Accumulation and the Two Halves of Calculus

Derivatives and integrals. One of them measures how fast something is changing. The other measures how much of something has piled up. Two completely different questions, and this whole lecture is about the fact that they are one question, asked from opposite ends. So let me start with a tank. Somebody has opened a tap, and I know the rate, at every instant, exactly how fast the water is arriving. The question is simple enough. How much water is in there? Let's draw the rate. Time runs along the bottom, and the height of this graph is how fast the water is arriving at that moment. This tap is being opened steadily, so the line climbs. So at time one, the water is arriving at one litre a second. Move out to time three, and it is arriving at three litres a second. Now hold that thought, and imagine a different tap for a moment. Suppose the rate were constant. Say two litres a second, flat, the whole way across. Held there for three seconds, the answer is easy. Two litres a second, for three seconds, is six litres. Rate times time. And on this picture, rate times time is the area of one rectangle. But our tap is not that tap. Our rate climbs the whole way, so there is no single rectangle to draw. So we chop the time up into slivers instead. Over a stretch short enough the rate barely changes, so each sliver is rate times time again. One thin rectangle. Then we add them all up. Now watch what happens as we use more and more of them. The staircase closes on the region underneath the curve, and in the limit the water that has arrived is exactly the area under the rate graph. So here is the object I care about. The amount of water that has arrived by time x is the area under the graph, from the start out to x. Give it a name. A of x, the accumulation function. And notice what kind of object A is. It is a function. You hand it a time, here, and it hands you back a number, the area up to there. Watch what happens when I slide that edge along. More of the region is swept in, and the number gets bigger. Every stopping point gives you one number, so this really is a function of x, built out of nothing but area. Hold on to that picture. The area under a rate graph is itself a function, and everything that follows comes from one question about it. So let me put that question on the board, in as many words. How fast does A grow? Answering that is the whole of the rest of this lecture.

A moment ago we ended with the accumulation function. A of x is the area under the rate graph, from zero up to x. For this particular tap we can work that area out exactly, and pleasingly, we need no calculus at all to do it. First, one word about the letters, because there are two of them and they are not the same letter. Along the bottom the clock is called t, since t is what sweeps across the region while we shade it in. The letter x is saved for where we stop. So stop the clock at x. There is the mark, on the axis the clock runs along. And look at what is shaded: it is a triangle. Its base runs from zero out to x, so the base is x. Its height is the rate at time x, and for this tap the rate at time x is x as well. The area of a triangle is a half, base, height. Here that is a half times x times x. Which comes to one half x squared. There it is. The accumulation function for this tap, in closed form. Now let us put the two of them side by side, because this is where it gets good. On the left, the rate. A straight line, climbing steadily. On the right, the accumulation. A parabola, bending upward. As the tap opens further, the water piles up faster and faster. Now a question about the curve on the right. How steep is it? The steepness of A at a point is exactly what the symbol A prime of x will mean, so take the tangent line at x equals two and read off its slope. Rise over run, the slope there is two. Now look across at the rate graph. At time two, the height of the rate line is, two. Coincidence? Let's slide x along and watch both pictures at once. Carry on out to x equals three. The tangent on the accumulation curve now has slope three. And the rate line at time three is sitting at height three. The slope of the accumulation is the height of the rate, every single time. That deserves a line of its own. Two things to write down, and the first one is what the symbol is going to mean. A is a function of x, so A prime of x is its rate of change: how fast the shaded area grows as x moves off to the right. And the second is what we just watched. That rate of change is the height of the rate graph at x. So A prime of x equals f of x. The slope of the accumulation curve is the height of the rate curve, at every single point. We got there from one triangle and one parabola, but it has nothing to do with triangles, and I want to show you why it has to be true for absolutely any rate at all.

That was one particular tap, and one particularly friendly shape. But the fact we landed on has nothing to do with triangles. So here is a rate that wanders. No nice formula, and no geometry that will help you. Same set-up as before. A of x is the area underneath it, from zero up to x. We stop the clock here, at x, and everything shaded to the left of that line is A of x. Now nudge it. Push x forward by a small amount, and give that amount a name: h. There is h, on the picture, the little step from x across to x plus h. The area grows by this sliver, the water that arrives between time x and time x plus h. Call that extra area delta A. And here is the whole trick, so let me put it on the screen rather than leave it in the air. That sliver is very nearly a rectangle. There it is, standing on the sliver it is pretending to be. Its width is h, the step we just marked. Its top sits level with the curve at x, at a height of f of x, because over a stretch that short the curve has not had time to change. So delta A is about f of x times h. Now divide both sides by h, and the left-hand side turns into something you have met before. Change in A, divided by change in x. The average rate of change of A. And now let h shrink. Watch the sliver, and watch the rectangle standing on it. The thinner they both get, the better the rectangle fits, and the closer those two sides come to being equal. In the limit, the left-hand side is the derivative of A at x. That is precisely the definition of a derivative. And the right-hand side never moved. So A prime of x equals f of x, for any rate you like. Said out loud, it is almost a tautology. How fast is the water in the tank going up? At the rate the water is coming in. That is the whole theorem. The surprise is not that it is true, the surprise is what it lets you do.

Before we cash this in, we need the notation everybody actually writes. The area under f, from a up to x, has a symbol of its own. A long, stretched S. The function goes in the middle, and dee t closes it off. Out loud: the integral of f of t, dee t, from a to x. And let me say what those two ends are, because nobody ever does. The little a at the bottom is where we start counting. It is the left-hand edge of the region, chosen once and then left exactly where it is. The x on the top is where we stop. So the whole symbol is a function of where you stop, which is exactly what the shaded area was. And the S is for sum, because that is precisely what it is. All those thin rectangles, added up, with dee t as the width of one of them. So in this language, what we proved with the thin sliver reads like this. Differentiate an integral with respect to its upper limit, and you get the integrand straight back. Now the payoff. Here is a rate again, and here is the job in front of us: the area underneath it, between two fixed times. The left-hand end we call a. That is where the counting starts. The right-hand end we call b, and that is where it stops. Everything shaded between those two lines is the area we are trying to find, and we have no formula for a shape like that. Now suppose somebody hands you a function big F whose derivative is little f. Any such function at all will do. There it is, written down: F prime of x equals f of x. And we already have one function like that. Our own area function, A. Last time we proved that A prime of x is f of x too. So big F and our area function A have exactly the same derivative, everywhere, whatever big F happens to be. Which raises a fair question. How different can two functions with the same derivative actually be? Let me draw one of them over here. This is big F, and its slope at every point is the height of the curve beside it. Now take a copy of it, and lift the whole copy upward. Every point goes up by the same amount, so the curve keeps its shape exactly. At every x it is still leaning the way it was leaning. Its slope has not changed anywhere. Push a copy downward and you get the same story again. All three of these are antiderivatives of the same little f. There is a whole family of them stacked up the board, and every one of them is parallel to the others. And here is where the constant comes from. Measure the gap between two of them at x equals a. Now measure it again over at x equals b. It is the same gap. It is the same gap everywhere, and that gap is the number we call C. So two antiderivatives of one function differ by a constant, and only by a constant. Big F of x is our area function A of x, plus some fixed number C. Now watch what that does the moment we subtract. Start from the line we just earned. Big F of x is A of x, plus C. That is true at every single x, so in particular it is true at b. Big F of b is A of b, plus C. And it is just as true at a. Big F of a is A of a, plus C. The very same C both times, because it is the very same big F both times. Now take the second of those two lines away from the first, exactly the way you would take one number off another. So rule a line underneath them both. And there is the whole trick, standing one above the other at the end of those two lines. Plus C on the top, plus C underneath. The same unknown number added on and then taken straight back off. So slash the pair of them out, and whatever C was, it has gone. So what survives on the right is A of b minus A of a. The constant has gone, and it was always going to go, whichever antiderivative you had picked up in the first place. And A of b minus A of a is exactly the area under f between a and b. Which, in the notation we built at the start of this scene, is the integral of f from a to b. So there is the bottom line. Let me write it out properly, because that is the theorem. So there it is, in a single line. The integral of f from a to b equals big F of b, minus big F of a, where big F is any antiderivative of f. That is the Fundamental Theorem of Calculus, and it is the reason integration is something you can actually sit down and do. To find an area, you no longer add up rectangles. You find a function whose derivative is the one you started with, and subtract at the two ends.

Let's spend it on a problem you cannot do with geometry. Here is y equals x squared, and here is the piece of it we want: everything under the curve between zero and two. The old route uses thin bars like these, then takes a limit. That is real work, because this is not a triangle, a rectangle, a circle, or any other shape with a school formula. By the new route you need exactly one thing. Some function whose derivative is x squared. So let us go and find one. Start by guessing. Differentiate x cubed, and you get three x squared. Write that down, because it is nearly right. It is three times too big. So divide the whole thing by three before you differentiate it. Now the three on top and the three underneath are the same three, and what comes out is exactly x squared. So there is our big F. F of x is x cubed over three, and its derivative is the x squared we started with. Now the theorem does the rest. The integral from zero to two of x squared, dee x, is F at two minus F at zero. F at two is two cubed over three. F at zero is zero. Take the second one away from the first. And what is left is eight thirds, which is about two point six seven. That is the exact area under the parabola, and we got it by differentiating backwards. No rectangles, no limits, three lines of algebra. Does that number look sensible? The region sits inside this two by four rectangle, which has area eight, and by eye the shaded part is about a third of it. Two point six seven out of eight. That checks out. So here are the two halves of one idea. On the left, differentiate an accumulation, and you get back the rate you were accumulating at. On the right, to total up a rate across an interval, find something that differentiates to it, and subtract at the two ends. Area and slope are not two separate subjects that happen to share a course. They are one machine, run in the two directions. That is the Fundamental Theorem of Calculus.