Welcome to Calculus, I'm Professor Greist. We're about to begin Lecture 17 on the Indefinite Integral. Welcome to Chapter 3 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 differentiated, 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 very easy to right down a list of integrals that we can compute easily. The integral of x to the k we know is x to the k plus 1 over k plus 1, unless k equals negative 1. In which case, you got 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 anti-derivative is highly non-trivial. 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 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 of 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 its derivatives. For example, the simplest differential equation is of the form dx dt equals f of t, for some f. This equation has as its solution facts given by the indefinite integral of f with respect to t. Now, what does that mean? How do 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 extra ordinary 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 minus g. Now, 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 equals 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 its solution, v as a function of time is the indefinite integral of negative g, with respect to t. What does this have as its solution? Well, of course, the anti-derivative of a constant. It's just that constant times T, and we have a plus C as our integration constant. What does that plus 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 v not, 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 of t dt. We already know what v of t is, and so we compute the anti-derivative to get negative 1 half t squared plus v not t plus not t plus another constant of integration. What is that plus C? Well, that is really an initial height, since that time zero all of the other terms vanish except for the plus C. Or, we can write that as negative 1 half g t square plus v not t plus x not. This should be a familiar formula to you from your basic physics. Whether is or not, it is transparent, now why? Projectiles move in a parabolic shape under the influence of gravity. We have a negative 1 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 equals f of x instead of f of 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 the dx dt equals 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 equals a times x. Let's simplify and say that a equals 1. We're looking for a function whose derivative is itself. You and I both know an excellent solution, that is x of 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 a t 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 x not, 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 plus c1t plus c2 t squared, etcetera. 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 a 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 a times c0, plus a times c1t, plus a times c2 t squared, etcetera. Now, here is the important step. If we have two series that are equal than their coefficient 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 is equal to a times c0. Well that's good. If we knew c0 we would know c1. What does the next equation, that of the first order coefficients tell us? Well, it tells us that c2 is equal to 1 half a times c1. But we already know what c1 is, c1 is a times c0, so that tells us that c2 is actually 1 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 1 3rd a c2. Using what we know about c2, we get 1 over 3 factorial. Times a cubed, c not. 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 is 1 over k times a times c sub k minus 1. And that is, 1 over k factorial, a to the k times C not. That means, when we look at our original series, x of t, it's the sum over k of ck, t to the k, that is c0 times 1 over k factorial times a to the k times t to the k. Now,, I'll let you verify that, that is really c not times the exponential function e to at. We obtain our solution, which we already knew was true, but from a more principled, serious 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 is dx dt times dt. Now, we notice something about dx dt since this is a solution to the differential 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, t terms, over to the other side. This gives us dx over x equals adt. And now, using the fact that the integral is an operator, we're going to apply it to both sides of this equation. Obtaining the integral of dx over x equals the integral adt. What is the integral of dx over x? Well, it's the anti-derivative. Natural log of x. What is the integral of a constant times dt? It's simply at 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. And 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 equals x. Or, as the series, sum of t to the k, or k factorial, or is 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. So, the question of why the plus 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.