Welcome to calculus. I'm professor Ghrist. We're about to begin lecture 17 on the indefinite integral. Welcome to chapter three on integration. In this chapter we're going to take what we've learned about differentiation and run it in reverse. In today's lesson we'll begin with an introduction to the indefinite integral and see its applications and motivations in differential equations. We begin with a definition. The indefinite integral of a function f of x is the anti-derivative of f. That means if you take the integral of f and differentiate it, plug that into the differentiation operator, then what you get back is f. You may think that this means that the integral is really the inverse of the differentiation operator, well that's almost true, but not quite. If you reverse the order of operations and first compute the derivative of f. Then what will you get when you integrate it? WIll you get f back again? Well not exactly. You may recall there is a constant of integration involved. And indeed, the indefinite integral of f is a class of functions, all equivalent up to a constant. It's fairly easy to write down a list of integrals that we can compute easily. The integral of x to the k, we know is x to the k +1 / k + 1, unless k equals negative one, in which case you get the log of x. Integrals of sines and cosines are easy enough, anti-differentiate and watch your minus signs. And wonderful to tell, the integral of e to the x is of course e to the x. Don't forget the constants. That much is simple, take the anti-derivative. On the other hand, there are some functions for which computing an antiderivative is highly nontrivial. Maybe with work you'll be able to compute an anti-derivative for log of x or secant of x. But no matter how hard you try, you are going to have an extremely difficult time finding an anti-derivative for something like e to the x squared. We're going to take some time a little bit later in this chapter to go over methods for computing these anti-derivatives. For the moment, and for the next couple of lectures, we're going to spend some time answering the question, why? Why do we care about the indefinite integral? One excellent answer to that question is to be found in the subject of differential equations, where a differential equation is simply an algebraic equation on a function x and it's derivatives. For example, the simplest differential equation is the form dx/dt = f(t) for some f. This equation has, as it's solution, x given by the indefinite integral of f with respect to t. Now what does that mean? How to we interpret it? Well one natural interpretation for the derivative is as the slope. So the differential equation is telling you something about the slope of the solution curve. It has to match this function f. Now of course, we can see the indefinite integral in here, we can see why there is a constant of integration. Because if I translate this solution curve up or down, I am not changing any of the slopes, and it still must be a solution. Thus, solutions to this simple differential equation come in the form of a family depending upon a constant. These are often called ODEs, which stand for ordinary differential equations. You will see some extraordinary differential equations later in your calculus career. For the moment, let's stick to a basic example and look at gravitation. As you know from your background in physics, when you drop an object, then it is acted on by gravity, which near the surface of the Earth, is modeled as a constant, g. Now, we can model the behavior of this falling object in terms of its height, x, its velocity, v, and its acceleration a. All of these are functions of time. Since acceleration is a constant we can write the simple differential equation dv/dt, that is acceleration, equals -g. Note we put the minus sign in because we orient things so that up is positive, down is negative. Likewise because we know the relationship between velocity and height, we can write a differential equation for x. dx/dt = v. Let's solve these one at a time both of these are simple differential equations. The first dv/dt equals minus g, has as it's solution v, as a function of time, is the indefinite integral of -g with respect to t. What does this have as it's solution? Well of course the anti-derivative of a constant is just that constant times t. And we have a + C as our integration constant. What does that + C really mean? It is really an initial velocity. Did we simply drop the object or did we toss it at some speed? Well in this case, if we plug in t equals 0, we see that v at time 0, otherwise known as v0, is equal to this integration constant. Now taking that solution for velocity and plugging that into the right hand side of our second differential equation, allows us to solve for the height, x, as the integral of v(t)dt. We already know what v(t) is, and so we compute the anti-derivative to get -1/2 g t squared + v0t + another constant of integration. What is that + C? Well, that is really an initial height, since at time 0, all of the other terms vanish except for the + C. We can write that as -1/2 gt squared + v0t + x0. This should be a familiar formula to you from your basic physics. Whether it is or not, it is transparent. Now why? Projectiles move in a parabolic shape under the influence of gravity. We have a negative one half G, T squared coming from the integral. So far so good these, simple differential equations have simple solutions in terms of integrals, but what happens when we look at the next simplest class namely dx/dt = f(x) instead of f(t)? How do you solve that for x as a function of time? Well, we're going to restrict to the simplest, the canonical differential equation of this form, one that is linear in x. This is a very important differential equation. It is dx / dt = ax, for a, a constant. We're going to think about this one over and over. How do we go about solving this equation? Well, the first thing that we're going to do is look at it and think dx/dt = ax. Let's simplify and say that a is equal to 1. We're looking for a function who's derivative is itself. You and I both know an excellent solution. That is, x(t) is e to the t. Now, what happens when we put the a back in there? Well, if we just think about it a little bit more, we see that putting an a up in the exponent gives us, as its derivative, e to the at times a. That is what we're looking for. We could, of course, put an arbitrary constant C out in front, and you may check that this is a solution to that differential equation. What is that constant? Well, we might, following our previous example, call it x0, the initial condition. What you get at time 0. This method of solution is sometimes called solution by ansatz, which is a fancy word for saying, we took a guess and it turned out to be correct. There is, of course, a more principled approach. One such approach is based on series, where we assume that our function, x, has a series expansion. c0 +c1t+c2t squared, etc. These constants, c sub k, are to us unknown. Now what are we going to do with this? Well the differential equation tells us something about the derivative of x with respect to t. We can differentiate the terms of this series very easily. What the differential equation tells us is that this is equal to a times x, which of course is ac0 + ac1 t + ac2 t squared, etc. Now, here is the important step. If we have two series that are equal, then their coefficients in front of the various terms must also be equal. And now we can work one step at a time. The constant terms equating them tell us that c1 = ac0. Well, that's good if we knew c0, we would know c1. What does the next equation that's of the first order of coefficients tell us? Well, it tells us that c2 is equal to one-half a times c1. But we already know what c1 is. c1 is ac0. So that tells us that c2 is actually one-half a squared times c0. What happens when we take the coefficients of the quadratic term and set them equal to each other? Well, 3 times c3 equals a times c2. That means c3 is one-third a. c2, using what we know about c2, we get 1 / 3 factorial times a cubed c0, I'm going to let you keep going with this, and see if you can find a pattern. That pattern through the method of induction can show that c sub k = 1 / k times a times ck-1 and that is one over k factorial a to the k times c0. That means when we look at our original series x(t), it's the sum over k of ck t to the k, that is c 0 times 1 / k factorial times a to the k times t to the k. Now I'll let you verify that that is really c0 times the exponential function e to the at. We obtain our solution, which we already knew was true, but from a more principled series approach. That, however, is a little complicated. And so, we are led to our last and best method of solution, that of integration. Beginning with x as a function of t, we differentiate using the chain rule to obtain dx = dx / dt dt. Now we we know something about dx dt since this is a solution to the deferential equation. It is a times x. And now we're going to use the method of separation. Which means we move all of the x terms over to one side of the equation. And we keep the terms over to the other side. This gives us dx/x = a dt. And now using the fact that the integral is an operator, we're going to apply it both sides of this equation, obtaining the integral of dx / x = the integral a dt. What is the integral of dx / x? Well, it's the anti-derivative, natural log of x. What is the integral of a constant times dt? It's simply a t plus an integration constant. Now to solve for x, we apply the exponentiation operator. e to the log of x is simply x, on the right hand side we get e to the at plus a constant splitting that up into a product. We obtain a new constant, we'll call that C as well, times e to the at. That is our solution, and it was obtained effortlessly. One of the things that you'll notice, is that e keeps coming up. It is not a coincidence that e to the t appears as the solution to the simple ODE dx/dt = x, or as the series, sum of t to the k over k factorial or as the indefinite integral as we just saw. In all three cases, you can interpret the constant of integration as an initial condition or the constant term in a series or the arbitrary constant of integration. And so the question of why the + C has many answers. And so we see that three of the main characters in our story cross paths. Integration, series, and solutions to differential equations. In our next three lessons we're going to focus on one of these characters. Solutions to differential equations and see how integration helps us to compute and understand them.