 With the fundamental theorem of calculus, we can define a function in terms of an integral. We can then ask all the questions we'd normally ask about the function. It helps to remember the two versions of the fundamental theorem of calculus. First, let F be continuous over some interval. Then the definite integral over the interval is going to be the difference between some function F at B minus F at A, where the derivative of our function is the integrand. Now there's a second version of the fundamental theorem of calculus, but in retrospect, we really don't need this and shouldn't rely on it. But you should always remember the chain rule. So, for example, let's define F of x to be the integral from 0 to x of cosine t squared. Let's find F of 0 and F prime of 0. So definitions are the whole of mathematics, all else is commentary. F of 0 is, according to the definition, the integral from 0 to 0 of cosine t squared dt. The fundamental theorem of calculus, one version, says that this is the difference between some function g evaluated at 0 minus g evaluated at 0, where g prime of x is cosine x squared. So let's see, we need to find some function whose derivative is cosine x squared. Actually, we don't. This is g of 0 minus itself, which is guaranteed to be 0. How about F prime of 0? We can still rely on this version of the fundamental theorem of calculus, since F of x is the integral from 0 to x of cosine t squared dt, we find that F of x is g of x minus g of 0, where g prime of x is cosine x squared. Now, remember g of 0 is a constant, it's some numerical value, and so when we differentiate F prime of x is g prime of x, if only there was some way of knowing what g prime of x was. Oh wait, here it is. And so we can find F prime of 0. Now, this version of the fundamental theorem of calculus is particularly useful because we don't always define our integral functions from 0 to x. So g of x might be defined as the integral from minus x to x of something. So again, g of x is going to be F of x minus F of negative x, where F prime of x is our integrand. And we differentiate. Now here we have to apply the chain rule because this is F of negative x. So g prime of x will be, and we know what F prime of x is, F prime of negative x will just substitute negative x in for x, and we'll do a little simplification. And so g prime of 3 is, or let's take another function, let L of x be the sine of h of x, and let's find L prime of 1. So again, chain rule, the derivative of L of x will be, now we do need to find h prime of x, so we'll pull in our fundamental theorem of calculus, h of x is going to be g of x minus g of 1, where g prime of x is the integrand. Again, since g of 1 is some particular constant value, the derivative will be g prime of x, and we know what that's equal to. And so we can substitute these values into our expression for L prime of x. And again, since h of x is the integral from 1 to x, then h of 1 is going to be 0. And that allows us to find L prime of 1.