So, or we could say, we could take, we could divide by the width of the interval, one over two minus one, and this all simplifies to one, times the definite integral over the interval. So one to two of a of t, which is, so this can be 12t to the negative four power dt. So what does this simplify to? Once again, this is one over one. That's just going to be one. If we take the antiderivative of this, and then we actually, well, let me just, so this is going to be equal to, the antiderivative of this is, so we're gonna go t to the negative three power, but then we divide by negative three. So an antiderivative of this is going to be, if we don't take the, well, an antiderivative is going to be negative four t to the negative three power, and we saw that over here.