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Main content
Current time:0:00Total duration:6:39
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)

Video transcript

imagine were asked to approximate the area between the x-axis and the graph of F from x equals 1 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 1 to x equals 10 using a right Riemann sum with three equal subdivisions so I'm assuming you 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 you'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 because whenever I do Riemann sums you can do them 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 1 to x equals 10 so this is 1 2 3 4 5 6 7 8 9 10 and so they give us the value of f of x when x equals 1 when x equals 2 3 4 when x equals 7 5 6 7 8 9 10 and x equals 10 and they tell us that when X is 1 we're at 6 and we go to 8 2 3 5 so let me mark these off so we're going to go up to 8 so 1 2 3 4 5 6 7 8 and so what we know when X is equal to 1 F of 1 is 6 so this is this is 1 2 3 4 5 6 7 8 so this point right over here is f of 1 is this is the point 1 comma 6 and then we have the point 4 comma 8 4 comma 8 we'll put us 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 ten comma five so ten comma five put us 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 1 to x equals 10 so let me make those boundaries clear so this is from x equals 1 to x equals 10 and what we what we want to 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 we do a Riemann sums you don't have to have equal subdivisions although that's what you'll often see so we've just divided going from one to ten to three equal sections that are three wide so that's three this is 3 and this is 3 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 4 for this first section what is f of 4 it's 8 so we're going to 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 7 so the value of the function is 3 so this would be our second rectangle our second division I guess I 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 5 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 4 which is 8 so this is going to be 24 square units whatever the units happen to be this is going to be 3 times the height here is 3 f of 7 is 3 so that is 9 square units and then here this is 3 the width is 3 times the height F of 10 is 5 so 3 times 5 which gets us 15 and so our approximation of the area would be summing these 3 values up so this would be 24 plus let's see 9 plus 15 would be 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 of an approximation is it depends on what the function is doing there's a world where it could be a 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 bad approximation but we can at least do the approximation using a right Riemann sum just using this table
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