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### Course: AP®︎/College Calculus BC > 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: Rewriting limit of Riemann sum as definite integral

When given a limit of Riemann sum with infinite rectangles, we can analyze the expression to find the corresponding definite integral.

## Want to join the conversation?

- If we are working with Riemann sum from the left side, should we write the definite integral from i=0 to n-1 ?(13 votes)
- Yes, this is correct. We are taking the value of the function on the left corner of the rectangle, so on the first time, we do not add a delta-x.(9 votes)

- At5:14, Sal says that it's a right Riemann sum. How do you know based on the formula that it is a right Riemann sum?(5 votes)
- Because the sum has i = 1 and not i = 0 you are starting on the right side of the first rectangle as the first side would be located at i = 0 and the sum Sal uses starts at i = 1.(6 votes)

- Could we not also define the integral as the limit as delta-x goes to 0 of the general Riemann sum (in addition to doing so by setting as the limit 'n goes to infinity')?(4 votes)
- How does he know it isn't an integral with a bound of [0, 5], and the function itself is ln(2+x)? In this case, delta x still equals 5/n, but xi = 5i/n. An integral calculator gives the same result for both of these. Does it not matter which one you pick?(4 votes)
- So... how do we solve definite integrals for actual values other than using anti-derivatives? (i.e. can we plug infinity in for n and solve for an actual value other than 0? )

Because n is in the denominator of a term, and n approaches infinity, that value would approach 0. But maybe I'm just confused, very possible.(3 votes)- Yes, that's the idea.

As 𝑛 approaches infinity, the number of terms also approaches infinity, while the value of each term approaches zero.(3 votes)

- What if delta x was (5/4n) would that affect the upper bound?

Also what if the function does not have an 'a'. what then? would the lower bound be zero or one?(2 votes)- ofcourse it'll affect the upper bound because "delta x = (b-a)/n ".

if delta x =(5/4n) then (5/4n) = (b-2)/n => b=(13/4)

Also if the function does not have an 'a'. Then the lower bound will be 'zero'.

f(a+ (delta)X_i) this will be written like this " f((delta)X_i) " where 'a' equal 'zero'

Hope i didn't Confuse you and Sorry for my bad English(4 votes)

- What is the area under that curve . How can i use all this to find the area ?(3 votes)
- The definite integral or the Riemann sum are both the area under the curve under specified intervals.(2 votes)

- What about if we didn't have something like 5/n as the width, where b-a is clearly 5.. What if we had 1/n as the width in this same problem, what would we do then?(3 votes)
- Width tell us about "how far the upper boundary to lower boundary" over as many as N rectangles we want to use from our integral; (b-a)/n. if we have 1/n as the width then b-a or the "distance" from our upper boundary to lower boundary should be equal to 1 or b-a=1(2 votes)

- Recreating this riemann sum in software like symbolab or volframalpha, it interprets the limit of this integral as [a,b] = [0,5] instead of [2,7]. Why is that?

The same problem occurs in my textbook where a very similar sum to the one in this video appears, and you are supposed to express it as a definite integral:

*lim(n->∞): Sigma[i=1, n] (2/n)(ln(1+(2i/n)))*

Here, following the steps in this video we should get the interval [a,b] = [1,3] right? However the correct answer according to the textbook is [a,b] = [0,2], using software also gives me this same result.

I have been struggling with this for very long and I feel very confused about what the correct method of doing this is. What should I do about this?(2 votes)- Hmm... I haven't used symbolab or volframalpha. Maybe you should ask your teacher or someone else, and if you still can't find out, I'd say to just skip ahead for now. Maybe later you'll find out the answer. If you do, please let me know, as I'm a little confused now to.

FYI: I know each way is different, but here on KA, I haven't had any trouble with it, so I'd say just to do it the way KA said to until you find a better way.(2 votes)

- at4:50why did Sal said "when N isn't approaching infinity"? N should be approaching infinity, right?(1 vote)
- If you remove the limit from the Riemann sum, you'll see that you're only left with a sum of n terms. This here would be an approximation of the area, as n here would indicate a finite amount of rectangles. This is what Sal was referring to. But yes, to get the exact area (which is what we want in most cases), you would take the limit as n tends to infinity.(2 votes)

## Video transcript

- [Instructor] So,
we've got a Riemann sum. We're gonna take the limit
as N approaches infinity and the goal of this video is to see if we can rewrite this as a definite integral. I encourage you to pause the video and see if you can work
through it on your own. So, let's remind ourselves how a definite integral can
relate to a Riemann sum. So, if I have the definite
integral from A to B of F of X, F of X, DX, we have seen in other videos this is going to be the limit as N approaches infinity
of the sum, capital sigma, going from I equals one to N and so, essentially
we're gonna sum the areas of a bunch of rectangles where the width of each
of those rectangles we can write as a delta X, so your width is going to be delta X of each of those rectangles and then your height is going to be the value of the function evaluated some place in that delta X. If we're doing a right Riemann sum we would do the right
end of that rectangle or of that sub interval and so, we would start
at our lower bound A and we would add as many delta
Xs as our index specifies. So, if I is equal to one, we add one delta X, so we would be at the right
of the first rectangle. If I is equal two, we add two delta Xs. So, this is going to be delta X times our index. So, this is the general form that we have seen before and so, one possibility, you
could even do a little bit of pattern matching right here, our function looks like
the natural log function, so that looks like our func F of X, it's the natural log function, so I could write that, so F of X looks like the natural log of X. What else do we see? Well, A, that looks like two. A is equal to two. What would our delta X be? Well, you can see this right over here, this thing that we're multiplying that just is divided by N and it's not multiplying by an I, this looks like our delta X and this right over here
looks like delta X times I. So, it looks like our delta
X is equal to five over N. So, what can we tell so far? Well, we could say that,
okay, this thing up here, up the original thing
is going to be equal to the definite integral, we know our lower bound is going from two to we haven't figured
out our upper bound yet, we haven't figured out our B yet but our function is the natural log of X and then I will just write a DX here. So, in order to complete writing this definite integral I need to be able to write the upper bound and the way to figure out the upper bound is by looking at our delta X because the way that we
would figure out a delta X for this Riemann sum here, we would say that delta X
is equal to the difference between our bounds divided
by how many sections we want to divide it in, divided by N. So, it's equals to B minus A, B minus A over N, over N and so, you can pattern match here. If this is delta X is
equal to B minus A over N. Let me write this down. So, this is going to be equal to B, B minus our A which is two, all of that over N, so B minus two is equal to five which would make B equal to seven. B is equal to seven. So, there you have it. We have our original
limit, our Riemann limit or our limit of our Riemann sum being rewritten as a definite integral. And once again, I want to emphasize why this makes sense. If we wanted to draw this it would look something like this, I'm gonna try to hand draw
the natural log function, it looks something like this and this right over here would be one and so, let's say this is two and so going from two to seven, this isn't exactly right and so, our definite integral
is concerned with the area under the curve from two until seven and so, this Riemann sum you can view as an approximation when N
isn't approaching infinity but what you're saying is look, when I is equal to one, your first one is going to
be of width five over N, so this is essentially
saying our difference between two and seven, we're taking that distance five, dividing it into N rectangles, and so, this first one is
going to have a width of five over N and then what's
the height gonna be? Well, it's a right Riemann sum, so we're using the value of
the function right over there, write it two plus five over N. So, this value right over here. This is the natural log, the natural log of two plus five over N, and since this is the first rectangle times one, times one. Now we could keep going. This one right over here the width is the same, five over N but what's the height? Well, the height here,
this height right over here is going to be the natural log of two plus five over
N times two, times two. This is for I is equal to two. This is I is equal to one. And so, hopefully you are
seeing that this makes sense. The area of this first rectangle is going to be natural log of two plus five over N times one times five over N and the second one over here, natural log of two plus
five over N times two times five over N and so, this is calculating the sum of the areas of these rectangles but then it's taking the
limit as N approaches infinity so we get better and better approximations going all the way to the exact area.