All right, so some of you might have been a little bit challenged by this notion of, hey, instead of an x on this upper bound, I now have a sine of x. If it was just an x, I could have used the fundamental theorem of calculus. Just to review that, if I had a function, let me call it h of x, if I have h of x that was defined as the definite integral from one to x of two t minus one dt, we know from the fundamental theorem of calculus that h prime of x would be simply this inner function with the t replaced by the x. It would just be two x minus one. Pretty straightforward. But this one isn't quite as straightforward. Instead of having an x up here, our upper bound is a sine of x.