Approximating Area Using Rectangles - Problem 3

Let's do another problem where we estimate area under a curve using rectangles. This time I want to estimate the area under the curve f(x) equals x² plus 1 from x equals 0 to x equals 2, with 20 left hand, and right hand rectangles.

So let's do the left hand first. The first thing I like to do is figure our what delta x, the width of the rectangles is. So delta x is going to be the total width of the interval 2 minus 0 over 20. So that's 2 over 20 which is 0.1. I'm going to have 20 rectangles in my sum. So I don't really want to do this by hand. I'm going to use my calculator for this. Let me show you how to set it up first.

Let me express the left hand sum as first the area of the first rectangle. If the width is 0.1, then this is 0.1, this is 0.2. I'll also need to know that the last rectangle has a height determined by 1.9. 0.1 before 2. So the first rectangle is going to be f(0) times delta x.

The second rectangle is f(0.1) so we advance by 0.1, because that's our delta x. F(0.1) times delta x plus. Then the third rectangle f(0.2) and so on. So the last term is going to be f(1.9) times delta x. There are going to be 20 terms and all 20 rectangles that divided the interval with the 20 sub intervals. I don't have to write this out in any more detail. I just need to know how to type it into my calculator.

Now if you're using a TI graphing calculator, you're going to want to learn these commands. Sum and sequence. We're going to use those two commands right now.

First, make sure that you enter this function under y1 in a graphing mode. Y1 equals x² plus 1. So I'll enter that function, and I'll use a sum sequence command. Now the sum sequence command works like this. You need to enter the quantities to be summed up. Each quantity is going to be of the type; function value times width. The function value is y1(x). This is what the times symbol looks like in your calculator. The width value is going to be 0.1.

Then you need to enter the variable x, that's the variable that you you'll be summing up by. Then you need to enter the first value that x is going to take on. That's 0 in our sum 0. Then you need the last value 1.9. Finally the step size. It's a lot of arguments, but this what you'll have to enter. These are the terms, the variable; start, stop, step. That's how I remember it.

So this is what I'm going to enter into my calculator right now. I'm going to come up with an answer and then we'll take it back to the board. So let's look at the TI-84 right now. I'm looking at the TI-84. I have already entered Y1 equals x² plus 1 you can see up here in this little screen. All I need to do is do my sum sequence command.

So to get the sum and sequence commands, you could go to the catalog, but you can also find them in second list. Under Math, 2 to the right, you've got the sum command under number 5. Second list again under OPs, you have the sequence command also number 5. Then you need y1. Y1 can be found in the VARS menu, so I don't need second. VARS, Y-VARS, function, I'll just use Y1. Y1(x) times 0.1, .

Now I have to put my variable x in, 0 the starting point, 1.9; the ending point, and the step size 0.1. Now I finish off with two parenthesis to finish off these starting parenthesis, hit enter, 4.47. This is our answer. So let's take it back to the board. That was easy. Imagine adding the area of 20 rectangles by hand. So our answer is 4.47. That's approximate, 4.47.

Now I want to do the same thing with the right hand sum. So let's take it over here. This is still f(x) equals x² plus 1, only I'm using a right hand sum to estimate the area under the curve. The right hand sum is different from our left hand sum. The rectangle reach up, and touch the curve in the upper right hand point.

Again I'm going to use the same number of rectangles, 20. So when n is 20, my delta x is 2 minus 0 over 20. So it's still 0.1. Our right hand sum is going to be a little different. The first rectangle first of all let me fill this in. This width is 0.1, this is 0.1, this is 0.21, this is 1.9. The first rectangle is going to depend on 0.1 not 0 for its height. The height is going to be f(0.1). The width will be delta x. So that's my first area. F(0.1) times delta x.

The second rectangle; its height will be f(0.2). Its width will be delta x. Then we keep going. The final rectangle will depend on 2 for its height. So f(2) times delta x. So you can see that this sum will be a little different in that the sum starts at 0.2, and ends at 2 whereas the left hand sum started at 0, and ended at 1.9. We're still incrementing up by 0.1.

How will this look in our calculator command? Well, we would use sum sequence. We still use y1(x) times 0.1. We still use x as our variable. Our starting point and our stopping point are different though. Starting point is 0.1. Stopping point is 2. Our step size is 0.1. So this is the part that we have to do differently this time. Starting point, and stopping point.

Just remember there are five arguments to the sequence command. The terms in the sequence, the variable of the sequence is the variable that I'm going to increment up through these values. X is going to start at 0.1, end at 2, and go up in increments of 0.1. Let's do that calculation on the TI calculator.

It looks like I cleared what I did before. So let me actually call it by doing second enter. Here is the command I entered before. So instead of actually retyping all this stuff, let me go back a row back. The only things I need to change are the starting value, and the stopping value. So let me just wait for the cursor to catch up. 1, 2, 3. So I want 0.1 as my starting point. I want 2 as my stopping point. That's it. So calculate that, 4.87. Done.

Let's take it back to the board. So we got the value for this sum to be 4.87. So that's our estimate for the right hand sum. Now that we have the right hand sun, and the left hand sum, we can say that the area is between those two values. The area for n equals 20 is between 4.47 and 4.87. So this is a decent approximation of the area. We know the area is somewhat between those numbers.

We wanted to get a better approximation we'd use still more rectangles. I took the liberty of making this calculation. For n equals 200, the area would be, if you'll excuse me I'll look at my values, 4.65 and 4.69. It's between those two numbers.

Now those two values would at least allow me to approximate the area to the nearest 10th. I could say that the area is approximately 4.7. So using Riemann sums with more, and more rectangles, you can approximate the area under a curve more, and more closely until you get the kind of accuracy you need. If needed only like accuracy to the nearest 10th, 4.7 would be my answer.