 A common practice in mathematics is we spend a lot of time learning how to go in one direction, and then once we've figured out how to go in one direction, we reverse it and try to figure out how to go in the opposite direction. So we've spent the first part of the semester looking at finding ways of finding and using derivatives, so now we're going to go backwards and find what are called antiderivatives. And the basic idea behind an antiderivative is fairly straightforward, suppose I have some function, then I can talk about the antiderivative of that function, it's going to be some other function whose derivative is what I start with. And so, well, I could say find a function whose derivative is this original function. The notation that we use for that is going to look something like this. And so this is for right now, you can read this as the antiderivative symbol of a function. And this dx here will play an important role. What this is telling us is that when I differentiate the function here capital F, I want to find the derivative with respect to x. The derivative of this with respect to x gives us the original function. And there's three important things to remember about finding antiderivatives. First off, one of the most important ideas, if you multiply by a function by a constant, then the derivative is going to be multiplied by the same constant, and we're going to be using that quite a bit. The second important thing you have to remember when finding antiderivatives is remember the chain rule. You can never go wrong by applying the chain rule. And the third important thing you have to remember about antiderivatives is also the chain rule. Again, it is very important to remember that when you differentiate you must apply the chain rule. And the most common mistake in finding antiderivatives is forgetting that the chain rule must be applied when we look at derivatives. So let's take a few examples of that. So to start off with, well let's take a function xq plus 4x minus 7, and in this case we want to show that this is an antiderivative of some function, and then while we're at it, let's see if we can find another antiderivative of the same function. So what does that mean? If this is an antiderivative, what that means is the derivative of xq plus 4x minus 7 should be 3x squared plus 4. And so I can verify that by just finding the derivative. Derivative of xq plus 4x minus 7 is 3x squared plus 4. So yes, in fact the derivative of xq plus 4x minus 7 is 3x squared plus 4, and so xq plus 4x minus 7 is an antiderivative of 3x squared plus 4. How about another antiderivative? The key to remember here is that if I differentiate this quantity here, this x cubed, the derivative gives me 3x squared, this 4x, the derivative gives me the 4, and this minus 7, the derivative doesn't show up any place over here, and that's because the derivative is 0. And what that means is if I change this constant, then when I differentiate, derivative of a constant is 0, and it's not going to appear over on the right-hand side. So that means if I change only the constant of my antiderivative, so let's change that to a plus 12 instead of a minus 7, then when I go and differentiate it, the derivative should be the same. Let's verify that, good practice to do so. Derivative of xq plus 4x plus 12 is in fact 3x squared plus 4, and so this is an antiderivative of 3x squared plus 4. And so is xq plus 4x plus 12, and that's going to be an important idea. In general, if I have an antiderivative of a function, I have an antiderivative of 3x squared plus 4, I can get a different antiderivative by changing what the constant is. Well, let's do an actual problem. Find the antiderivative of x squared, where we will be differentiating with respect to the variable x. So I want to solve this problem. It helps if I remember how to go forward and how to find a derivative. And a good place to start is to ask yourself what type of function am I dealing with. So this is an x squared function. That's a function of the form x to the n, and I might think about that and think, well, I know how to find the derivative of x to the n. It's nx to the power n minus 1. And we might notice that the exponent has gone from n to 1 less. So since I want a function whose derivative is x to the 2, I want to start with a function x to the 3. So let's try it out. What's the derivative of x cubed? So I'll differentiate that. That's going to be 3x squared, almost what we want, but not quite. So I can definitely say that x cubed is not the antiderivative of x squared. However, it is close. This is 3x squared, that's x squared. And as the saying goes, close only counts in horseshoes and calculus. And the reason that that's useful is that while I have this 3x squared here, I actually want an x squared. What I can do is I can then multiply both sides because it is an equality. I want to maintain the equality. I can multiply both sides by 1 third. And over on the right-hand side, I get 1 third times 3x squared, that is x squared. And over on the left-hand side, I get 1 third times the derivative of something. And at this point, we use that very useful rule that we have, which is that the derivative of a constant times a function is the same as the constant times the derivative of a function. And again, the equal sign says that if you see the one side, you can immediately and always replace it with the other. So here, I see constant times derivative, and I can immediately replace it with derivative of constant times function. I can move that constant inside the differentiation operator, in other words. Over on the right-hand side, again, after some horribly tedious algebra, I simplify 1 third 3x squared, that gets me x squared. And over on the left-hand side, I get derivative of 1 third x cubed. And it's worth trying to evaluate the derivative just from the beginning. The derivative 1 third x cubed 3 comes out front cancels. x to power 2 does in fact give us x squared. Now, one last step here. The derivative of 1 third x cubed is x squared. So one of the anti-derivatives of x squared is 1 third x cubed. And if I want to indicate all of the possible anti-derivatives, I want to write down 1 third x cubed plus any constant c that I care to add. And so my final statement of the answer here, anti-derivative of x squared is 1 third x cubed plus some constant c which is going to be determined or not as the case may be. It depends on the context of the problem. Now, in the first part of calculus, the difficulty was in the algebra. You'll find that the difficulty in finding anti-derivatives more often than not is actually in the arithmetic. So let's take a look at this anti-derivative of square root of x. And again, what this dx tells us is that whatever function we find, when we differentiate it with respect to x, we should get square root of x, at least after all the dust settles. So it's worth remembering that square root of x is x to power 1 half. So this is another function of the form x to the n. And again, if I differentiate x to the n, then the exponent drops by 1. So anti-differentiating should cause the exponent to increase by 1. So let's take a look at that derivative of, well, let's try this out. There's our square root power 1 half. I'm going to raise it by 1. And if I differentiate that, that's x to the 3 halves. What I should get, 3 halves x to power 1 half. And here we at least have our square root of x. I do have this extra coefficient 3 halves that I don't want to have. Well, close only counts in horseshoes and calculus. So let's multiply that by 2 thirds so that I can get rid of that coefficient. And over on the left-hand side, I now have constant times derivative. So here's constant times derivative. And I can move the constant underneath the differentiation operator. So now derivative of 2 thirds x to power 3 halves, check it out. This is x to the n, exponent 3 halves comes out front. 3 halves times 2 thirds cancels. x to power 1 half does give me what I'm looking for, x to power 1 half. And so I can make my final conclusion. The anti-derivative is this function here plus any constant that we want. Because when I differentiate, that constant goes away. And the other thing I'm left with is going to be the x to power 1 half.