Area Under a Curve - Online Tutor, Practice Problems & Exam Prep

Finding the area under a function involves using familiar shapes like rectangles and triangles. For linear functions, the area can be calculated directly using geometric formulas. For curvy functions, approximations using rectangles are employed, with the base calculated as $(b-a)/n$. The exact area can be determined using limits, allowing for infinite rectangles, leading to precise calculations of the area under the curve.

Area Under Linear Functions

Welcome back, everyone. So over the course of these next couple of videos, we're going to be learning an important skill to have in this course, which is being able to find the area underneath a function. Now for this video, we're specifically going to be focusing on linear functions, and if all of this sounds scary, don't sweat because it turns out finding the area under linear functions is very straightforward because we can use shapes that we're already familiar with, like the area of rectangles or triangles. So let's go ahead and just jump right into an example of this to see what these types of problems look like.

In this example, we are given the function \( f(x) = x \), and we are asked to find the area under the curve from \( x = 0 \) to \( x = 5 \). Now I can see 0 is right about here on the x-axis, and 5 is right there. If I want to find the area underneath this curve, all I'm really needing to do is figure out what the familiar shape is that this makes. And I can see that we clearly have a triangle which forms right here, and the area of a triangle is \( \frac{1}{2} \text{ base } \times \text{ height} \). So the area is going to be \( \frac{1}{2} \) the base of this triangle, which I can see we go from 0 to 5 times the height which goes from 0 to 5 on the y-axis. So we're gonna have \( \frac{1}{2} \times 5 \times 5 \), which is \( \frac{1}{2} \) of 25, and \( \frac{1}{2} \) of 25 is 12.5. So this right here is going to be the area underneath this function curve, and that's all there is to it.

As you can see, whenever you're asked to find the area under a curve, what you're really being asked to do is find the area between the function's graph and the x-axis, and that's exactly what we did. Now it's also possible you'll see situations where rather than just using one familiar shape, we actually have to use a combination of multiple familiar shapes. And to make sure we can understand some more of these complicated problems, well, let's just try a couple more examples.

Now for each of these examples, we need to calculate the area underneath the curve for the interval that we're given. Now we'll go ahead and start with example a. Example a asks us to find the area underneath this function from \( x = -8 \) to \( x = 4 \). I can see -8 is right here on the x-axis, and I can see that positive 4 is up here. So this is the area that we're trying to calculate. Now to find this area, I could try to find a trapezoid or some kind of familiar shape here, but the more simple thing to do is going to be to split this up into two shapes that we're familiar with. Notice if I divide the shape in this fashion, we get two areas, area 1 and area 2. Area 1 is a triangle and area 2 is a rectangle. So all we need to do is add these areas together, and that's going to give us the total area underneath this curve for the interval we have. So the total area is going to be the area of a triangle, which is \( \frac{1}{2} \) base times height, plus the area of a rectangle, which is just base times height.

Looking at this, I can see that the base of our triangle is going to be this distance right here. We go from -8 to 0, so this is a distance of 8. And the height is going to be up here on the y-axis which is also 8. Now what I'm going to do is add this to the base of this rectangle which goes from 0 to 4 times the height which goes from 0 to 8. So we have this as our math right here, and I just need to multiply and add everything together. Now one half of 8 is 4, and 4 times 8 is 32. Then we're going to add this to 4 times 8 which is 32, and 32 + 32 is 64. So this right here is the area underneath this curve, and that is the solution to example a. But now let's go ahead and try example b.

Example b asks us to find the area underneath the curve from \( x = -6 \), which is right there, to \( x = 8 \), which is right here. So this is the interval that we're given in this problem, but notice when we try to find this area, we're actually going to get an area above the curve if we're looking between the function and the x-axis. So is this actually something that's possible to calculate? Well, it turns out that it is. The statement about being under the curve can actually be misleading, because when the function goes below the x-axis, the area is going to be above the x-axis, meaning the area we calculate is going to be negative. Let's see if we can understand why this area becomes negative. We have a rectangle that forms here, and the area of a rectangle is base times height. Now what I first need to do is calculate the base of this rectangle. The base is going to be this distance, which goes from -6 to 0, so that's a distance of 6, plus this distance, which goes from 0 to 8. \( 6 + 8 = 14 \), so that is going to be our base. Now the height is going to be from here down there, but notice we're actually going down to find our height. We haven't had to do that yet. So since our height goes from 0 to -6, that means the height is going to be -6. So we have \( 14 \times (-6) \), which gives us an area of -84. This is how you can calculate the area underneath this function curve, or in this case, it would be the area above the function curve. So this is the strategy for finding the area between a linear function and the x-axis. Hope you found this video helpful, and let's try getting some more practice.

Find the area under the curve in the function below from $x=-4$ to $x=8$.

• A. 16
• B. 32
• C. 24
• D. 40

Find the area under the curve in the function below from $x=-2$ to $x=8$.

• A. 34
• B. 38
• C. 50
• D. 42

Approximating Area Under the Curve

Hey, everyone. So in a recent video, we talked about how you could find the area under the curve of a function if you have some kind of linear function. Well, here's my question for you. How would you go about calculating the area underneath a function curve if you have a more curvy graph like this one? Well, one of the things that might be tempting is to try and draw some kind of familiar shape, but there really isn't a familiar shape that's going to work. Because if I try drawing a triangle, notice how this is not going to capture the area here, because this is not the exact shape we're dealing with. If I try drawing a rectangle, well, this shape is not even close. So how can I go about doing this? Because it's clear that the familiar shape strategy does not work for these more curvy graphs. Well, what I'm going to be discussing with you in this video is our strategy for approximating the area underneath these more curvy functions. So without further ado, let's get right into things.

Now the strategy for approximating this area is by taking it and cutting it up into rectangles. Now rectangles are a familiar shape, and we know that the area of a rectangle is pretty straightforward: base times height. So if you take this shape here, this curvy graph, and you fill it with rectangles, you can find the base and the height of each rectangle, and then add everything together, and doing that will give you an approximation for the area underneath the function.

So let's just go ahead and try this example here to see what this looks like. We're asked to use 5 rectangles to approximate the area underneath the function f(x)=x2, and we're asked to do this on the interval from x=0 to x=5 with left and right endpoints. Now we're going to start with the left endpoints. And all this means when you're dealing with left endpoints is it means you want to start with the upper left corner, the top left corner of each rectangle and touch it to the function curve. This will make sense as we go ahead and draw these rectangles.

Now my first step before actually drawing anything on our graph though, is going to be to figure out what the base is, and the base of each rectangle can be found with this equation right here. So it's going to be that our base is equal to the high point on the x-axis, which we can see is b or 5, minus the lowest point on the x-axis, which we can see here is 0, all divided by n, which is the number of rectangles, and we can see from our example that that's 5. So we have 5-0/5, which is equal to 1. So that means that our base is 1. And this actually makes sense because if we're filling our curve here with 5 rectangles, and we want to find the width or the base of each of those rectangles, it makes sense that it would be 1, since we need to evenly split it up from 0 to 5 on the x-axis. So let's go ahead and now draw in these rectangles. But remember, we're using the left endpoints or the top left corners. So my first rectangle is gonna look something like this. I'll touch it to the top left corner here, and it's going to go right there. So it's really just kind of a line touching the x-axis because we're starting with left endpoints. But if I draw my second rectangle, again, these all have a base of 1, so we're going to start here on our curve and it's going to look like that. That's our second rectangle. Then our 3rd rectangle is going to touch at its top left corner right there. Our 4th rectangle is going to touch at the top left corner right about there. And then our 5th rectangle is going to touch at the top left corner here. So this is going to be what 5 rectangles underneath this curve looks like. And what I need to do is find the area for each one of these rectangles, Because what I can do here is if I'm able to find the area, which we know is just base times height, I could find the base times the height of the first rectangle plus base times the height of the second rectangle, and I could just keep adding these on until I get all these rectangles. So let's go ahead and do this.

Now we can already see here that the base is 1, so we're gonna have a base of 1, but what's the height of this first rectangle? Well, the height is going to be this point right about there, and we can see if we look here we're at 0. 02 on the y-axis here would just be 0, so that means our height is 0. Now let's move on to the second rectangle. We can see that the base is 1, that's going to be the same for all these rectangles, but the height here, well the height is going to be right about at that point. And I can see it's going to be 1 squared and 12 is 1, so that's going to be our height. Now let's move to the 3rd rectangle. The 3rd rectangle has a base of 1 and then the height, well, we can just look right about here. For the 3rd rectangle, it's going to be 22 which we know is 4, so that's our height. So now we'll move to the 4th rectangle. The 4th rectangle has a base of 1 and the height we can see is going to be 32, which is 9. Now let's add the 5th rectangle. The 5th rectangle is right there, it has a base of 1, and the height is going to be 42, which is 16. So this right here is the approximate area underneath our function. And if you go ahead and do the math here, where you multiply all these numbers and add them together, you should get that your area is approximately equal to 30. So this is how you can approximate the area underneath a function curve using these rectangles.

Now we talked about how this was a left endpoint approximation, but the question becomes, what would happen if we were dealing with right endpoints? Well, for a right endpoint approximation, it's the same exact idea, except rather than starting at the top left corner, you're now going to start at the top right corner, which we can see is already the case on this graph here. So what we need to do is find the area of each one of these rectangles to do our right endpoints. So let's go ahead and do this. Well, we know that our base is already equal to 1, and this is gonna be the same in both these cases since we're using 5 rectangles and our interval goes from 0 to 5. So our base is going to be 1, but the height is going to be a little bit different, because notice that our height is going to start on the right side this time. So height is going to be 12, which is 1. Now we're going to add this to our next rectangle, the base is always 1, but the height, you can see here, is going to be 22, which is 4. Now we can go ahead and add our next rectangle. The base is going to be 1, but the height of this next rectangle is going to be 32, which is 9. Now we'll add our next rectangle, the base is 1, but the height is going to be 42, which is 16, and then we'll go ahead and add our last rectangle. The base is 1, and the height is going to be 52, which is 25. So this is what we end up getting when we use right endpoints, and if you go ahead and multiply out these numbers and add everything together, you should get that your area is approximately equal to 55. So this right here is the approximate area underneath the curve using right endpoints, and notice the two answers we got here. One of these answers gave us a lower number, but that makes sense because this is actually an under approximation. Notice there's a bunch of areas here that we actually haven't accounted for, so that's why we got this lower number, whereas for this case, we got a higher number, but we actually did an over approximation because notice these rectangles have some extra area that comes over the curve that we actually don't want to account for if we're trying to get an accurate approximation. So a lot of times people will try to take these low and these high numbers and kind of average them, or figure out some number between to see how accurate they can get this area. But either way you wanna go about it, this is the strategy for approximating the area underneath a function curve when you don't have a nice linear function like a line or a constant. So hope you found this video helpful. Let me know if you have any questions, and let's move on.

Use 3 rectangles to approximate the area under $f\left(x\right)=x^2$ from $x=2$ to $x=5$ with right endpoints. 

• A. 50
• B. 40
• C. 30
• D. 35

Use 6 rectangles to approximate the area under $f\left(x\right)=x^3$ from $x=0$ to $x=3$ with left endpoints. 

• A. $\frac{225}{16}$
• B. $\frac{225}{8}$
• C. $\frac{81}{4}$
• D. 81

Approximating Area Algebraically

Everyone, you may recall in a recent video, we talked about how to approximate the area underneath a curvy function like this by filling it with a bunch of rectangles and adding the area of each individual rectangle. Well, in this video, we're going to learn how to solve these same types of problems, which you're going to come across in this course, except the difference is we're not going to be given a nice graph. And it's very common that you're not going to have a graph to be able to solve these problems, but rather you're just going to need to be able to use the algebra and the equations to do it. Now if that sounds kind of scary, don't sweat it, because it turns out that this process is going to be very familiar to what we've already done. We're just using simply numbers this time instead of a graph. So let's just go ahead and get right into an example of what these problems would look like.

So here we're asked to use 4 rectangles to approximate the area underneath this function f(x)=x2, from the interval x=1 to x=5 using right endpoints. So this is the problem we have. Now when doing these types of problems, this is the only equation that you need. So if you plug in the variables and use this equation properly, you can get your answer. Now in this equation, Δx here is the base, and recall that we've talked about how the base of each rectangle is the same. So once we calculate Δx, that's going to stay constant throughout this problem. Now next, we need to find the height, f(x_k). And the height is going to be each individual rectangle's height, because recall that not all rectangles are going to have the same height in these types of problems, but that's why we have this x_k here, Because k here is this counter, 1, 2, 3, 4, that counts how many rectangles you have, and it counts for the height of each individual one so you can find the area.

Now once you do this, it's base times height, the area of each rectangle. You're just adding this all up for however many rectangles you have, which in this case is 4 rectangles. So as you can see, all this equation is doing is telling us to do the same thing we did before, where we add a bunch of rectangles, but we just use a formula rather than a graph. Now the way that I'm going to solve this problem is by first calculating the base. And to calculate the base, well, I can use this familiar equation here, b-an. Now b, recall that that's going to be the end of the interval on the x axis, which in this case is 5, and a is going to be the start of the interval, which is 1. Now for n, that's the number of rectangles. We can see in this problem, we're asked to approximate using 4 rectangles. So to calculate Δx, we're going to have 5-14 is 1. So that's going to be the base of each rectangle that we have.

Now at this point, I can set up our equation. So we're going to have our base of 1 multiplied by our first rectangle, the height f(x)=x₁(2). Then we'll have the base of 1 multiplied by the height of our next rectangle, f(x)=x₂(3). So, basically, we have the first rectangle, then the second rectangle, then we're going to have one times the height of our 3rd rectangle, f(x)=x₃(4). And then I think you're seeing the pattern here. Then we'll have one times f(x)=x₄(5), which is going to be the height of our 4th rectangle. So this right here would be the area of each rectangle we have in this problem, and since we have 4 rectangles, we have 4 areas that we're adding up.

Now of course the next piece here is going to be to figure out what all of these heights are, And to calculate the height of each rectangle, well, I first need to calculate what these x ones, x,x two, and all these x's are inside. But it turns out there actually is a pattern for this. If you want to calculate the x1 here for left endpoints, you're going to have the start of your interval a, plus 0 times Δx. And then from here, you're gonna keep incrementing up, so it's gonna be 0 Δx, and it's gonna be a+1Δx, and a+2Δx, and that's gonna keep going on. Now if you have right endpoints, you're going to start with a+1Δx, so you're gonna go 2, 3,...

Approximating Area Algebraically Example 1

Let's try solving this example. So here we're asked to use 4 rectangles to approximate the area under this function, f(x)=4x2 from x=0 to x=8. And they want us to use left endpoints to solve this problem. So how can we do this? Well, whenever you are not given a graph for these situations where you need to use rectangle approximations, you're going to need to use this equation right here to solve the problem. But thankfully, it's pretty straightforward, or at least very related to what we've already done, where we just add a bunch of rectangles. So let's see what this looks like.

Now what I'm first going to do is calculate the base. I think this is always a good starting point with these types of problems since the base stays constant for every rectangle that you have. Now the base is going to be this equation right here, (b-a)/n. Now b, I can see is the end of the interval, which is 8. a is the start of the interval, which is 0, and n is the number of rectangles, which is 4. So we're going to have 8-0=2. So this is going to be the base for every rectangle that we have.

Now, at this point, what I'm trying to do is set up this equation. So I can do this by recognizing this is the sum of every rectangle we have. We have 4 rectangles, So our area is going to be this base of 2 times the height of each rectangle. So we'll have base×height, which is 2×f(x1), that's our first rectangle, plus 2×f(x2), our second rectangle, plus 2×f(x3) of our third rectangle, plus 2×f(x4) of our fourth rectangle. So these are all of our rectangles added up.

Now remember, there's actually a pattern when it comes to these types of rectangles. When you're using left endpoints, x1 is going to be the a value, which I can see is 0, plus 0 times delta x. Now delta x is equal to 2, but anything times 0 is just 0, so x1 is just going to be 0. Then we'll have x2, which is 0, the a value, plus 1 times delta x. Now delta x is equal to 2, so we'll have 1 times 2, which is 2. Then we'll have x3. x3 is going to be 0 plus 2 times delta x. Now 2 times delta x, well, that's going to be 2 times 2, which is 4. And then we'll have x4, which is 0 plus 3 times delta x. 3 times 2 is 6, so we're going to have 0 plus 6, which is just 6. So this right here is all of the x values that we have. We have x1, x2, x3, and x4. So now that we have all of these calculated over here, well, now I can apply them to this equation.

So we're going to have 2, the space stays the same, 2 times f(x1), which is that x1 was 0. So we have 2 times f(0), plus 2 times f(2), plus 2 times f(4) I'm just using the x1,x2,x3,x4 we calculated it over here. Plus 2 times f(6). Now at this point, what we need to do is take all of these inputs, plug them into our function, and figure out what the output is going to be. So we're going to have 2 times 0 plugged into our function. So it's going to be 0 squared times 4, which is 0. Then we're going to have plus 2 times 2 plugged into our function, so that's 2 squared, which is 4. 4 times 4 is 16. Now we're going to have 4 plugged into our function. So that's going to be 2 times 4 squared, which is which is 16 times 4, which is 64. And then we're going to have 6 plugged in to our function. So we're going to have 6 squared, which is 36 times 4, which is 144, so we're going to have plus 2 times 144. So this is what we end up getting. Now at this point, all you need to do is multiply out all the numbers and add everything together, and you should be able to do this. You could check this on a calculator or do this by hand, and you should get that your area is equal to 448. So this is the approximate area underneath the curve using left endpoints. So this is how you can solve these types of problems where you don't have a graph, but need to use the equation to approximate the area under a curve. Hope you found this video helpful, let's move on.

Using Limits to Find Area Under the Curve

Everyone, and welcome back. Up to this point, we've spent a lot of time figuring out how we can approximate the area underneath the curve of a function by filling it with rectangles and adding the area of each rectangle. Now, in this video, we're going to learn how we can find the exact area underneath the curve. So we're no longer approximating. Now the results we get are going to be exact using this process of filling it with a bunch of rectangles. And this might sound like it's impossible to do, but it turns out there is a process that allows us to do this with what we've learned. So pay close attention, because I'm going to walk you through these steps and show you how to solve these types of problems in this course. Let's get right into things. The main thing we need to think about is how exactly we could use rectangles to find the exact area. Well, for approximating with rectangles, we're in essence just filling this up and calculating the area of each rectangle, base times height. Suppose I were to try to do an approximation with 2 rectangles. It would look something like this, where we'd have one rectangle right here and another rectangle right there, using right endpoints. Notice, if I were to try and find this area, we would get a severe over-approximation because we have all this extra area above the curve that's sticking out, and this would give us an overapproximation. But now let's say I wanted to fill this with more rectangles, so rather than using 2, we use 4. So four rectangles would look something like this. Again, I'll use these right endpoints to approximate. We'd have one rectangle here, one there, one right about there, and one right here. Notice, when doing this, we still get an over-approximation. Because if you look, we have all this extra area that's still above the curve, but this is a better approximation than the area we had before. As we draw more rectangles as n gets bigger, the width or the base of these rectangles, I should say, gets smaller, and we get a more accurate approximation. We need to increase the number of rectangles to get the most accurate approximation possible. Let's say that we had 100 or 1000 or even 1000000 rectangles. This would give us a very accurate approximation. We would have very little error, very little extra area popping up above the curve. So if we could somehow find a way to get this n value, this number of rectangles to approach infinity, that would allow us to get a truly accurate approximation or the exact amount. And in this case, we actually can do that because we can't necessarily just say, oh, we have an infinite number of rectangles, but what we can do is apply a lim n → ∞ to this equation that we've already been using, and say, what happens if this limit approaches infinity, the number of rectangles? Well, in that case, we're going to get the most accurate approximation we can, meaning that that's going to be the exact area underneath the curve. So let's actually see how we can use this equation to solve the type of problem where we want to find the exact area. In this example down here, it says find the exact area underneath the function f ( x ) = 1/2 x from x = 0 to x = 6 , that's our interval, using the limit definition. Now we have this argument out in front right here. This argument says we need to add a certain number of rectangles, and this time we're going to an infinite number of rectangles. What I need to find is delta x to go in here. We know how to calculate delta x; it's this equation that we've been using, b - a / n . So delta x is going to be b = 6 minus a = 0 , all divided by the number of rectangles, which in this case, we have an infinite number of rectangles. But I can't just put infinity in the bottom of the fraction. It's not really a number, so I have to leave this as just n. So what we end up getting is that delta x is 6-0 n = 6 n . Now, here we have delta x calculated, so I can go ahead and put 6 n into this summation. But now we need to find f of x k. f of x k is the height because remember this is the base, that's the height, but how do I find the height of each of these rectangles? What you first need to do is find what this x of k inside the function is, and x of k is going to be a, the start of your interval, plus k times delta x. This might look kind of confusing, but all this is saying is that if we're using, say, right endpoints to approximate, well, in this case, we would just keep counting up for each rectangle we have. And this k would be like, well, we have 1 delta x or 2 delta x or 3 delta x. We've already seen that before, so that's all this is saying. We're just using variables to describe that. For x of k, what I need to do is take a, which is the start of our interval 0, and then add this to k times delta x, which we figured out was 6 n . So what we're going to end up having for x of k is that this is all equal to 6 k n . Now that we have delta x and x of k calculated, what I can now do is plug in this function here f of x k, which is going to be f of this value 6 k n . So now we have everything plugged into our equation, and at this point, I just need to do some algebra to get this figured out. Now, this argument right here that you see, going to want to keep this consistent every time you are doing a step here. But because I want to save some space here, I'm just going to say, remember this argument is out in front for each step that we do because I'm mostly just going to focus on simplifying this part right here. So we have 6 n , that's our delta x. And to find f of 6 k n , well, that means I need to take this argument right here and plug it into our function. Our function is 1/2 x, so that means we're going to have 1/2 of this value 6 k n . Now what I can do is take this 6 right here and divide it by this 2, that's going to give us 3. So we're going to have 3 n , and keep in mind this is all going to be what this area is equal to, so we're going to have 3 n and that's going to be times 6 k n . Now 3 times 6, that's going to give us 18. So we're going to have 18 k, and n times n is n squared. So this is what everything simplifies to when we go ahead and plug this into the function, and we reduce this down as far as we can get. Now keep in mind this argument we're matching is still out in front here. And what I can actually do at this point is I'm trying to get this into a certain form with my summation because this summation stays in the argument. And if I can somehow get just the k that we have right here to be by itself, well, that's going to allow me to get a form of a summation that I can simplify with the n's that I have right here. So what I'm going to do is take this 18 n 2 and I'm going to bring it outside of the summation. So I'm going to have 18 n n 2 on the outside, meaning all we're going to have left over is this k on the inside. And this is perfect right here because remember this summation that we have there. We know this is a familiar form; it's n × ( n

Find the area under the curve, of the function $f\left(x\right)=3x+2$ from $x=0$ to $x=4$ using limits.

• A. 32
• B. 56
• C. 26
• D. 50

Find the area under the curve, of the function $f\left(x\right)=x^2$ from $x=0$ to $x=8$ using limits.

• A. 256
• B. $512$
• C. $\frac{1}{3}$
• D. $\frac{512}{3}$