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Current time:0:00Total duration:4:01

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)

- [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.