Topic 8.1: Finding the Average Value of a Function on an Interval

The ghost booster

What’s the average of a velocity that never sits still? Our booster falls from twenty-three hundred and four meters to zero in twelve seconds, and its velocity is different at every instant, all the way from negative three sixty up to zero. The most natural answer is its average velocity, the change in height divided by the time, and we know both of those, twenty-three oh four meters down in twelve seconds.

To see what that number really means, let’s give the booster a ghost, a second booster that falls at one steady speed the whole way down and never slows down at all. Set the ghost’s speed so that it reaches the deck at exactly the same moment as the booster, and then let them both go.

The average velocity gets you to the right place at the right time, and it’s still a terrible way to land. On the velocity graph, the ghost is this flat line. Now let the curved region settle like water. It levels out into a rectangle exactly twelve seconds wide and one ninety-two deep, and the area never changes on the way.

The change in height is the integral of the velocity, so the average velocity is that integral divided by the length of the interval, one over twelve times the integral from zero to twelve. Nothing about that is special to boosters. For any continuous function f on an interval from a to b, the average value of f is one over b minus a, times the integral of f from a to b. In the picture, it’s the height of the flat line whose rectangle has the same area as the region under f.

Let’s make sure this works for things that aren’t motion at all. Water flows into a tank at six t minus t squared liters per minute for six minutes, so what’s the average rate of flow over those six minutes?

Back to the booster. The tempting shortcut is to average the first and last velocities, negative three sixty and zero, which gives negative one eighty, and that’s wrong here. Look what that number does, too. One eighty times twelve seconds is twenty-one sixty meters, the old height that crashed.

That’s exactly what we did in precalc, a hundred eighty times twelve. For a straight line the region is a triangle or a trapezoid, and its area really is the width times the average of the two ends. Our curve bends, so that shortcut misses.

The AP exam loves a question like this one. Is there a moment during the burn when the booster’s velocity is exactly equal to its average velocity? A full-credit answer has four pieces. The velocity is a polynomial, so it’s continuous on the interval from zero to twelve, and negative one ninety-two is between v of zero and v of twelve.

So by the Intermediate Value Theorem, there’s a time c between zero and twelve when the velocity is exactly negative one ninety-two meters per second. On the graph you can see it at about six point two seconds. That’s the moment the booster and its ghost are falling at exactly the same speed.

Real readings

Now let’s do one the way the exam does it, with real data. These speed readings come from the webcast of SpaceX’s CRS-12 mission in August 2017. We start a few seconds into the landing burn, with the booster about three point eight kilometers up. Notice that the readings aren’t evenly spaced. The gaps are four seconds, then eight, eight and eight, and then four, and the exam does that on purpose to catch anyone who assumes they’re all the same.

Speed is the absolute value of the velocity, so the integral of the speed from zero to thirty-two is the total distance the booster traveled in those thirty-two seconds. With readings like these, we estimate it with trapezoids.

Use a trapezoidal sum with these five intervals to estimate that integral, to the nearest meter. The webcast’s altitude readout said three point eight kilometers at the first reading, and we got just under four kilometers from speed alone. The little extra is sideways motion plus the error in our five trapezoids. If the exam asked you to explain what that integral means in context, which of these sentences would earn the point?

If they ask for the average speed over these thirty-two seconds, you already have it. One over thirty-two times that integral is about a hundred twenty-four meters per second, the average value formula with a trapezoid sum inside it.

Land it

One more booster, and this one’s all yours. Its flight computer has planned a burn that brakes hard for twelve seconds, down to sixty meters per second, and then eases off for a gentle final eight seconds.

You’ve got its velocity graph and nothing else, so work out how high it has to light its engines, set that height, and press light. That’s a landing. You took a velocity graph, found the area under it, and turned that area into the one height where a booster can touch down on a ship in the middle of the ocean.

Let’s put the whole lesson in one place. The integral of velocity is displacement, and adding the starting position tells you where you are. The integral of speed is the total distance, and an integral divided by the length of its interval is the average value, which for our booster was exactly the ghost’s steady velocity. Next time, the things we integrate will be rates of water and people flowing in and out, and you’ll find it’s the same idea all over again.