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.