The two properties that let an integral be broken apart and rescaled

0:00You are about to be handed five statements about integrals. Two of them you will use on every question you ever answer. Two more you will probably never use directly, and they are the reason the first two are true. The fifth is the first two written out over a longer list. So the honest count is not five things to remember. It is two tools and the argument that earns them.

0:27But the item that pays for all of it is not numbered at all. It is one unremarkable sentence about what an equals sign between two integrals is claiming, and almost nobody reads it. Every liberty taken with constants after that point is drawn on that one account, and we are going to open it. Start with the two halves of the first statement, because they do not match and the mismatch is the whole lesson.

0:58Differentiate an integral and you get the integrand back, bare. Integrate a derivative and you get the function back with a constant attached. That is usually read as one slogan: the two operations cancel. They do not cancel in the same way, and here is the difference as a count. Eight integrands were each handed an answer, and each answer put through one measured door: does the rate of change this rule actually settles on, read from both sides at five inputs, equal the integrand?

1:35All eight pass. Now go downward. Take seventeen different members of each family, a quarter apart, and differentiate every one of them at the same input. Each of the eight comes back with exactly one value. Seventeen different functions, one rate. The parameter dies on the way down. Now go upward. Ask which functions have that integrand as their rate, and put all forty-one members of the grid through the same door.

2:07All forty-one pass. One value down. Forty-one up. That is not a slogan about cancelling; it is a count, and the two directions are not each other's mirror. So the second statement has work to do. It says that if two integrals have the same rate of change, they name the same family. The argument is three lines. Subtract them. The difference has a rate of nought everywhere. A function whose rate is nought on an unbroken stretch is constant. So the two differ by a number.

2:46That is a claim about two sets, so a set is what gets built here. A family is a rule plus a grid of parameters, and a member is that rule with one of them added. Two families are equal when every member of the first is a member of the second, and every member of the second is a member of the first. Not when the two rules look alike. Membership, both directions, counted.

3:15Here is why that matters, and it is the moment most students first think they have made a mistake. Take half the square of the sine. Take minus half the square of the cosine. Those are not the same expression. They are not even the same sign. Both were put through the rate door on the same integrand, at five inputs, and both pass. Now read their difference across the inputs. It takes exactly one value.

3:45And read the two rules as numbers. They are different at every one of the five inputs. Five out of five. Never once equal. Two rules that are never equal, naming one family. The same thing happens with half the square of the tangent against half the square of the secant. For both of those pairs the membership search runs in both directions and finds every one of the seventeen inner members in the other family. Seventeen for seventeen, each way.

4:19If you have ever written an answer that differed from somebody else's by a constant and assumed you were wrong, this is the sentence you were missing. A third pair belongs here, and it is the one that went wrong. The inverse sine and minus the inverse cosine also answer one integrand, and their difference also takes exactly one value. But the membership search came back with nought of seventeen. In both directions.

4:51That is a failure worth sitting with, because it is not the families disagreeing. The number between those two answers is a quarter turn, and no fraction is a quarter turn. The grid of parameters was laid out a quarter apart, so it contains no name for the number that separates them. The search had nowhere to point. Slide the same grid along by that quarter turn, found by halving and carried as a bracket rather than rounded, and all seventeen are found at once.

5:26The families were identical the whole time. What was missing was a name for the gap. That is the difference between a test failing and a claim being false, and it is worth knowing which one you are looking at. Now the unnumbered sentence. When you write one indefinite integral equal to another, you are not saying two numbers are equal. You are saying two families coincide, and the parameter is simply left off the page.

5:58That is why the constants can be shuffled, merged, renamed and dropped without anybody objecting. It is one permission, granted once. Read that sign as an equality of numbers and half of what follows looks like a sequence of small favours the writer is granting themself. Read it as an equality of sets and they are all the same move. Everything left in this video is a consequence of that reading, and we are going to watch it pay for itself four times.

6:34First payment. An integral splits across a sum. The proof is short and it has a shape worth stealing. Do not compare the two integrals. Compare their rates, and then appeal to the family statement. Twenty-one pairs were drawn from seven integrands. For each pair the two answers were added, and the sum put through the rate door against the sum of the two integrands. All twenty-one pass. Now try the comparison the proof refuses to make.

7:07Take that same sum of answers, and take another member of its own family, three quarters along. Their difference takes one value, so they are in one family. But as rules, do they agree? At nought of the twenty-one pairs. Two things in one family need not be one thing. That is exactly why the proof goes the long way round, and why the statement that seemed like scaffolding turns out to be load-bearing.

7:39Second payment. A constant factor comes out through the integral sign. The word doing the work in that sentence is constant, and it can be measured rather than emphasised. Six real numbers were put in front of each of the eight integrands, and in front of each of their answers. All forty-eight pass the door. Then four functions were put in the same two places: the input, the square, the sine, the exponential.

8:09Thirty-two attempts. Nought pass. And every one of them was measured and refused, not skipped for want of a value. Forty-eight against nought. A factor that stays still may leave. A factor that moves is part of the integrand and stays where it is. And now the awkward case, which is the sharpest test of what we agreed an integral names. The statement admits any real number. So it admits nought.

8:40Put nought in front, and on the left you have the integral of the zero function. Every one of the forty-one constants on the grid passes the rate door for it. Forty-one answers. On the right you have nought times a family. Read that as the single number nought and exactly one of those forty-one survives. Forty of the forty-one answers are lost by that reading, which is why students who reach for it conclude that the integral of nothing is nothing.

9:13It is not. It is every constant there is. Read the equals sign the way that one sentence told us to, as two families, and the two sides match member for member. Seventeen for seventeen, both directions. This is the smallest and the loudest illustration of what that sentence is for. Third payment. Both moves at once, over any finite list. This one is usually presented as a fifth result, and it is not a result at all. It is the two tools applied repeatedly.

9:51Twelve finite lists were made, each of two to four rules with a real coefficient in front of each. Every answer was assembled by scaling each piece and adding, using nothing but the two tools. All twelve pass the rate door. Now let one of those coefficients vary with the input, which is the one thing the statement forbids. Replace the first coefficient by the input itself and change nothing else.

10:22Every one of the twelve fails. So the generalisation is real, and the restriction in it is real, and neither had to be taken on trust. Fourth payment, and this one is a habit rather than a statement. When you split an integral you get two constants. When you write the answer you write one. Everybody does it, and it looks like tolerated sloppiness. It is not. It is a fact about parameters, and it can be counted.

10:54Carry the two constants separately through a worked split, over seventeen values each. Two hundred and eighty-nine pairings. All two hundred and eighty-nine pass the rate door. So nothing you write with two constants is wrong. Now the merge. Every one of the forty-one constants a single symbol can take is reached by some pair from that grid. So the one-constant answer names nothing the two-constant answer did not. And no target is reached by fewer than seventeen different pairs, which is the other half: the two-constant answer names nothing extra either. The pairs were never independent.

11:39One symbol loses nothing and invents nothing. That is the whole justification for the habit, and it is a measurement. Here is the split those constants came out of. A cube less one, all over the square. Nothing on the standard list matches that. Divide through and it is the input, less one over the square. Two pieces, each of which is on the list. Added back together the two pieces return the original integrand at every input tried. Apart at nought of five.

12:16Integrate each and the answer is half the square, plus the reciprocal. It passes the door. Two constants arose. One is written. You now know exactly what that costs, which is nothing. The splitting statement earns its keep in a way that is easy to miss, because the split is usually invisible until you open a bracket. Three worked integrands were taken. Each was put to a list of eleven standard rows, with a stated set of constant multiples in front of each row, exactly as it stands.

12:56The first matches nothing. The second matches nothing. The third matches nothing. Nought, nought and nought. The second of them is the clearest: a cosecant multiplied by a bracket holding a cosecant and a cotangent. As written, it is on no list anywhere. Expand it and it is two standard rows sitting side by side. Do that for all three, and each becomes two pieces that add back to the integrand at every input tried, and every one of those six pieces is on the list. One row each.

13:34Nought matches became six. That is not tidiness. That is the entire value of the property, and it is why expanding a bracket is worth the trouble. All three assembled answers pass the rate door. Now the two things that are not being claimed, because this is where the errors come from. Nothing here says anything about a product. Nothing here says anything about a quotient. Those are not warnings. They were searched for.

14:07Twenty-eight pairs were drawn from the eight rows. Multiply the two answers together and ask whether the product answers the product of the two integrands. Nought of twenty-eight pass. Divide them instead and ask the same. Nought of twenty-eight pass. And that search is capable of saying yes: hand it the zero rule against itself, where the product of two constants really is an answer, and the same door accepts it at once.

14:39So the nought is a measurement and not a broken loop. Products are handled, but by a different technique entirely, and it is some way off. Until then, a product in an integrand is a reason to rewrite, not a reason to split. One last thing, about what survives when this is condensed. A summary of all this keeps the sum, keeps the constant factor, keeps the version over a finite list.

15:10It drops the two statements that prove them. That is a fair trade for a revision sheet. Those two really are never used directly on a question. Fifteen items of the shape a question actually takes were built, each a finite list of two to four standard rows with a real coefficient in front of each. Put to the eleven-row list as they stand, not one of the fifteen matches a single row. Nought out of fifteen.

15:42Assembled by scaling each piece and adding, all fifteen pass. Nought to fifteen. That is the whole distance the two tools carry you, and it is why a revision sheet keeps them. But keep the two dropped statements somewhere too. The first one is where the constant comes from, and losing it is how people lose the constant permanently. The second one is the only reason you are allowed to compare two integrals at all, and it is the sentence that says an equals sign between them is about two families rather than two numbers.

16:20Keep the tools by all means. Just do not throw away the licence you are using them under.