Estimating Area with Finite Sums: Study with Video Lessons, Practice Problems & Examples

To estimate the area under a curve, we can divide it into rectangles, using left, right, or midpoint approximations. The width of each rectangle, denoted as Δx, is calculated using the formula $(b-a)/n$. The height of each rectangle varies based on the chosen endpoint method. Increasing the number of rectangles improves accuracy, reducing overestimation or underestimation of the area, which is crucial for understanding integrals and calculus concepts.

Estimating the Area Under a Curve Using Left Endpoints

Hey, everyone, and welcome back. So, one of the concepts we're going to need to be very familiar with in this course is the process for finding the area underneath the curve of a function. Now, this might sound pretty simple since all we're trying to do is figure out what the area is underneath these functions that we've already seen before. However, here's where things get a bit difficult. Notice we don't usually have a simple shape that forms.

It's not like we're dealing with a triangle or a square where we know what the area of those shapes is. We're dealing with a shape that we don't really know. But that's what we're going to be talking about in this video is how we can estimate the area underneath these shapes. I think you're going to find this is pretty straightforward, a concept we've already learned, so let's just jump right into things. Now, the process for estimating the area under a curve of a function is to take whatever your function curve is and break it up into a certain number of rectangles.

Now, what you'll likely see in your textbooks or class is that n is going to represent the number of rectangles that you'll have. And this n, we also call the number of subintervals. So subintervals or the number of rectangles mean the same thing. And what we need to do is take whatever sheet that we have, fill it with these subintervals or rectangles, and then add all of their areas together. So let's say, for example, we're dealing with this case over here.

For this situation, what I want to do is fill this with rectangles. And it says in the example here that we're dealing with to approximate the area under the curve, which is this region r that we see. And what we want to do is we want to approximate this region for two rectangles and four rectangles. Now we'll start by putting only two rectangles underneath the curve of our function. And notice that we've had these rectangles so that the upper left corners are touching our function.

This is what we call left endpoints for our approximation here. So since we're focusing on the left endpoints, we're always going to be focusing on these values within our function. Now, I can see that I'm dealing with two rectangles, and I need to add up the areas for each of these rectangles. Now we know the area for a rectangle is width times height. So we need to find the width and height for each rectangle, and add it all up.

Now, thankfully, the width is going to be the same for each rectangle that we have. We can clearly see that on our graph here. We call this width delta x. So delta x, which I can see on our graph here, goes from zero to one. So we have that that's a width of one because one to two would also be a distance of one.

Delta x is going to be our width, and that's going to be multiplied by the heights. Now the way that we find these heights is by using the function values because the height of each rectangle is going to be the function value at the left endpoint. So if we go over to our left endpoint, I can see that that's zero for this first rectangle. So what that means is that I'm going to have f of zero as my first height. Now next, we're going to add up our next rectangle, which is going to be the width delta x times the next height, which is going to have one plugged into our function.

Because I can see that that's going to be the left side of the second rectangle, so that's the next input I need to plug in. Now we already figured out that delta x here is one. And then what I'm going to do is multiply that by our height f of zero. Now to find f of zero, there are two ways I could do this. One way is to just look at my graph and I can see when we're at an x value of zero, we're at a y value of four.

But another way that I can find these heights here is by simply plugging the function value in the function that I have. So I can see that the x value that we have is one. And if I put one into this function right here, well, that's going to give me three as an output. So our area is going to be one times four plus one times three, which all comes out to seven. And this right here is the approximate area underneath the curve of this function.

Now, this is not a truly accurate answer for the area, because notice we have some extra area that peeks over these curves. So using these rectangles is really just a way of estimating about what this area is. But this is the strategy we use for finding these regions where we're trying to estimate the area under a function. Now the question, of course, becomes, is there another way to do this? Are there other strategies we can use or possibly ways to get more accurate answers?

Well, what we could try doing is simply adding more rectangles. So we dealt with two rectangles or subintervals in this example, but let's try now doing four rectangles and see the result that we get. So if I add two more rectangles to this, giving us four rectangles underneath this curve, well, again, notice we're using these left endpoints here, and I'm going to use the exact same strategy for calculating this area. Now this one's going to be a little bit more tedious because notice there are four areas I have to add up since we're dealing with four rectangles. Now, each of these rectangles will be width times height that we're adding up.

And what I need to do is go through and figure out the function value to get the height, and then figure out the width, which is going to be the same for all these rectangles. And this is going to allow me to calculate or estimate the area under this curve. Now to find the width, what I could do is look at my graph here, but it turns out there is an equation you can use for calculating width. The equation that we use looks like this, it's going to be b minus a over n up here. So what I can do is use this and recognize that our b value is going to be this high point of two minus a, which is the low value of zero, all divided by the number of rectangles we have, which is four.

Now two minus zero over four is one half or 0.5, so that is going to be the width of each rectangle. So that's going to be our delta x value. And notice we can clearly see just looking at our graph that each rectangle has a distance of 0.5. Now next, what I need to do is figure out what the heights of these rectangles are. And to do this, well, I just plug in my left most endpoint for each rectangle and put it into this function.

So we're going to first have f of zero, that's this endpoint. Then we're going to have f of 0.5. Then we're going to have f of one, then we're going to have f of 1.5. So I've gone ahead and plugged in these values here, and now I need to just go ahead and write what these numbers are. So delta x, we calculated to be 0.5.

That's going to go on the outside here. I was able to factor this delta x out since we know that this width is always the same for each rectangle. Now next, we're going to have f of zero. Now f of zero, well, what you need to do is just take zero and plug it into this function, which is going to give you four. Then we're going to have plus f of 0.5, which you can plug into this function to give you 3.75.

Next, we'll add f of one, which if you plug one into this function, you'll get three. And then you will have f of 1.5, which if you plug 1.5 into this function, you're going to get 1.75. So this is what happens when you plug in all these x values into your function. And if you go ahead and multiply this out and add everything up, you should get 6.25 as your result. Now notice how we did get a different result even though we had the same function in both cases.

So we had the same function and we were going from zero to two in both situations. The only difference was the number of rectangles that we had. Now 6.25 is also not going to be an accurate answer since we have some extra area peeking over this curve. But the question becomes, which one of these is more accurate, the 6.25 or the seven? Well, it turns out the more rectangles or subintervals that we add, the more accurate our result is going to be.

And this is because notice how if we added more rectangles, we actually got a little bit less area peeking over this curve. We still got some, but it wasn't as much as when we had two rectangles. So this 6.25 is actually more accurate than the seven was. Now this still doesn't mean that you're going to get a perfectly accurate answer, but it does mean you will get a more accurate answer for the more rectangles. So if we were to add eight rectangles in here, then we would get an even more accurate answer than the 6.25.

However, if we only did one big rectangle, then we would get a much less accurate answer than either of the results we had here. So this is the process for estimating the area under a curve. As you can see, it is a little bit tedious, but hopefully, it makes sense that all we're trying to do is figure out what the area is for each rectangle that we put underneath the curve of our function, and we're just adding it all up. So hope you found this video helpful, and let's try getting some more practice with this concept. See you in the next video.

Use five rectangles to estimate the area under the curve of $f\left(x\right)=x^2$ from $x=0$ to $x=5$ using left endpoints.

• A. $A=30$
• B. $A=55$
• C. $A=41.25$
• D. $A=41.67$

Answer: (A)

Use two rectangles to estimate the area under the curve of $f\left(x\right)=\frac{1}{2}{x}^2$ from $x=0$ to $x=3$ using left endpoints.

• A. $A=1.6875$
• B. $A=4.21875$
• C. $A=8.4375$
• D. $A=4.5$

Answer: (A)

Use four rectangles to estimate the area under the curve of $f\left(x\right)=x^2+2$ from $x=0$ to $x=2$ using left endpoints.

• A. $A=5.00$
• B. $A=6.67$
• C. $A=5.75$
• D. $A=7.75$

Answer: (C)

Estimating the Area Under a Curve with Right Endpoints & Midpoint

So in the previous video, I want you to recall how we talked about being able to estimate the area under the curve of a function by breaking it into many rectangles or subintervals. We talked about how we could do this calculation using left endpoints. Now what this meant is that we would take the curve of some function and we would fill it with rectangles, a certain number of rectangles, we call it subintervals. And what happened is these rectangles would all have the same width, but the height would vary depending on where the left side of the rectangle was. Doing this would give us an overestimation because we'd have this area peeking over the curve, but it turns out there are other ways we could do estimations as well.

There are certain ways that we can draw these rectangles to get an underestimation or even some kind of mix of an over and underestimation that might allow us to get a more accurate answer. So let's take a look at how we can do some of these other calculations to approximate the area. Now since we were able to draw these function values and find them at the left endpoints, it turns out that we can also do this when dealing with the right endpoints of our rectangle. We could even do this if we went somewhere directly down the middle, which we call the midpoint approximation. These are two other ways that we can do approximations to find the area under a curve.

Now, it's going to be the same process that we saw in the previous video for just adding up the area of these rectangles. The main difference when doing either right endpoints or midpoints is to focus on the heights, because the widths are always going to be the same for these rectangles. We'll have the same widths here and the same widths over there. So that's not going to change. What is going to change is the heights. Because notice how this height on the left side, when we did the left endpoint approximation, is going to be different than this height when we do the right endpoint approximation. So let's see what this all turns out to be. Well, if I'm doing a right endpoint approximation, there's a few things I know. I know that I'm going to be filling the inside of this curve with rectangles. Notice this rectangle is touching the x-axis here.

It's pretty small because we're touching this rightmost point right there on this interval. Now it's going to be the same process of me adding up the areas, but you may recall previously when we did this, we got this relationship here where we had these specific inputs. Well, I'm still going to have the area of rectangles, so we're still going to have width times height, but the inputs are going to change. Because since we're now looking at the rightmost endpoints, that means we're going to focus on the right x value. So if I go to the right side of this rectangle, this value I can see is one.

So we're going to have our width delta x times f of one. Now likewise, if I go to the right endpoint of this other rectangle right here, notice that we're at two. So that means we're going to have f of two. So this is how we could estimate the area using right endpoints. And notice how we are going to get different values. Because we have f of one and f of two, that means that we're looking at this rightmost side, meaning that we're going to get a different area that we calculate. So it's going to vary on the outputs we have here and there in order to estimate about the area under the curve. So this is a right endpoint approximation. This is still an estimation. It's not going to give us the true or accurate answer for what the area really is under this curve, but it's just another way that we can try approximating this area to figure this out.

Now the third strategy is going to be to use a midpoint approximation. And oftentimes, the midpoint approximation is actually the most accurate. Because rather than getting a drastic overapproximation or a drastic underapproximation, the midpoint kinda gives us this balance between some peeking under the curve and some peeking over the curve. Again, this is not going to be a completely accurate answer, but it is going to give us another way to estimate about what the area is. So it's going to be the same process where we go ahead and find the area for each rectangle.

It's going to be width times height. That's always the area of a rectangle. But now, once again, it's going to be the x inputs that change. But how exactly could we find the x inputs for these rectangles? I mean, for the right endpoints, we're able to just look at where these right values were. What can we do for midpoints? Well, notice here how we're going right down the middle of each of our rectangles. So if we're trying to find these midpoints, we need to find what these x values right in between would be. Well, what I can do is I can, in essence, just average these two numbers or think about what goes between zero and one exactly. And that would be 0.5, which is the same thing as one half.

If you were to take one plus zero and then divide it by two, that would give you one half. That's just averaging those two numbers, but you can also see it's directly in the middle of zero and one. So we're first going to have f of 0.5. Now I want to emphasize again, the width is always the same. For all of these rectangles, and for all of these estimations, this delta x doesn't change.

And again, you could use this calculation over here to find the width exactly. But, what is changing is what the inputs are into our function, and therefore, it's going to change the outputs as well. So you get some output that's right about there. Now next, I can go to this next midpoint here, and I can see it's between one and two. Directly between one and two is 1.5.

So my next input for our function is going to be 1.5. And again, this is going to give me another output, which is right about over here. So this is how we can find these function values, and what we can do is multiply them by the width, and that's going to give us the area, and that would be our way of calculating the approximate area underneath the curve of the function. Now, again, this can often give us a more accurate calculation, but it's still not going to be the true area under the curve because these are all just estimations whenever you do a rectangle approximation. And the width stayed the same in all of these examples because we had the same number of rectangles under the same function curve. And so because of that, we were able to get the same width for each rectangle. The only times the width is going to change is if you add more rectangles under the curve or you change the interval in some way that you're looking at. But for this example, as you can see, the widths were the same. It was really just the heights that were changing.

And because the heights were changing, it was the inputs inside of our function that was changing these outputs, thus changing the heights of the rectangles. So this is how you can do approximations when it comes to trying to find the area underneath the curve of a function. And I encourage you to check out some of the practice and examples after this video so we can see directly what some of the problems are going to look like as we get farther into this course and have to solve different situations where we're needing to estimate this area. See you in the next video.

Use three rectangles to estimate the area under the curve of $f\left(x\right)=\frac{1}{3}{x}^3$ from $x=0$ to $x=3$ using the right endpoints.

• A. $A=3$
• B. $A=12$
• C. $A=6.5$
• D. $A=5.3$

Answer: (B)

Use three rectangles to approximate the area under the curve of $f\left(x\right)=3{(x-2)}^2$ from $x=0$ to $x=3$ using the midpoint rule.

• A. $A=9.00$
• B. $A=6.00$
• C. $A=8.25$
• D. $A=15.0$

Answer: (C)

Estimating the Area Under a Curve with Right Endpoints & Midpoint Example 1

So let's try solving a problem like this. Here we're asked to approximate the area under the curve of \( f(x) = x^3 \) on the interval from zero to three using three rectangles. Now notice this problem asks us to use three different methods: the left endpoints, the right endpoints, and the midpoints. So we're really trying to see how accurate of an answer we can get to find the curve underneath this function \( f(x) \). Now we're not given a graph of this function, but what I'm going to do is draw kind of a mock-up graph right here.

Now this doesn't have to look exactly like the function does. I'm just gonna say it looks something like this. But what this does is tells me how I can draw this curve, what I can make my rectangles look like, and gives me kind of a reference as to what I'm trying to get out of this problem. So let's go ahead and get into things.

Now, what I need to do is fill up the underneath of this curve with rectangles, and that's going to give me the approximate area that we have by calculating the area of each rectangle individually. Now, what I'm first going to do right off the bat is calculate the width of these rectangles because we know that \( \Delta x \), the width, is \( b - a \) over \( n \). Now, \( b \), that's going to be the high point of our rectangles, which is three. \( a \) is going to be the low point of our interval, which is zero, and then it's going to be divided by \( n \), which is the number of rectangles we have, and in this problem that's going to be three rectangles. So we have three minus zero over three, which all comes out to one.

So one is going to be the width for each rectangle. Now, it turns out this width that we have built to \( x \), that's going to stay the same for every type of approximation we use because notice we have the same interval and the same number of rectangles every single time. So that's why this width is going to stay the same. So this isn't going to change in this problem. But what is going to change is the area for each rectangle because the heights of each rectangle are going to change depending on whether we start from the leftmost side, whether we go to the rightmost side, or we go somewhere in the middle, that the heights are going to change each time.

So that's what we need to figure out. Now we're going to start by using a left endpoint approximation. And I'm going to write this as \( A_L \), for area from the left side. Now, to do this, well, it's going to be the area of the first rectangle, plus the area of the second rectangle, plus the area of the third rectangle. There are always three rectangles in this problem.

Now, to write these, I'm starting from the leftmost side. So I'm going to touch the leftmost side of my graph here, and that's going to be my first rectangle. I will then touch the left side again, that's going to be the second rectangle, and then the third rectangle is going to go somewhere right about there. Again, I'm not writing this exactly as it would be, but this is just kind of a reference here so we can solve this problem. So we're trying to find the area right here.

Now to find this area, well, what I'm going to do is figure out where exactly the leftmost sides are. Now this left side would be on the leftmost side at the beginning of my interval, which would be zero. Then we would go one width over, and that would give me one. Then we would go another width over, and that would give me two. The zero plus one is one, one plus one is two, and then we just keep on going, but we're only focusing on the left side.

So those are all the numbers we care about. Now I know the area of a rectangle is width times height. The width is something that we already calculated. It's \( \Delta x \). It's the same for every approximation that we do.

Now the heights are going to depend on the inputs we have into our function. So we're going to have \( f(0) \), and it's going to be plus \( f(1) \) plus \( f(2) \), since we're only looking at the left endpoints of these rectangles. Notice how the way I did this is I just did width times height for each rectangle, width times height, width times height, and that's going to give me the approximate area under the curve of this function. So let's go ahead and plug in these numbers. Now we're going to have \( \Delta x \), which we already know is one.

We calculated this already. And that's going to be multiplied by \( f(0) \). Now to find \( f(0) \), just take zero and plug it into our original function. So we'd have zero cubed, which is zero, plus, and then we're going to have one cubed, which comes out to one, plus then we're going to have two cubed, which comes out to eight, because two times two times two is eight. So we're going to have zero plus one plus eight, and all of that's going to come out to nine.

So nine is going to be the result when we use left endpoints. So this is the process for solving these types of problems. In this case, we wanted left endpoints, so that's what we went ahead and used. But let's go ahead and see some other examples where we're using different endpoints. Like, let's say, instead, we start from the right endpoints rather than the left endpoints.

Well, to do this, I'm going to fill the area under my curve with rectangles, but this time I'm going to start on the rightmost side and fill it with rectangles like this. So I'll have a right endpoint there or another right side here, then another right endpoint rectangle there. And now we're trying to calculate the area of these rectangles to get the approximate area under the curve. Now I noticed right off the bat for this function, what's going to happen is you're going to get a little bit of extra area up here. So we should get we should get a more, a larger number for the results of our functions

We have some extra area. And, again, this is all just an approximation for the area about under the curve of our function. And since I know that the \( x^3 \) function does look something kinda like this, then I know we are gonna get this approximation. If it was something different, we might get an under-approximation. But here, it looks like it's going to be over.

So to calculate this area, well, let's go ahead and use the same method we used before. I'm gonna take the area from the right side this time, and it's going to be area one plus area two plus area three. Now again, for each of these areas, the \( \Delta x \), the width is always gonna stay the same. So we just need to find the heights of each point. But how can we find these?

Well, to find these, all we need to do is look at our rightmost endpoints for our rectangle. Now notice that we have the start of our interval here. Our right endpoint is gonna be one width over. So we start with one, then we go another width over, which is gonna be two, then we go another width over, which is gonna be three. So our inputs to our function are going to be \( f(1) \) starting here, plus \( f(2) \) going there, then plus \( f(3) \).

So this is going to be how we can calculate the area approximately under our curve from the right endpoints. Now doing this, \( \Delta x \), that's one. It stays the same in this whole problem. But to find \( f(1) \), \( f(2) \), and \( f(3) \), we'll just take one, plug it into our function here. That gives us one cubed, or one.

We take two. We plug it into our function. It gives us two cubed, or eight. And we take three. We plug it into our function.

It gives us three cubed, or 27. So we get one times one plus eight plus 27, which all comes out to 36 for the right endpoints, and that is the solution to part b of this problem. Now we've gone ahead and taken a look at the left endpoints, and we've taken a look at the right endpoints. But there's one more approximation that we are asked to do. Notice we are also asked to find the midpoints.

So to find the midpoints of these rectangles, well, I'm just gonna go ahead and redraw my graph again or redraw the rectangles underneath the curve of this function. So doing that, we're going to start from the middle points of each piece, which can oftentimes be the trickiest part when dealing with these. But if you think about it, this would be the start of our interval zero. We would go over here to one, and then between zero and one would be 0.5. So this would be the height of this first rectangle, and that's or the 0.5 put into our function would be the height of our first rectangle, and that's going to be this first rectangle we're dealing with.

And next, I would go somewhere in between in the middle right here to get my next rectangle, which would look something like that. Then I'll go somewhere in the middle right here to get my next rectangle, which looks something like that. And each of these midpoints, well, that's just gonna be midway between the intervals that we saw before for the left and right endpoints. So we have a zero and then a one here that give us the halfway point 0.5. Then we have a one and a two there.

The halfway point is 1.5. Then we have a two and a three here. The halfway point would be 2.5. And these would be the midpoints of these rectangles. This is now the area we're calculating for the midpoints.

So to find the area of these midpoints now, well, it's just going to be area one plus area two plus area three. And I can go ahead and do this the same way. I can take this \( \Delta x \). I can factor it out of these areas. But now I need to find the heights of each rectangle.

Well, the heights, well, we're just going to plug in these points that we found. So that's going to be \( f(0.5) \) plus \( f(1.5) \) plus \( f(2.5) \). So this is going to be the numbers that we plug into our function that we have right here. So we're going to have \( 0.5^3 \). And I'm going to go ahead to so I have a bit more space here.

I'm going to move this over to the left a little bit. So, we'll take this whole piece here. We'll move it over to the left. And now to find that \( \Delta x \), well, we already know that's one. Then we have \( f(0.5) \).

And what you wanna do is take 0.5 and plug it into your function \( x^3 \). \( 0.5^3 \) is going to come out to 0.125. Then I need to add this to 1.5 plugged into \( x^3 \). \( 1.5^3 \) comes out to 3.375. Then we need to take 2.5, plug it in for \( x^3 \), and that's going to give us 15.625.

And these are going to be the numbers right here. So we need to take one and multiply it by all these numbers added up here, and that's going to give me 19.125 as my result, and that is the solution to this problem. So this is how you can solve problems where you're dealing with approximations, whether it be from the left side or right side or midpoints, and that is going to be the solution right there. We've done all three approximations. And as you can see, we got three different results.

Notice when we did a left endpoint approximation, we got the small number of nine because we had an underapproximation that we did. When we did the right endpoint approximation, we got 36, which was an overapproximation. And then we got 19.125, which would be somewhere in between those two. Again, it's still not an accurate number, but this would be the estimation with the midpoints. So these are the three ways that you can approximate areas under the curves of functions.

Hope you found this video helpful, and let's try getting some more practice.