So let's take the definite integral from negative one to zero of x plus one dx. Well, let's see, the antiderivative of x plus one is, antiderivative of x is x squared over two. I'm just incrementing the exponent and then dividing by that value. And then plus x, and you could view it as I'm doing the same thing. If this is x to the zero, it'll be x to the first, x to the first over one, which is just x. And I'm gonna evaluate that at zero and subtract from that it evaluated at one, sorry, it evaluated at negative one. And so this is going to be equal to, if I evaluate it at zero, let me do this in another color.