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. I should just copy and paste that. So I'll just copy and paste. Let me paste it. So times the derivative with respect to x of all of that. And now the fundamental theorem of calculus directly applies. This is going to be equal to, we deserve a drum roll now, this is going to be equal to the negative.