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## Calculus 2

### Unit 1: Lesson 2

Approximation 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
- Motion problem with Riemann sum approximation

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# Worked example: finding a Riemann sum using a table

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)

When we have a table of values of a function, we can use it to find a Riemann sum approximation of that function.

## Want to join the conversation?

- If a rectangle is under x axis , should I subtract its area instead of adding ? Why ?(5 votes)
- Yes, even though an area can never be negative you will always want the "Net Area." So, when you have your areas of the rectangles for this instance you would subtract any area that is found below the x-axis.(5 votes)

- What if we are asked to do this same example but with unequal divisions?(4 votes)
- I may be misunderstanding, but if you're talking about the distance between two adjacent x values being different, then the answer is you won't be asked that. As shown in previous videos, you set each Δx values equal to one another in order to approximate. In the unlikely case you are actually asked to evaluate with different Δx values, you would simply take each value of Δx and multiply it with its corresponding f(x) value and add them all together. For example, if you had a table that listed several x values such as 1, 3, 7 and 10 as well as their respective f(x) values, say, 6, 7, 3 and 5, you would take Δ of the first two values and multiply it by the left or right side, like this: (3-1)(6) if you're taking the left side or (3-1)(7) if you're taking the right. then you move on to the next, which would be (7-3)(7) [for left] or (7-3)(3) for the right, and so on and so forth. Once you have all values, you would add them together.

Sorry this is long, didn't want to underexplain.(1 vote)

- are upper values = right endpoint values & lower values = lefthand values?(2 votes)
- If your talking about in terms of the type of Riemann sum [left/right], yes then the lower values are the left side and upper values are the right side.(2 votes)

- What if we are asked to find the reimann sums without an interval?(2 votes)
- What would be the proper sigma notation for a riemann sum, like the example sal did here, for instance?(1 vote)
- The sigma notation for left and right Riemann sums can be found in this Khan Academy article.

https://www.khanacademy.org/math/ap-calculus-ab/ab-integration-new/ab-6-3/a/riemann-sums-with-summation-notation(3 votes)

- A function T(x) is continuous and smooth with values given in the table below. Use the values in the table to estimate each of the following. Use best estimates and round your answers to two decimal places.(1 vote)
- What would one do if the intervals indicated by the table are different lengths? In this case it is always three but the problem I am working on it changes for every interval.(1 vote)
- With non-equal subdivisions the width of each rectangle will be different, but the method is still the same: multiply the height and width of each rectangle and add the results together.(1 vote)

## Video transcript

- [Tutor] Imagine we're
asked to approximate the area between the x-axis and the graph of f from x equals one to x equals 10 using a right Riemann sum with three equal subdivisions. To do that, we are given
a table of values for f, so I encourage you to pause the video and see if you can come
up with an approximation for the area between
the x axis and the graph from x equals one to x equals 10 using a right Riemann sum
with three equal subdivisions, so I'm assuming you've had a go at it, so now let's try to do that together and this is interesting, because we don't have a
graph of the entire function, but we just have the value of the function at certain points, but as we'll see, this is all we need in order to get an
approximation for the area, we don't know how close it is to the actual area with just these points, but it'll give us at least a right Riemann sum for the approximation, or an approximation using
a right Riemann sum. So let me just draw some axes here, 'cause whenever I do Riemann sums, you can do 'em without graphs, but it helps to think
about what's going on if you can visualize it graphically, so let's see, we are going from
x equals one to x equals 10, so this is one, two, three, four, five, six, seven, eight, nine, 10 and so they give us the value
of f of x when x equals one, when x equals two, three, four, when x equals seven, five,
six, seven, eight, nine, 10 and x equals 10 and they
tell us that when x is one, we're at six, then we go
to eight, to three, five, so let me mark these off, so
we're gonna go up to eight, so one, two, three, four, five, six, seven, eight and so what we know, when x is equal to one, f of one is six, so this is one, two, three, four, five, six, seven, eight, so
this point right over here is f of one, this is
the point one comma six and then we have the
point four comma eight, four comma eight, we'll
put this right about there and then we have seven
comma three is on our graph, y equals x f of x, so seven comma three would put us right over there and then we have 10 comma
five, so 10 comma five, put this right over there. That's all we know about the function, we don't know exactly what it looks like, our function might look like this, it might do something like this, whoops, I drew a part that
didn't look like a function, it might do something like this and oscillate really quickly, it might do, it might be nice and smooth and just kind of go and do
something just like that, kind of a connect the dots, we don't know, but we can still do the approximation using a right Riemann sum
with three equal subdivisions, how do we do that? Well, we're thinking about the area from x equals one to x equals 10, so let me make those boundaries clear, so this is from x equals
one to x equals 10 and what we wanna do is have
three equal subdivisions and there's three very
natural subdivisions here, if we make each of our
subdivisions three-wide, so this could be a subdivision and then this is another subdivision and when you do Riemann sums, you don't have to have
three equal subdivisions, although that's what you'll often see, so we've just divided going from one to 10 to three equal sections,
that are three-wide, so that's three, this is
three and this is three and so the question is how do we define the height of these subdivisions, which are going to end up being rectangles and that's where the
right Riemann sum applies, if we were doing a left Riemann sum, we would use the left boundary
of each of the subdivisions and the value of the function there to define the height of the rectangle, so this would be doing a left Riemann sum, but we're doing a right Riemann sum, so we use the right
boundary of each of these subdivisions to define the height, so our right boundary
is when x equals four for this first section, what
is f of four, it's eight, so we're gonna use that as the height of our, of this first rectangle, that's approximating the area
for this part of the curve. Similarly, for this second one, since we're using a right Riemann sum, we use the value of the
function at the right boundary, the right boundary is seven, so the value of the function is three, so this would be our second rectangle, our second division, I guess,
used to approximate the area and then last but not least, we would use the right boundary
of this third subdivision when x equals 10, f of 10 is five, so just like that and so then our right
Riemann approximation, using our right Riemann sum
with three equal subdivisions to approximate the area, we would just add the area of these rectangles,
so this first rectangle, let's see, it is three
wide and how high is it? Well, the height here is
f of four, which is eight, so this is going to be 24 square units, whatever the units happen to be, this is going to be three times
the height, here is three, f of seven is three, so
that is nine square units and then here, this is
three, the width is three times the height f of 10 is five, so three times five, which gets us 15 and so our approximation of the area would be summing these three values up, so this would be 24 plus, let's see, nine plus 15 would give us
another 24, so it's 24 plus 24, it gets us to 48, there you go, we just
using that table of values, we've been able to find an approximation. Now once again, we don't know
how good our approximation is, it depends on what the function is doing, there is a world where it could
be very good approximation, maybe the function does
something like this, maybe the function just happens, let me make it a little bit, maybe the function does
something like this, where in this case, what we just did would be a very good approximation or maybe the function
does something like this, where in this situation, this might be a very bad approximation, but we can at least do the approximation using a right Riemann sum
just using this table.