So let's start off, let's evaluate g at negative 4, at kind of the lowest end or the starting point of our interval. So g of negative 4 is equal to 2 times negative 4, plus the integral from 0 to negative 4 f of t dt. The first part is very easy, 2 times negative 4 is negative 8. Let me do it over here so I have some real estate. So this is equal to negative 8, and instead of leaving this as 0 to negative 4 f of t dt, let's change the bounds of integration here, especially so that we can get the lower number as the lower bound. And that way it becomes a little bit more natural to think of it in terms of areas. So this expression right here can be rewritten as the negative of the integral between negative 4 and 0 f of t dt.