝐹
→
· 𝑠
→
for any two piecewise-smooth curves C1 and C2 with the same start and end points. It is important that these
curves are defined on the domain of .
𝐹
→
Defining the Fundamental Theorem of Line Integrals
Let and = ∇ , where is continuous on a connected, open region in . On any
𝑓: ℝ
𝑛
→ ℝ 𝐹
→
𝑓 𝐹
→
ℝ
𝑛
piecewise-smooth curve from A to B,
𝐶
.
𝐶
∫ 𝐹
→
· 𝑑𝑠
→
= 𝑓(𝐵
→
) − 𝑓(𝐴
→
)

3. 3
Proving the Fundamental Theorem of Line Integrals
Now we will prove the Fundamental Theorem of Line Integrals. Let and . We can use the
𝐴 = 𝑥
→
(α) 𝐵 = 𝑥
→
(β)
chain rule to obtain,
𝐶
∫ 𝐹
→
· 𝑑𝑠
→
=
α
β
∫ ∇𝑓(𝑥
→
(𝑡)) · 𝑥
→
'(𝑡)𝑑𝑡
=
α
β
∫
𝑑
𝑑𝑡
[𝑓(𝑥
→
(𝑡))]𝑑𝑡
= 𝑓(𝑥
→
(β)) − 𝑓(𝑥
→
(α))
= 𝑓(𝐵
→
) − 𝑓(𝐴
→
).
This shows that the line integral only depends on the endpoints, making it path-independent. Path independence
is a corollary of the FTLI.
Section 1.2
Finding the Potential Function
Option 1 - The “Safe Method”
We begin by using the “Safe Method.” Define = for some and . Our goal is
𝐹
→ ∂𝑓
∂𝑥
,
∂𝑓
∂𝑦
( ) 𝐹
→
: ℝ
2
→ ℝ
2
𝑓: ℝ
2
→ ℝ
to find the function such that ∇ = . Take the antiderivative of with respect to .
𝑓 𝑓 𝐹
→ ∂𝑓
∂𝑥
𝑥
∫
∂𝑓
∂𝑥
𝑑𝑥 = 𝑓(𝑥, 𝑦) = 𝑔(𝑥, 𝑦) + 𝐶(𝑦)

4. 4
In the equation above, is a function of and . This “integration function” is the most important
𝐶 𝑦, 𝑔: ℝ
2
→ ℝ
part of this process and can not be overlooked. Seeing that this is a function of y, if we took the derivative of
+ with respect to , would disappear. Let’s move on to the next step. Take the derivative of
𝑔 𝐶(𝑦) 𝑥 𝐶(𝑦)
+ with respect to . We can cross reference it with the component from the vector field to obtain
𝑔 𝐶(𝑦) 𝑦
∂𝑓
∂𝑦
= = .
∂
∂𝑦
(𝑔 + 𝐶(𝑦)) ⇒ 𝑔𝑦
+ 𝐶'(𝑦)
∂𝑓
∂𝑦
⇒ 𝐶'(𝑦)
∂𝑓
∂𝑦
− 𝑔𝑦
Now, we take the antiderivative of with respect to , obtaining
𝐶'(𝑦) 𝑦
.
∫𝐶'(𝑦) 𝑑𝑦 = 𝐶(𝑦)
Recall back to the beginning of our process when we took the antiderivative of with respect to and got
∂𝑓
∂𝑥
𝑥
+ . We just discovered what is, so we’ve filled in the missing piece. At this point, we have
𝑔 𝐶(𝑦) = 𝑓 𝐶(𝑦)
fully assembled the potential function. Here is an example with a defined vector field .
𝐹
→
= =
𝐹
→ ∂𝑓
∂𝑥
,
∂𝑓
∂𝑦
( ) 𝑦 cos(𝑥) + 3𝑥
2
+ 1, sin(𝑥) + 4𝑒
4𝑦
( )
Integrate with respect to .
∂𝑓
∂𝑥
𝑥
=
∫
∂𝑓
∂𝑥
𝑑𝑥 ∫(𝑦 cos(𝑥) + 3𝑥
2
+ 1) 𝑑𝑥
= 𝑦 sin(𝑥) + 𝑥
3
+ 𝑥 + 𝐶(𝑦)
= 𝑔(𝑥, 𝑦)
Differentiate with respect to .
𝑔(𝑥, 𝑦) 𝑦
=
∂
∂𝑦
(𝑔(𝑥, 𝑦))
∂
∂𝑦
(𝑦 sin(𝑥) + 𝑥
3
+ 𝑥 + 𝐶(𝑦))

5. 5
= sin(𝑥) + 𝐶'(𝑦)
Compare the result above to .
∂𝑓
∂𝑦
=
sin(𝑥) + 𝐶'(𝑦)
∂𝑓
∂𝑦
= +
sin(𝑥) + 𝐶'(𝑦) sin(𝑥) 4𝑒
4𝑦
=
𝐶'(𝑦) 4𝑒
4𝑦
Integrate with respect to to find .
𝐶'(𝑦) = 4𝑒
4𝑦
𝑦 𝐶(𝑦)
∫4𝑒
4𝑦
𝑑𝑦 = 𝑒
4𝑦
= 𝐶(𝑦)
Put together the scalar potential function.
𝑓 = 𝑦 sin(𝑥) + 𝑥
3
+ 𝑥 + 𝑒
4𝑦
Option 2 - The “Dangerous Method”
We can also use the “Dangerous Method” to find the potential function. We’ll explain this method using an
example vector field. Define .
𝐹
→
(𝑥, 𝑦) = 2 cos(2𝑥) cos(𝑦) − 1 + 𝑦𝑒
𝑥𝑦
, − sin(2𝑥) sin(𝑦) + 𝑥𝑒
𝑥𝑦
+ 3𝑦
2
( )
First, we take the antiderivative of with respect to and of with respect to .
∂𝑓
∂𝑥
𝑥
∂𝑓
∂𝑦
𝑦
∫(2 cos(2𝑥) cos(𝑦) − 1 + 𝑦𝑒
𝑥𝑦
) 𝑑𝑥 = sin(2𝑥) cos(𝑦) − 𝑥 + 𝑒
𝑥𝑦
+ 𝐶(𝑦)
∫(− sin(2𝑥) sin(𝑦) + 𝑥𝑒
𝑥𝑦
+ 3𝑦
2
)𝑑𝑦 = sin(2𝑥) cos(𝑦) + 𝑒
𝑥𝑦
+ 𝑦
3
+ 𝐶(𝑥)
Now, we compare the two antiderivatives to see what they have in common. We do this to synthesize a “guess”
as to what the scalar potential function might be by writing down function components that both
𝑓
antiderivatives exhibited, followed by each unique component seen in either antiderivative. In this case, both

6. 6
antiderivatives had and with unique components being - and . Combing these gives
𝑠𝑖𝑛(2𝑥)𝑐𝑜𝑠(𝑦) 𝑒
𝑥𝑦
𝑥 𝑦
3
us
.
𝑓 = sin(2𝑥) cos(𝑦) + 𝑒
𝑥𝑦
− 𝑥 + 𝑦
3
Next, we verify we got the correct function by confirming = .
𝑓 ∇𝑓 𝐹
→
∇ = , )
𝑓 (
∂
∂𝑥
(sin(2𝑥) cos(𝑦) + 𝑒
𝑥𝑦
− 𝑥 + 𝑦
3
)
∂
∂𝑦
(sin(2𝑥) cos(𝑦) + 𝑒
𝑥𝑦
− 𝑥 + 𝑦
3
)
= ( )
2 cos(2𝑥) cos(𝑦) + 𝑦𝑒
𝑥𝑦
− 1, − sin(2𝑥) sin(𝑦) + 𝑥𝑒
𝑥𝑦
+ 3𝑦
2
= 𝐹
→
Section 1.3
FTLI Examples
A “Mundane” Example
Suppose we have the vector field = We wish to compute where C is any piecewise smooth
𝐹
→
(4𝑥
3
, 2).
𝐶
∫𝐹
→
·𝑑𝑠
→
,
curve from the point (0,0) to (3,4). The first step for us is to integrate with respect to . We solve to obtain
4𝑥
3
𝑥
=
∫4𝑥
3
𝑑𝑥 𝑥
4
+ 𝐶(𝑦).
The next step is to integrate 2 with respect to so we then obtain
𝑦,
=
∫ 2 𝑑𝑦 2𝑦 + 𝐶(𝑥).

7. 7
Combining our two results, we see that We can double check our answer by computing the
𝑓(𝑥, 𝑦) = 𝑥
4
+ 2𝑦.
gradient of .
𝑓
∇𝑓 = (
∂𝑓
∂𝑥
,
∂𝑓
∂𝑦
) = (4𝑥
3
, 2)
Now that we have checked our function we are ready to apply the Fundamental Theorem of Line Integrals
𝑓,
and compute our integral.
𝐶
∫𝐹
→
·𝑑𝑠
→
= 𝑓(3, 4) − 𝑓(0, 0)
= 3
4
+ 2 · 4 − 0
4
− 0 · 8
= 81 + 8
= 89
For any paths from (0,0) to (3,4) to which the FTLI is applicable, for , the line integral will always be 89
𝐹
→
because of path-independence.
To further our understanding, let's compute an integral using the definition of line integral over a path to show
that the FTLI works and is indeed equal to what we computed. Let our first curve be where
𝑥(𝑡) = (3𝑡, 4𝑡),
Plugging everything in,
𝑡 ϵ [0, 1].
0
1
∫ (4 · (3𝑡)
3
, 2) · (3, 4) 𝑑𝑡 =
0
1
∫ 324𝑡
3
+ 8 𝑑𝑡
= 81 (1)
3
+ 8 − 81 (0)
3
+ 8 · 0
= 81 + 8
.
= 89

8. 8
Thus, the Fundamental of Line Integrals indeed works as both of our computations were equal, showing that we
may use this shortcut.
Solving the Circulation Integral with FTLI
Define a gradient field on an open, connected space in , with a scalar-valued potential function . Consider
𝐹
→
ℝ
𝑛
𝑓
C, a closed curve with a counterclockwise parameterization defined on the same space. Using the Fundamental
Theorem of Line Integrals, we can prove that d = 0. Points A and B lie on curve C, dividing it into two
𝐶
∮𝐹
→
· 𝑠
→
open curves. We label the curve traveling from A to B as C1, and the curve traveling from B to A as C2, as shown
in Fig. 1.
Since both C1 and C2 are open paths with initial and terminal points, we can apply the Fundamental Theorem of
Line Integrals to calculate the following line integrals:

9. 9
d = (B) - (A)
𝐶1
∫ 𝐹
→
· 𝑠
→
𝑓 𝑓
d = (A) - (B)
𝐶2
∫ 𝐹
→
· 𝑠
→
𝑓 𝑓
Our objective is to calculate the circulation integral over C. Seeing that C1 and C2 combined form C, we
accomplish our objective by finding the sum of the line integrals over C1 and C2. Thus, we obtain
d = d + d
𝐶
∮ 𝐹
→
· 𝑠
→
𝐶1
∫𝐹
→
· 𝑠
→
𝐶2
∫𝐹
→
· 𝑠
→
= (B) - (A) + (A) - (B)
𝑓 𝑓 𝑓 𝑓
= 0.
The relationship between the circulation integral and the FTLI can be conceptualized using path independence.
Since we know that the line integral over a single point is zero, we can deduce that the line integral over a curve
with the same initial and terminal point is also zero, regardless of the curve’s length. Thus, d = 0 for any
𝐶
∮𝐹
→
· 𝑠
→
closed curve C.