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Oil leaked from a tank at a rate of $ r(t) $ liters per hour. The rate decreased as time passed and values of the rate at two-hour time intervals are shown in the table. Find lower and upper estimates for the total amount of oil that leaked out.

lower $2[7.6+6.8+6.2+5.7+5.3]L$upper $2[8.7+7.6+6.8+6.2+5.7]L$

01:24

Frank L.

Calculus 1 / AB

Chapter 5

Integrals

Section 1

Areas and Distances

Integration

Oregon State University

University of Nottingham

Idaho State University

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Yeah. Yeah. So problem. 5 15. We have oil leaking from a tanker and they show us a table here. So our F. T. This is my rate. So what I have here in this column is the rate of the oil leaking. And the units are leaders for our and then they also give me at a moment in time what the rate is. So the top row that you see here, this is my time given in hours. So I don't see this is not the volume of oil is the rate at which is leaking. So at zero hours I'm losing 8.7 liters per hour. At eight hours I'm losing 5.3. So we're asked to find the lower and upper some estimates for the amount of oil. So in order to do that, I've got to figure out, okay, you have to know a little bit about the shape of the curve anytime you start dealing things with with area here. So if I look at these numbers 8.77 point six it looks like every time these numbers are going down. So if I look at, if I could construct what's going on here, this is a function. If it were continuous, it would be a function that would be always decreasing. So what that would mean as if I drew a rectangle and use the left end point. So if I use the left endpoint, that's how I would come up with an upper some. If I use the right side of a rectangle there, you can see that. That would give me a lower some. Okay. If the function changed from increasing the decreasing, I would have to go and reflect that. But in this one it makes it easy, it's always decreasing. So I know that my upper some is going to be my left some. So my if I look at my upper some it's just going to use all the left sides of the rectangles. So what that means is I'm going to use um it's gonna be what zero times eight point seven. Um And if you look at the distance between each one so the width of the rectangle so that first rectangle. Sorry though I need to choose the the width of the rectangle. So the first one not zero back up we just erase that Each of these is 02468. The width is always too. So what we're going to have here is if you look at this real quick the width of each rectangle is too so you're going to have to as the width of each rectangle and then the heights of each rectangle. If I'm using the left side It's going to be the numbers 8.7 Plus 7.6 Plus 6.8 plus six point two. And that real quick. Mhm. And sorry I left out a number on here, let me back up. Um I left off the number, There was one extra number here, so sorry about that. Um the number at 10 and it still is decreasing at 5.3. Yeah, and The one that you have right before it is 5.7. Mhm. So again, if I'm using the left side, Just double check that the left side is going to be (877 668 62 57 60 82 57. And that should be it. So 1234566 endpoints. So that should give me five uh intervals for that. So 8776 um 6862 and 5.7. This turns out to be 70, So this is going to be 70 leaders. So that is my approximation as an upper some. Now if I do my approximation as a right some, so that's going to be the sum for the lower. And we established that that's just gonna be using the right side of each rectangle. So still the width is still too, But the first rectangle starts at 7.6, so you get 7.6 Plus 6.8 Plus 6.2 Plus 5.7 Plus 5.3. And that's how I get the lower some. And this turns out to be Um, 62 663 0.2 leaders as my approximation there. Yeah. So you can see that if you have an upper some, you expected to be higher as an estimate than your lower some. And that is what I see reflected here. So again, I don't have a continuous curve here. In previous problems, you had to plug into a function here. I have discrete data, so I'm limited here. And what I could do in terms of I can only evaluate the data at the discrete places that I have. And so there is the difference with the upper and the lower some.

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