Riemann Sums: Study with Video Lessons, Practice Problems & Examples

Understanding sigma notation is essential for evaluating finite sums efficiently. It allows for compact representation of summations, such as $\sum k_1{k}^2$. Key rules include the sum and difference rule, which separates terms, and the constant multiple rule, allowing constants to be factored out. Riemann sums, which estimate the area under curves using rectangles, can be calculated using left, right, or midpoint evaluations, enhancing understanding of integration concepts.

Sigma Notation

So for this video, I want you to imagine that I gave you this problem, where I asked you to find one plus two plus three plus four. Now I'm sure most of you could probably solve this, especially in your heads, when doing these types of problems, because it's pretty straightforward. But what if I kept on adding numbers, plus five, plus six, plus seven, all the way up to a hundred? Well, doing this in our heads would be a lot more difficult, and trying to write out all these numbers would be a complete pain. But in this video, what I'm going to show you is there's a way that we can actually write this in a more simple and compact way.

And it turns out that doing this is going to be very helpful when it comes to approximating the area under curves with rectangles, like we've talked about in recent videos. So let's just go ahead and get right into things. Now the notation we're going to learn about is called sigma notation. And as I mentioned, it's a way that when you have the sum of many terms, you can write them in a more compact way. And doing sigma notation looks something like this.

Now what you see below here, this kind of squiggly line here, this is the Sigma notation. And this means that you're adding up a bunch of individual pieces. Now what we do with Sigma notation is we start on the bottom here with this number. This number is what we call the index of summation. So that just tells us where we start when it comes to our summation.

Now what we're going to do is finish up here. This is where the index ends. And to do this, well, that just means we're going to take whatever our value is that we start with and go to where we end. So the index starts right here and finishes up there where we see on top. And what each of these terms are is the equation that you're plugging things into.

So you're going to take whatever this value is, and you're going to plug it into whatever equation or function this represents, And then you're going to keep putting discrete values in for each value until you get all the way up to this value here. So that's the idea of sigma notation, and I think it's going to help if we look at an example to see what this is all about. So let's try the example down here, which asks us to evaluate the following finite sums. Now to evaluate a finite sum, what you want to do is plug in the integer values from the start of the index, which we discussed was this bottom number here, to the end of the index, which we discussed was this top value. So let's go ahead and try that, and then we're going to add all these terms together.

So starting right here, notice that our equation is k2. So that means from one to three, we're going to plug in values all the way up to k2. And they're discrete values. So it's like one, two, three, all the way up until we get to this end of the index here. So let's go ahead and try this.

So we start with one down here. And so we're going to have one squared since this is k2. We're going to have plus, and we're going to go up another number, which is going to be two squared. And we're going to have plus three squared. And we're going to stop there since we stop at three.

So we have one squared plus two squared plus three squared, which is the same thing as one plus four plus nine. And then we're going to have five plus nine, which turns out to be 14. So 14 is going to be the result of this summation. So that's how you can evaluate a summation. As you can see, it's quite straightforward.

You're just taking these values and you're plugging them into this equation all the way up until you reach whatever this number is here. Now something I'm going to mention is that when it comes to the index of summation, we've seen that this is k in the examples so far. We saw that up here as well as down there, but it turns out that this k value is not always going to be this letter k. You could see any letter here that represents the start of our index. So it could be k, you could also see something like i, or maybe that's going to be a j or an h or something else.

It could be any letter on the bottom, but that just always represents the start of the index if it's on the bottom of the summation. So we'll try this example down here where we start from an i value of zero, and we go all the way up to four. You may see it written like this in your textbook instead, but you solve this the exact same way. This is the equation that you reference, and you just plug in the numbers all the way until you get to the end of your index. So we'll start with zero.

That's where we're starting. So we're going to have zero plus three. That's going to be plus, and then we're going to have one plus three plus we're just replacing this i here, so it's two plus three plus, and then we're going to have three plus three, and then we're going to have plus, and then we're going to finish off with our index here, which is four plus three. So this is what we get. Now adding up all these numbers here, you should finish with 25.

So 25 is the solution to this summation. So that is how you solve these problems where you're dealing with the sigma notation. As you can see, once you know how to read it and you understand what the values are, it becomes pretty straightforward. I hope you found this video helpful, and let's try getting a bit more practice to r

Evaluate the following summation: 
$\sum_{i=1}^4i^2-3i+8$

• A. 64
• B. 18
• C. 32
• D. 20

Answer: (C)

Evaluate the following summation: 
$\sum_{k=1}^4\left(\frac{k}{_{}2}\right)^2$

• A. $\frac{15}{2}$
• B. $15$
• C. $\frac{17}{2}$
• D. $\frac{17}{4}$

Answer: (A)

Evaluate the following summation: 
$\sum_{i=0}^2\frac{2i}{3}$

• A. $6$
• B. $\frac23$
• C. $2$
• D. $\frac43$

Answer: (C)

Sigma Notation Example 1

This is an interesting problem. So here we are given these kinds of repetitive operations happening to numbers, and what we're asked to do is write the following expanded sums in sigma notation. Now you may recall that we've learned about how the summation notation, what that looks like, and sigma notation in general. What we're now asked to do is basically go backwards. So rather than expanding out the whole sum like we've done before, we're now given an expanded sum, and we're asked to go backwards and write the summation for it.

So let's see if we can do that for both of these problems. Now we'll start with example a right here. Now what I need to do is pattern recognition. I need to see what's happening and what's staying the same and what's changing. Now, noticing this pattern here, I can see that we have a three that constantly stays in this situation.

So it's clear that three is going to be something that is inside of our sum here. Now, something else that I notice is we have this number right here, which is being squared. It's always going to be squared. So something's going to be squared here. And notice that we go one, two, three, four, five.

It keeps going up by one. So that tells me that we're going to start at k equals one, because that's where we're starting. And then that's going to go all the way up to five. And the k is going to be right here, but we need to also square that since we can see all of these numbers are squared. So that's going to be the pattern, and this right here would be a sum that describes what we have.

So it's going to be from one to five, and it's going to be 3kk2. And that would be the solution for this sum. So this is how we can take this repeated notation here, and we can write it in this sigma notation. Now we'll move on to example b. Example b looks interesting because notice we have a bunch of fractions that we're dealing with, so this looks a little more complicated.

But I'm going to write this a little bit differently. Notice that we have three fourths that happens first, and then we have three halves. Now, I'm going to try to see if I can get common denominators here. Now, to do this, I would multiply the top and bottom of this fraction by two to get four on the bottom. So it's going to give me six over four as my next fraction.

Now next, we have nine over four. There's already a four on the bottom there. And then to get three over a four, what I need to do is multiply and divide by four. So doing this is going to give me four times three, which is 12 all divided by four. And that's in a sense the pattern that we have.

So how exactly could I write this in our sigma notation? Well, notice that we have this three right here, and we have this four in the denominator. So we're starting with a three. We have a four in the denominator. So what I'm just going to do is put a three-fourths on the outside here.

Now it looks like something is changing this three up here, but let's see what we can figure out. So what I'm going to do is say that our k is going to start at a value of one. And if I go ahead and slap a k right about there, well, three times one is three. And notice if I if k were equal to two, when we go up to two, that's going to give me three times two, which is six over four. Then if this were three, if we go up to a value of three, we get to nine over four, because three times three is nine.

And then if we got four, we get 12 over four. So notice that this is the pattern that we need. So all I need to do is put this all the way up to four because that's how many places we go when we do this sum. So this right here will be the summation, for example, b, and that is the solution to this problem. So this is how you can take expanded sums and condense them using the Sigma notation.

Hope you found this video helpful, and let's move on.

Algebra Rules for Finite Sums

In the last video, we were introduced to sigma notation and solving these types of summation problems that look like this. Now, in this video, we are going to take a look at some of the rules that apply to these finite sums. Just like we saw for derivatives and indefinite integrals, we've seen things like sum and difference rules and constant multiple rules, and it turns out these same rules will actually apply when it comes to summations in a very similar fashion. So let's just dive right into it. First off, we are going to take a look at the sum and difference rule when it comes to dealing with these summations.

We could have a situation like where we have the summation, a finite sum of these two pieces right here, a(k)+−b(k). It turns out if you're dealing with plus or minus, we can expand this and separate it into two individual sums. So it would be the summation of this piece plus or minus the summation of that piece. We did the same thing where we separated out into two individual derivatives or two individual integrals.

Solving an example of this, let's say we have the summation from k1to3, it's going to be these two values right here. Doing this, we're going to have the same summation of these same values for this piece, and it's going to be the summation of k2 that we're going to have or the summation of k. This would separate these out into two different sums. However, doing this example, we would have ∑k1to3k2.

Another rule that we need to be familiar with is the constant multiple rule. This rule says that we can take some constant being multiplied by this piece a(k) and pull it out to the front. Therefore, we're just going to have the sum from k1ton of our equation a(k), and multiply by our constant to get the result. An example would be like this, we have this sum right here from k1to3 of 2k2. What I can do is take this constant two and pull it outside of the summation.

This sums up to the front, meaning all we are going to have is the summation from k1to3 of k2. Now doing this, we already know one to three for k squared, that's going to give us 14. So ultimately we'll have is two times 14. If we were to run the summation here, and two times 14 comes out to 28, which is the result for this sum here. As you can see again, we could have just taken the numbers and plugged them in, but it was much simpler to pull out this constant from the front, and then go ahead and multiply everything out.

Next, let's consider whenever we have the summation of a constant value. An example of this would be the case where we have the summation from k1ton of some constant c. If we ever have this, you only need to know that this is going to be the same as taking your constant c and multiplying it by the end of the index, which is n. For instance, the summation from k1to18 of three. This is going to be three times the end of our summation, which is 18. So we have three times 18, which is 54, and that's the solution there.

For all these summations, as you can see, we could just take the values and plug them into our equations and solve that way up to whatever end of the index we have, but that could be really lengthy and quite tedious. It's far better to use these rules that we know about. In most problems that we're going to encounter, we'll need to apply multiple rules together to find the summation we're aiming for. For example, the given sum needs evaluations using the rules that we've learned. Recognizing that we have two portions that are being subtracted, we can apply the sum and difference rule. To do so, we'll split this into two separate sums, meaning we will have ∑6k2-9k1to3.

Recognizing the constant six in front of k2, we pull it outside of the summation, resulting in 6∑k2k1to3 and factor in the constant sum -9∑ck1to3. Evaluating further, the result simplifies further down. Six times the sum of the squares minus nine times three yields a result of 57. This is the solution to this problem, showing how rules can simplify solving situations with sigma notation.

I hope you found this video helpful, and let's take a look at some more examples and practice to see how we can apply these rules and solve various sums. See you in the next video.

Evaluate the following summation: 
$\sum_{t=2}^65t+\frac{4}{t}$

• A. 94.2
• B. 105.8
• C. 8.67
• D. 7.33

Answer: (B)

Evaluate the following summation: 
$\sum_{i=0}^42i+1$

• A. 9
• B. 20
• C. 21
• D. 25

Answer: (D)

Evaluate the following summation: 
$\sum_{i=1}^24i^3-3i^2+2i-1$

• A. 6
• B. 25
• C. 4
• D. 13

Answer: (B)

Introduction to Riemann Sums

So in recent videos, we've been talking about how we can solve finite sums, which look something like this. We've talked about what these summations are in the general notation, and what we're going to learn in this video is a way that we can combine this idea of finite sums with this idea we've already learned about estimating the area underneath the curve of a function. So this should be pretty direct since we've learned both of these concepts already. So let's just go ahead and jump right into this. Now when you combine the summation notation with estimating the area under a curve, we call these Riemann sums.

So that's what we're going to be learning about. Now I want you to recall that when it comes to estimating the area under the curve of a function, we can do this by breaking it into a certain number of n rectangles that we call subintervals. And this is the equation for finding the width of a subinterval. So going down to this example here, this would be what would happen if we estimated the area underneath this function curve, which is this function here. And notice how we fill it up with four rectangles.

Now the general strategy is always going to be adding up the area of each individual rectangle. But since we now know what summations are, there's a more compact way that we can write this, rather than having to just do area one plus area two plus area three and so on. So the way that we can do this is by using sigma notation. And what we can do is use this notation to represent the area, which we call the Riemann sum. Now how exactly would sigma notation look for this example right here?

Well, the way that we could do this is we could say our notation is going to go from k equals one at the start of our index to n, and that's going to be for f(x_k-1). And that's going to be multiplied by Δx, and this would be our notation. Now the reason we have this x_k-1, this just tells us that we're looking at the leftmost endpoint. So let's say that we're looking at a certain subinterval, we'll call this the kth subinterval, and we go to the leftmost side of this interval. Well, the leftmost side right here of this rectangle would be x_k-1.

And we've talked about how we would look at the left endpoints or the left sides. So if we're looking at the left side of each rectangle, we're always going to look to the x_k-1 side. So we take whatever this x value is, we plug it into our function, and then we multiply that by Δx, and then we just keep going up by one increment, and that's going to get us the area of each rectangle because this is the height of each rectangle looking at the left endpoints, and then this would be the width. So to go ahead and expand out this example a little bit more, let's see what we're working with. We're told to estimate the area under the curve of our function using left endpoints, and we're trying to do this Riemann sum with four subintervals, which is four rectangles.

Now to do this, I'm going to set up my sum over here. So to set up this sum, what I can do is I can take a look at how many rectangles we have. In this case, I can see we have four rectangles or four subintervals. So that means my summation is going to go from k equals one all the way up to four. Now what I'm going to have inside here is this function f(x_k-1) since we're looking at the left endpoints.

We're asked to do a left endpoints sum. Now something I noticed about what we just found is we have this Δx right here, and recall that Δx is the width of each rectangle or subinterval, and that's going to be constant. Because Δx, the width is always the same, we can take that Δx and pull it outside of our summation. So we're going to have Δx, and that's going to be our summation from k equals one to four of this function f(x_k-1). So that is another way that we can write this.

Now, the real trick here is going to be figuring out what this summation is right here. Now looking at this function that I have, I can see that we start on this leftmost side, and this point right here would be the start of our interval, which is going to be x of zero, because that's going to be the very start of the leftmost side of this first subinterval. And then we finish all the way over here at this right side, which we'll call x of n. That's going to be the highest point that we go to on our curve. So if I want to write out this function notation here, well, what I can do is think about what each of these points are going to be in between.

And so we're in essence just going to be doing the same thing that we did with estimating the area with just adding these rectangles. So we just need to evaluate the sum in the same way. So doing this, we're going to have f of zero, that's this leftmost side. It's going to be plus f of 0.5. Notice that we're going to be right here on our interval.

Then we're going to go next to f of one, which is going to be this point right here. Then we're going to finish at f of 1.5, which is going to be on this point right here. Notice that we don't go all the way to two because that's on the right side of the sub interval, and we're only looking at the left endpoints. So we've gone ahead and taken a look at the leftmost side of each subinterval. So that's what happens when we add up this summation.

That's what we get. Notice it's the exact same thing that we got when dealing with just filling this with rectangles and finding the leftmost endpoints. But now we have a more compact way of writing this. So when we go ahead and do the math here, these would be at the heights of each rectangle, and then the width, Δx, we could find it using the equation that we talked about up here. So we have the Δx is b - a / n.

So if we want to calculate Δx, it would be the b value of two minus the a value of zero divided by the number of rectangles, which is going to be four.

For the following graph, write a Reimann sum using left endpoints to approximate the area under the curve over [0,5] with 5 subintervals.

• A. $5$
• B. $\frac{15}{2}$
• C. $6.25$
• D. 10

Answer: (A)

For the following graph, write a Reimann sum using left endpoints to approximate the area under the curve over [0,6] with 6 subintervals.

• A. 9.62
• B. 6.62
• C. 7.16
• D. 8.15

Answer: (A)

Write the Riemann sum that would approximate the area of the following graph over the interval [0,3] using 3 subintervals.

• A. 8
• B. 12
• C. 24
• D. 6

Answer: (D)

Left, Right, & Midpoint Riemann Sums

So in the last video, we got introduced to Riemann sums. And recall that this gave us a way of describing the area underneath the curve of a functions when we approximate with rectangles. This gave us a more compact equation, which allowed us to find this area. Now what we're going to learn in this video is that just like we saw in the last video for left endpoints when it came to doing Riemann sums, we can also do Riemann sums with right endpoints and midpoints using the same notation. So I'm going to walk you through this notation because it oftentimes seems challenging to students at first, but don't sweat it because I'm going to show you what this all looks like and how our equation is going to change based on what we've already learned.

So let's just go ahead and get right into things. Now you may recall that when estimating the area under the curve of a function, we actually can use function values at the left endpoints, right endpoints, or midpoints of the rectangles we draw inside the function. And Riemann sums are no exception to this. Let's say we had an example where we were asked to set up the left, right, and midpoint Riemann sums for this function right here given that interval, zero to two.

Now, at first, what I would do is I would set up the left endpoint Riemann sum because this is what we're familiar with. We know that inside of the function is going to be this x_k-1, where x_k-1 is going to start on the leftmost side of each rectangle. So we're looking at this rectangle, we start on the left side. If we look at this rectangle, we start on the left. And this x_k-1 that goes inside the function, well, what this tells us is where we're starting and how we're going to increment each rectangle.

So in this case, what it's going to be is (k - 1) times Δx. And the reason of this (k - 1) here is because we're starting on the leftmost side, then we're going to go over to a whole width Δx, then we're gonna travel another width Δx, then another width to get every single rectangle within the function that we drew. So if I wanna set up this notation down here, it's gonna look like this. The left endpoint Riemann sum, ℓ(n), is gonna go from k = 1 to n, which is the number of rectangles. Now in this case, I can see we have four rectangles, so n is going to be four.

Now inside of here, what I'm going to do is put in f of x_k-1 times Δx. We have the height times the width right there. So the height f of x_k-1, well, that's going to be this function with k - 1 plugged in. So we're going to have four minus, and then we're going to have x_k-1. That's going to be squared. This is all going to be multiplied by Δx. Now, what is Δx? Well, the way that I can find Δx is using this equation we've talked about in the previous videos, which is going to be the high point of our interval, which I can see here is two minus the low point, which I can see starts all the way over at zero divided by the number of rectangles that we have, which is four. Two minus zero over four is 0.5.

So the width here is going to be 0.5. And that right there would be how we could set up the Riemann sum for the left endpoint. Now this should be pretty straightforward to us because we've talked about this in the previous videos. But things might get a little more challenging if we try to look at other sides of the rectangles. Since we're looking at the left endpoints, we know how to solve this.

But what if we shifted gears here and took a look at the right side of the rectangles instead? So notice we drew these rectangles on the right endpoints, but it turns out we can use a similar process when it comes to finding this Riemann sum. So when it comes to a right endpoint approximation, this equation is going to change. But the way that we can change this equation is by thinking about what exactly is going to happen if we look at the right sides. Now when we look at the right sides, this x_k^* that goes inside of our function is going to change.

Rather than having x_k-1, we're just going to have x_k since we're starting on the right side of these rectangles instead of the left side. So our function is going to be f of x_k times Δx is going to go inside of our summation, and then the rest of the summation is going to be the same as we've already learned about. So doing a problem like this is going to look something where we have this summation here. It goes from k = 1 to 4, since we have four rectangles. And what we're also going to do is change this function here.

So we're now plugging in x_k instead of x_k-1. So we're going to have four minus, and that's going to be x_k squared. So this right here would be our function, and then that's going to be multiplied by Δx. And since we still have four rectangles and we're going to the same interval, that's going to still be a width of 0.5. You could use the equation to check, but we can clearly see there are four rectangles and it's the same interval zero to two.

So this is going to be right endpoint Riemann sums. Now when it comes to this x_k value that we put inside the function, what we do is we say that this is going to be a + k times Δx. And I want to make something clear here. This is just some notation you might see in your textbook, so it's not critically important that you memorize this. But the reason that I'm showing this to you is because you may come across it, and all this is really telling you is where you start when it comes to your rectangles.

Since we have a + k times Δx, that means we start at the beginning of our interval. Then we're going to start one width over since it's k times Δx rather than k - 1 times Δx. So we're starting one width over, then we travel another width, then we travel another width. And notice how we're just finding the right endpoints for this rectangle as opposed to the left. So now let's take a look at our final example here, which is the midpoint approximation.

Now the midpoint can oftentimes be the trickiest. Because notice for the midpoint approximation, we're not starting from the left or right side. We now need to think of some kind of notation that will get us to the middle point of each of our rectangles. Now the way that we deal with midpoint approximations is like this. The x_k^* that's going to go inside your function is going to be the left side of each rectangle.

So we look at a rectangle, we're going to start with the left side, which is x_k-1, and then we're going to add that to the right side of that same rectangle, which we learned was x_k. So we just have the left endpoint and then the right endpoint, then we're going to divide this by two. Because in essence, what we're doing is we're kind of averaging these two numbers right here to find this middle point in between the rectangle. Right? So that's why we add them up and divide by two, because we're getting this middle point, and that's going to give us the inside of our function.

The inside of our function is going to be x_k-1 + x_k all divi

Approximate the area under the curve $f\left(x\right)=\sqrt{x+3}$ over the interval $\left\lbrack1,5\right\rbrack$ using the Right Riemann sum with 8 subintervals.

• A. 9.75
• B. 9.96
• C. 9.54
• D. 9.72

Answer: (B)

Approximate the area under the curve $f\left(x\right)=\frac{2}{3x+5}$ over the interval $\left\lbrack4,10\right\rbrack$ using the Midpoint Riemann sum with $n=3$ subintervals.

• A. 0.55
• B. 0.48
• C. 0.43
• D. 0.61

Answer: (B)

Left, Right, & Midpoint Riemann Sums Example 1

So in this problem, we're asked to set up and solve the left, right, and midpoint Riemann sums given this function, \( f(x) = x^3 \) on the interval from zero to four. We're asked to use four subintervals to solve this. We're also asked to tell what is an overestimate and what is an underestimate based on the results we get. This is an interesting example where we're asked to apply all of these techniques for Riemann sums that we've learned about. But let's just go ahead and go right through them to see what all of these look like.

The main function we're going to be dealing with is \( f(x) = x^3 \). The next thing I'm going to calculate is the width of every subinterval that we're dealing with. Recall that the width of each subinterval, which is essentially the width of every rectangle, is \( b - a \div n \). \( b \) is the high point of our interval, which is four, and \( a \) is the low point of our interval, which is zero, then it's going to be divided by \( n \), the number of subintervals, in this case, four. So, \( 4 - 0 \div 4 \) comes out to one. I am calculating this width because if we are dealing with a function plotted on a graph \( x \) and \( y \), and we have the curve of our function on the interval from zero to four, my strategy would be to fill this area with a certain number of rectangles, approximate the area of each rectangle, and add them all up using Riemann sums.

We don't necessarily need a graph to do this; knowing the width of each rectangle, which we call \( \Delta x \), is crucial. This value is the same for every single subinterval. If I take this value and multiply it by the function value, which would be the height, it allows me to compute width times height, essentially estimating the area under the curve of a function. Now, let's start with the left endpoint approximation. The equation for dealing with left endpoints is \( L(n) = \sum_{k=1}^n f(x_{k-1}) \Delta x \). We already have \( \Delta x \) calculated as one. This summation goes from \( k = 1 \) to \( n \), where \( n \) is four in this case, and the function will take \( x_{k-1} \) and plug it into \( x^3 \) since that's our function.

Evaluating this sum, we go from \( k = 1 \) to four, and because we start with \( k - 1 \), our values will be zero cubed, one cubed, two cubed, and three cubed since we start from zero, which is the leftmost side of our interval. Summing these, you should get 36 as the approximate area under the curve of our function, which is the result for the left endpoints.

For the right endpoints, the equation looks like \( R(n) = \sum_{k=1}^n f(x_k) \Delta x \). Unlike the left endpoint, we start from one and increment to the right, thus covering one cubed, two cubed, three cubed, and four cubed, totaling 100. This sum would represent an overestimate as it's higher than the actual area under the curve.

The midpoint rule is calculated using the equation \( M(n) = \sum_{k=1}^n f \left( \frac{x_{k-1} + x_k}{2} \right) \Delta x \). Here, we average the left and right sides for each subinterval and plug this into our function. The midpoints being 0.5, 1.5, 2.5, and 3.5 give us a total area of 62 for the summed cubed values, providing a more accurate but not exact approximation.

From these calculations, the left endpoints underestimate the area (36), the right endpoints overestimate (100), and the midpoints provide a result closer to the actual but not exactly (62). The gerencial strategy would be to visualize these approximations by considering how filling these rectangles might look under the actual curve plotted from these sums results. This helps understand the character of over and underestimation in Riemann sums. If you have any questions, let's discuss and clarify further.