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The Fundamental Theorem of Line Integrals
Definition If $C$ is a smooth curve given by the vector function $\vec r(t)$, $a\leq t\leq b$ and $f$ is a differentiable function whose gradient vector $\vec\nabla f$ is continuous on $C$. Then- $$\int_C \vec\nabla f \cdot \mathrm{d}\vec r=f(\vec r(b))-f(\vec r(a))$$^[1] This is also known as "Gradient Theorem"^[3] 1
Motivation This is a beautiful theorem which allows us to calculate the value of line integral without even doing the integral under some specific conditions. By this theorem line integral value along any curve over a vector field(not every) depends only on the endpoints of the curve viz. the line of integration. This indirectly implies that we can encode the information of a vector field (vectors associated with each point in space) into a scalar field(values associated with points in space) which is indeed spectacular^[3]. Example of such fields are gravitational field, electromagnetic field. 2
Bird's Eye View In the the beginning of integral calculus one may find Fundamental Theorem of Calculus written as : If $f^\prime(x)$ is derivative of $f(x)$ continuous on $[a,b]$ then - $$\int_a^b f^\prime (x)\, \mathrm{d}x= f(b)-f(a)$$ The extension of that theorem to multiple variables is the current theorem if we think $\vec\nabla f$ as some sort of derivative of the multivariable function $f$.^[1] Although $f(x,y,z)$ is a scalar function $\vec\nabla f(x,y,z)$ is a vector valued function, \begin{align}\vec\nabla f(x,y,z) &=\frac{\partial f}{\partial x}\hat i+\frac{\partial f}{\partial y}\hat j+\frac{\partial f}{\partial y}\hat k\\ \implies\vec F(x,y,z) &=F_x(x,y,z)\hat i +F_y(x,y,z)\hat j+F_z(x,y,z)\hat k \end{align} For good illustration in 3-dimension function of two variables are used. 3
3
Animation 1: Gradient of a scalar function $f(x,y)=xy$ is a vector function $\vec F=y\hat i+x\hat j$.
So, we can restate the theorem as- Line integral in a vector field $\vec F$ can be evaluated as the difference of $f$-value at the final and initial points of the path of integration $\vec r(t)$. $$\int_C \vec F \cdot \mathrm{d}\vec r=f(\vec r(b))-f(\vec r(a))$$ Once we have found the proper $f$ for the field $\vec F$ the line integrals become this easy in that field. $enlightened$ Remarks: We can calculate the value of the line integral in a vector field ($\vec F$) by this theorem if and only if, that vector field can be represented as the gradient of some scalar function ($f$). Such kind of vector fields are called conservative vector fields. 3
3
Animation 2: Line integral is independent of path in the conservative field.
Context of the Definition We are going to see the proof of the theorem first, The parametric equation of the curve along which we want to do the line integral is smoothly defined as $\vec r(t)= x(t)\hat i+y(t)\hat j+z(t)\hat k,\, (a\leq t\leq b)$. and, $$\vec F(x,y,z)=\vec\nabla f(x,y,z)=\dfrac{\partial f}{\partial x}\hat i+\dfrac{\partial f}{\partial y}\hat j+\dfrac{\partial f}{\partial z}\hat k$$ Now the line integral, \begin{align} \int_C \vec F(x,y,z)\cdot \mathrm{d}\vec r &=\int_C \dfrac{\partial f}{\partial x}dx+\dfrac{\partial f}{\partial y}dy+\dfrac{\partial f}{\partial z}dz\qquad & \\ &=\int_a^b \left(\dfrac{\partial f}{\partial x}\dfrac{dx} {dt}+\dfrac{\partial f}{\partial y}\dfrac{dy}{dt}+\dfrac{\partial f}{\partial z}\dfrac{dz}{dt}\right)\, dt \\ \text{using the chain rule..}\\ &=\int_a^b \dfrac{d}{dt} f(\vec r(t))\, dt\\ \text{from Fundamental Theorem of Calculus.. }\\ &=f(\vec r(b))-f(\vec r(a)) \end{align} The proof is based on the fact that the vector field $\vec F$ can be represented as a gradient($\vec\nabla$) of some scalar function $f$. Now the question arises, how do we know that a vector field has this property(called conservativeness) and how to get that scalar function? So that we can use this theorem to calculate line integrals easily. 4
$yes$ Test the field for conservativenss If two points are fixed in the vector field and line integral along any path between the two points gives the same answer, the field has conservative nature. Example: A Vector field is given as $\vec F(x,y) =y\hat i+x\hat j$ and two points are $(0,0), (1,1)$, Find $\int_{(0,0)}^{(1,1)} \vec F \cdot d\vec r$ along two different paths. 1. integral along the path $(0,0)\rightarrow (0,1)\rightarrow (1,1)$ \begin{align} & \int_{(0,0)}^{(0,1)} \vec F(0,y) \cdot (0\hat i+dy\,\hat j)+ \int_{(0,1)}^{(1,1)} \vec F(x,1) \cdot (dx\,\hat i+0\hat j) \\ &=\int_0^1 0\,dy + \int_0^1 1\,dx\\ &= 0 + 1 =1 \end{align} 2. integral along the path $(0,0)\rightarrow (1,0)\rightarrow (1,1)$ \begin{align} & \int_{(0,0)}^{(1,0)} \vec F(x,0) \cdot dx\,\hat i+ \int_{(1,0)}^{(1,1)} \vec F(1,y) \cdot dy\,\hat j \\ &=\int_0^1 0\,dx + \int_0^1 1\,dy\\ &= 0 + 1 =1 \end{align} The two answer is same indicating a conservative field. Furthermore we can deduce another theorem : If $C$ is a simple close curve and $\oint_C\vec F \cdot d\vec r = 0$ anywhere in the field $\vec F$ then it is a conservative field and there is a scalar potential $f$ associated with it. 4
5
Animation 3: Line Integration of the previous example
The most consistent way to prove a field is conservative is to see if the curl^[5] of the field is $0$. $${\mathop{\rm curl}\nolimits} \vec F = \vec\nabla \times \vec F = \left\| {\begin{array}{{20}{c}}{\vec i}&{\vec j}&{\vec k}\\{ \dfrac{\partial }{{\partial x}}}&{ \dfrac{\partial }{{\partial y}}}&{ \dfrac{\partial }{{\partial z}}}\\y&x&0\end{array}} \right\|=0$$ In the previous example observing the plot of the vector field ( in animation 3 ) you can also guess this answer. $yes$ Find the scalar potential function if the field is conservative. An example will be best for it rather than saying too much. Example :* Show that $\vec F =(2xy+z^3)\hat i+ x^2\hat j +3xz^2\hat k$ is conservative and find its scalar potential $f$.^[6] $$ \vec\nabla \times \vec F = \left\| {\begin{array}{*{20}{c}}{\vec i}&{\vec j}&{\vec k}\\{ \dfrac{\partial }{{\partial x}}}&{ \dfrac{\partial }{{\partial y}}}&{ \dfrac{\partial }{{\partial z}}}\\2xy+z^3 & x^2 & 3xz^2 \end{array}} \right\|=0$$ so $\vec F$ is conservative. now we have to find $f$ such that, $\vec\nabla f =\vec F$ or ,$$\dfrac{\partial f}{\partial x}\hat i+\dfrac{\partial f}{\partial y}\hat j+\dfrac{\partial f}{\partial y}\hat k=(2xy+z^3)\hat i+ x^2\hat j +3xz^2\hat k$$ The three equations - (1) $\dfrac{\partial f}{\partial x}=2xy+z^3$ ,(2) $\dfrac{\partial f}{\partial y}=x^2$, (3) $\dfrac{\partial f}{\partial z}=3xz^2$ Integrating (1),(2),(3) respectively , \begin{align} &f &= &x^2y &+xz^2 &+g(y,z) \\ &f &= &x^2y & &+g(x,z) \\ &f &= & &+xz^3 &+g(x,y) \end{align} if we choose $f(y,z) = 0, g(x,z) = xz^3, h(x,y) = x^2y$ we get , $f=x^2y + xz^3 + k$, where $k$ is any arbitrary constant. this is the scalar potiential of the given field $\vec F$ and now we can use this "Fundamental theorem of line integral" for calculating any line integral in this field. 6
Applications We have already seen that if we are doing a line integral in a vector field, we can use this theorem to reduce a significant amount of work by introducing the scalar(potential) function associated with it. Then every integral in that field become merely finding values at the endpoints and subtract them. You have heard about gravitational potential, it is the scalar function related to the gravitational field (a force-vector field). You have used this theorem maybe without knowing it. Like if you are asked what amount of work is needed to lift a $15$ kg load of bricks to a roof of $10$ meter height against the gravitational field, you instantly calculate it using the $W=mgh$ formula, $15\times 9.8\times 10=1470\, Joule$ even without the information in which path you are going to do it i.e. using a pully or along the stairway. Generally, if you want to calculate the work done in a force field you need to do a line integral along the path from the initial point to the endpoint. But as gravitational field is a conservative field you are easily doing it using the difference of potential between the two points(note that $mgh$ is an approximated formula). Similar applications are found in electromagnetism to calculate work done on charges. 11
History During the decade 1660–1670 Isaac Barrow, James Gregory, and Isaac Newton independently developed the fundamental theorem of calculus. Further development of mathematics in need of describing certain physical theories in a concise way lead to the formulation of Fundamental Theorem for Line Integrals. 11
Pause and Ponder This theorem is a really a "calculation saver"; from the next time when you are going to do a really involving line integral in a vector field, check if it is conservative viz. the $(\vec\nabla \times)$ is $0$, then the rest will 'boring' (obviously not the potential finding part !). The conservation formulas are the result of this potential formulation of the vector fields and that's why these fields are called conservative. 12
References James Stewart. Calculus : Concepts and Contexts, chapter Vector Calculus.The Fundamental Theorem for Line Integrals, page 925. Brooks/Cole, Cengage Learning, 4th edition. Howard Anton, Irl bivens, and Stephen Davis. Calculus: Early Transcendentals, chapter 15.Topics in vector calculus, page 1112. 10th edition Weisstein, Eric W. "Gradient Theorem." From MathWorld--A Wolfram Web Resource. https://mathworld.wolfram.com/GradientTheorem.html https://mathinsight.org/gradient_theorem_line_integrals Weisstein, Eric W. "Curl." From MathWorld--A Wolfram Web Resource. https://mathworld.wolfram.com/Curl.html Murray R. Spiegel. In Vector Analysis and an introduction to Tensor Analysis, Schaum’s Outline, chapter 5. Vector Integration, page 91. McGraw-Hill,1959. Further Reading 1. For more detailed study: James Stewart. Calculus : Concepts and Contexts, chapter Vector Calculus.The Fundamental Theorem for Line Integrals, Brooks/Cole, Cengage Learning, 4th edition. Howard Anton, Irl bivens, and Stephen Davis. Calculus Early Transcendentals, chapter 15.Topics in vector calculus, page 1111. 10th edition. George B.Thomas and Ross L. Finney. Calculus and Analytic Geometry,chapter 14.3 'Path Independence,potenial Funcions, and conservative Fields'. Addison-Wesley, 9th edition, 1998. 2. A large collection of problem and solutions: Murray R. Spiegel. In Vector Analysis and an introduction to Tensor Analysis, Schaum’s Outline, chapter 5. Vector Integration. Line Integrals. McGraw-Hill,1959. 3. Theorems of vector calculus : https://www.math.ucla.edu/~josephbreen/The_Theorems_of_Vector_Calculus.pdf 13

The Fundamental Theorem of Line Integrals

Definition

If $C$ is a smooth curve given by the vector function $\vec r(t)$, $a\leq t\leq b$ and $f$ is a differentiable function whose gradient vector $\vec\nabla f$ is continuous on $C$. Then-
$$\int_C \vec\nabla f \cdot \mathrm{d}\vec r=f(\vec r(b))-f(\vec r(a))$$^[1]

This is also known as "Gradient Theorem"^[3]

Motivation

This is a beautiful theorem which allows us to calculate the value of line integral without even doing the integral under some specific conditions. By this theorem line integral value along any curve over a vector field(not every) depends only on the endpoints of the curve viz. the line of integration. This indirectly implies that we can encode the information of a vector field (vectors associated with each point in space) into a scalar field(values associated with points in space) which is indeed spectacular^[3]. Example of such fields are gravitational field, electromagnetic field.

Bird's Eye View

In the the beginning of integral calculus one may find Fundamental Theorem of Calculus written as : If $f^\prime(x)$ is derivative of $f(x)$ continuous on $[a,b]$ then -
$$\int_a^b f^\prime (x)\, \mathrm{d}x= f(b)-f(a)$$

The extension of that theorem to multiple variables is the current theorem if we think $\vec\nabla f$ as some sort of derivative of the multivariable function $f$.^[1]

Although $f(x,y,z)$ is a scalar function $\vec\nabla f(x,y,z)$ is a vector valued function,
\begin{align}\vec\nabla f(x,y,z) &=\frac{\partial f}{\partial x}\hat i+\frac{\partial f}{\partial y}\hat j+\frac{\partial f}{\partial y}\hat k\\
\implies\vec F(x,y,z) &=F_x(x,y,z)\hat i +F_y(x,y,z)\hat j+F_z(x,y,z)\hat k
\end{align}

For good illustration in 3-dimension function of two variables are used.

Animation 1: Gradient of a scalar function $f(x,y)=xy$ is a vector function $\vec F=y\hat i+x\hat j$.

So, we can restate the theorem as- Line integral in a vector field $\vec F$ can be evaluated as the difference of $f$-value at the final and initial points of the path of integration $\vec r(t)$.
$$\int_C \vec F \cdot \mathrm{d}\vec r=f(\vec r(b))-f(\vec r(a))$$
Once we have found the proper $f$ for the field $\vec F$ the line integrals become this easy in that field.

$enlightened$ Remarks: We can calculate the value of the line integral in a vector field ($\vec F$) by this theorem if and only if, that vector field can be represented as the gradient of some scalar function ($f$). Such kind of vector fields are called conservative vector fields.

Animation 2: Line integral is independent of path in the conservative field.

Context of the Definition

We are going to see the proof of the theorem first,

The parametric equation of the curve along which we want to do the line integral is smoothly defined as $\vec r(t)= x(t)\hat i+y(t)\hat j+z(t)\hat k,\, (a\leq t\leq b)$. and, $$\vec F(x,y,z)=\vec\nabla f(x,y,z)=\dfrac{\partial f}{\partial x}\hat i+\dfrac{\partial f}{\partial y}\hat j+\dfrac{\partial f}{\partial z}\hat k$$
Now the line integral,

\begin{align}
\int_C \vec F(x,y,z)\cdot \mathrm{d}\vec r &=\int_C \dfrac{\partial f}{\partial x}dx+\dfrac{\partial f}{\partial y}dy+\dfrac{\partial f}{\partial z}dz\qquad & \\
&=\int_a^b \left(\dfrac{\partial f}{\partial x}\dfrac{dx} {dt}+\dfrac{\partial f}{\partial y}\dfrac{dy}{dt}+\dfrac{\partial f}{\partial z}\dfrac{dz}{dt}\right)\, dt \\
\text{using the chain rule..}\\
&=\int_a^b \dfrac{d}{dt} f(\vec r(t))\, dt\\
\text{from Fundamental Theorem of Calculus.. }\\
&=f(\vec r(b))-f(\vec r(a))
\end{align}

The proof is based on the fact that the vector field $\vec F$ can be represented as a gradient($\vec\nabla$) of some scalar function $f$.
Now the question arises, how do we know that a vector field has this property(called conservativeness) and how to get that scalar function? So that we can use this theorem to calculate line integrals easily.

$yes$ Test the field for conservativenss

If two points are fixed in the vector field and line integral along any path between the two points gives the same answer, the field has conservative nature.

Example:
A Vector field is given as $\vec F(x,y) =y\hat i+x\hat j$ and two points are $(0,0), (1,1)$, Find $\int_{(0,0)}^{(1,1)} \vec F \cdot d\vec r$ along two different paths.

1. integral along the path $(0,0)\rightarrow (0,1)\rightarrow (1,1)$
\begin{align}
& \int_{(0,0)}^{(0,1)} \vec F(0,y) \cdot (0\hat i+dy\,\hat j)+ \int_{(0,1)}^{(1,1)} \vec F(x,1) \cdot (dx\,\hat i+0\hat j) \\
&=\int_0^1 0\,dy + \int_0^1 1\,dx\\
&= 0 + 1 =1
\end{align}

2. integral along the path $(0,0)\rightarrow (1,0)\rightarrow (1,1)$
\begin{align}
& \int_{(0,0)}^{(1,0)} \vec F(x,0) \cdot dx\,\hat i+ \int_{(1,0)}^{(1,1)} \vec F(1,y) \cdot dy\,\hat j \\
&=\int_0^1 0\,dx + \int_0^1 1\,dy\\
&= 0 + 1 =1
\end{align}

The two answer is same indicating a conservative field.

Furthermore we can deduce another theorem : If $C$ is a simple close curve and $\oint_C\vec F \cdot d\vec r = 0$ anywhere in the field $\vec F$ then it is a conservative field and there is a scalar potential $f$ associated with it.

Animation 3: Line Integration of the previous example

The most consistent way to prove a field is conservative is to see if the curl^[5] of the field is $0$.
$${\mathop{\rm curl}\nolimits} \vec F = \vec\nabla \times \vec F = \left| {\begin{array}{*{20}{c}}{\vec i}&{\vec j}&{\vec k}\\{ \dfrac{\partial }{{\partial x}}}&{ \dfrac{\partial }{{\partial y}}}&{ \dfrac{\partial }{{\partial z}}}\\y&x&0\end{array}} \right|=0$$

In the previous example observing the plot of the vector field ( in animation 3 ) you can also guess this answer.

$yes$ Find the scalar potential function if the field is conservative.

An example will be best for it rather than saying too much.

Example : Show that $\vec F =(2xy+z^3)\hat i+ x^2\hat j +3xz^2\hat k$ is conservative and find its scalar potential $f$.^[6]

$$ \vec\nabla \times \vec F = \left| {\begin{array}{*{20}{c}}{\vec i}&{\vec j}&{\vec k}\\{ \dfrac{\partial }{{\partial x}}}&{ \dfrac{\partial }{{\partial y}}}&{ \dfrac{\partial }{{\partial z}}}\\2xy+z^3 & x^2 & 3xz^2 \end{array}} \right|=0$$

so $\vec F$ is conservative.

now we have to find $f$ such that, $\vec\nabla f =\vec F$ or ,$$\dfrac{\partial f}{\partial x}\hat i+\dfrac{\partial f}{\partial y}\hat j+\dfrac{\partial f}{\partial y}\hat k=(2xy+z^3)\hat i+ x^2\hat j +3xz^2\hat k$$
The three equations - (1) $\dfrac{\partial f}{\partial x}=2xy+z^3$ ,(2) $\dfrac{\partial f}{\partial y}=x^2$, (3) $\dfrac{\partial f}{\partial z}=3xz^2$

Integrating (1),(2),(3) respectively ,
\begin{align}
&f &= &x^2y &+xz^2 &+g(y,z) \\
&f &= &x^2y & &+g(x,z) \\
&f &= & &+xz^3 &+g(x,y)
\end{align}

if we choose $f(y,z) = 0, g(x,z) = xz^3, h(x,y) = x^2y$ we get , $f=x^2y + xz^3 + k$, where $k$ is any arbitrary constant.
this is the scalar potiential of the given field $\vec F$ and now we can use this "Fundamental theorem of line integral" for calculating any line integral in this field.

Applications

We have already seen that if we are doing a line integral in a vector field, we can use this theorem to reduce a significant amount of work by introducing the scalar(potential) function associated with it. Then every integral in that field become merely finding values at the endpoints and subtract them.

You have heard about gravitational potential, it is the scalar function related to the gravitational field (a force-vector field). You have used this theorem maybe without knowing it.

Like if you are asked what amount of work is needed to lift a $15$ kg load of bricks to a roof of $10$ meter height against the gravitational field, you instantly calculate it using the $W=mgh$ formula, $15\times 9.8\times 10=1470\, Joule$ even without the information in which path you are going to do it i.e. using a pully or along the stairway.

Generally, if you want to calculate the work done in a force field you need to do a line integral along the path from the initial point to the endpoint. But as gravitational field is a conservative field you are easily doing it using the difference of potential between the two points(note that $mgh$ is an approximated formula).

Similar applications are found in electromagnetism to calculate work done on charges.

History

During the decade 1660–1670 Isaac Barrow, James Gregory, and Isaac Newton independently developed the fundamental theorem of calculus. Further development of mathematics in need of describing certain physical theories in a concise way lead to the formulation of Fundamental Theorem for Line Integrals.

Pause and Ponder

This theorem is a really a "calculation saver"; from the next time when you are going to do a really involving line integral in a vector field, check if it is conservative viz. the $(\vec\nabla \times)$ is $0$, then the rest will 'boring' (obviously not the potential finding part !). The conservation formulas are the result of this potential formulation of the vector fields and that's why these fields are called conservative.

References

James Stewart. Calculus : Concepts and Contexts, chapter Vector Calculus.The Fundamental Theorem for Line Integrals, page 925. Brooks/Cole, Cengage Learning, 4th edition.
Howard Anton, Irl bivens, and Stephen Davis. Calculus: Early Transcendentals, chapter 15.Topics in vector calculus, page 1112. 10th edition
Weisstein, Eric W. "Gradient Theorem." From MathWorld--A Wolfram Web Resource. https://mathworld.wolfram.com/GradientTheorem.html
https://mathinsight.org/gradient_theorem_line_integrals
Weisstein, Eric W. "Curl." From MathWorld--A Wolfram Web Resource. https://mathworld.wolfram.com/Curl.html
Murray R. Spiegel. In Vector Analysis and an introduction to Tensor Analysis, Schaum’s Outline, chapter 5. Vector Integration, page 91. McGraw-Hill,1959.