 What if our definite integral doesn't start at 0? So I suppose we want to find the integral from a to b. Our fundamental theorem tells us something about the integral from 0 to x. So let's start with what we know and define capital F of x to be this integral. So this tells us that the derivative of capital F of x is F of x. Since we'd like to find the integral from a to b, but we only know about the integral from 0 to someplace, let's see if we can incorporate that using our theorem that we can split an integral. So we'd like to be able to write an equation that includes this integral from a to b. So let's see what we can build around it. Since we know something about the integral from 0 to someplace, let's start with an integral from 0 to a, and this gives us the integral from 0 to b. But we know the integral from 0 to a and the integral from 0 to b, so we can substitute those in and solve. And that gives us the value of the integral from a to b. And this gives us a new version of the fundamental theorem of calculus. To find the definite integral from a to b, find an antiderivative of F of t, and evaluate it at the end point and find the difference. Of course, this leaves the problem of finding the correct antiderivative. So let's consider another problem. So to find the definite integral from 3 to 5 of 2t dt, the revised version of the fundamental theorem says that this is going to be capital F of 5 minus capital F of 3, where the derivative of capital F is 2x. So we find a function whose derivative is 2x. Now, ordinarily at this point we try to find what c is. But let's be a little bit lazy and not worry about the value of c just yet. So I can apply the fundamental theorem. Capital F of 5 is going to be, and capital F of 3 is going to be, and notice that when we subtract them, the value of c drops out and we just get the number 16. And this suggests that the value of c is irrelevant, and so we have a new version of the fundamental theorem. As before, the definite integral is the difference between some function evaluated at b minus some function evaluated at a, where this time our function is any antiderivative of our integrand. So let's evaluate the integral from 1 to 6 of cosine x dx. Our new version of the fundamental theorem says we can find any function whose derivative is cosine of x. So capital F of x is going to be an antiderivative of cosine. And so that'll be sine x plus any constant we want. Well, let's make that constant zero. So we need to evaluate our function at 6 and at 1, and the difference between the two is going to be the value of the definite integral.