Think about it. You're gonna take the antiderivative of f prime of x, which is going to be f of x, and you're going to evaluate it at b, f of b, and then from that, you're going to subtract it, evaluate it at a, minus f of a. These two things are identical, and then you can, of course, divide by b minus a. Now this is starting to get interesting. One way to think about it is there must be a c, if we, there must be a c that takes on the average value of, there must be a c that when you evaluate the derivative at c, it takes on the average value of the derivative. Or another way to think about it, another way to think about it, if we were to just write, if we were to just write, let me just say g of x is equal to f prime of x, then we get very close to what we have over here, because this right over here is going to be g of c. Remember, f prime of c is the same thing as g of c, is equal to one over, is equal to one over b minus a, one over b minus a. So there exists a c, where g of c is equal to one over b minus a, times the definite integral from a to b of g of x, g of x, dx.