Roger's Equations

Integrals are Summations

The gravitational force between two masses is described by the equation:

What if there is more than two masses? What if there were many masses named m₁, m₂, m₃, m₄,... m_n and we wanted to know the force at m₁? What equation do we use then? Luckily for us, we can take the vector sum of forces due to the pairs of masses, add them up and we get the total force on m_1. So we would find the force due to the pairs m₁ and m₂ (F₁), m₁ and m₃ (F₂), m₁ and m₄ (F₃), etc., and add them together like so:

In the equation above, the right hand side can be represented by the symbol:

which means "the sum of the forces" Σ, the greek letter "sigma", being the mathematical symbol for summation. So the net force is simply the sum of the individual forces.

So what happens if instead of masses, we have a continuous line of mass instead? We can still take the same approach but we quickly run into a serious problem, take a look at the image below:

Lets say we want to find the gravitational force at P due to a line of mass of length L. If we take the approach above and try to express the total force as a summation of forces from point masses on the line, we run into a serious problem, there are an infinite number of point masses. That would imply an infinite net force (if you add them up), which is clearly not correct so obviously this approach won't work. We need a new approach that can sum the forces due to an infinite number of points on a continuous line of length L. The answer is the definite integral:

.

So what is a definite integral? A definite integral is the sum of an infinite number of infinitely small continuous "elements" over a specified range. In our line example above, dx would be a tiny (infinitely small (1/∞)) line segment with mass of λdx (lamda is the mass/distance), f(x) describes how each infinitely small point mass (dm=λdx) effects P (the gravitational equation from above describes this) and a and b would be the length of the continuous line of point masses we are summing over, in this case L₁ to L₂. Notice the integral sign:

Looks like a stretched out S. This is no accident, since the stretched out S is meant to represent a continuous sum, just as the greek ∑ is meant to represent a discrete sum.

Area Under The Curve

The value of an integral is often described as "the area under the curve". The curve in this expression is the function f(x) being integrated. I said above that the Integral is a summation, so why should a summation be equal to the area under a curve? To answer this, lets take a look at a simple function.

For a simple function, say f(x)=3, we would would have the following values:

f(x)=3
f(1)=3
f(2)=3
f(3)=3
f(4)=3
f(5)=3
etc.

Graphically we can represent this with a straight line like the one below:

So lets say we're interested in summing f(x) over the range x=0 to x=5 above. We see that the value of f(x) = 3 for all values of x, so the sum is 3 + 3 + 3 + 3 + 3 = 15. So "area under the curve" means the same thing as summing since the "height" of the function f(x) is simply it's value at a particular point.

Unfortunately, most functions f(x) aren't so easy to add up since they are varying continuously. Take a look at this example:

The curve in the example above is changing continuously, so for any given x we will have a slightly different f(x). This makes adding the values of the function up very difficult, even for the smallest of ranges. So we use a clever trick. Since we can add up discrete values, like in our f(x)=3 function above, why don't we just chop up the curve into discrete values that we can add up? The more we chop it up, the closer we'll get to an accurate sum. So the most accurate sum would be if we chopped our function up an infinite number of times, which would require infintely small cuts. You can see in the example above, using limits as a way of getting to "infinitely small cuts" that's exactly what Integrals do.

An Example

Take a look at the example below. The relation between Electric Potential and charge is:

V = k Q / r

where V is the Electric Potential, k is 1/4πε₀ (the electrostatic constant), Q is charge, and r is distance from the charge. That's for a point charge, but what if you have a line charge like below?

The trick is to divide the line into little "charge elements" dq and add them up from -a to b. We can express dq = λ dx. The lamda (λ) is simply charge per length, which, when multiplied by a infinitely small length (dx) will give an infinitely small charge (dq). By summing up all the contributions of those small charges we can get the total potential due to the line of charges.

Special Thanks to Hyperphysics for a bunch of the formulas used above. Also thanks as always to Wikipedia.