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# Over- and under-estimation of Riemann sums

AP.CALC:
LIM‑5 (EU)
,
LIM‑5.A (LO)
,
LIM‑5.A.1 (EK)
,
LIM‑5.A.2 (EK)
,
LIM‑5.A.3 (EK)
,
LIM‑5.A.4 (EK)
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?
• Interesting question! Not exactly. The average of the right and left Riemann sums of a function actually gives you the same result as if you had used a trapezoidal approximation (instead of rectangular). This approximation is closer to the actual area of the function though!
• 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.
• 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.
• 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!
• Is it possible to calculate the trapezoid instead of the rectangles ?
(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.