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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).

## Want to join the conversation?

• Would the average of the right and left Riemann sums get you the actual area under the curve?
• 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?
• 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.
• If you take an average of a right Riemann Sum and Left Riemann Sum do you get the exact area
• Nope, but you will get the trapezoidal Riemann Sum, which is a better estimation.
• 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!
• 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.
• Is it possible to calculate the trapezoid instead of the rectangles ?
(1 vote)
• At and 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?
• 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.
• 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.
• Is it possible to use a mix of both left and right bounded rectangles so that the over approximations and the under approximations might "balance" each other out?
• Absolutely! The average of the left hand rectangle approximations and right hand rectangle approximations is called the trapezoidal approximation. This is because when you average the left and right sums, you get the same formula as if you used trapezoids to approximate the area. The trapezoidal sum is more accurate than the rectangle approximations. The midpoint rule and Simpson’s rule, however, give even more accurate approximations than the trapezoidal rule.
(1 vote)
• what if the function is decreasing and in the third/fourth quadrant? because then it would be upside down?
• 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.