Your experience with the power of integrals has been expanding since you first learned about them. You started with evaluating the area under a curve and then quickly went to evaluating area between curves. Then you were able to calculate volumes of revolution. Double and triple integrals expanded your ability to solve all kinds of problems in two and three dimensions. Now you are about to embark on another integration adventure involving line integrals.
If you want complete lectures on this topic, here are some videos.
video by Prof Leonard 

video by Prof Leonard 

Basically, the idea is that you can evaluate an integral along a specific path. This allows you to solve all kinds of problems involving vector fields in physics and engineering. Before you jump into this technique, you need an important skill that is usually not covered very well as a separate topic. Many instructors just assume you can pick it up as you go. However, it does help to have a bit of explanation and some examples. This skill is the ability to parameterize a smooth line or curve, i.e. to be able to describe a line or curve using parametric equations. Make sure you understand the material on this other page before going on.
Line integrals and path integrals are two different things, even though some instructors don't separate the two ideas.
A path integral, also called a scalar line integral, is an integral along a certain path and applies to scalar functions. 
A line integral is also an integral along a certain path but we require the integral to be evaluated in a specific direction, i.e. we need to take into account the direction in which the curve is traced, and applies to vector fields. 
We start by discussing the more basic idea of path integrals and then discuss how line integrals are different.
Path Integrals 

Let's start off with a video. This is a great video which includes a complete introduction, with examples, and explanation of applications of path integrals. It is well worth your time to watch it.
video by Dr Chris Tisdell 

The equations from the last video are important for evaluating path integrals. Here is a summary.
requirements 
result  

continuous scalar function \(f(x,y,z)\) 
path integral of \(f\) over \( \mathcal{C} \)  
curve \( \mathcal{C} \) parameterized as \(\vec{c}(t)\) on an interval \(a \leq t \leq b\) 
\(\displaystyle{ \int_{\mathcal{C}}{ f(x,y,z)~ds} = }\) \(\displaystyle{ \int_{a}^{b}{ f(\vec{c}(t)) ~ \ \vec{c}'(t) \ ~dt } }\)  
continuous \(\vec{c}'(t)\) [derivative of \(\vec{c}(t)\) with respect to \(t\)] 
Most of the above equations should be familiar to you. However, one comment is in order about the term \(f(\vec{c}(t))\). How do you substitute a vector into a function?
If we write the vector as \( \vec{c}(t) = \langle X(t), Y(t), Z(t) \rangle = X(t) \hat{i} + Y(t) \hat{j} + Z(t)\hat{k} \) then we substitute the vector components into \(f(x,y,z)\) as \(x=X(t)\), \(y=Y(t)\) and \(z=Z(t)\). Another way to write this is
\(f(x,y,z) = f(X,Y,Z) = f(X(t), Y(t), Z(t))\). The result is a function of \(t\) only. All \(x\)'s, \(y\)'s and \(z\)'s will be gone, leaving only an integral in \(t\).
A very interesting comment he makes in the above video is that the direction of the curve does not affect the answer, i.e. integrating in the opposite direction will give the same answer. This is true because we use only the magnitude of the derivative of the parameterized curve, i.e. \(\ \vec{c}'(t) \\), which is the same regardless of the direction. However, the path integral is dependent on the chosen path, i.e. different paths will, in general, give different results. So path integrals are sensitive to the paths.
Okay, let's try some practice problems.
Unless otherwise instructed, evaluate the path integral over the curve \(\mathcal{C}\), giving your answers in exact terms.
Problem Statement 

Evaluate the path integral \( \int_{\mathcal{C}} x+y ~ds \)
\( \mathcal{C}: \) straight line from \((0,0)\) to \((1,1)\)
Solution 

video by Dr Chris Tisdell 

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Problem Statement 

Evaluate the path integral \( \int_{\mathcal{C}}{x+y+z ~ ds } \)
\( \mathcal{C}: \langle \cos(t), \sin(t), t \rangle; 0 \leq t \leq 2\pi \)
Solution 

video by Dr Chris Tisdell 

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Problem Statement 

Evaluate the path integral \( \int_{\mathcal{C}}{x+y+z ~ ds } \)
\( \mathcal{C}: \) straight line \( (1,0,0) \) to \( ( 1,0, 2\pi) \)
Solution 

video by Dr Chris Tisdell 

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Problem Statement 

Evaluate the path integral \( f(x,y,z) = x^2+y^21+z \)
\( \mathcal{C}: \langle \cos(t), \sin(t), t \rangle; 0 \leq t \leq 3\pi \)
Solution 

video by Dr Chris Tisdell 

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Problem Statement 

Evaluate the path integral \( \int_{\mathcal{C}}{ x+y^2 ~ds }\)
\( \mathcal{C}: \) 2 lines \((0,0)\)  \((1,1)\) and \((1,1)\)  \((1,0)\)
Solution 

video by Dr Chris Tisdell 

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Problem Statement 

Evaluate the path integral \( \int_{\mathcal{C}}{xy^4 ~ds } \)
\( \mathcal{C}: \) right half of the circle \( x^2+y^2 = 16 \)
Solution 

video by PatrickJMT 

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Problem Statement 

Evaluate \( \int_{\mathcal{C}}{ 2x ~ds } \) where \(\mathcal{C} = \mathcal{C_1} + \mathcal{C_2}\); \(\mathcal{C_1}\) goes from \((0,0)\) to \((1,1)\) along \(y=x^2\); \(\mathcal{C_2}\) goes from \((1,1)\) to \((1,2)\) along \(x=1\).
Final Answer 

Problem Statement 

Evaluate \( \int_{\mathcal{C}}{ 2x ~ds } \) where \(\mathcal{C} = \mathcal{C_1} + \mathcal{C_2}\); \(\mathcal{C_1}\) goes from \((0,0)\) to \((1,1)\) along \(y=x^2\); \(\mathcal{C_2}\) goes from \((1,1)\) to \((1,2)\) along \(x=1\).
Solution 

video by Michel vanBiezen 

Final Answer 

\((5\sqrt{5}+11)/6\) 
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Problem Statement 

Find \(\int_{\mathcal{C}}{x^2z~ds}\), where \(\mathcal{C}\) is the line segment from \((0,6,2)\) to \((1,8,7)\).
Hint 

There are an infinite number of ways to parameterize this curve. This instructor uses the equations
\(x=(1t)x_1+tx_2\)
\(y=(1t)y_1+ty_2\)
\(z=(1t)z_1+tz_2\)
\(0 \leq t \leq 1\).
Problem Statement 

Find \(\int_{\mathcal{C}}{x^2z~ds}\), where \(\mathcal{C}\) is the line segment from \((0,6,2)\) to \((1,8,7)\).
Hint 

There are an infinite number of ways to parameterize this curve. This instructor uses the equations
\(x=(1t)x_1+tx_2\)
\(y=(1t)y_1+ty_2\)
\(z=(1t)z_1+tz_2\)
\(0 \leq t \leq 1\).
Solution 

video by MIP4U 

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Line Integrals 

Here is an important introduction to line integrals that explains very well the difference between line integrals and path integrals. He also explains the notation and how to calculate them. Notation is critical at this point since line integrals can be written several different ways. So be especially aware of that section of this video.
video by Dr Chris Tisdell 

Similar to the path integrals video, this video gives the equations used to calculate the line integral, which are summarized below.
\(\displaystyle{ \int_{\mathcal{C}}{\vec{F} \cdot d\vec{s}} = \int_{a}^{b}{ \vec{F}(\vec{c}(t)) \cdot \vec{c}'(t) ~dt } }\) 
This line integral form is called the vector form. Make sure you see the difference between this equation for the line integral and the equation given earlier for the path integral.
Line Integrals Using The Unit Tangent Vector 

For a parameterized vector function \(\vec{c}(t)\), it's unit tangent vector is \(\displaystyle{ \hat{T} = \frac{\vec{c}'(t)}{\ \vec{c}'(t) \} }\). Since we are dividing by \(\ \vec{c}'(t) \\), we require that \(\vec{c}'(t) \neq \vec{0}\). Notice that we are dividing the vector by it's length, giving us a unit vector, i.e. \(\ \hat{T} \ = 1\).
Since the parameterized vector function \(\vec{c}(t)\) sweeps out the curve in a certain direction, the unit tangent vector will be the tangent vector in the forward direction, i.e. in the same direction that the curve is being swept out.
The unit tangent vector comes into the equation for the line integral given above. Here are the intermediate equations.
\( \int_{\mathcal{C}}{\vec{F} \cdot d\vec{s}} \) 
\( \int_{\mathcal{C}}{\vec{F} \cdot \hat{T} ds} \) 
\( \int_{a}^{b}{ \vec{F}(\vec{c}(t)) \cdot \hat{T}(\vec{c}(t)) \ \vec{c}'(t) \~dt } \) 
\( \int_{a}^{b}{ \vec{F}(\vec{c}(t)) \cdot \displaystyle{ \frac{\vec{c}'(t)}{\ \vec{c}'(t) \} } \ \vec{c}'(t) \~dt } \) 
\( \int_{a}^{b}{ \vec{F}(\vec{c}(t)) \cdot \vec{c}'(t) ~dt } \) 
These equations came from the last video above. We highly recommend that you download the notes for this topic from
Dr Chris Tisdell's site. The video also describes the geometric meaning of the unit tangent vector. If you haven't watched it yet, now would be a good time.
You will probably find in your class, textbook and some of these videos, the boundary between path integrals and line integrals is difficult to see. So don't be too worried. The techniques are very similar. You just need to get some practice with the notation.
Problem Statement 

Evaluate the line integral \( \vec{F} = \langle y/(x^2+y^2), x/(x^2+y^2) \rangle \)
\( \mathcal{C}: \langle \cos(t), \sin(t) \rangle; 0 \leq t \leq 2\pi \)
Solution 

video by Dr Chris Tisdell 

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Problem Statement 

Evaluate the line integral \(\displaystyle{ \vec{F} = zx\hat{i} + zy\hat{j} + x\hat{k} }\)
\( \mathcal{C}: \langle \cos(t),\sin(t),t \rangle; 0 \leq t \leq \pi \)
Solution 

video by MIT OCW 

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Problem Statement 

Determine the work done by the conservative vector field \(\displaystyle{ \vec{F} = yz\hat{i} + xz\hat{j} + xy\hat{k} }\) along any smooth curve from \(A(0,0,0)\) to \(B(1,1,1)\).
Solution 

video by Dr Chris Tisdell 

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Problem Statement 

Evaluate the line integral \( \vec{F} = \langle x^3, z, 2xy \rangle \)
\( \mathcal{C}: \langle t^2, \sqrt{t}, \sqrt{t} \rangle; 2 \leq t \leq 4 \)
Solution 

video by Dr Chris Tisdell 

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Problem Statement 

Consider the force field \(\vec{F}(x,y) = \langle y^2x^3, x^5 \rangle\) Newtons, where \(x\) and \(y\) are in meters. Let \(C_1\) be the oriented curve from \((0,0)\) to \((1,1)\), along the graph of \(y=x\). Let \(C_2\) be the oriented curve along \((0,0)\) to \((1,1)\), along the graph of \(y=x^2\). Calculate the work done by \(\vec{F}\) along \(C_1\) and \(C_2\) and show that they are not equal.
Solution 

He has several mistakes when writing this problem.
1. He says the curves start at \((0,1)\) but they actually start at \((0,0)\). He works the problem as if they start at \((0,0)\).
2. His second integral should be \(\int{ F(\hat{r}(t)) \cdot \hat{r}â€™(t)~dt }\)
Here is the link to the blog entry for this problem: blog entry
video by World Wide Center of Mathematics 

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Problem Statement 

Calculate the work done by the vector field \( \vec{F}(x,y) = xy\hat{i} + 3y^2\hat{j} \) along the path \(x=11t^4\), \(y=t^3\) by evaluating the integral \( W = \int_{\mathcal{C}}{ \vec{F} \cdot d\vec{r} } \) on \( 0 \leq t \leq 1 \).
Final Answer 

Problem Statement 

Calculate the work done by the vector field \( \vec{F}(x,y) = xy\hat{i} + 3y^2\hat{j} \) along the path \(x=11t^4\), \(y=t^3\) by evaluating the integral \( W = \int_{\mathcal{C}}{ \vec{F} \cdot d\vec{r} } \) on \( 0 \leq t \leq 1 \).
Solution 

video by Michel vanBiezen 

Final Answer 

45 
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Problem Statement 

The force exerted by an electric charge at the origin on a charged particle at the point \((x,y,z)\) with position vector \(\vec{r} = \langle x,y,z \rangle\) is \(\displaystyle{ \vec{F}(\vec{r}) = \frac{K\vec{r}}{\vec{r}^3}}\), where K is constant. Assume \(K=10\). Find the work done as the particle moves along a straight line from \((2,0,0)\) to \((2,2,3)\).
Solution 

video by MIP4U 

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For these next 3 problems, notice that the vector function \(\vec{F}\) and the beginning and ending points are the same in all 3 problems. However, the line integrals are along 3 different paths.
Problem Statement 

Calculate the work done by the vector field \(\vec{F}(x,y)=(x+y)\hat{i} + (yx)\hat{j}\) along the path \(x=y^2\) from \((1,1)\) to \((4,2)\) by evaluating the integral \(W=\int_{\mathcal{C}}{ \vec{F} \cdot d\vec{r} }\).
Solution 

video by Michel vanBiezen 

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Problem Statement 

Calculate the work done by the vector field \(\vec{F}(x,y)=(x+y)\hat{i} + (yx)\hat{j}\) along the path \(x=2t^2+t+1\), \(y=1+t^2\) from \((1,1)\) to \((4,2)\) by evaluating the integral \(W=\int_{\mathcal{C}}{ \vec{F} \cdot d\vec{r} }\).
Solution 

video by Michel vanBiezen 

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Problem Statement 

Calculate the work done by the vector field \( \vec{F}(x,y) = (x+y)\hat{i} + (yx)\hat{j} \) along \(y=1\) from \((1,1)\) to \((4,1)\) then along \(x=4\) from \((4,1)\) to \((4,2)\) by evaluating the integral \( W = \int_{\mathcal{C}}{ \vec{F} \cdot d\vec{r} } \).
Solution 

video by Michel vanBiezen 

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Line Integrals in Differential Form 

Another way to write line integrals of vector fields is in differential form.
If we are given a vector field \(\vec{F}(x,y,z) = M(x,y,z)\hat{i} + N(x,y,z)\hat{j} + P(x,y,z)\hat{k} \) and we need to find the line integral of this vector field over the curve \( \vec{r}(t) = x(t)\hat{i} + y(t)\hat{j} + z(t)\hat{k} \), we can do the following, where \( \vec{r}'(t) = x'(t)\hat{i} + y'(t)\hat{j} + z'(t)\hat{k} \).
\(\displaystyle{ \int_{\mathcal{C}}{ \vec{F} \cdot d\vec{r}} }\) 
\(\displaystyle{ \int_{\mathcal{C}}{ \vec{F} \cdot \frac{d\vec{r}} {dt}~dt } }\) 
\(\displaystyle{ \int_{a}^{b}{ \left( M\hat{i} + N\hat{j} + P\hat{k} \right) \cdot \vec{r}'(t) ~dt } }\) 
\(\displaystyle{ \int_{a}^{b}{ \left[ M\frac{dx}{dt} + N\frac{dy}{dt} + P\frac{dz}{dt} \right] ~dt } }\) 
\(\displaystyle{ \int_{\mathcal{C}}{ M~dx + N~dy + P~dz } }\) 
Some general comments are in order.
1. The above equations relate to vector fields in (3dim) space. For vector fields in the plane, we just drop the \(dz\) term to get \(\displaystyle{ \int_{\mathcal{C}}{ \vec{F} \cdot d\vec{r}} = \int_{\mathcal{C}}{ M~dx + N~dy } }\).
2. The final equation that we use to calculate the line integral does not seem to involve a vector field. However, you can see that M, N and P are the components of the vector field \(\vec{F}\). So keep in mind that this is a line integral of a vector field (not a path integral of a scalar function).
Problem Statement 

Evaluate the line integral \( \int_{\mathcal{C}}{ xy~dx+y~dy } \)
\( \mathcal{C}: \vec{r}(t)=2t\vec{i} +10t\vec{j}; 0 \leq t \leq 1 \)
Final Answer 

Problem Statement 

Evaluate the line integral \( \int_{\mathcal{C}}{ xy~dx+y~dy } \)
\( \mathcal{C}: \vec{r}(t)=2t\vec{i} +10t\vec{j}; 0 \leq t \leq 1 \)
Solution 

\(x=2t\)  \(y=10t\)  
\(dx=2~dt\)  \(dy=10~dt\) 
\(\displaystyle{ \int_{\mathcal{C}}{ xy~dx+y~dy } }\) 
\(\displaystyle{ \int_{0}^{1}{ (2t)(10t)2~dt + 10t(10~dt) } }\) 
\(\displaystyle{ \int_{0}^{1}{ 40t^2 + 100t ~ dt } }\) 
\(\displaystyle{ \left[ \frac{40t^3}{3} + 100\frac{t^2}{2} \right]_{0}^{1} }\) 
\(\displaystyle{ \frac{40}{3} +50 = \frac{190}{3} }\) 
Final Answer 

\(\displaystyle{\frac{190}{3}}\) 
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Problem Statement 

Evaluate the line integral \( \int_{\mathcal{C}}{ (2xy)~dx+(x+3y)~dy } \)
\( \mathcal{C}: \) along the xaxis from 0 to 5
Final Answer 

Problem Statement 

Evaluate the line integral \( \int_{\mathcal{C}}{ (2xy)~dx+(x+3y)~dy } \)
\( \mathcal{C}: \) along the xaxis from 0 to 5
Solution 

To parameterize the curve, we let \(x=t\) which gives us the interval \(0\leq t\leq 5\). The yvalue is just zero since the line is along the xaxis. So we have \(\vec{r}(t)=t\vec{i}+0\vec{j}\)
\(x=t\)  \(y=0\)  
\(dx=dt\)  \(dy=0~dt\) 
\(\displaystyle{ \int_{\mathcal{C}}{ (2xy)~dx+(x+3y)~dy } }\) 
\(\displaystyle{ \int_{0}^{5}{ 2t~dt } }\) 
\(\displaystyle{ \left[ \frac{2t^2}{2} \right]_{0}^{5} = 25 }\) 
Final Answer 

25 
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Problem Statement 

Evaluate the line integral \( \int_{\mathcal{C}}{ 9x~dy9y~dx }\)
\( \mathcal{C}: \langle 3\cos(t), 2\sin(t) \rangle; 0 \leq t \leq \pi \)
Solution 

video by Dr Chris Tisdell 

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Problem Statement 

Evaluate the line integral \( \int_{\mathcal{C}}{(2x+yz)dx +xz dy + xy dz} \)
\( \mathcal{C}: \) straight line \( (0,0,0) \) to \( (1,2,1) \)
Solution 

video by PatrickJMT 

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Problem Statement 

Compute \(\displaystyle{ \int_{(1,1)}^{(4,2)}{ (x+y)dx+(yx)dy } }\) along the curve \(x=y^2\).
Solution 

video by Dr Chris Tisdell 

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Fundamental Theorem of Line Integrals 

Similar to the Fundamental Theorem of Calculus we have the Fundamental Theorem of Line Integrals. Basically, this says, that for a conservative vector field, the line integral is independent of the path and depends only on the endpoints.
Here is a great video that explains this theorem in detail and includes examples.
video by Dr Chris Tisdell 

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