So this area over here is 12, but the integral is going to be a negative because once again, our function is below the x-axis. So this integral, this right over here is going to be a negative 12. This is negative 12. So negative 12 plus three is negative nine. All right, now let's evaluate f of negative two. f of negative two is equal to the integral from one to negative two of f prime of u du plus three. 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.