2015 AP Calculus AB/BC 3b

- [Voiceover] Part B, using correct units, explain the meaning of the definite integral, so it's definite integral from zero from T equals zero to T equals 40 of the absolute value of V of T DT in the context of the problem. Approximate the value of that integral using a right Riemann sum with the four subintervals indicated in the table. All right so let's first tackle the first part. So what's the meaning of this definite integral? So the velocity, velocity is a vector, it has direction, and you can see when you look up here that sometimes the velocity is positive and sometimes it is negative, and a reasonable interpretation of that is well this is when she's walking in one direction and then that's where whatever direction we assume is the positive direction and this is when she's walking the other way, and so she has a positive velocity here and then she has a negative velocity. She has some speed in the other direction and then she has a positive velocity again. So if you were to take the absolute value of this, if you just cared about the magnitude of the velocity, well we're talking about, we're talking about speed, but we're not just talking about the absolute value of velocity, we're integrating the absolute value of velocity over time from time equals zero to time equals 40. So if you integrate speed, if you're saying okay for every for every little amount of time DT she has some speed, well that's going to give you the distance that she travels. So the distance remember, is the difference between distance and displacement is displacement you can think of as the net distance it takes if you go back and forth a bunch of times it's going to add up to the distance, but they're going to net out with respect to each other in displacement. So this right over is going to give us the total distance, not displacement, that she travels over those 40 minutes. If there was no absolute value sign here then we'd be talking about the displacement and so let me write that. So this integral from zero to 40 absolute value V of T DT is the total distance distance she travels she travels over the 40 minutes over the 40, over the 40 minutes, remember our unit of time here is in minutes. All right. So using correct units and so this is and so the units, let me write it this way. Is a total distance she travels over the 40 minutes. The units are meters. The units are meters, are meters. All right, now this says approximate the value of that integral using a right Riemann sum with the four subintervals indicated in the table. So what do they mean by a right Riemann sum? Well actually, let me just, let me just draw, let me draw the absolute value of the V of T and I already took the trouble of the last part of drawing the axis and you wouldn't actually have to do this on the test, but I'm doing this to just help you understand what we're actually going to do when we talk about a right Riemann sum. So if we, so in orange I have V of T, but if I want to do the absolute value of V of T, I'll do that in magenta, so the absolute value of that is that, absolute value of that is that, absolute value of that is that, absolute value of negative 220 is going to be positive 220. So we saw here when time, when T is equal to 24 V of T is negative 220, she's running in the opposite direction or jogging on the other direction, but we're going to take the absolute value of that. So when T is equal to 24, her speed, we could say, is going to be 220, we're going to take the absolute value of it. So it's going to be right around, right around there and then the absolute value of this is going to be that point again and so we don't know the actual velocity function. The actual velocity function, I don't know, it might look something like, it might look something like, like this. It goes like that, maybe it dips down. I should say the speed function. We don't know the actual speed function. So this is the absolute value of V of T, which is the speed function you could say. Speed, speed function, and so if we wanted to actually integrate it, so if we wanted to integrate it, well we don't know how to do it if we don't know, we don't know how to do it exactly without knowing the exact function, but we can approximate what the integral is going to be with a right Riemann sum and a Riemann sum is just breaking up this area into rectangles and then finding the total area of those rectangles. And then there's a couple of ways you could do it, you could use, you could use for each rectangle you could use the left boundary as the height or you could use the right boundary as the height and so they tell us to use a right Riemann sum. A right Riemann sum, so for each of the rectangles, for each of the intervals we're going to use the value of our function on the right side to approximate the area. So let's do that. So let me divide this into four rectangles and they give us the four intervals. So one interval goes from zero to T equals 12, we see that there. The other integral goes from 12 to 20, the next integral goes from 20 to 24, the next integral goes from 24 to 40, and if the whole notation of a Riemann sum is completely foreign to you, I encourage you to watch the videos on Khan Academy on Riemann sums. And so here we are. So the first interval's from zero to 12 and so we want to use the value of our function on the right side to define the height of the rectangle. So let me define. So that rectangle is going to look, is going to look like this, so that's a rectangle right over there. Then the next interval goes from 12 to 20. The next rectangle goes from 12 to 20, we see that there. From 12 to 20. So let's use a value of our function at 20 to define the height of this rectangle. So just like that. Look like, let me do it, I'll just hand draw that. It would look like that and then the next rectangle goes from T equals 20 to T equals 24 and we want to use the right hand side to define the height of that rectangle. You can see it's a little irregular and that's okay, they all don't have to have the same width. So the height is something like this, and then the last interval is T equals 24 to T equals 40 and so we use the right side as the height. So the interval, so this was the last interval we did, the next interval is going to be that and then the height we use the right hand side to define the height that's why it's a right Riemann sum and so our approximation of the integral of the absolute value of V of T is going to be the area of these rectangles over here, and so we could write, we could write the integral from zero to 40 of the absolute value of V of T DT is approximately, well it's going to be this first change in time times this height, so this first change in time this is a change in time of 12 times, so it's going to be 12 times, what's the height? Well V of 12 we knew was 200, V of 12 is 200. So it's going to be 12 times, 12 times 200 plus, okay the height here or the next interval goes from 12 to 20, so it's going to be eight times and then the height here is going to be what is V of 20, I think that was V of 20 was 240. V of 20 was 240, and then the next interval goes from 20 to 24 so that only has a change in time of four and then what is the height at 24? The height at 24, and remember we took the absolute value of V of T so it's 220. So four times 220 and then our last rectangle, our change in time we go from 24 to 40, so our change in time is 16 times the height at 40. What was that? That was 150 I think. Yup, that was a 150. So times 150. 150, and so, just to make clear what I just did, I'm just adding up the sums of the areas of these rectangles. The area of this rectangle is 12 times 200, area of this rectangle's eight times 240, the area of this rectangle's four times 220, area of this rectangle's 16 times 150, and then let's see if we could calculate that, we're not allowed to use our calculator for this part so we're going to do it by hand. 12 times 200 is going to be 2400, plus eight times 240, that's going to be 1600, 1600 plus 3200 that is 1920. Did I do that right? Eight times 40 is 3200, or sorry eight times 40 is 320 plus 1600, yup that's right. Plus, this is going to be 880 plus, let's see 15 times 15 is going to be, would be 2250 that would be 15 times 150, and so we're going to have another 150 and so this is going to be 2400. 2400, whoops, it's 2400. Then we want to add all these together, let me just do it the way that, the easiest way. 2400, plus 1920, plus 880, plus, plus 2400 again, plus 2400 again will give us, will give us, we get a zero, we get a 10, so let's see we have nine plus nine is 18, plus eight is 26 and then we have seven. So we get 7,600, so is equal to 7,600 and just to be clear, this would be in meters. In, in meters. In meters. Now you might be saying wait, wait, wait didn't we just find an area, wouldn't areas be in meters squared? But remember, we're finding the area under the velocity, under the velocity graph, and so if you look at our units, say for this rectangle right here you're multiplying 12 minutes, you're multiplying 12 minutes times 200, times 200 meters per minute, times 200 meters per minute. So when you multiply them out, the minutes the cancel with the minutes and you're left with 2400 meters. So even though it's area, if you look at the units, it's going to be in meters not meters squared.