Well, this is going to be, by the same logic, the antiderivative of this evaluated at x, so f of x minus f of a. Minus f of a. Or if we wanted to solve for f of x, we could add f of a to both sides, and we would get f of x is equal to the integral from a to x of f prime of u du, and once again, why did I pick u? Well, I just needed some letter other than x, since I already used x as one of my bounds of integration. And I'm adding f of a to both sides. I swapped the sides, too. So f of x is going to be equal to this, plus f of a.