The area function, and the two theorems that tie area to antiderivative

0:00Here is an integral with two numbers attached to it. A lower end, an upper end, and a single number as the answer. Now do one small thing to it. Take the number at the top and replace it with a letter. That letter is not inside the integrand. It is sitting on the end of the integral, where a limit goes. And that changes what kind of object you are looking at. Before, you had a number. Now, for every place you might put that upper end, there is a number -- so what you have is a function.

0:36It is called the area function, and it is the one genuinely new construction in this part of the subject. It is easy to miss, because nothing on the board moved very far. But everything else in this video is a consequence of that one move. First, what the number actually measures. With the curve sitting above the axis, a definite integral measures a region. And that region has four boundaries, which is worth counting out loud, because a question will usually hand you two of them and expect you to supply the other two yourself.

1:17Above, the curve. Below, the horizontal axis. On the left, a vertical line standing at the lower end. On the right, a vertical line standing at the upper end. Four edges. The curve, the axis, and the two verticals. The two verticals are the numbers written on the integral sign. The curve is the integrand. The axis is not written down anywhere, and you are expected to know it is there.

1:48A region enclosed on all four sides is a bounded region, and that is the thing whose area this is. Now put the sliding edge somewhere in the middle, and the region falls into two strips. The left strip runs from the fixed lower end across to the sliding edge. That strip, and only that strip, is the area function's value. The right strip runs from the sliding edge out to the far end. It carries no label, because it is not the area function. It is what is left to sweep.

2:25This is the commonest thing to get wrong about the picture, so it is worth measuring rather than warning about. Across seven integrands, with twenty-one different places to put the dividing line on each, the strip up to the line plus the strip beyond it comes to the whole region in all one hundred and forty-seven cases, and misses in none. Now take that second strip from the fixed left edge instead of from the dividing line, which is exactly what reading both strips as the area function amounts to. Only ten of the hundred and forty-seven still come out right. The other hundred and thirty-seven are simply the wrong number.

3:13So let us actually slide it. Twenty-one places, from the fixed left edge across to the far one, on four integrands whose curve stays above the axis. Eighty-four different numbers in all -- and of the eighty steps between them, eighty take the value up, and none take it down. Which is what you would expect. Sliding the edge to the right only ever adds area, as long as the curve is above the axis.

3:43But a loop that reports eighty rises might be a loop that is not sliding anything, so here is the control. Freeze the sliding edge at the far end and run the identical loop. It returns four numbers, one per integrand, and all eighty steps are flat. That is the difference between a reading and a description. The rises are there because the edge moved. There is one more thing to say about the moment before it starts.

4:16When the sliding edge has not moved yet, the left edge and the right edge are in the same place. The region has no width at all. So the area function is nought there. That holds on all seven of the integrands used here. And that is exactly what fixing one end buys you. One end is a setting -- you choose it once and leave it. The other end is the argument, the thing the function is a function of.

4:46Swap which one you fix and you get a different function. Fix both and you are back to a single number, which is where we came in. Now the assumption in that picture, and what it quietly does to you. I drew the curve above the axis throughout. That is why the strips are areas in the ordinary sense of the word -- pieces of the plane with a size.

5:12But that assumption is a courtesy to the picture. It is not part of the result, and you must not carry it away as a fact about integrals. Take three integrands that dip below the axis. Slide the edge across each of them, sixty steps in all: thirty take the value up and thirty take it down. The area function is under no obligation to rise. And the sixty-three readings hold only forty-nine different numbers rather than sixty-three, because two of those three area functions come back to values they had already taken. It goes up, and then it comes back down through the same numbers.

5:57Now put those numbers against a summed reading of the integrand's modulus -- the ordinary unsigned area, with nothing counted negative. The four rows above the axis agree in all four cases, and differ in none. The three that dip below agree in none, and differ in all three. So below the axis the number and the ordinary area part company. They are not the same quantity, and the picture cannot tell you that, because the picture was drawn above the axis.

6:34Now the first of the two results this whole construction exists for. Take an integrand that is continuous throughout the closed interval. Then the area function's derivative is the integrand itself, at every point of that interval. Read it slowly, because it is doing something strange. The slope of the area function at a place is the height of the curve at that place. One of those is about how fast an accumulated area is growing. The other is just how tall the curve happens to be. The theorem says they are the same number.

7:14It is believable if you watch the edge move. Push the sliding edge a hair further to the right, and the extra area you sweep is a thin strip -- height the curve, width the hair. Divide the extra area by the hair, and the width cancels, leaving the height. That is a reason to believe it. It is not a proof, and I am not going to give you one. I am going to measure it instead.

7:43That distinction is the whole of this scene, so here is what measuring means here. Every value of the area function used in this video was built by summing. Cut the stretch into a million pieces, read the integrand once in the middle of each piece, multiply each reading by the width of a piece, and add them all up. There is no anti derivative anywhere in that. Then take the slope of that summed function the honest way. Read it a thousandth to the left of a point, read it a thousandth to the right, subtract, and divide by the gap.

8:23At five inputs on each of the four rows above the axis, that slope lands on the integrand's own value in twenty cases out of twenty. None miss, and none are unreadable. Ask the same test to land on the integrand plus one instead, and it lands nought times out of twenty. So the test can fail, and it does fail when it should. Run it again on the three rows that dip below the axis -- the ones whose area function spends half its steps falling -- and the slope lands on the integrand in all fifteen cases there too.

9:03The theorem never needed the curve above the axis. That was only ever the picture being polite. Landing on a number is still not the same as a reading that is settling on it, so here is the sharper version. Halve the step and divide the two errors into each other. The error of a symmetric quotient is governed by the square of the step, so halving the step should quarter the error.

9:33It does, on three of the four rows above the axis. On the fourth there is nothing to divide, because a straight-line integrand makes the quotient exact and the error is nought at both steps. And an error is a size, not a direction. Some of these readings sit above the true rate and some sit below it, and both are equally wrong. Halve the step on all thirty-five readings. Twenty-nine shrink. Five were already exact and cannot shrink. And one does not shrink -- the one whose error had already fallen to two parts in a million million, which is the floor left by the million pieces the sum underneath it was cut into.

10:20There is also somewhere the quotient simply refuses to answer. Ask for it at the two ends of the interval themselves, and half of the neighbourhood it needs is off the end, where the integrand was never declared. It comes back as nothing at all in all eight cases. Not a hit, not a miss -- nothing. A test that returns nothing where it cannot see is worth more than one that guesses.

10:50It is easy to undersell what that theorem hands you. Before it, an anti derivative was something you found by recognising the integrand. A guess, with a check attached. And there are perfectly ordinary continuous functions nobody can guess one for, because it cannot be written with the symbols we have. After it, every continuous integrand has an anti derivative, and you have been handed one. It is the area function.

11:22So an area -- a piece of the plane -- and an anti derivative -- a function whose slope you were hunting for -- turn out to be the same object. That is the bridge. The first theorem builds it out of an area. The second theorem spends what the first one earned. Take an integrand continuous on the closed interval, and take any anti derivative of it. Then the definite integral is that anti derivative's value at the upper end, less its value at the lower end.

11:57That is the two-step recipe. Find an anti derivative, then subtract two of its values. Every evaluation of a definite integral you will ever carry out is this sentence. Measured against the summed numbers, on all seven rows the summed area equals the anti derivative's value at the far edge less its value at the fixed one. None miss. Now hand the same evaluation the integrand's own rate of change in place of its anti derivative -- differentiate where you should have integrated -- and six of the seven fail.

12:37The one survivor is the single row whose integral happens to be nought, where subtracting a thing from itself gives the right answer for the wrong reason. There is exactly one such row, and it is worth naming rather than hiding, because a test that passes for the wrong reason is the one that will fool you later. The words any anti derivative in that statement mean what they say, and that is worth showing rather than asserting.

13:10Every integrand has infinitely many anti derivatives, differing by a constant. The theorem lets you use whichever one you like. So take forty-one different constants and carry each of them through each row's evaluation. Every one of the seven rows comes back with exactly one value. Forty-one starting points, one answer. Which happens because the constant is added at the upper end and added again at the lower end, and the subtraction takes it straight back out.

13:46The control makes that mechanism visible. Add the constant at the far end only, so there is nothing to cancel against, and the same loop returns forty-one different values on every single row. So the cancellation is doing real work. It is not that the constant was never there. Now put the first theorem to work on the kind of question it actually answers. Suppose a function is defined as an integral from nought up to the variable, and you are asked for its derivative.

14:23It looks like a hard question and it is a one-line question, but only if you notice which theorem it is. Here are four things a reader might reasonably write down. The integrand at the input. The integrand at the fixed end. The integral itself. The integrand's own rate of change. Put all four to the measured slope, on all seven rows. The integrand at the input passes seven times out of seven. The integrand at the fixed end passes nought. The integral itself passes nought. The integrand's own rate of change passes nought.

15:05So the answer is the integrand, with the variable written in where the upper end was. Nothing is integrated and nothing is evaluated. It is the first theorem, wearing a disguise. And it is worth being able to recognise it in the middle of an exercise that is nominally about something else, because that is where it will turn up. One last distinction, and it is exactly the sort that gets lost when a result is restated from memory.

15:40The first theorem puts its hypothesis on a closed interval -- continuous there -- and states its conclusion on that same closed interval. Both ends, and everything between. Restate that conclusion for every point at or beyond the lower end, and you have quietly claimed more than you were given. Beyond the far end there is no hypothesis at all. And there is nothing there to read, either. Take the area function at ten inputs past the far edge of each row, where the integrand was never declared: nought of the seventy can be read. Every one of them comes back as nothing at all.

16:24On the interval itself, all one hundred and forty-seven readings are there, and none are missing. A conclusion stated on a wider set than the hypothesis covers is not the same conclusion. Usually nothing goes wrong, because nobody asks outside. But it is a habit worth not having. So: two theorems. Neither of them was proved here, and that is worth saying plainly rather than gliding past. The proofs need machinery that sits beyond this stage, and stating a result you cannot yet prove is a normal and honest thing to do -- as long as you say that is what you are doing.

17:10What was done instead was to measure them. The area function was built by summing and never by an anti derivative. Its slope was taken from those summed values. Only then were the two theorems tested against it, and every count in this video came out of that testing. And all of it rests on the small move at the start. The variable came out of the integrand and sat down on the end of the integral.

17:41A number became a function. Its slope turned out to be the curve. And that function, being an anti derivative, could be spent on any definite integral you like. One letter, moved a couple of centimetres to the right.