 One application of the integral is to find the center of mass of an extended object. So to begin with suppose you have several point masses. These are masses that only have location. The center of mass is the average distance from the origin. That's the sum of the products of the masses times their distances divided by the total mass. Now even though we're actually summing up masses, remember that in physics weight is proportional to mass. And so what we're finding is very literally the weighted average of the distances from the origin. Now for future reference this formula has two parts. We'll call the numerator this sum of the products of the masses and their distances the moment. And if we need to distinguish these distances and we will, we'll specify the axis. So this is mx. Meanwhile the denominator, the sum of the individual masses, is well, that's just the total mass. Which we can call m. For example suppose you want to find the center of mass of a system with a mass of 5 kilograms. Located 3 meters to the right of the origin and a 2 kilogram mass located 2 meters to the left of the origin. So we have our formula for the center of mass and a useful reminder from physics. Units are your best friends. It's very important to keep track of the units of our numbers and of our final answer. So we want to add up the products of the units and their distance from the origin. So a 5 kilogram mass 3 meters to the right of the origin, that's 5 kilograms times 3 meters. A 2 kilogram mass 2 meters to the left of the origin, well that's 2 kilograms and our distance is going to be negative 2 meters because we're to the left of the origin. Meanwhile the masses themselves are 5 kilograms and 2 kilograms. So the thing to remember is that units are just like algebraic variables. So in the denominator, this is like 5x plus 2x or 7x, 7 kilograms. In the numerator, this 5 kilograms times 3 meters, we could multiply the numbers to get 15 and the units to get kilogram meters. Similarly 2 kilograms times minus 2 meters, that's minus 4 kilogram meters. Our numerator units are the same so we can combine like terms and get 11 kilogram meters and remember this dash when we're talking units really represents a multiplication. So here we have a common factor of kilograms, which we can remove giving us our final answer and remember this is giving us the location of the center of mass, 11 sevenths of a meter and since this is positive, it's going to be to the right of the origin. Now that's a discrete case. If our mass is continuous, we can find the center of mass by using integration and that's because every sum is an integral. Now we have our formula for center of mass. We have to sum the products of the masses and their distances and we also have to sum the masses. The second sum isn't a problem. This is just the integral of our differential of mass. The first sum is a little more complicated. Do we find the integral of m dx or the integral of x dm? To decide this, note that in the discrete case we're summing the products of the actual distance xi with the actual mass mi. So m dx would be the entire mass at a small distance from the origin. Meanwhile, x dm would be a small part of the mass at its distance x from the origin. And since we want to sum the masses times their actual distance from the origin, we want to find the integral of x dm. And so this sum is the integral of x dm and the center of mass is the integral of x dm divided by the integral of dm. And an important thing you should keep in mind every time you get a formula in mathematics or physics or chemistry or biology or anything else really is don't memorize formulas, understand concepts. In this case, what we're doing is we're finding the sum of the individual moments and we're dividing by the sum of the masses. So for example, a three meter long bar has a density of rho of x equal x kilograms per meter. Let's find the center of mass. So remember, units are your best friend. And we want to find the differential of mass dm. And remember, the differential of a variable has the same units as the variable. So this differential of mass should have units of kilograms. Now the only thing in sight that actually includes a unit of kilograms is this density, rho of x. So let's start with rho of x and that's in kilograms per meter. And note that if we want to get units of kilograms, we need to multiply by meters. We need to multiply by some distance. Well, it seems that x should be the distance. The problem is that because our density does change depending on what x is, we need to multiply by a small bit of x dx over which our density is essentially constant. So rho of x dx will have units of kilograms and we can exchange rho of x for x. And so that says x dx will have units of kilograms. And since we want to find the differential of mass, which should also be measured in kilograms, this suggests that dm is equal to x dx. So remember that the moment about x is the weighted sum of the mass dm and its distance x from the origin. dm is x dx. And so this integral becomes. Now we do need to find the limits of integration. So remember, the differential variable is controlling. So x is going to go from zero to three because the bar is three meters in length. And we compute. Now we don't always have to check the units on our work. In fact, the only time you should check the units on your work is when the correct answer matters. Well, actually, I guess that's always so we should always check the units on our work. So the important thing here is to recognize that x dx actually has units of kilograms. That's because it's our differential of mass. And so this summation x times x dx, well, x is measured in meters. x dx is measured in kilograms. And so the units are meters times kilograms or a kilogram meter. We'll also sum to find the actual mass. So again, our differential of mass is x dx and the differential variable is controlling. x goes from zero to three. And so our actual mass will be. And remember, the differential of a variable has the same units as the variable. And so dm, the differential of mass, has units of kilograms. So our final answer here, nine halves, is also in kilograms. And finally, we have the moment of x. We have the mass, and so our center of mass is the quotient. And again, kilograms is a common factor, so we can remove that and get our final answer to meters, which will be the location of the center of mass.