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Use the form of the definition of the integral given in Theorem 4 to evaluate the integral.

$ \displaystyle \int^5_2 (4 - 2x) \, dx $

-9

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Okay, so this problem 5 to 21. So we have a definite integral from 2 to 5 for minus two X. And were asked to use this theorem four we saw in this section which is the limit definition of the definite integral to figure this out exactly. So what I will do when I start out one way to make this easier is to break this definite interrupt. So The integral from 2 to 5 of 4 -2 x. Dx is the integral from 2- five of 4 DX Minour the integral from 2 to 5 of two X. Dx. The reason I do that is this first one is very easy. Okay, this first one is just for times five minus two, And so that was just turns out to be this one is equal to 12. Okay, and so we'll come back there in just a moment. So if I can just figure out what the next interval is, that was gonna be, where more of the work is involved. So let's see if we can figure out how do we figure out the integral From 2 to 5 of two? X. Dx Ok, so in this case delta X is going to be 5 -2 over in Which is three over in. So this is going to be the limit. Yeah, as in approaches infinity. Yeah, of the some yeah, I Equal 1 to end. The width of each rectangle is three over in and then the height of each rectangle is going to be dysfunction evaluated. So if you look at it, so it's the function evaluated at um a plus I, delta X. So that is going to be um two Plus 3 I over in. So that is going to be um So you've got a two Yeah. And then X is going to be when you substitute an X. That is gonna be two plus Hi Tom So three I over and so all of this, so it's going to be the limit as n approaches infinity. Ah The some I got one to end and you've got to two plus three items three over in um things we can do out front, we can bring the two and the three out front. So this is the limit as in Approaches Infinity. Um I can have a six and then I'm left with the some I equals one to end and I'm left with one over in and two plus three I over in. Yeah. And so now I just go and evaluate um um each of these. Okay, so now what you're keying in on is things that I need to know. The sum I equal one to end of two is going to be two in and the sum I equal one to end of, I Thanks to Gaza's role that is going to be in times in plus 1/2. So those are the pieces of information that are going to get me where I need to be on this particular problem. So if I look at this, this is going to be yeah, the limit as in approaches infinity of six and then now let's just take those thumbs away, so you're gonna have one over in. Mhm. And then the sum for two is just going to be two in and then you're going to have plus three over in and then the sum of I that's just going to be in Times in plus one over two. So now I just need to do some simplification, so if I look at this um I can have some cancellations with this end, this end this in, so this is going to be equal to What is it? six times the limit in approaches infinity. And so what I'm left with here is to uh huh Plus Mhm 3/2. Yeah, and then I'm left with N plus one. Yeah, over in. Yeah. Mhm. And I can also write this Mhm As six times the limit. Mhm In approaches infinity of two. Okay, Plus three halves and I can just write this as one plus one over in. Now as N goes to infinity then this term right here the term is going to go to zero, So this is going to be six times two plus three halves which is going to be um That is going to be six times, that's four plus three, that is six times seven halves, which is 21. So now what I've proven is that I know that this is the integral from 2 to 5 of two, X. D. X. is 21. So if I go back to where I started all of this This integral from 2 to 5 to excess 21 so 12 -21 is minus nine. So the big key here is like, well in reality, yeah, the calculus ended about the time you got two right here and set up that limit. And then everything else was just you know, there's a lot of sequence in series and limits from freak out that are coming in. But places that mistakes can be made for sure.

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