Adding two limits turns a family of functions into a single number

0:00Up to now an integral has been a family. You find one anti derivative and then you write plus a constant, because every other anti derivative is that one shifted up or down, and there are infinitely many of them. Now attach two numbers to the integral sign -- one below it, one above it -- and the object stops being a family. It becomes a single number. That is not a decoration. It is a change of kind.

0:31And almost everything that follows in this part of the subject is a consequence of it. The constant stops mattering. The answer can be checked against arithmetic instead of against another anti derivative. And a question can have a wrong answer in a way an indefinite integral never could. So it is worth being slow about what those two numbers actually do. The notation first, because the names are asked for directly.

1:05The number written below the integral sign is the lower limit. The number written above it is the upper limit. Together they are the limits of integration. The word limit is doing double duty here and it is worth flagging. These are not limits in the sense of something a sequence approaches. They are just the two ends of the stretch you are integrating across. And an integral carrying them is a definite integral, as against the indefinite one, which carries none and is a family.

1:40One number below, one number above, and one number as the answer. Now, that object has two different definitions, and both are worth knowing about. The first is the one you will use. Take any anti derivative of the integrand, read its value at the upper limit, read its value at the lower limit, and subtract the second from the first. The second builds the number directly, with no anti derivative anywhere in it. Cut the stretch into pieces. Read the function once in each piece. Multiply each reading by the width of its piece, and add them all up. Then let the pieces get finer without bound and see what the running total settles on.

2:30Both are real definitions, and both give the same number. But look at what the second one is asking of you. Not a limit you can write down in closed form -- a limit of sums with more and more terms in them, where the number of terms runs away to infinity while every single term shrinks to nought. You cannot execute that with a pen for anything but the very simplest integrands. Which is why every evaluation you will ever be asked to carry out goes through the first route, and why all that work on finding anti derivatives exists at all.

3:11So the phrase is worth knowing, and the route is not one you will be running by hand. It would be easy to leave it at that. I would rather show you the thing exists. So here is the second construction, carried out. Take five integrals with rational ends. A square from two to three. A root over a squared bracket, from four to nine. A fraction needing a decomposition, from one to two. One over one plus a square, from nought to one. And one over the variable, from one to two.

3:48For each one: cut the stretch into eighty equal pieces, read the integrand once in the middle of each piece, add the eighty readings, and multiply by the width of a piece. All five land within a thousandth of the number the anti derivative gives. None misses. Two controls, because landing near the right answer is easy to arrange by accident. Leave the width factor off -- add the readings and stop -- and none of the five lands. A heap of readings is not an integral.

4:24And put each row's sum against the other four rows' numbers, twenty comparisons, and none of the twenty lands either. The sums are not simply landing near everything. But landing near a number is still not the same as converging on it, so here is the sharper reading. Double the number of pieces from forty to eighty, and divide the two errors into each other. Reading the middle of each piece, the error goes down by a factor of four, on every one of the five rows. Double the pieces, quarter the error.

5:03Reading the left-hand end of each piece instead, it goes down by a factor of two on four of the five rows. Double the pieces, halve the error. Reading the right-hand end, the same. So the midpoint reading is not just more accurate, it is a different order of accurate, and that is a measurement rather than a slogan. There is one row where the endpoint readings also quarter their error rather than halving it, and it is not a fluke.

5:35Exactly one of the five integrands takes the same value at both ends -- the fraction, which comes out a sixth at one end and a sixth at the other. The leading part of the endpoint error is proportional to the difference between those two values, and here that difference is nought, so the leading part vanishes and what is left behind is the smaller one. I would rather report that than hide it, because a rule of thumb that has an exception you can explain is worth more than one that has none.

6:10And one more distinction, which took a second measurement. Getting the whole number four out of a division is not the same as the rate settling on four. So compare the ratio at ten and twenty pieces with the ratio at forty and eighty: the midpoint rule is nearer four at the fine end on four of the five rows, and on the fifth -- the square -- it is exactly four at both ends, because there the error is a fixed multiple of the width squared and nothing else.

6:43Right. Back to the route you actually have, which is two lines long. Step one: find the indefinite integral. Any anti derivative of the integrand. Step two: evaluate it at the upper limit, evaluate it at the lower limit, and subtract the second from the first. That is the whole of it. Every definite integral you will be asked for is those two lines. There is a notation for the second step: the anti derivative written inside a bracket, with the two limits stacked at its right-hand edge. That bracket is an instruction, not a value -- it says substitute both and subtract.

7:27And the thing to notice about step two is that it is the same two operations no matter what the integrand was. Two readings and one subtraction. Which means all the difficulty of a definite integral lives in step one. That is a useful thing to know when you are stuck: if you are stuck, you are stuck on an anti derivative, and everything from the previous module is what gets you out.

7:58Now the constant, and why you are allowed to drop it. You are not allowed to drop it because someone said so. You are allowed to drop it because it cancels, and you should watch it cancel once. Carry the constant through. At the upper limit you get the anti derivative there plus the constant. At the lower limit you get the anti derivative there plus the same constant. Subtract, and the constant is subtracted from itself.

8:29It contributes nothing. Not approximately nothing -- nothing. That is easy to assert, so it was measured instead. Forty-one different constants were attached to each of six anti derivatives and each evaluation run again, forty-one times over. Every one of the six rows comes back with exactly one value. The family really has collapsed. And the control is the part that makes that mean something. Put the constant at the upper end only -- which is what you have effectively done if you add it after substituting one end and forget the other -- and the same loop comes back with forty-one values instead of one. All forty-one, one per constant.

9:17So exactly one is a reading about the arithmetic, not an artefact of the loop. One more thing worth checking, because one value could be a suspiciously easy result. The six numbers themselves are six different numbers, and none of them is nought. Four worked evaluations, and the point of the group is not the definite step. It is where the work is. The first: a square, from two to three. The anti derivative is the cube over three, straight off the power rule, and the value is nineteen thirds.

9:57The second: a root over the square of a bracket, from four to nine. Here the anti derivative needs a substitution before anything else can happen. Substitute for the bracket, and you get two thirds of its reciprocal. Then the limits go in, and the value is nineteen ninety-ninths. The third: the variable over a product of two linear factors, from one to two. The anti derivative needs a partial-fraction decomposition first -- the coefficients come out minus one and two -- and then the value is the logarithm of thirty-two over twenty-seven.

10:39The fourth: a cubed sine of twice the variable, times the cosine of twice it, from nought to a quarter turn. Another substitution, an eighth of the fourth power of the sine, and the value is one eighth. Three rational answers, three that are not, and the routine that produced them does not know which is which. Every one of those six anti derivatives, and the two more I used for the sums, was put to the same test used everywhere in this series: its measured rate of change has to settle on the integrand from both sides, at inputs inside the interval being integrated over. Six out of six pass. With every answer's sign turned over, none of the six passes.

11:30Read the same six pairs a long way outside their own intervals and one of them comes back unreadable rather than as a verdict, because one of the rules does not exist out there. That is what makes none unreadable a real reading rather than a bucket nothing ever falls into. Look at the four again. One of them needed no preparation. Three of them needed a technique from the previous module before the definite step could even begin. And the definite step was the same two lines in all four.

12:09A short one, because it comes back later with a name attached. The lower limit goes below the sign, the upper limit goes above it, and swapping them is not free. Attach the two numbers the other way round and add the two readings together. Across all six rows the sum takes exactly one value, and that value is nought in every one of the six. Their difference is not nought anywhere: six rows, six distinct differences, none of them nought.

12:42So swapping the two limits changes the sign of the answer. It does not leave it alone, and it does not do something unpredictable either. Right now that is a fact to file. Later it becomes a numbered property with a sign in front of it. Now a condition that usually turns up as an afterthought, and belongs at the front. The integrand has to be well defined and continuous all the way across the closed interval. Not at the ends. Across.

13:17And it is worth looking at an integral that gets rejected outright by that condition, because there is one that is famous for it. The integrand is the variable, times the half power of one less than its square. The interval runs from minus two to three. One less than the square is negative whenever the variable sits strictly between minus one and one. A half power of a negative number is not a real number. So the integrand does not exist on that stretch.

13:51That was measured rather than argued. Over fifty-one evenly spaced inputs of the interval from minus two to three, the integrand is not a real number at nineteen of them and is at the other thirty-two. Nineteen out of fifty-one. That is a whole stretch missing from the middle of the interval, not one awkward point. And the control: over the twenty-one inputs of the interval from one to three, it is a real number at all twenty-one and fails at none. So the machinery is not simply refusing everything it is shown.

14:32Here is the part I think is actually worth your attention. Nothing stops you. Take that same rejected integrand. Find an anti derivative for it -- a third of the cube of the root. Substitute three. Substitute minus two. Subtract. The recipe does not protest. It does not refuse. It hands you an ordinary number, a little over five and eight tenths, and it looks exactly like every other answer you have written this week.

15:06That number is not the value of anything, because the thing it claims to be the value of cannot be written down. Which tells you where the domain question has to go. Not after the recipe, as a check. Before it, because the recipe will not take it for you. And there is a distinction underneath that worth keeping separate. Put that same pair to the test at inputs where the integrand does exist, and it passes -- it genuinely is an anti derivative there. Put it to the test at inputs inside the missing stretch and the test returns nothing at all, rather than a verdict.

15:48This is not an anti derivative and there is nothing here to read are two different answers, and a machine that reports the second as a pass is telling you nothing. Two multiple-choice items, and one of them carries a small trap that has nothing to do with integration. The first integrates one over one plus a square, from one to the root of three. The anti derivative is the inverse tangent. The value is the inverse tangent of the root of three, less the inverse tangent of one -- which is a twelfth of a half turn.

16:29To settle that without agreeing with anybody else's answer, both readings were put against a grid of twenty-four simple parts of a half turn, fixed in advance. The value lands on exactly one of the twenty-four, and it is the second: a twelfth. Now read the upper limit as a plain three instead of the root of three -- which is exactly what happens when a radical is lost. The value lands on none of the twenty-four. The item stops having an answer at all.

17:04A radical is an easy thing to lose in transcription. If you ever meet this question with a bare three on top and none of the offered answers fits, the question has been miscopied -- you have not misjudged it. The second item integrates one over four plus nine times a square, from nought to two thirds. The constants come out as a sixth outside and three halves inside, the inverse tangent evaluates at one, and the value is a twenty-fourth of a half turn -- the first entry on that same grid.

17:42And neither value rests on its anti derivative alone. Both were read a second time as a sum over eighty pieces, and both times the two readings agree to within a thousandth. The control sums the other item's integrand over the same interval, where neither reading agrees with anything. One thing to head off before the next topic arrives. You will shortly be told that a definite integral is an area. That identification comes with conditions, and it arrives for a function that stays above the axis.

18:22Nothing said so far forbids a negative answer, and nothing should. Over six fixed pairs of an integrand and an interval -- three different powers over three different intervals, all six anti derivatives having passed the same test as everything else here -- two of the values come out negative, two come out nought, and two come out positive. The variable itself, from minus two to one, comes out minus three halves. The same integrand from minus one to one comes out nought, because the two halves cancel exactly.

19:01So a definite integral is a number. It can be negative, it can be nought, and neither of those is a mistake. Hold the word area until it is offered to you with its conditions attached. Back to the second definition, because I want to leave you with the right sentence rather than a vague unease. You have seen the sums converge. They are real. The two definitions do give the same number, and the rate at which the sums close is measurable and behaves exactly the way it should.

19:38What that does not hand you is a way to answer a question. Eighty pieces got us to within a thousandth. A question wants the exact value, and the exact value is the limit, not any one of those sums -- and a limit of sums whose number of terms is running away is not something you evaluate on paper. So the sentence that is safe to write down is this: the definite integral is defined as the limit of a sum, and it is evaluated as the difference of an anti derivative's values at the two ends.

20:16The sentence that is not safe is: I will compute this one as a limit of a sum. For a square you could manage it. For almost anything else you could not, and an examination is not where you want to find out which case you are in. The two definitions are equals in mathematics. They are not equals at a desk. Quote the definition; use the recipe. What is worth carrying out of here.

20:46Two numbers attached to an integral sign change the object. A family of functions becomes one number, and that is why the constant stops mattering. The constant stops mattering because it is subtracted from itself, and you should have watched that happen once rather than been told. Across forty-one constants each row gives exactly one value; put the constant at one end only and the same loop gives forty-one. The recipe is two lines: find any anti derivative, then subtract its value at the lower limit from its value at the upper. The same two lines every time.

21:31So when a definite integral is hard, it is never the definite part that is hard. Of four worked evaluations, three needed a substitution or a decomposition before the limits could go in at all. Ask the domain question first, not last. Some integrals must not be written down at all, and the one we looked at has an integrand that fails at nineteen of fifty-one inputs across its stated interval -- and the recipe will still hand you a number if you let it.

22:05The order of the limits is a sign. Swap them and the two readings sum to nought, every time. A definite integral is a number, and a number may be negative. And the other definition is real -- the sums close, and they quarter their error every time you double the pieces -- but it is not something you can run by hand. Quote it as the definition; evaluate with the anti derivative.