In recent videos, we've spent some time talking about definite integrals. We've talked about what they represent and the various rules that apply. Now, we're going to take a look at a very important use for definite integrals, which is known as the fundamental theorem of calculus part one. If that sounds super important, it is. But don't sweat it because in this video, I'm going to walk you through some examples that we will likely see in this course, and I'm also going to show you how surprisingly simple this equation and theorem is once you get the hang of it.
So without further ado, let's get right into things. I want you to recall something. We've talked about antiderivatives in the past, and recall that the function big F of x is the antiderivative of the function little f of x. The reason I'm asking you to recall this is because it's going to be important for the fundamental theorem of calculus to understand this relationship. For the fundamental theorem of calculus part one to work, a couple of things need to be true.
We need to have a function that is continuous on some interval a to b. So some sort of interval that we see right here, we need our function continuous so there are no breaks or jumps. And we also need this integral to exist. If this integral exists, there's something we can say about its derivative, because that's what the fundamental theorem of calculus part one is all about. This theorem shows us the connection between derivatives and definite integrals.
So going down here to our equation, the main fundamental theorem says this. If we take the derivative of big F of x, which we recall big F of x in the past has been the antiderivative, then what we're in essence doing is taking the derivative of this definite integral. So we have this definite integral that goes from some number a, that could be any number, to x of this function right here. What we're going to say is that this derivative is going to come out to little f of x. If this looks confusing right at first, I want you to focus on a couple of things.
Focus on this side and that side of the equation. Notice something. What we're literally saying is, if we take the derivative of the antiderivative, we get just the function. And that should make sense, because basically all this is saying is that the derivative and antiderivative will cancel each other, and that's basically exactly what happens. So we say that this will cancel with our antiderivative here, giving us just the function itself.
Now take a look at this portion of the equation right here. Notice we're taking the derivative of an integral. But we've seen with indefinite integrals that in essence, the process for this integral is just like the reverse or the antiderivative. So what we can see here is that this derivative is basically canceling this integral and giving us just the function. One main thing that does change here is this variable.
Notice we had a t inside, and now we have an x. The reason for that is because we're taking this function here, and we're bounding it from some random number a all the way up to x. So it changes everything to x, but that's okay because when we take this derivative, we just have the same function. We just switch up the t's with x's. To make sure we understand this equation, let's take a look at a couple of examples.
We'll start with example a. Notice for example a, we have this equation right here, y is equal to this integral. For both of these examples, we're asked to use the fundamental theorem to find dy/dx for the following, which is the derivative. So if I want to find the derivative of this integral, that's going to give me the solution to my problem. If I went ahead and tried to integrate all of this and then take its derivative, that would be a long and tedious process.
But what I can do is use the fundamental theorem. That says that I can just take these t's and replace them with x's because this is just going to be the same function we have. So dy/dx is going to be x3 - 3x2 + 4, and that right there would be the solution. Notice how quick it is to use the fundamental theorem of calculus. We didn't have to do this long and painful process here to find this derivative.
We simply just switched the variable t with x. And remember, that's all our fundamental theorem says. We just take our function t that's inside the integral, and we replace it with x, and that would give us the solution. But now let's try example b. And example b is a little bit more tricky because notice that we're going from some value of five all the way up to x squared.
We no longer have an x on top of our integral. Now we have an x squared. So what exactly do we do when this happens? Well, whenever we have a situation where the upper bound of our integral is a function of x, what we want to do is use the fundamental theorem of calculus with the chain rule. Recall that the chain rule is taking the derivative of the outside and multiplying it by the derivative of the inside to get the total derivative for our function.
If we want to use the chain rule when dealing with the fundamental theorem, well, it looks like this. What we do is we take whatever this t is, and we replace it with this function of x on top, just like we had up here. The main difference, though, is now you have to take this entire thing and multiply it by the derivative of the top function. So using the rule down here, what I need to do first to find dy/dx is I first need to take this t and replace it with x squared. That's taking this function and just putting it into what we have.
So that's going to give us two over 1 + and replace this t with x squared. Now what I need to do from here is multiply this by the derivative of the top function, and the derivative of x squared with respect to x is two x. Now from here, I can multiply these twos to give me a four x on top and then one + x squared on bottom, and that is the solution to example b. As you can see, it's pretty quick to solve these problems when we use the fundamental theorem of calculus. The main thing you should remember from this video, if you're just looking at these equations, is to remember how to use this one.
Because notice here that I could use this exact same equation for the first example that we tried. Because if I went ahead and took this function of x and replaced it for t, that would give me what we had down here, and then I could multiply this by the derivative of the outside function or the top function here. But notice that would just be the derivative of x, which is one. So we would just get the exact same thing that we did if we used the initial theorem that we learned about. But the main thing that we need to remember is this scenario where if you take the derivative of an integral, you're just replacing the inside of that function with the top here, and you may need to apply the chain rule if there's a function of x on top that is not just x.
That is the idea for the fundamental theorem of calculus part one. I highly encourage you to check out the practice and examples after this video to really make sure we have this concept down. See you in the next one.
