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We begin with a definition. The indefinite integral of a function, f

Â of x is. The anti-derivative of f, that means if

Â you take the integral of f, and differentiated, plug that into the

Â differentiation operator, then what you get back is f.

Â You may think that this means that the integral is really the inverse of the

Â differentiation operator. Well, that's almost true, but not quite.

Â If you reverse the order of operations, and first, compute the derivative, of f,

Â then what will you get when you integrate it?

Â Will you get f back again? Well, not exactly.

Â You may recall there is a constant of integration involved, and indeed, the

Â indefinite integral of f is a class of functions all equivalent up to a

Â constant. It's very easy to right down a list of

Â integrals that we can compute easily. The integral of x to the k we know is x

Â to the k plus 1 over k plus 1, unless k equals negative 1.

Â In which case, you got the log of x. Integrals of sines and cosines are easy

Â enough. Anti-differentiate, and watch your minus

Â signs. And wonderful to tell the integral of e

Â to the x is, of course, e to the x. Don't forget the constants.

Â That much is simple. Take the anti-derivative.

Â On the other hand, there are some functions for which computing an

Â anti-derivative is highly non-trivial. Maybe with work you'll be able to compute

Â an anti derivative for log of x or secant of x.

Â But no matter how hard you try, you are going to have an extremely difficult time

Â finding an anti-derivative for something like e to x squared.

Â We're going to take some time, a little bit later in this chapter, to go over

Â methods for computing these anti-derivatives.

Â For the moment, and for the next couple of lectures, we're going to spend some

Â time answering the question of why. Why do we care about the indefinite

Â integral? One excellent answer to that question is

Â to be found in the subject of differential equations.

Â Where a differential equation is simply an algebraic equation on a function x and

Â its derivatives. For example, the simplest differential

Â equation is of the form dx dt equals f of t, for some f.

Â This equation has as its solution facts given by the indefinite integral of f

Â with respect to t. Now, what does that mean?

Â How do we interpret it? Well, one natural interpretation for the

Â derivative is as the slope. So, the differential equation is telling

Â you something about the slope of the solution curve.

Â It has to match this function f. Now, of course, we can see the indefinite

Â integral in here. We can see why there is a constant of

Â integration, because if I translate this solution curve up or down, I am not

Â changing any of the slopes and it still must be a solution.

Â Thus, solutions to this simple differential equation come in the form of

Â a family depending upon a constant. These are often called ODEs, which stand

Â for ordinary differential equations. You will see some extra ordinary

Â differential equations later in your calculus career.

Â 5:27

All of these are functions of time. Since acceleration is a constant, we can

Â write the simple differential equation, dv dt, that is acceleration, equals minus

Â g. Now, we put the minus sign in because we

Â orient things so that up is positive, down is negative.

Â Likewise, because we know the relationship between velocity and height,

Â we can write a differential equation for x, dx dt equals v.

Â Let's solve these one at a time. Both of these are simple differential

Â equations. The first dv, dt equals minus g, has as

Â its solution, v as a function of time is the indefinite integral of negative g,

Â with respect to t. What does this have as its solution?

Â Well, of course, the anti-derivative of a constant.

Â It's just that constant times T, and we have a plus C as our integration

Â constant. What does that plus C really mean?

Â It is really an initial velocity. Did we simply drop the object or did we

Â toss it at some speed? Well, in this case if we plug in t equals

Â 0, we see that v at time 0, otherwise known as v not, is equal to this

Â integration constant. Now, taking that solution for velocity

Â and plugging that into the right-hand side of our second differential equation

Â allows us to solve for the height, x, as the integral of v of t dt.

Â We already know what v of t is, and so we compute the anti-derivative to get

Â negative 1 half t squared plus v not t plus not t plus another constant of

Â integration. What is that plus C?

Â Well, that is really an initial height, since that time zero all of the other

Â terms vanish except for the plus C. Or, we can write that as negative 1 half

Â g t square plus v not t plus x not. This should be a familiar formula to you

Â from your basic physics. Whether is or not, it is transparent, now

Â why? Projectiles move in a parabolic shape

Â under the influence of gravity. We have a negative 1 half g t squared

Â coming from the integral. So far so good.

Â These simple differential equations have simple solutions in terms of integrals.

Â But what happens when we look at the next simplest class?

Â Namely, dx, dt equals f of x instead of f of t.

Â How do you solve that for x as a function of time?

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How do we go about solving this equation? Well, the first thing that we're going to

Â do is look at it and think. dx dt equals a times x.

Â Let's simplify and say that a equals 1. We're looking for a function whose

Â derivative is itself. You and I both know an excellent

Â solution, that is x of t is e to the t. Now, what happens when we put the a back

Â in there? Well, if we just think about it a little

Â bit more, we see that putting an a up in the exponent gives us, as its derivative,

Â e to the a t times a. That is what we're looking for.

Â We could, of course, put an arbitrary constant, C out in front.

Â And you may check that this is a solution to that differential equation.

Â What is that constant? Well, we might, following our previous

Â example, call it x not, the initial condition.

Â What you get at time 0. This method of solution is sometimes

Â called solution by ansatz. Which is a fancy word for saying, we took

Â a guess, and it turned out to be correct. There is, of course, a more principled

Â approach. One such approach is based on series

Â where we assume that our function x has a series expansion c0 plus c1t plus c2 t

Â squared, etcetera. These constants c sub k are, to us,

Â unknown. Now, what are we going to do with this?

Â Well, the differential equation tells us something about a derivative of x with

Â respect to t. We can differentiate the terms of this

Â series very easily. What the differential equation tells us

Â is that this is equal to a times x, which of course is a times c0, plus a times

Â c1t, plus a times c2 t squared, etcetera. Now, here is the important step.

Â If we have two series that are equal than their coefficient in front of the various

Â terms must also be equal, and now we can work one step at a time.

Â The constant terms equating them, tell us that c1 is equal to a times c0.

Â Well that's good. If we knew c0 we would know c1.

Â What does the next equation, that of the first order coefficients tell us?

Â Well, it tells us that c2 is equal to 1 half a times c1.

Â But we already know what c1 is, c1 is a times c0, so that tells us that c2 is

Â actually 1 half a squared times c0. What happens when we take the

Â coefficients of the quadratic term and set them equal to each other?

Â Well, 3 times c3 equals a times c2. That means c3 is 1 3rd a c2.

Â Using what we know about c2, we get 1 over 3 factorial.

Â Times a cubed, c not. I'm going to let you keep going with

Â this, and see if you can find a pattern. That pattern through the method of

Â induction, can show that c sub k is 1 over k times a times c sub k minus 1.

Â And that is, 1 over k factorial, a to the k times C not.

Â That means, when we look at our original series, x of t, it's the sum over k of

Â ck, t to the k, that is c0 times 1 over k factorial times a to the k times t to the

Â k. Now,, I'll let you verify that, that is

Â really c not times the exponential function e to at.

Â We obtain our solution, which we already knew was true, but from a more

Â principled, serious approach. That however, is a little complicated,

Â and so we are led to our last and best method of solution, that of integration.

Â Beginning with x as a function of t, we differentiate using the chain rule to

Â obtain dx is dx dt times dt. Now, we notice something about dx dt

Â since this is a solution to the differential equation.

Â It is a times x. And now, we're going to use the method of

Â separation, which means we move all of the x terms over to one side of the

Â equation, and we keep the, t terms, over to the other side.

Â This gives us dx over x equals adt. And now, using the fact that the integral

Â is an operator, we're going to apply it to both sides of this equation.

Â Obtaining the integral of dx over x equals the integral adt.

Â What is the integral of dx over x? Well, it's the anti-derivative.

Â Natural log of x. What is the integral of a constant times

Â dt? It's simply at plus an integration

Â constant. Now, to solve for x, we apply the

Â exponentiation operator. E to the log of x is simply x.

Â On the right hand side, we get e to the at plus a constant.

Â Splitting that up into a product, we obtain a new constant.

Â And we'll call that c as well, times e to the at.

Â That is our solution and it was obtained effortlessly.

Â 15:57

One of the things that you'll notice is that e keeps coming up.

Â It is not a coincidence that e to the t appears as the solution to the simple ODE

Â dx dt equals x. Or, as the series, sum of t to the k, or

Â k factorial, or is the indefinite integral, as we just saw.

Â In all three cases, you can interpret the constant of integration as an initial

Â condition or the constant term in a series, or the arbitrary constant of

Â integration. So, the question of why the plus c has

Â many answers. And so, we see that three of the main

Â characters in our story cross paths. Integration, series and solutions to

Â differential equations. In our next three lessons, we're going to

Â focus on one of these characters. Solutions, to differential equations, and

Â see how integration, helps us to compute, and understand them.

Â