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

Understanding sigma notation is essential for evaluating finite sums, which can simplify complex calculations. The sum and difference rule allows separation of terms, while the constant multiple rule enables pulling constants outside the summation. Riemann sums, which estimate the area under curves, can be calculated using left, right, or midpoint approximations, with the width of rectangles defined as $\Delta =\frac{b-a}{n}$. Mastering these concepts enhances problem-solving efficiency in calculus.

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 Afks are are 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 a 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 with 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 turned 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 gonna 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. So I hope you found this video helpful, and let's try getting a bit more practice to really make sure we have this down. See you in the next video.

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

• 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. $\frac{2}{3}$
• C. $2$
• D. $\frac{4}{3}$

Answer: (C)

Evaluate the following summation (make sure your calculator is in radian mode): 
${\sum }_{i=1}^4\sin \left(\frac{\pi i}{8}\right)$

• A. 1.00 rad
• B. 1.38 rad
• C. 3.01 rad
• D. 0.62 rad

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 gonna 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 gonna start at k = 1, because that's where we're starting. And then that's gonna go all the way up to 5. And the k is gonna 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 gonna be from one to five, and it's gonna be ∑ k = 1 5 3 k 2 . And that would be the solution for this sum. So this is how we can take this repeated notation here, 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 gonna 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 gonna 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 gonna 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 gonna 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 k were equal to two, when we go up to two, that's gonna 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 got introduced to sigma notation and solving these types of summation problems that look like this. Now, in this video, what we're going to do is take a look at some of the rules that apply to these finite sums. Because 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 go ahead and jump right into things. First off, we're going to take a look at the sum and difference rule when it comes to dealing with these summations.

Now we could have a situation like this where we have the summation, a finite sum of these two pieces right here, a_k plus or minus b_k. Now it turns out if you're dealing with plus or minus, what we can do is we can expand this out and separate it into two individual sums. So it would be the summation of this piece plus or minus the summation of that piece. As you can see, it's very similar to what we've seen for derivatives and integrals. 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 k equals one to three, and it's going to be these two values right here. Well, doing this, we're going to have this 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 minus the summation of k. And this would be how you could separate these out into two different sums. Now doing this example, we would have 1² + 2² + 3², and all of that would come out to 14. Then going to this piece, we would have 1 + 2 + 3, which comes out to six.

And if we subtract these values, we get eight, meaning eight would be the solution to this sum. Now we could get the same result if we just took these values and directly plugged them in here. But notice how I have to write out 1² - 1 + 2² - 2 + 3² - 3. That might actually take a little bit longer, so it's easier to split this out into two different sums and solve it this way. Now another rule that we need to be familiar with is the constant multiple rule.

The constant multiple rule says that we can take some constant being multiplied by this piece a_k and pull it out to the front. So that means we're just going to have the sum from k equals one to n of our equation a_k, and we multiply by our constant, that will give us the result. So an example would be something like this. We have this sum right here from k equals one to three as of 2k². What I can do is take this constant two and pull it outside of the summation.

So this comes out to the front, meaning all we're going to have is the summation k equals one to three of k². Now doing this, we already know one to three for k², that's going to give us 14. So what we're going to ultimately 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. So as you can see again, we could have just taken the numbers and plugged them in here, but it was way more simple to pull out this constant front, and then go ahead and multiply everything out.

So that is the process for dealing with these summations. Another thing we need to take a look at is whenever we have the summation of a constant value. So an example like this would be this case, where we have the summation from k equals one to n of some constant c. If we ever have this, all you need to know is that this is going to be the same thing as taking your constant c and multiplying it by the end of the index, which is n. So an example of this would be the summation from k equals one to 18 of three.

Well, this is going to be the same thing as 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. Now for all these summations, as you could see, we could just take the values and plug them into our equations and solve that way and go all the way up to whatever end of the index we have, but that would take a really long time and probably be quite tedious. So it's a lot better to just use these rules that we know about. As you can see, they really follow the same process when it comes to these rules we learned for derivatives and integrals.

Now in most problems that we're going to see, we'll need to use multiple rules together to find the summation we're looking for. But that's okay because with this practice up here, I think we should be able to do that in the example we have. So it's here, it says to evaluate the given sum, and this is the sum that we're dealing with. Now the way that I'm going to do this is by using the rules that we've learned. So what I'm first going to do is recognize that we have these two portions that are being subtracted.

And since we have a subtraction here, I can use the sum and difference rule. So to do that, I'm going to split this into two separate sums. So you're going to have the summation of 6k², and then that's going to be minus the summation of nine. So we'll have from k equals one to three here and k equals one to three there. Next, what I'm going to do is recognize something.

We have this six here, which is a constant in front of k². So what I'm going to do is pull that outside of the summation. So it's going to be six, and that's going to be the summation from k equals one to three of k². Then we're going to have minus, and then we have this constant nine here. So we have from k equals one to three.

Now evaluating this, we'll have six, and then we have k equals one to three for k². And we've already done this multiple times already. That's just 1² + 2² + 3², which all comes out to 14. So that's what we get for that summation, and that's going to be minus. Then that's going to be minus the sum from k equals one to three of nine.

But remember, whenever we have this constant here, we see to take whatever this value is and multiply it by the front of this number. So this is just going to be nine times three. We'll have six times 14 minus nine times three, and all of this just comes out to 57. So 57 is the result of the sum and the solution to this problem. So this is how you can use rules to solve these situations where we're dealing with this 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 these various sums. See you in the next video.

Evaluate the following summation: 
${\sum }_{t=2}^{6}5t+\frac{4}{t}$

• A. $\frac{471}{5}$
• B. $\frac{529}{5}$
• C. $\frac{26}{3}$
• D. $\frac{22}{3}$

Answer: (B)

Evaluate the following summation: 
${\sum }_{i=0}^{4}2i+1$

• A. 9
• B. 20
• C. 21
• D. 25

Answer: (D)

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

• A. 6
• B. 25
• C. 4
• D. 13

Answer: (B)

Evaluate the following summation (make sure your solution is in radians):
${\sum }_{k=0}^{5}2\tan \left(\frac{\pi k}{3}\right)-\pi k$

• A. 0 rad
• B. -43.65
• C. -47.12 rad
• D. -50.58

Answer: (C)

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. 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 k1=1 at the start of our index to n, and that's going to be for f(x(k−1))⁢⁢∙⁢⁢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, well, what I can do is I can take a look at how many rectangles that 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 k1=1 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 k1=1 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 x0, 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 xn. That's going to be the highest point that we go to on our curve. So if I wanna 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(0), that's this leftmost side. It's going to be plus f(0.5). Notice that we're going to be right here on our interval.

Then we're going to go next to f(1), which is going to be this point right here. Then we're going to finish at f(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 over n.

So if we wanna 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. Now doing this math here would give us 0.5 or one half. So if you took 0.5 and put it in here, and then you evaluated the function at each of these values that we see for the function that we have, that's going to give you 6.25 or the same result that we would get if we just fill this with a bunch of rectangles and added them all up. So this is how we can use our Riemann sums for this new notation to figure out what the approximate area is underneath the curve of a function. And this was for left endpoints, which we saw in this case, but it turns out we are going to see examples where we're dealing with right endpoints or midpoints, so we need a more general rule.

And the general rule states that our approximate area is going to be from k1=1 to n, is going to be f(xk*). So this star right here just tells us that we could be starting from any side. We could be starting from the leftmost side. We could be starting from the right, somewhere in the middle. This is just the general notation for any type of Riemann sum that you could have left, right, midpoint.

And then this is the notation for specifically the left endpoints that we did in this video. So hope you found this video helpful, and let's try getting some more practice with these Riemann sums. See you in the next one.

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: (B)

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 function 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 often 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 left, right, and midpoint Riemann sums for this function right here given that interval, zero to two.

At first, what I would do is 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.

In this case, what it's going to be is (k - 1) × Δx. The reason for 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 going to travel another width Δx, then another width to get every single rectangle within the function that we drew. So if I want to set up this notation down here, it's going to look like this. The left endpoint Riemann sum, L_n, is going to 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(xk-1)×Δx. We have the height f(x_k-1), well, that's going to be this function with k-1 plugged in. So we're going to have 4 - (x_k-1)² all 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(xk)×Δx 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 four, 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 4 - (x_k)². 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 × 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 × Δx, that means we start at the beginning of our interval. Then we're going to start one width over since it's k × Δx rather than k - 1 × 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 often 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 of 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)/2. So to solve an example using the midpoint approximation, well, it's going to go from a k value of one, the start of our index, all the way up to the number of rectangles we have, which is four. So that's our subinterval right there, four rectangles. And then we're going to go ahead and take this function and replace it with the inside of the function we figured out here. So we're going to have this 4, and it's going to be minus this value.

So we're going to have (x_k-1 + x_k)/2, and then that whole thing is going to be squared. Since we have an x squared there, we're plugging in the inside of this function. And then it's going to be multiplied by our width Δx, which again, we have the same interval and the same number of rectangles. So this is just going to be 0.5. So this right here would be the solution for midpoints,

Approximate the area under the curve $f\left(x\right)=\sqrt{x+3}$ over the interval $[1,5]$ 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 $[4,10]$ 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) = x³ on the interval from zero to four. We're asked to use four subintervals to solve this. We're asked to tell what is an overestimate and what is an underestimate based on the results we get. So 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 looks like.

So what we're dealing with is this function ultimately, f(x) = x³. That's the main function we're going to be dealing with. And the next thing I'm gonna calculate is the width of every subinterval that we're dealing with. Because recall that the width of each subinterval would be basically the width of every rectangle, which is b minus a over n. B is going to be the high point of our interval, which is four.

And we're gonna have minus a, which is the low point of our interval zero, and then it's gonna be divided by n, which is the number of subintervals. In this case, that's going to be four. So we have four minus zero over four, which all comes out to one. Now, why exactly am I calculating this width here? Well, because my ultimate goal is this, if we are dealing with some kind of function, let's say that it's plotted on a graph x and y, and we have some kind of curve of our function.

I'm just gonna write something like this. This is f(x). And what we're doing is we're looking on some interval from zero all the way to four, and we're trying to find this area. My strategy would be to take whatever area I have and fill it with a certain number of rectangles and approximate the area of each of these rectangles and add them all up. Now in this case, we're using Riemann sums.

We don't even necessarily need a graph to do this. What I would need to know is the width of each rectangle that we would have. And the width of each rectangle would, in essence, be the width of the subinterval, which is what we call Δx. So this is gonna be the same for every single subinterval we have. Every single piece that we're estimating the area is going to be this value right here.

And if I go ahead and take this value, and then multiply it by the function value, the function output, which would be the height, that allows me to get a width times height, which is basically the area of a rectangle, thus estimating the area under the curve of a function. Now we're gonna start by dealing with the left endpoint approximation. And we actually have an equation when it comes to dealing with left endpoints. The left endpoints are going to be L_n = ∑ (k = 1 to n, f(x_k-1) × Δx).

Now at this point we already have Δx calculated, we said that that's one. So I can see that this whole Riemann sum is equal to one multiplied by this summation. Now the summation goes from k equals one to n. Now what exactly is n? Well, n is the number of subintervals, which in this case is four.

So we go from k equals one to four, and then we have this function. Now this function is going to take x_k-1 and plug it in to the function we have up here. Since I can see our function is clearly x cubed, that means we're gonna have x_k-1³. And this is going to be the function we have, and one multiplied by anything is just gonna be that number or that value, so this sum I'm just gonna write by itself. So, we'll have ∑ (k = 1 to 4, x_k-1³).

Now how exactly could we evaluate this sum? Well, what I need to do is look at these numbers I have here. We go from one to four, but notice that our Δx here is one, and we're starting from the left endpoints. Because we're dealing with the left endpoints k minus one, we're going to start at an x of zero value which would be right at the start of our interval of zero. So doing this sum, what we're going to have first is we're going to have zero cubed.

Then we're going to have plus and then we need to go up by a width of one. So we're going to have zero plus one, which would be one cubed. Then we're going to go up by another width of one, we're going to increment up by one. So then we're going to have two cubed, and then we're going to increment up by one again. That's gonna give us three cubed.

And this right here is where we would stop. Because notice we have one, two, three, four pieces that we're adding together, and notice that this goes from one to four. So this is good. The reason we don't actually go all the way up to four is because we're dealing with left endpoints. So what we've done here is we've gone all the way up to three, because if we're looking at the leftmost side of each rectangle we would stop at three and not go up to four.

So this right here would be our sum, and if you go ahead and crunch all these numbers, you should get 36 as the approximate area underneath the curve of our function. So 36 would be this result, and that would be the solution for this first situation, the left endpoints. Now what if we take a look at the right endpoints? Well, the right endpoints would look like this. It's R_n = ∑ (k = 1 to n, f(x_k) × Δx), and then this is going to be within a sum.

So notice we have the height times the width for our rectangles, but then this is gonna be the sum from k equals one to n. So this would be the Riemann sum from the right side, which is gonna be the next piece that we take a look at. Notice it's very similar to the left side, except rather than having k minus one, we have k. So rather than starting here at the zero like we would before, we would actually start at one this time, because we're incrementing up by one, and then starting our count to the right sides of the rectangle. So going to this sum, we're going to go ahead and have this Δx just be one multiplied by the front of this whole piece, but I can just write this as just the sum by itself because one multiplied by anything is just that number.

So we're gonna have ∑ (k = 1 to 4, f(x_k)). Now for f(x_k), I need to take x_k and plug it into this function that we have, that's gonna give me x_k³. So this is the result that we get for our sum, and now I just need to evaluate this. Now doing this, remember, we're going to start one width over from zero. So if we're looking at an x axis for example or some number line, if we started here at zero, we're going to actually go over here to one to start, and then we're going to increment all the way up to four.

So that's gonna be the result that we do. So going through this process, well, let's just go ahead and do that. So I'm going to go ahead and increment up based on what we have down here. So we're going to first have one cubed. Right?

And keep in mind, we're looking at this interval up here, zero to four. We're gonna have one cubed first, and then if we go one unit over, then we're gonna have two cubed. Then we could go one unit over, and that's going to give us three cubed. Then we could go one unit over, and that's going to give us four cubed. So we have one cubed plus two cubed plus three cubed plus four cubed, and this entire thing is going to come out to 100.

So 100 would be the area when we use right endpoints. So we now have the left endpoints, and we have the right endpoints. But what about if we took a look at the midpoints? Well, the midpoint is going to be this equation that I write down. It's gonna be that M_n = ∑ (k = 1 to n, f((x_k-1 + x_k) / 2) × Δx). Now for this sum, Δx, we already figured out is one. So I'm not even gonna write it down here, since I know it's just gonna be multiplied by this whole sum. So what we're gonna have is the sum from k equals one to the number of subintervals, which we said already was four. And then within our function here, x³, we're going to plug in this quantity.

So what that's going to give me is (x_k-1 + x_k) / 2, and then this whole thing is going to be cubed. So this is the result. Now how exactly do we find these midpoints? Well, keep in mind for the midpoint, this is what we're plugging in. And if I go over from zero to four, well, if we think about this on an x axis, we would have zero right about there.

And our first increment up would be one since we have a width of one. We're trying to go to the midpoint, trying to take the leftmost side, we're trying to take the rightmost side, and we're trying to average those. So we'd go right here in the middle between zero and one, which is 0.5. So that would be the midpoint that we would be dealing with. So if we want to do the midpoint rule, we'd first start with 0.5.

So writing out this sum, we're going to have 0.5 cubed. And then we're going to have plus, and we're going to go up by a width of one. So it's gonna be 1.5 cubed plus, and then we're gonna go by width of one. It's gonna be 2.5 cubed, then we'll have plus, and then width of one is gonna be 3.5 cubed. So this is what you need to evaluate right here when doing this sum.

If you go ahead and plug this all into a calculator or solve this out by hand, you should get 62 as your result. So this would be the solution with the midpoints, and then we have the left and the right endpoints as well. And that is how we deal with Riemann sums from all three cases left, right, and midpoints. Now a final piece of this question asks us to tell which is an overestimate and which is an underestimate. Well, by looking at these numbers here, I can see that we got a low number when we did the left endpoints.

So I can see that this right here would be an underestimate for our results. So that means when we did this estimation right here, we got a lesser result. We got less area than we truly have for the curve of this function underneath. But if I go to the right endpoints, notice I got a big number of a hundred. So what I could say is this right here would be an overestimate.

We call this over, and the reason we're saying that this is over is because in this case, we got a larger number that we were dealing with. We got a hundred. This is clearly over the estimate. And then 62, this midpoint, this is a much closer to accurate result in the left or the right, but it's still not necessarily gonna be fully accurate. This was just a way of kind of averaging things out by taking some of the left and some of the right and going right down the middle, but this still is not a truly accurate answer.

But this gives us a more accurate answer than the direct left or right endpoints did. But we can clearly see based on what the midpoints are that the left endpoints would be under this time and the right endpoints would be over. This is not always consistent. Sometimes it's the other way around. Sometimes the right endpoints are going to give you an under approximation and the left endpoints will give you an over.

But in this case, this is just how it happens to work out. So this would be the solution to our problem, and that is how you can deal with left, right, and midpoints when it comes to Riemann sums, as well as figure out what's an under approximation versus an over approximation. And if you ever have any doubts when solving these types of problems, feel free to draw up a little mock graph. So feel free to kinda, like, draw some kind of random arbitrary curve for your function, and then fill this with endpoints from the left side or from the right side. And then go ahead and check your area and how much is gonna peak over the curve based on what you calculate, how much peaks over, or how much goes under.

And that's going to tell you what your approximation is and where you end up when it comes to your calculations. So, I hope you found this video helpful. That is how we can do these Riemann sum problems. Let me know if you have any questions, and let's move on.