Substituting inside a definite integral, and why the limits have to move with it

0:00Substitution is a technique you already have. Rename a piece of the integrand, convert the differential, integrate in the new letter. Attach two numbers to the integral sign and almost nothing about that changes. There is one thing, and it is the whole of this video. The two numbers are not decoration on the integral sign. They are two values of the variable. So the moment you rename the variable, they stop meaning what they meant.

0:32That is the entire idea. Everything else is a consequence of it, and so is the one mistake everybody makes here. And that mistake is quiet. It does not throw an error. It hands you a perfectly ordinary number that happens to be the wrong one. Here is the safe route, in four steps. One. Ignore the two numbers for a moment. Make the substitution and convert the differential. Two. Integrate in the new letter. And do not write a constant -- there is nothing for it to do here, because it is going to be subtracted away.

1:14Three. Put the original letter back. You now have an expression in the letter you started with. Four. Only now, put the two numbers in. Evaluate at the upper one, evaluate at the lower one, subtract. Look at where the limits are for three of those four steps. They are sitting off to one side, untouched, waiting. That is the safety of this route: the limits never have to move, because you always come back to the letter they belong to.

1:47There is a second route, and it is a trade. Delete step three -- do not put the original letter back at all. Stay in the new letter and evaluate there. You pay for that by moving the limits. If the integral is now in a new letter, the two numbers attached to it have to be two values of that new letter. That is not optional and it is not a shortcut for the confident. It is what skipping step three costs.

2:17Neither route is more correct than the other. They are the same piece of arithmetic in two different orders, and which one is shorter depends entirely on the integral in front of you. So where do the two new numbers come from? You feed the old limits to the substitution. To the substitution, and to nothing else. That sounds obvious written down and it is the commonest place to slip, because by the time you need the new limits there are three or four expressions on the page and only one of them is the substitution.

2:56Put the old limits into the anti derivative instead -- a different expression, sitting right next to it -- and here is what happens across seven worked substitutions. It lands on the right answer nought times out of seven. It returns a wrong number three times, and four times it cannot return anything at all. So it is not a slip that sometimes works. It is a slip that never works.

3:26Before we run any of this, there is a factor that has to be right, and it is worth showing rather than quoting. When you rename the variable, the differential changes too, and the thing it changes by is the substitution's own rate of change. I am not going to write that rate down from a table. It is measured, off the substitution, by the same difference quotient that judges every anti derivative in this video.

3:57Then the test is this: is the original integrand equal to the new integrand, composed with the substitution, times that measured rate? At five inputs inside each of seven intervals -- thirty-five readings -- it holds all thirty-five times, fails none, and is unreadable on none. Now drop that factor. Rename the variable and carry the old differential across unchanged, which is exactly the mistake. It holds at seven of the thirty-five.

4:33And those seven are not luck. They are exactly the seven inputs where the substitution's measured rate happens to be one -- the places where there was nothing to drop. Now the two routes, run against each other. Except that is the one comparison that would prove nothing. The two routes are the same arithmetic in a different order, so of course they agree. If both were built on a wrong anti derivative, they would agree on the wrong number.

5:06So both are checked against something that knows nothing about any substitution: a midpoint sum of the original integrand, over the original limits, cut into two hundred pieces. The four-step route lands on that sum, within a thousandth, on all seven rows. It misses none and is silent on none. The route that moves the limits does the same. Seven of seven. And their difference, taken on all seven rows, takes exactly one value, and that value is nought every time.

5:43Put the route that leaves the limits behind in the second place instead, and the same subtraction is available on five rows, takes five different values, and is nought on none of them. So the nought above is a reading about the two routes and not a fact about subtraction. The new limits can come out in either order, and this is where confidence goes. Of the seven substitutions here, six send the lower limit to a smaller number than the upper one -- the ordinary case, and everything looks familiar.

6:21One does not. Its upper limit lands below its lower one. Nothing has gone wrong. The substitution is decreasing across that interval, so it turns the two ends round. And that is a reading rather than a coincidence: in all seven the direction the ends move agrees with the sign of the substitution's own measured rate across the interval. The control is a substitution that turns round inside its own interval. Its two ends come back equal, so it is neither rising nor falling, and there the direction agrees with the rate on none of them.

7:04There is one more shape of surprise, and it is worth being ready for. Take an inverse tangent divided by one plus a square, from nought to one. The substitution is the inverse tangent itself. The lower limit is nought and the inverse tangent of nought is nought, so that end is quiet. The upper limit is one, and the inverse tangent of one is a quarter of a half turn. A number that looks nothing whatever like the one it came from.

7:37The value comes out at the square of a half turn, over thirty-two. If you were expecting the answer to look like the limits you started with, this is where you would stop and check for a mistake -- and there is no mistake. Now the failure mode itself, carried out rather than warned about. Rename the variable. Integrate in the new letter. Then evaluate between the two old numbers, because they are still written on the integral sign and nothing shouted at you.

8:11Across the seven substitutions here, that lands on the right answer nought times. Five times out of seven it hands back an ordinary number. Not an error message, not a refusal -- a number that looks exactly like an answer. Only twice does it decline to answer at all, and that is luck: it happens when the old limits fall somewhere the new expression cannot be read. So five times out of seven you would walk away with a wrong answer and no reason to suspect it. That is why the check has to be part of the habit rather than something an alarm sets off for you.

8:53And here is the control that says this is about the substitution and not about the route. Take a substitution that renames the variable to itself. Leaving the limits alone is then exactly right, and it lands on the sum. When there is nothing to move, moving nothing is correct. Sometimes a question arrives with a hint that names the substitution for you. It is easy to read that as a courtesy. It is not. It is a statement about which of two workable substitutions is shorter.

9:32Take the input times the square root of two more than it. The obvious move is to set the bracket -- two more than the input -- equal to a new letter. That works. The hint says something else: set the bracket equal to the square of a new letter. Both reach the same number. Their difference takes one value and that value is nought. So what separates them? What is left to integrate. And that can be measured rather than asserted: fit a polynomial through five values of each transformed integrand, then try it at twenty different inputs.

10:11The squared route survives all twenty and misses none, because it genuinely is a polynomial -- the root is gone. The plain route survives none of the twenty and misses all twenty, because a half power is still standing in it. The hint was telling you which one clears the root. Here is something two attached numbers can do that no amount of algebra can. One item asks for the integral of an inverse sine, applied to twice the variable over one more than its square, from nought to one.

10:49Every solver reaches for the same rewriting: that expression is twice the inverse tangent of the variable. It makes the item easy. But that rewriting is not always true. It is true on part of the line and false elsewhere. Read at twenty-one inputs across the stated interval, it holds twenty times, fails none, and refuses once -- at the far end itself, where the inverse sine is right at the edge of what it is defined on. Read across an interval three times as long, it holds seven times and fails fourteen.

11:29In every one of those runs, the inputs it survives are exactly the inputs whose modulus is under one. Not under two, not under a half -- both of those were put to the same test and both came back false. The two numbers on the integral sign are what keep the item inside the region where the rewriting is honest. Strip the limits off and that guarantee goes with them.

11:58I want to come back to something from earlier, because it is the most useful habit in this whole video. If you work an integral two ways and the two ways agree, you have not checked your answer. Here is the demonstration. Take an integrand whose anti derivative is nearly right -- a cube over two where a cube over three was wanted. Run both routes on it. Resubstitute and evaluate at the old limits; or move the limits and evaluate in the new letter. The two answers agree exactly. Their difference is nought.

12:36And both of them are wrong. The sum catches it immediately: the shorter route lands on the sum nought times and misses once. Two routes agreeing is not two answers agreeing. If you want a check, it has to come from somewhere else -- differentiate your answer, or estimate the integral numerically, or read the sign and rough size off the picture. One last practical thing: how to pick the substitution in the first place.

13:09Look for a piece of the integrand whose rate of change is also sitting there, up to a constant factor. That is the whole heuristic, and it is why substitution works at all. A fifth power under a root, with a fourth power outside it. A bracket cubed, with the input outside it. An inverse tangent, with one over one plus a square beside it. In each case one part is the other part's rate of change wearing a disguise.

13:42If a root is in the way, ask what would clear it. Setting the bracket to a square rather than to a plain letter is a standard move for exactly that. And be ready for items filed under substitution that are not substitutions at all. Some complete a square and land on a formula you already know. Some are a shape you recognise on sight, where the anti derivative is spotted rather than derived.

14:12The heading on an exercise is a hint about what was on the author's mind, not a promise about every item under it. So, the short version. The two numbers on an integral sign are two values of the variable. Rename the variable and they have to be renamed too. You have two routes. Come back to the old letter and use the old numbers, or stay in the new letter and convert them. Both are correct and neither is the clever one.

14:46The new numbers come from the substitution and from nothing else. Put the old limits into anything else on the page and, across seven substitutions here, it is right nought times. And the failure is silent. Leave the limits behind and five times out of seven you get a number that looks exactly like an answer. So convert the limits at the moment you convert the differential, in the same breath, every time. Not because you might forget -- because if you do forget, nothing is going to tell you.