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# The Definite Integral - Problem 3

###### Norm Prokup

###### Norm Prokup

**Cornell University**

PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

Estimating the area under a curve can be done by adding areas of rectangles. The area of a rectangle is A=hw, where h is height and w is width. To find the width, divide the area being integrated by the number of rectangles n (so, if finding the area under a curve from x=0 to x=6, w = 6-0/n = 6/n.

The height of the rectangle will be f(a) at whatever number a the rectangle is starting. This depends on whether you are taking a left-hand or right-hand sum. For a left hand sum, the height will be taken from the leftmost part of the rectangle. So, if starting at x=0, the height of the first rectangle will be f(0). The height of the next rectangle will be f(0+w), and so on. For a right hand sum, the height will first be taken from the right side of the rectangle. Going back to the example, the height of the first rectangle in a right-hand sum will be f(0+w).

The area under the curve can be approximated by adding the areas of the rectangles. The left-hand and right-hand sums may be different.

This approximation can be found using a graphing calculator.

Let's do another problem with definite integrals. This times we'll sort of examine how the definition of the definite integral works, and how we can use it to approximate a definite integral. Let's interpret the integral from 0 to 2 of x² plus 1 as an area. Estimate the value of this integral to the nearest 100th using left hand sums. So we'll do both of these things.

First, let's interpret the integral as an area. Now this function x² plus 1, I'll graph it as y equals x² plus 1. This is what it looks like. The interval 0 and 2, these limits of integrations as they're called, these determine the interval that you're integrating over. So the left limit is 0, the right limit is 2.

Since this function is non-negative, on this interval, the value of the integrals can exactly equal the area under this curve. So let me just represent that area. So this integral for this area equals the integral from 0 to 2 of x² plus 1 dx. The area under the curve above the x axis to the right of x equals 0, and to the left of x equals 2.

Now let's get to approximating this value. I want to approximate the value of this integral with left hand sums. Now we've done this before. The way we did it was we divided the interval from 0 to 2 into four equal subintervals. So we would be using four rectangles.

Let's start with a nice small number like 4. If we do this, what's the width of each rectangle? Now let's make the assumption that the rectangle are all the same width. So the width is going to be the total width of the interval which is 2 minus 0 divided by the number of rectangles, 4. So I divide by 4. I get 2 over 4 which is 0.5. So each of these rectangles is 0.5 units wide. Each of them.

Now I have to calculate the area of each rectangle. Now remember that I calculate the area using the width, and the height. The height of this first rectangle is going to be f(0). So the left hand sum is going to start with f(0) times 0.5. This is the height, this is the width.

The second rectangle is going to be, this will be 0.5 based on the width, f(0.5) times 0.5 plus. Then the next one, this is 1, this is 1.5. This will be f(1) times 0.5. F(1) is the height of the third rectangle, 0.5 is the width. The final rectangle f(1.5) times 0.5. Notice that in the left hand sum, the 2 doesn't get used at all. So if you consider each of these four subintervals, we're only using the left end point of each subinterval. So this right point never gets used.

So how do we do this on our calculator, because we're going to have to approximate this integral to the nearest 100th. We're going to use a lot more than four rectangles. So I want to get into the calculator commands for how to do this.

The first thing I want to do is set y1 on your calculator so that it equals x² plus 1. So it equals our function. Then to annotate this summation, we're going to use the sum sequence command. It's actually two different commands. We'll put in y1 of x times the width. This is going to represent what kind of a thing we're adding up. Y1 of x represents this thing. The width will have to be for n equals 4, 0.5.

Now there are five arguments. This is the first argument of the sum sequence command. The second one is just x. X is the variable that we're going to be incrementing up. The third, fourth, and fifth are the starting value, the stopping value, and the step size. So we're going to start incrementing x up from 0. We're going to come up by 0.5. These are the numbers we have to go through, 0.5, 1, 1.5.

So we'll start at 0, we'll end at 1.5, and we'll step up in increments of 0.5. That's how much we're increasing each time. So 0.5. You'll notice that this number will tend to be the same as this number. It's the width of the rectangles. So we actually calculated this in a previous exercise. We found it to be 3.75. So this is clearly an underestimate of the area under the curve.

Now I want to keep a running tally of what these estimates are. So I have a table set up here. These are left hand sums. For n equals 4, I got 3.75. I want to do n equals 20 next. So for 20 rectangles, I have to go through this process again to figure out what the width is, what the numbers are that I'm going to be plugging into f, and so on. You can actually do that without drawing the graph all over again. I do recommend at least setting up a number line from 0 to 2.

Imagine if you're setting up from 0 to 2, the width for n equals 20 is going to be 2 minus 0 this length divided by 20. That's point 1. 2 over 20, point 1. That means that we're going to start at 0, and we're going to increment up by point 1. So 0.1, 0.2, 0.3, and so on. This is not just a scale clearly.

Then the last triangle is going to have a left end point of 1.9. Remember it's the left end point that count in the left hand sum. So x will be going from 0 to 1.9. It will be stepping up by increments of point 1. So this is point 1. So let's look at our sum sequence command.

So the sum sequence commands has to have the heights which are y1(x) times the widths, which in this case are all 0.1. We'll still use x here as our counter. X is going to start at 0 just like before, because it's a left hand sum. It's going to finish at 1.9, then it's going to step up by 0.1. Again this width is the same as this.

So let's go on our calculator and figure out what this value is going to be. Then we can try it for n equals 200. Here we're looking at the TI-SmartView. I need to enter my sum sequence command. You'll notice they've already set 1 equals to x² plus 1. So I need the sum sequence command. That comes from second list. Under Math, you'll find the sum command, number 5. Then second list again to get the sequence command, that's under OPs also number 5. Sum sequence.

Now I need y1(x). So I go into the VARS menu. I go Y-VARS then function; Y1. Y1(x). I need to multiply this by, so times the width, 0.1 comma. That's the first argument. Then I need x, comma I need the starting point which is 0. The stopping point, 1.9. The step size, 0.1. I need two parenthesis, because each command; sum and sequence had a starting parenthesis.

Let's execute this to see what we get. 4.47 that was pretty quick. 4.47 is our left hand sum for n equals 20. Now let's see what it would be for 200. I can make this adjustment really easily right here now. If I do second entry, I get the whole command I just executed. I can go back and edit some things here.

First of all, if I use n equals 200, it's pretty easy to see that the widths are going to be 1/10 of what they were for n equals 20. So let me go back. Let me insert an extra 0 here. So second, insert, and then 0. So now my widths are 0.01. Then I have to start from 0, it's a left hand sum. I have to end up at 0.01 to the left of 2, so 1.99. So let me cursor over there. Insert again, second insert 9. I have 1.99, then I have changed this width as well. So second insert, 0.

So I'm ready to go. These are my rectangles y1(x) times 0.01. This is my variable x. My starting value 0, that's left most end point. This is the last left end point representing the last rectangle; 1.99. This is the step size. So this will give me 200 rectangles. So I execute it. Pretty quick. 4.6467.

Let's take these results to the board. So before had for n equals 4, 3.75. We just calculated for n equals 20, the left hand sum is going to be 4.47. For n equals 200, it was 4.6467. I took the liberty of calculating this on my own time. For n equals 200 we get 4.664667. So it looks like this thing is a approaching 4.66 possibly 4.67.

The only way to be sure would be to take it maybe a little further than this which I did take the liberty of doing. This thing actually rounds to 4.67. Let me put this value in here 4.47. So to the nearest 100th, the definite integral seems to be going to 4.67. This is as n is going to infinity.

These top values represent numbers of rectangles in my left hand sum. As those numbers increase, the value of the left hand sum approaches 4.67. Because of the way the definite integral is defined, this becomes the value of my definite integral. So I conclude that the definite integral from 0 to 2, of x² plus 1 dx is approximately 4.67.

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###### Norm Prokup

PhD. in Mathematics, University of Rhode Island

B.S. in Mechanical Engineering, Cornell University

He uses really creative examples for explaining tough concepts and illustrates them perfectly on the whiteboard. It's impossible to get lost during his lessons.

Thiswas EXCELLENT! I am a math teacher and have been looking for an easy/logical way to explain the lateral area of a cone to my students and this was incredibly helpful, thank you very much!”

I just learned more In 3 minutes of polygons here than I do in 3 weeks in my math class”

Hahaha, his examples are the same problems of my math HW!”

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