Well, it's kind of, feels a little strange to have the upper bound being lower than the lower bound, so we can swap the bounds and then add a negative out here. So this is going to be equal to the negative of, if we swap the bounds here, so from negative two to one, f prime of u du plus three. So what's the integral from negative two to one? Well, they give us that. From negative two to one, they told us that this area over here is nine, but once again, because it's below the horizontal axis and above the curve, we would say that the, or the integral will evaluate to be negative nine. So the area is nine, but once again, the curve is below the x-axis, so the integral would give us negative nine. So this would evaluate to negative nine.