2015 AP Calculus AB/BC 3b | AP Calculus AB solved exams | AP Calculus AB | Khan Academy

Part B using correct units, explain the meaning of the definite integral.

So, it's the definite integral from zero to tal 40 of the absolute value 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? The velocity is a vector; it has direction. You could see when you look up here that sometimes the velocity is positive and sometimes it is negative.

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. 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.

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 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 0 to time equals 40.

So, if you integrate speed, if you're saying, okay, for every little amount of time DT, she has some speed, well, that's going to give you the distance that she travels. Remember, 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 here 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.

Let me write that, so this integral from 0 to 40 of the absolute value V of T DT is the total distance she travels over the 40 minutes. Remember, our unit of time here is in minutes.

All right, so using correct units, the total distance she travels over the 40 minutes, the units are meters.

Now, they say 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 draw the absolute value of V of T.

I already took the trouble in 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; the absolute value of that is that; the absolute value of that is that; the absolute value of -220 is going to be positive 220.

So, we saw here when time—when T is equal to 24, V of T is -220; she's running in the opposite direction or jogging in 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'll take the absolute value of it, so it's going to be right around there. And then the absolute value of this is just going to be that point again.

So, we don't know the actual velocity function; the actual velocity function might look something like this. It might look something like this; it goes like that, maybe it dips down, or 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.

If we wanted to actually integrate it, 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.

A Riemann sum is just breaking up this area into rectangles and then finding the total area of those rectangles. There are a couple of ways you could do it; you could use the left boundary as the height, or you could use the right boundary as the height.

So, they tell us to use a right Riemann sum. 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. Let me divide this into four rectangles, and they give us the four intervals. One interval goes from 0 to 12; you see that there. The other interval goes from 12 to 20. The next interval goes from 20 to 24, and the next interval goes from 24 to 40.

If the whole notion of a Riemann sum is completely foreign to you, I encourage you to watch the videos on Khan Academy on Riemann sums.

So, here we are. The first interval is from 0 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 like this; 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 the value of our function at 20 to define the height of this rectangle. Just like that; look like— let me hand-draw that; it would look like that.

Then the next rectangle goes from T = 20 to tal 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.

Then the last interval is tal 24 to tal 40, and 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.

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.

We could write— we could write the integral from 0 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 * 200.

Plus okay, the height here, or the next interval, goes from 12 to 20. So, it's going to be 8 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.

Then the next interval goes from 20 to 24, so that only has a change in time of four. 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, 4 * 220.

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. Yep, that was 150, so times 150.

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 * 200; the area of this rectangle is 8 * 240; the area of this rectangle is 4 * 220; the area of this rectangle is 16 * 150.

Now, let's see—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 * 200 is going to be 2400. Plus 8 * 240, that's going to be 1920.

1,600 + 3200, that is 1920—did I do that right? 8 * 40 is 320. Oh, sorry, 8 * 240 is 1920, yep, that's right, plus— this is going to be 880. Plus, let’s see, 16 * 150— would be 2400.

15 * 15 is going to be— would be 2250; that would be 15 * 150. And so, we're going to have another 150.

So, this is going to be 2400. Oops, is 2400. Now we want to add all these together.

Let me just do it the way— the easiest way, 2400 plus 1920 plus 880 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 9 + 9 is 18 + 8 is 26, and then we have 7. So we get 7600, so the total is equal to 7600. And just to be clear, this would be in meters.

Now, you might be saying, 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 graph.

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 m per minute.

So when you multiply them out, the minutes cancel with the minutes, and you're left with 2400 m. So, even though it's area, if you look at the units, it's going to be in meters, not meters squared.