 This lesson will be on interpretations of the definite integral. In former lessons, we were working on RAM, LRAM, MRAM, RRAM, in order to find area under curves. Well, this time we're just going to start counting some squares or rectangles. Sometimes that is easier than working with the RAM idea. The other idea we will be working on in this lesson will be average value. In counting squares, I want you to look at the example on page 240, and that's example number 5, which reads, a car starts at noon and travels with the velocity shown in the graph. A truck starts at 1pm from the same place and travels with a constant velocity of 50 miles per hour. The first question asks, how far away is the car when the truck starts? Well, let's look at this graph. This is the graph of the car, and you'll see the velocity starts at zero and increases, and keeps increasing until this point, and then it begins to decrease, which means the car is going further and further away from the point of starting. Now, when we include the truck, which starts at 1pm, going 50 miles an hour. It is going constantly at 50 miles per hour. The two is moving away from the starting point, or it was the same starting point, but at that constant rate. So at some point, the car and the truck will get closer and closer together, and that's what we have to start thinking about. But our first question asks, how far away is the car when the truck starts? Well, we see the truck starts at 1, so the car has gone all this distance in here. We know that if we have a time velocity graph, we can add up the area under the curve in order to find the distance. Well, a simple way to do this is to add up all the rectangles, these little rectangles that you see, and when we add them all up, we find out we have 14 of them. But each rectangle is half an hour wide and five miles high. And when we put all of that together, we get 35 miles. So the car has traveled 35 miles before the truck even starts. Second question, during the period when the car is ahead of the truck, when is the distance between them the greatest and what is the greatest distance? Well, we see the car is moving along again, velocity increasing, increasing, increasing, and then it begins to decrease. Now, when is the distance between them the greatest? Well, remember, the car is still going faster than the truck until it reaches the time when their velocities intersect. So again, let's put in that velocity for the truck. The car is pulling further and further away from the truck all along here, but when it reaches this point, their distance is beginning to close up. So the car is furthest away from the truck at this point. Now, let's find out what the distance between them is. We know this distance right in here because we've just figured that out to be 35 miles. This distance where both of them are overlapping, we don't have to worry about. We don't want total. We only want a difference. So the only thing we have to worry about is this distance in here. Again, we are going to add up all the squares, and this time we find out there are approximately 20 squares. Again, multiply it by the one-half and the 2.5, and we get, adding that to 35, we get a total of 85 miles. That's when they are furthest apart. And then the third question asks, when does the truck overtake the car? Again, let's put the truck in. So the car is ahead during all of this time here. So what we need to do is find out how much area this takes in here because all the rest of the area below this graph and below that graph, they're doing the same amount of distance. So once we do all of that, we realize that we need 34 squares in here to count up. And if we start counting them vertically, we come to about this point in here, which is about 8.25 hours that we will find that the truck will begin to overtake the car. So we want to do this problem as simply as possible. So we don't ever need this common area under the curve or distance in our case as we do the problem. We just look for what is above or below the curve we need. The next topic we will go over is average value. First, let us get a function y equals 5x e to the negative x. And we're going to sketch this over the interval 0 to 3. So it looks something like this on that interval from 0 to 3. The next thing we want to find is the area under the curve. So the area under this curve is found by a is equal to the integral from 0 to 3 of 5x e to the negative x dx. I put this on my calculator and you can see I have the curve. So if I want to find the area, I'll do second calc 7. And the lower limit is 0. The upper limit is 3. And we will get that calculation to be 4.0042586. I want to keep as many digits as possible in order to keep this accurate. Now to determine an average value, what we are looking for is a rectangle who has the same area as the area under the curve. So if we go from A to B where A is our 0 and B is 3, we can determine an area of a rectangle that fits this. Well, the area of that rectangle is 4.0042586. We also know the width of the rectangle, which is 3. So if we want the average value, it will be as you can figure out from what I'm saying here, it will be the height of the rectangle. And that will be the area over the width. And if I do that computation, I will get 1.335. And that will give me the height of the rectangle, which is indeed the average value. What is the average value? Well, we already know the height that we're supposed to get for the area of that rectangle. And that happens to be the average value. But how does this work in theory? Well, we know the area under the curve is the integral from A to B of f of x dx. We know the area of the rectangle to be the B minus A, which is the length times the height, which we'll call h of our rectangle. So the average value now is this h. So in order to determine it, we write it out as h equals the integral from A to B of f of x dx. We multiply that by 1 over B minus A. And this is the formula for average value. It's nothing more than the height of the rectangle that we use that is the exact same area as the area under the curve. How is this used in problems? Well, we have a question here. Determine the average value of the integral from 0 to pi of x sin x dx. Well, we know the average value is the height and that's equal to 1 over B minus A. Well, in this case, it's pi minus 0 or pi integral from 0 to pi x sin x dx. We can go to our calculator and type in 1 divided by pi times the integral of our y1, x, 0, pi. And we get the average value is 1. So in this case, the height of our rectangle is equal to 1. Well, let's look at this on our calculator. We have our function. Let's get a window that we need, 0, pi, and the negative 2 to 2 is probably good. So let's graph that. There's our function. And if we draw in the line y is equal to 1, that will give us the height of the rectangle we need. And if we look at the area under the curve and then the area of that rectangle, we see that the top piece that's above the rectangle can kind of fit in the piece that's under the top piece of that rectangle yet above the curve. So the areas are the same and we have found the height or average value of the rectangle that gives you the same value as the area under the curve. This concludes our lesson on interpretations of the definite integral.