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.