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## AP®︎/College Calculus AB

### Unit 6: Lesson 3

Riemann sums, summation notation, and definite integral notation- Summation notation
- Summation notation
- Worked examples: Summation notation
- Summation notation
- Riemann sums in summation notation
- Riemann sums in summation notation
- Worked example: Riemann sums in summation notation
- Riemann sums in summation notation
- Definite integral as the limit of a Riemann sum
- Definite integral as the limit of a Riemann sum
- Worked example: Rewriting definite integral as limit of Riemann sum
- Worked example: Rewriting limit of Riemann sum as definite integral
- Definite integral as the limit of a Riemann sum

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# Worked example: Riemann sums in summation notation

AP.CALC:

LIM‑5 (EU)

, LIM‑5.B (LO)

, LIM‑5.B.2 (EK)

Here we express the approximation of the area under a curve in sigma notation. Created by Sal Khan.

## Want to join the conversation?

- How do you determine the accuracy of the estimations you make when using this method?(14 votes)
- Well, an easy way to do that is to compare the area of the Riemann approximation with the area that using the indefinite integral gives you. So to calculate it, it would be R/I, with 'R' being the area given to you through your Riemann approximation, and 'I' being the area that the integral gives for the same area.

For example, suppose we want to check the accuracy of our Riemann approximation for the function x^2 in the section 0-3. We used right-hand rectangles, so we already know this is an over estimation. We decide to use three rectangles in this calculation. That gives us 1+4+9, or 14 un^2. We then integrate the function x^2. That is (x^3)/3, in order the calculate the true area under the section of the curve we take F(3)-F(0), or (3^3)/3-(0^3)/3, which is equal to 3^2, or 9 un^2. We find that we were way off, but how off? 14/9 = 1.555... So we overshot it by about 56 percent.

However, should be mentioned that the accuracy of your approximation is subject to change, because your accuracy will vary depending on how many rectangles you make, and how the function itself behaves. You see, in the function x^2 our approximations will get less and less accurate as we go up the x-axis because the curve is getting steeper and steeper, whereas in the function f(x) = ln(x) we would find that we get more accurate as we go up the x-axis because the rate of change is getting smaller and smaller (approaching zero, in fact).(44 votes)

- why do we need to use riemann's approximation when we can just use the indefinite integral?(5 votes)
- Because the indefinite integral doesn't tell you anything about the area, its just another function. But if you are referring to the definite integral (which is basically just the indefinite integral evaluated at two endpoints), the reason we have to learn Riemann Sums is that this technique is actually the rigorous foundation of the definite integral. The definite integral is more or less defined as the limit of a Riemann Sum, taking the limit as the number of rectangles goes to infinity (or equivalently, as the width of each rectangle goes to zero).

So, its important to understand where the definite integral comes from. In the same way that its important to realize that the*true*(if you will) definition of the derivative is the limit as h goes to zero of (f(x+h)-f(x))/h, its important to realize that the whole concept of the definite integral is valid because its defined, at its very foundation, as a limiting case of a Riemann sum.

Hope that helps.(46 votes)

- Hey, I'm not quite sure if I understand where you got the formula f(2n-1) from. I do understand that that's the height of any rectangle. But how did u find this function from just looking at it?

Best regards(9 votes)- In the video, he says that it seems like the heights of the rectangles are based on the midpoints of the rectangles. The midpoints of the rectangles are located at the odd numbers
`1, 3, 5, 7`

. A general formula for these odd numbers is`2n - 1`

, where`n`

goes from`1`

to`4`

(inclusive). Therefore, the height of the`n`

th rectangle is`ƒ(2n - 1)`

.

If`2n - 1`

still seems mysterious, try plugging in some values for`n`

; it will become apparent that you get odd integers. The ability to recognise such formulas comes with experience.(8 votes)

- in the sigma notation where does he get the value n=1?(10 votes)
- We are just counting rectangles and counting begins at 1. In general, In general, there are, there are n rectangles. In this case there are 4, so we sum from the first, that is, n = 1 to the last, that is, n=4. The first rectangle is n=1, the second n=2 etc. It is just a coincidence that the center of the first rectangle also happens to be centered on the point x=1.

Hope that helped!(8 votes)

- Can you just put the two out in front of the sigma? I mean, Sal kind of ends up doing that anyways...(6 votes)
- Yes, absolutely. You're just factoring it out of each term in the sum.(12 votes)

- Um, there was a mistake in the video. He put a times sign when he needed to use a plus sign between f(5) and 2*f(7)(10 votes)
- correct, you can put it on
*tips & thanks*section so KA could see it. Good eyes!!(4 votes)

- It is true that more the number of rectangles we choose, better is the accuracy we get. However, this curve has something interesting. It appears that accuracy is also achieved, rather better, by choosing less number of rectangles (thus having wider rectangles) and height as the mid boundary (At1:10) of rectangle, because overestimating the area at the left half of rectangle is compensated by underestimating the area above the right half of that rectangle. Can someone explain to me that this advantage is specific to these types of curves and does not work in general.(5 votes)
- Are you talking about the fact that it looks like the part of the rectangle that protrudes above the curve on the left seems to fill the piece missing underneath the curve? I was also wondering about this. If true, then wouldn't taking the area of the rectangle work?(4 votes)

- Why is Sigma being used what is the significance of sigma?(2 votes)
- It is common practice in mathematics, for some reason, to use greek letters to denote various quantities. A capital sigma,
`∑`

, happens to be used to denote sums and series. There is no real significance of the symbol`∑`

; you could come up with your own notation for sums, something very different, but it would be annoying if everyone did so. Having a convention is nice.`∑`

happens to be the convention.(8 votes)

- Would this be an underestimate or an overestimate?(4 votes)
- The function is concave up in the interval we are interested in, that means that in any part segment, the left half of the segment will grow less than the right half of the same segment (you can see it quite clearly on the first rectangle). With this in mind we can say that the area we got from the rectangles will be a underestimate of the area under the function.(2 votes)

- I know that derivative is defined the difference between a point and another point infinitely close to the point divided by infinitesimal distance, but what is the definition of integral ?

Also if you take the left-most point instead of mid-point for Riemmann Sum, it would be f(n) dx instead right? And if right-most then f(2n)? How do we usually get a formula to represent all points used for the function(3 votes)- To describe the definition of the integral in the same terms you used to describe the derivative it would be something like: given a region* that has been partitioned into an infinite number of slices, each slice with an infinitesimal width, the integral, or (total area) is the sum of the areas of the infinitesimal slices, giving the total area of the region*.

*Region: area defined by two end points [a, b], the x-axis and the value f(x) for each x in [a, b](3 votes)

## Video transcript

Voiceover: What I want
to do in this video, is get a little bit of
practice trying to approximate the area under curves,
and also get a little bit more familiarity with the
sigma notation in this context. So what we have here, we have the graph of f of x is equal to one plus 0.1 x squared, that's this curve right over here, and then we have these rectangles that are trying to approximate
the area under the curve, the area under the function
f, between x equals zero, and x equals 8, and the
way that this diagram, or the way that we are
attempting to do it, is by splitting it into four rectangles, and so we can call this rectangle one, this is rectangle two, rectangle
three, and rectangle four. And each of their heights,
let's see the interval, it looks like they each
have a width of two, so they are equally spaced,
so we go from zero to eight, and we split into four sections,
so each has a width of two, so they're each going to be two wide, so that's two, that's two,
that's two, that's two. And their height seems to
be based on the midpoint, so between the start,
between the left side and the right side of the rectangle, you take the value of the
function at the middle value, right over here, so for example, this height right over
here looks like f of 1, this height right over
here looks like f of 3, this height of this rectangle is f of 5, this height right over here is f of 7. So given the way that
this has been constructed, and we want to take the sum of the areas of these rectangles as an
approximation as the area under this curve, how would we
write that as sigma notation? And I'll get us started,
and then I encourage you to pause the video and try to finish it. So the sum of these rectangles, we could say it's the
sum of, so we'll have n equals one to four, 'cause
we have four rectangles. And I encourage you to finish this up. Actually, just write it
in terms of the function, use function notation, you
don't have to write it out as one plus .01 times something squared. So I'm assuming you've had a go at it. So for each of these, so for
the first rectangle over here, We're gonna multiply two times the height, so the height right over
here is one, and this is the first rectangle, so
you might be tempted to say times f of n, but then that breaks down as we go into the second rectangle. The second rectangle,
the two still applies. This two is the width of the rectangle, but now we want to multiply
it times f of three, not f of two, so this f of n
isn't going to pass muster. And so let's see how we
want to think about it. So when n is one, two, three,
four, we're gonna take f of... So f of n, I shouldn't say
f of n, we're gonna take f of something, so here this
first one, we're gonna take f of one, then here, for
the second rectangle, we're taking f of three for the height, for the third rectangle
we're taking f of five, and then for the fourth rectangle,
we're taking f of seven. So what's the relationship over here? Let's see, it looks like
if you multiply by two and subtract one, so two
times one minus one is one, two times two minus one is three, two times three minus one is five, two times four minus one is seven. So this is two n minus one, so the area of each of these rectangles, the base is two, and the height is f of two and minus one. So that hopefully makes
it a little bit clear, kind of mapping between the sigma notation and what we're actually trying to do. And now let's just for fun, let's actually try to evaluate this thing. What is this thing going to evaluate to? Well, this is going to
evaluate to two times f of, when n is equal to one
this is one, f of 1, plus two times, when n
is two, this is gonna be f of, two times two minus
one is three, f of 3, when n is three, this is
gonna be two times f of 5, when n is four, this is going
to be two times f of seven, four times two minus one
is seven, f of seven. And so that is going to be,
well we're gonna have to evaluate a bunch of
these things over here, so let me actually, let me clear this out so I have a little bit more real estate. I'm feeling this might get
a little bit messy now. So this is going to be, actually
we could factor out a two. So this is going to be
equal to two times f of one is one plus 0.1 times one squared. So it's one plus 0.1, let me
color-code it a little bit so we can keep track of things. So this right over here is
1.1, so one plus 0.1 is 1.1, this right over here, f of
3, so that's one plus .1 times three squared, nine,
so one plus .9, so it's 1.9. And then, let's see, this
one right over here, f of 5, is gonna be one plus,
see five squared is 25, times .1 is 2.5, so one
plus 2.5 is gonna be 3.5. And then finally, f of
seven, is going to be one plus .1 times seven squared, so this is 49 times .1 is
4.9 plus one, so plus 5.9. And so what is this going to be equal to? So let's see, 1.1 plus 1.9, these two are going to sum up to be equal to three, and then these two are
going to sum up to be, let's see, if we add
the five, we get to 8.5, and then we add the .9, we
get to 9.4, so plus 9.4. Did I do that right -
three plus five is eight, .5 plus .9 is 1.4, yep,
and so this is going to be, so once again, we have
the two times it all, so this is going to be
equal to two times 12.4, which is equal to 24.8,
which is our approximation. Once again, this is just an approximation using these rectangles of
the area under the curve between x equals zero, and x equals eight.