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Worked example: over- and under-estimation of Riemann sums

Ordering different areas from least to greatest.

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  • blobby green style avatar for user ztbahbaz
    Would it be possible to better approximate integrals by doing both a left and right Riemann sum and then averaging the areas together?
    (7 votes)
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  • starky tree style avatar for user {Rayeed}^3
    Can it be possible that both reimman sums are greater (or smaller) than the actual area ? Or is the actual area always between the left and right reimman sums .Please give examples. Thanks
    (3 votes)
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  • blobby green style avatar for user Zaratustra
    Good day. Could somebody explain the meaning of question ''ordering from least (on top) to greatest (on bottom)''. That's mean when, for example, y coordinate goes down(decreasing) we should take least(less(underastimate)) area then at the bottom we trying to get more(greatest(overestimate))area? And when the curve increasing we need to do the opposite?
    (0 votes)
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    • aqualine ultimate style avatar for user sbb illusionist
      ordering from least (on top) to greatest (on bottom) - Meaning that when we take Left Riemman Sum, Right Riemman Sum and the actual area. One of them is greater than the actual area, one is the area itself, and the other lesser than the actual area. Ordering from the least to greatest is when you draw the bars for the sums and the one which covers less area than the shaded region(under the graph) is the least and the one that also covers the white part (unshaded region above the graph) is the greatest.
      (2 votes)

Video transcript

- [Instructor] The continuous function g is graphed. We're interested in the area under the curve between x equals negative seven and x equals seven, and we're considering using Riemann sums to approximate it. So this is the area that thinking about in this light blue color. Order the areas from least on top to greatest on bottom. So this is a screenshot from a Khan Academy exercise where you would be expected to actually click and drag these around. But it's just a screenshot. So what I'm gonna do instead of dragging them around, I'm just gonna write numbers ordering them from least to greatest where one would be the least and then three would be the greatest. So pause this video and try to think about these. Which of these is the least, which is in the middle, and which is the greatest? So let's just draw out what a left Riemann sum, a right Riemann sum would actually look like, and compare it to the actual area. And we could do an arbitrary number of subdivisions. I would encourage us to do fewer because we're just trying to get a general sense of things. And they don't even have to be equal subdivisions. So let's start with the left Riemann sum. So we wanna start at x equals negative seven, and we wanna go to x equals seven. Well, let's say that this is the first rectangle right over here. So this is our first subdivision. And it's a left Riemann sum, so we would use the value of the function at the left end of that subdivision, which is negative seven, x equals negative seven. The value of the function there is 12. And so this would be our first rectangle. You already get a sense that this is gonna be an overestimate relative to the actual area. And so the next subdivision would start here, so this would be our height of our rectangle. And once again, they don't have to be equal subdivisions. They often are, but I'm just show you unequal subdivision just to show you that this is still a valid Riemann sum. And once again, this is an overestimate where the actual area that we're trying to approximate is smaller than the area of this rectangle. And then let's say the third subdivision right over here starts right over there at x equals three. And we use the left end of the subdivision, the value of the function there, to define the height of the rectangle. And once again, you see it is an overestimate. So the left Riemann sum is clearly an overestimate. And it's pretty clear why. This function never increases. It's either decreasing, or it looks like it stays flat at certain points. And so for a function like that, the left edge, the value of the function at the left edge is going to be just as high or higher than any other value the function takes on over that interval for the subdivision. And so you get left with all of this extra area that is part of the overestimate. Or this area that is larger than the actual area that you're trying to approximate. Now let's think about a right Riemann sum. And I'll do different subdivisions. Let's say the first subdivision goes from negative seven to negative five. And here we would use the right edge to define the height. So f of negative five, or g of negative five, I should say. So that's right over there. That's our first rectangle. Maybe our next rectangle, the right edge is zero. So this would be it right over there. And then maybe we'll do four rectangles. Maybe our third subdivision, the right edge is at x is equal to three. So it would be right over there. And then our fourth subdivision, let's just do it x equals seven. And we're using the right edge of the subdivisions. Remember, this is a right Riemann sum, so we use the right edge. The value of the function there is just like that. And now you can see for any one of these subdivisions, our rectangles are under estimates of the area under the curve. Under, underestimate. And that's because once again, in this particular case, the function never increases. It's either decreasing or staying flat. So if you use the value of the function at the right edge, it's going to be smaller, it's never going to be large than the value that the function takes on in the rest of that subdivision. And so we are continuously underestimating. We're missing this. All of this area right over there is not being included. So we have an underestimate. So if wanna rank these from least to greatest, well, the right Riemann sum is the least. It is underestimating it. Then you have the actual area of the curve, which is just the area of the curve. (laughs) And then you have the left Riemann sum, which is the overestimate.