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(a) Find an approximation to the integral $ \displaystyle \int^4_0 (x^2 - 3x) \, dx $ using a Riemann sum with right endpoints and $ n = 8 $.

(b) Draw a diagram like Figure 3 to illustrate the approximation in part (a).

(c) Use Theorem 4 to evaluate $ \displaystyle \int^4_0 (x^2 - 3x) \, dx $.

(d) Interpret the integral in part (c) as a difference of areas and illustrate with a diagram like Figure 4.

(a) -1.5

(b) See graph

(c) $-\frac{8}{3}$

(d) See Diagram

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Harvey Mudd College

University of Michigan - Ann Arbor

University of Nottingham

Boston College

if we first want to get a response, I'm using eight rectangles. Um, let's go ahead and write everything. So the way I like to do it is first to come up with my partitioning set. And then from there, we can just kind of plug things in as needed. So, um, first, I need to figure out what is going to be A and B Well, a is going to be my lower balance zero, uh, b is going to be my upper bound for and then in is going to be eight. So we're going to use Delta X is equal to be my say over in, which is for minus 0/8. So that would be four rate or one half now to get X, I this is going to be a plus. I times Delta tax Another way we can think about it as we just start from a so zero. And then I'm going to keep adding Delta X, which is one half up until I get to four, which in this case is just going to be one half, so it would be zero. Then we have 0.5 and then 11.52 2.533 point five and then four. So we've reached beats, so we're going to stop now. Since we're doing the right room on some, we're getting essentially just throw zero out and then just use all of these other ones. Um and then we can go ahead and plug everything in so our eight is going to be equal to the sum. Um, actually, over here I'll write the equation in general first. So are in is equal to the some from my as you go to 12 in oh, Delta x times f of x I, and in these excise again is are set down here. So this will just be X one next to next three dot to dot all the way up two x eight. So over here there's going to be I is equal to 1 to 8 of Delta X is always a half. So I'll just write a half there and then we just have f of X. I now we can go ahead and plug in 0.5 all the way up to four into here and doing that should give us well first. Since we have one half being multiplied by everything. We could go ahead. In fact, that out, it would be one half times and then f of X I of 0.5. So here I'll just go ahead and write this out first. So F 0.5 plus f of one plus f of 1.5 plus half of two plus f of 2.5 plus f of three plus f of 3.5 and then actually let me skip this over a little bit and then plus f of four. And now we just need to take all of these and plug it into X squared minus three X Um, so let's see what we get when we do that. So we're going to get one half times negative 1.25 and then plus negative two, um, plus negative. 2.5 plus negative. Two plus negative. 1.25 plus zero plus 1.75 and then plus four. And then if we add everything up there on the inside, we should get one half times negative three, which is going to be negative 1.5. So our eight is negative 1.5. Okay. Now, for part be actually, maybe I should say this is the solution department right here. Now for part B. They want us to draw the rectangles of for this. So since we're doing right, Riemann Sums we're going to start on the rights of the right side of our interval and then go towards the left early. So that's the way I like to do. So let me go ahead and just plot these points first, Um, or at least the X value. So I can kind of see where I'll be starting and stopping. So 4, 3.5 3. 2.5 to 1.5 1.5. Well, then I guess I can go ahead and put zero also. So what I'm gonna do is just go start here on the right and go all the way up until I hit my function. So I hit it here, and I'm just going to go to the right until I hit 3.5, which is around here, and then we're just gonna go straight back down. Okay, So that's our first rectangle then. Our second rectangle is just going to be wherever we go from 3.5 up. So it'll be here. Come over to three, uh, three senses at zero is just gonna go straight across like that. Now, since we're below the X axis, we're going to go all the way down until we had it and then go over the same manner as we were doing before two. Go down until we hit function, go to the left, go all the way back down to 1.5 1.5 again. Go up, go over to one, go down like that one, go over, then go down. And then, lastly, from half to zero and then go down. So we go ahead and just draw the face of these. So what we found is all of this area right here. This is what we found in part a. Okay. And at least for right now, I'm going to skip, um, part C because that's actually going to take a lot of work. But we will come back to that after we do deep. So let's go ahead and come down to D. So they want us to interpret, um what this Integral was supposed to look like, um, from the areas here. So let me go ahead and write it down again to the 024 x squared minus three x dx. So notice how we have, um, two sections. So let me go ahead and draw what we have. So this is all of our area on 0 to 4 of the function that is above the X axis, and then this is all of our area on zero before that is below the X access. So the way we would interpret this is this should be equal to the green area minus whatever the red area is. So this is kind of the graphical interpretation of this here and now, for part C. Let me go ahead and just pull this other page up. They want us to use this definition to actually evaluate this, um, so we'll just go ahead and use these to help us out. So remember, we're doing this in general, so and it's just some arbitrary number, so let's just go ahead and start. So first Delta X is going to be so before minus zero over, and so that would just be for over in now, X, I is going to be what we start from zero, and then we have plus I times Delta X, which is going to just be times for over end. So that means normally over here, X, I is going to be four I over and all right. And now that we have all these, we can come up here and just plug everything. So there's going to be the limit, as in approaches infinity of the some from in as you go to 12 and of now, Delta X is supposed to be four I over in, So four I over in and then f of X, I Well, let's go ahead and see. Well, this should be, um Well, actually, sorry. This isn't for I over and this is just for over in. I was looking at X I as opposed to Delta X. Yeah. Um, but now we're going to plug in for I over end of four. I over n squared. Ah, and then minus three for I over in. Now, remember, this in here is a constant, so we can go ahead and just pull that out. Actually, before we do that, let's go ahead and just distribute everything. Ah, and then we can try to simplify later. So if you limit as n approaches infinity of So the sum from N Z could have one to not in. Uh, I don't know why I root ends here. This should have been I I'm used to using my index is on, uh, and as opposed to I, uh, Everything else looks good, though. Okay, Um, so this would be 64. I squared over insects jobs going and write this out first. I'm not trying to do a ton of things in my head. So 16, I squared over and squared and then minus, uh, 12 I over in. So just writing all this down again. So now we can go ahead and distribute the four in. So that should give us 64. I squared over and cubed, and then minus 48 I over in squared. Right now, we can use the fact that the sum is supposed to be linear. That means we can distribute it. Let me move this down here, actually, so I can first rewrite this some in the following way. So this would be the limit as n approaches infinity of So the only thing I have to keep on the inside of the limit or of the sum is going to be the eyes. So we can rewrite this as 64 over and cute some from my is he going to want to in of I squared and then minus 48 over in squared, some from eyes go to one to end of I. And now both of these limits here are these sums should say, um, they actually give these to us on page 3 81. So on page 3 81. So we don't really need to, like, derive these at all. So this should give us the following. So this first one, uh, for the cube is going to be in n plus one, two n plus one, um, divided by six. And then we have minus 48 over in squared and then the sum of Isaac would want to end up. I, um, should be in in plus one over. Ah, to Okay, so we have these now. Well, one thing we can do is go ahead and apply these limits because now we just have of two rational expressions. And actually, I'll go ahead and simplify some of this first, um, so we can do 64 divided by six. Ah, which is going to be 32 3rd. So we have 32 3rd times, and then, well, one of these ends and that in castles would be squared. So this is going to give us n plus 1/2 n plus one all over and squared. And then we have minus 24 divided by 28 so that would just be 48. Divided by two is 24. And then that would be, well, one of these ends. Cancel out with that. So it's going to be in plus one over in. Yeah. And again, both of these are rational functions where they have the same power, the numerator and the denominator. So we can just go ahead and divide their leading coefficient. So this is going to be wolf. We look, the leading coefficient here is going to be to because it would be too and squared in the denominator is just in squared. So it would be one, so that would be two or one, and then over here is just 1/1. So this is going to be two thirds 32 3rd times two minus 24. So that would be 64 3rd minus 20 for which would give us negative eight thirds. So this is going to be the integral from 0 to 4 of X squared minus three x DX. So this is our answer to part. I should let me write this to make it look more like a negative.

University of North Texas