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Course: AP®︎/College Calculus AB > Unit 6
Lesson 2: Approximating areas with Riemann sums- Riemann approximation introduction
- Over- and under-estimation of Riemann sums
- Left & right Riemann sums
- Worked example: finding a Riemann sum using a table
- Left & right Riemann sums
- Worked example: over- and under-estimation of Riemann sums
- Over- and under-estimation of Riemann sums
- Midpoint sums
- Trapezoidal sums
- Understanding the trapezoidal rule
- Midpoint & trapezoidal sums
- Riemann sums review
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Over- and under-estimation of Riemann sums
Riemann sums are approximations of area, so usually they aren't equal to the exact area. Sometimes they are larger than the exact area (this is called overestimation) and sometimes they are smaller (this is called underestimation).
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Would the average of the right and left Riemann sums get you the actual area under the curve?(20 votes)
Is there a general rule when RRAM is greater than LRAM? Is there also a rule to determine which (RRAM MRAM or LRAM) is the most accurate depending on the situation?(3 votes)
RRAM will always be greater than LRAM when the function is increasing, and LRAM will always be greater than RRAM when the function is decreasing.
LRAM/RRAM in my opinion is horrible way of estimation, but for AP they will tell you when to use LRAM/RRAM in the problem, and later MRAM as well.(16 votes)
If you take an average of a right Riemann Sum and Left Riemann Sum do you get the exact area(4 votes)
Nope, but you will get the trapezoidal Riemann Sum, which is a better estimation.(10 votes)
Hi,
For right hand rule:
For example:
My interval is from 1-9, the function is 1/x and lastly I need to take four intervals. Would I take my first point as f(1) and my second as f(3)? If yes, why? And if not, why?
Thank you!(4 votes)
Yes, you would. By doing 9-1 divided by the 4 intervals, you would get a step size of 2. The question also says that your first interval is 1 so you would start there and then increase by 2 each time until you reach 9.(2 votes)
Is it possible to calculate the trapezoid instead of the rectangles ?(1 vote)
Yes, it is possible. Sal talks about trapezoidal sums here:
https://www.khanacademy.org/math/ap-calculus-ab/ab-integration-new/ab-6-2/v/trapezoidal-approximation-of-area-under-curve
Note that a trapezoidal sum is exactly the average between a left Riemann sum and a right Riemann sum.(4 votes)
At3:41and onward, a rectangle seems to form between the right-edge rectangle and left-edge rectangle. The function also seems to cut through them, very close to halfway cut. Could this affect how we use Riemann sums?(2 votes)
Not exactly. Though for many equations if you average out the left and right rectangle sums you usually get closer to the true area under a graph. Of course there are some graph shapes where this is untrue, or the size of the rectangles makes it untrue.
With using a riemann sum though it doesn't matter if you do left or right in the end, since eventually riemann sums as you to divide the graph into infinitely many rectangles, or other shapes. This makes it so it doesn't matter where you start, the infinitely many rectangles will fit perfectly under the graph.
I'm not sureif you knew that yet or where this video is on the playlist, so apologies if that is confusing.(2 votes)
- Why dont we use triangles on top of the underestimations to get a more accurate estimation?(1 vote)
What you are referring to is trapezoidal rule.(3 votes)
- what if the function is decreasing and in the third/fourth quadrant? because then it would be upside down?(2 votes)
Howdy Patrick,
Good question! But the key here is that we usually don't talk about "negative area". Instead, we would say that it estimates the area under the x-axis. Thus we are still having less area.
Note that sometimes we want to calculate the net area, where we subtract the area below the x-axis from the area above the x-axis. In this case, you would be right that the left Riemann sum would be underestimating the amount that should be subtracted, and thus is overestimate the overall sum (provided that there is more area below the x-axis than above the x-axis: otherwise the underestimation from above the x-axis might cancel out the overestimation from below the x-axis).
Yeah... a bit of a tongue twister there, but I hope you get the point.(1 vote)
why not take a triangle and subtract it from the left Riemann sum? The triangle would be ((left side function value - right side function value)*change of x) / 2. That would be a slight under approximation but would be far more accurate.(1 vote)- That's equivalent to the trapezoid approximation, which is discussed later.(2 votes)
if we do the riemann approximation over a symetrical limit eg. ∫ with the top number being 5, and the bottom being -5. Could you do the riemann sum over this, and the overestimation and the underestimation cancel out?
because if you use the riemann approx,and use the left riemann approx, there would be a underestimation on the left side of the curve, but a overestimation on the right. Would these two cancel out in a symetrical graph such as y=x^2?(1 vote)- Your instinct is good but getting an exact cancellation would be unusual. If you graph this out you will notice that on the negative side of the graph you're underestimate is at a different point of the curve than its corresponding overestimate on the positive side. The areas won't exactly cancel. But conceptually, you're cancellation estimate is better than the positive side alone.(1 vote)
3:41Video transcript
- [Narrator] Consider the left
and right Riemann sums that would approximate the area
under y is equal to g of x between x equals two and x equals eight. We want to approximate this
light blue area right here. Are the approximations over
estimations or underestimations? So, let's just think about each of them. Let's consider the left
and right Riemann sums. First the left. I'm just gonna write left for
short but I'm talking about the left Riemann sum. They don't tell us how many
subdivisions to make for our approximation so
that's up to us to decide. Let's say we went with three subdivisions. Let's say we want to make them equal. They don't have to be,
but let's say we do. The first one would go from two to four, the next one would go from four to six, and the next one would
go from six to eight. If we do a left Riemann sum, you use the left side of
each of these subdivisions in order to find the height. You evaluate the function at
the left end of each of those subdivisions for the height of
our approximating rectangles. We would use g of two to
set the height of our first approximating rectangle, just like that. Then we would use g of four
for the next rectangle. We would be right over there. Then you would use g of six
to represent the height of our third and our final
rectangle, right over there. Now, when it's drawn out like
this, it's pretty clear that our left Riemann sum is going
to be an overestimation. Why do we know that? Because these rectangles, the
area that they're trying to approximate is contained
in the rectangles. And these rectangles have
this surplus area so they're always going to be larger than the areas that they're trying to approximate. And in general, if you have a function that's
decreasing over the interval that we care about right over here, strictly decreasing the entire time. If you use the left edge of
each subdivision to approximate, you're going to have an overestimate. Because the left edge, the value of the function there, is going to be higher than
the value of the function at any of the point in the subdivision. That's why for decreasing function, the left Riemann sum is going
to be an overestimation. Now let's think about the
right Riemann sum and you might already guess that's
going to be the opposite but let's visualize that. Let's just go with the
same three subdivisions. But now let's use the
right side of each of these subdivisions to define the height. For this first rectangle the
height is going to be defined by g of four. That's right over there. And then for the second one
it's going to be g of six. That is right over there. For the third one it's
going to be g of eight. Let me shade these in to make
it clear which rectangles we're talking about. This would be the right Riemann
sum to approximate the area. It's very clear here that this is going to be an underestimate. Underestimate because we see
in each of these intervals, the right Riemann sum or the
rectangle that we're using for the right Riemann sum
is a subset of the area that its trying to estimate. We're not able to, it doesn't
capture this extra area right over there. And once again, that is because this is a
strictly decreasing function. So if you use the right end
point of any one of these or the right side of any of
these subdivisions in order to define the height, that right value of g is going
to be the lowest value of g in that subdivision so it's
going to be a lower height than what you could say is even the
average height of the value of the function over that interval. So you're going to have an
underestimate in this situation. Now, if your function
was strictly increasing, then these two things would be
swapped around and of course, there are many functions
that are neither strictly increasing or decreasing and then it would depend on the function. Sometimes even, it would depend on the type
of subdivisions you choose to decide whether you have an
overestimate or an underestimate.