All I did is distribute the negative sign and then switch the two terms. But this right over here is equal to the definite integral from, instead of a to b, but from b to a of F of t dt. So notice, when you put a negative, that's just like switching the signs or switching the boundaries. Or if you switch the boundaries, they're the negatives of each other. So we can go back to our original problem. We can rewrite this as being equal to the derivative with respect to x of, instead of this, it'll be the negative of the same definite integral but with the boundaries switched. The negative of x, with the upper boundary is x, the lower boundary is 3, of the square root of the absolute value of cosine t dt, which is equal to, we can take the negative out front, negative times negative times the derivative with respect to x of all of this business.