 The most important application of calculus is, well, actually there's quite a lot to choose from, but one of the earliest uses of calculus was to solve problems in physics. A key concept in physics is the idea of work. Work is the product of force and distance. Now it's important to remember we should always include unit, so let's talk a little bit about work. The measure of force is known as the Newton, named after Isaac something. I forget the last name. Meanwhile the measure of work is the, well, there's some debate over exactly how to pronounce it. The standard accepted pronunciation is jewel. However, like the Newton, it is a person's name and there's some evidence that the person called himself James Prescott Jowell. For example, suppose it requires 15 Newtons of force to push a box across a floor. How much work is done moving the box 10 meters? So remember work is the product of force and distance, and so the amount of work done will be the force, 50 Newtons, times the distance, 10 meters, and remember units act like algebraic variables, so when we multiply these two together, we can multiply the coefficients 50 and 10 to get 500, and the units multiply to get Newton meters. And note that since we're measuring force in Newton and distance in meters, this Newton meter is the same as a jewel. If the force or distance varies, we can find the work done by summing the work done over a very small interval over which we can treat the force as essentially constant. And so one important idea to keep in mind is that if an amount varies, don't substitute a value until the calculus is done. For example, suppose the gravitational force in a spacecraft is given by 5000 divided by r squared Newtons, where r is the distance from the object in meters. How much work is required to move the spacecraft from a distance of 1000 meters to a distance of 2000 meters? So again, work is the product of force and distance, but because the force varies, we have to find the amount of work done over a small distance and sum the result. So suppose we're at some distance r, and we move out a small bit, dr. The amount of work we do is going to be the force 5000 over r squared times the distance dr. And now we want to sum this work done, and since our differential variable is r, we want that sum to go from 1000 meters to 2000 meters. So our limits of integration will be from r equals 1000 to r equals 2000. And so this is a nice simple definite integral, and we can evaluate it. And if force is measured in Newtons and distance in meters, then work will be measured in joules. So here our force is measured in Newtons, our distance is measured in meters, and so our final answer has units of joules. Or how about another example? A spring resist stretching by a force equal to 100 times x minus 0.2 Newtons, where x is the current length of the spring in meters. So how much work is required to stretch the spring from a resting length of 0.2 meters to a length of 0.5 meters? So again, since the force is a function of the length, we can find the work by summing the force multiplied by a small distance dx. So let's suppose we stretch the spring to a length of x. If we stretch the spring a little bit more, that's distance dx, so work is force times distance. If the spring's length is x and the force is 100 times x minus 0.2, and we stretch the spring a distance dx, sum that over our distance from 0.2 meters to 0.5 meters. And we can evaluate that integral. And again, since our units are Newtons and meters, the work is measured in joules.