Okay, So this question we're finding the limit as one approaches infinity for a midpoint room on some with and sub intervals on the function that were given is Kovacs vehicle. The 12 minus four x on the interval that we're working with is from 2 to 6 s o. The first step is to write out what the formula is for a midpoint remind some with ends up intervals just kind of just grew out what we're gonna be plugging into So men point reruns on with inseventy rules is equal to Delta Axe times. The sum from J is equal to zero too, and minus one of F uh x sub j plus X sub j plus one over, too. And just as a reminder of Delta X is equal to be minus a over end on B and I are the limits of the interval of an ex subject is equal to a plus J. Delta X and therefore X subject was one just plug in J plus one for Jang right here. Okay, so now we can start figuring out what each of the components is for this problem. So first, let's you Delta X Delta X is going to equal to you might say over end in this case, be a six and A is too so six minus two over end is equal to four over end. So that's where Delta X and our except J is gonna be equal to a, which is to plus J times, Delta X And were you handle two exits four over end. So four J over end Um can actually write X subject as well, So x So Jay pulls one's ari. So rewrite ex subject pulls one, um And that's going to be equal to two plus. So he just distribute this forward to the J please when you get four j plus four over and okay, so now that we have all of our components ready to plug in and figure out what our midpoint months, um, with ends up intervals is going to look like so this is gonna be equal to Delta X. So that means for over end times of some from J is equal to zero. The end minus one. Um, and here we're gonna have s of A except J. So two plus for J over end plus two plus for J plus four overrun. Um, and this is all divided by two. Okay, so now let's simplify this a little bit. So for this inside part here, we just do it all the one step. If you add the tunes, you get four. Um, and these have the same denominators. And just get eight j plus for on. And then we can just divide by two right away on you. Right out. You get four over end times. A sum from J is equal to zero. Toe end minus one from. And here relax. Aft with s of two plus for J over end Closed two over. And okay, so now that we've done this, let me go to the next page and we'll plug that into our function. This is gonna be equal to a four over end times. A son from J is equal to zero and minus one, and our function was 12 minus four x. So we're gonna have 12 minus four times that we had in the parentheses, which is two plus for J over end plus two over. And okay, so now we can distribute this four and subtract from 12 on and When you do that, you get the same well done by the same thing. So I must find my four over and And we have the sum from J is equal to zero and minus one. Um, and here we're gonna be left with four minus 16 J over end, minus eight over end. Okay, so now we can do is we can use the rules of summations and split this up into three parts to just take the sum of each of these components. So that's gonna be equal to four over end times the sum of each. So this same some from J equals zero too. And minus one of four minus on. And we can actually write this coefficient in front, so I'll just put that in front. So 16 over end times, times of some from J is equal to zero toe end minus one of J. Um, And again here we can write it as a coefficient out in front. So minus one over end crimes is some from J is equal to zero Thio end minus one of eight. So now we can again apply our rules of summations. So here, we're gonna be left with four end, uh, times four times and minus one. When you have a constant, you just multiplied by whatever you have on top here. So times so four times and minus one. Um, now we're gonna ply our power some formula. So we have minus 16 over end times and minus one, um, times and over, too. And so I'm just gonna go on to the next part. I ran out of room. Um, and here again, we just have a constant, but we're multiplying by negative one over end. Some of me left with minus eat over end times and minus one. Okay, so now when we simplify this, it's just a little bit of algebra. We can get that this is all equal to negative. 16 minus 16 over end. Plus 32 over and squared. Okay, So this just as a reminder, this is our midpoint, Ramon. Some friends of intervals. So we figured out what that is and the simplest form. And now we're ready to actually take the limit because that's what we're looking for Here were approximating the integral using the summation notation. So let me go to the next page. The limit as an approaches infinity, who were a midpoint. Three months I'm with and some intervals eyes gonna be equal to the limit as a purse. Infinity of the answer that we got on the previous page. So as an approaches, infinity of negative 16 a minus 16 over end, plus 32 over and squared. So now the limit as university new for negative 16 is just negative 16. Um, And here we have an end in the denominator. And when it gets really, really big, this is going to go to zero. Same thing for this right here. So that's just negative. 16 plus zero zero on and I just gives us negative 16. So this is actually our final answer. This is our, uh, limit for this question.