PrepShorts · Study sheet · Class 12 Mathematics · Chapter 7, Integrals
Chapter 7 · Integrals
Properties of the definite integral, and using symmetry to kill half the work
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The idea
Eight numbered rules arrive on one page and look like eight things to memorise. They are not. Two are restatements, one exists only to prove its neighbour, and what remains is a single technique with a single move behind it: fold the interval back on itself, get a second expression for the same number, add the two and watch the hard part cancel. Four of the seven worked examples are that move and nothing else, and once a learner sees it once they can do all four. Beside it sits parity, which is not a technique at all but a glance — test the negated input, and on a balanced interval the answer is either zero or twice a half. Two things an explanation should not let slide. This is the one place in the chapter where every claim on the page is proved, two folios after a section that proved neither of its theorems and said so; the contrast is worth naming. And the closing Summary stops before this section starts — the properties, the substituted limits, and thirty-one exercise items between them get no line at the end of the chapter. A student revising from the Summary alone will revise none of this.
What you should be able to do
- State each of the eight numbered properties and say what each one changes about an integral
- Identify which properties are essentially bookkeeping and which do real work
- Apply the reflection property to an integral that reappears after reflection, and solve for it by adding the integral to itself
- Recognise an even or odd integrand on a symmetric interval and write the answer down without integrating
- Split an interval at a sign change to remove a modulus
- Follow the chapter's longest example, including the step it marks with a question
- Choose a property from the shape of an integral rather than by trying them in turn
- State the general result that the last multiple-choice item of the chapter is a case of
Words to know
| Term | Definition in one line | First introduced |
|---|---|---|
| even function | one that is unchanged when the input is negated | printed in this chapter (§7.10, Part II p. 274) |
| odd function | one that changes sign when the input is negated | printed in this chapter (§7.10, Part II p. 274) |
| anti derivative | the function whose derivative is the integrand, spelt as two words here | printed in this chapter (§7.2, Part II p. 226, and throughout the §7.10 proofs, Part II p. 274) |
| definite integral | an integral carrying two limits, so its value is a number | printed in this chapter (§7.7, Part II p. 267) |
| limits of integration | the pair of numbers the integral runs between | printed in this chapter (Summary, Part II p. 291) |
| substitution | changing the variable, which is how most of these proofs are built | printed in this chapter (§7.3, Part II p. 235, and in the §7.10 proofs, Part II p. 274) |
| second fundamental theorem of calculus | the result the first two proofs invoke by name | printed in this chapter (§7.8.3, Part II p. 268, cited again at Part II p. 274) |
| property | the chapter's own label for each numbered rule | printed in this chapter (§7.2.1, Part II p. 230, and §7.10, Part II p. 274) |
| symmetry | the shared idea behind the reflection and even-odd properties | an added term; the word does not occur anywhere in this chapter |
| modulus | the bars that make the sign of the integrand change partway | an added term; neither modulus nor absolute value occurs anywhere in this chapter, though four exercise items and one example use the bars |
| sub-interval | a piece of the range produced by splitting | an added term, not printed in this chapter |
| periodic | repeating after a fixed step, which one exercise item quietly relies on | an added term; the chapter prints neither periodic nor period |
Where people slip up
- "The properties are shortcuts for the exam." Three of them are needed to prove the others, and one exists for nothing else. Present the list as a structure with a shape, not as eight independent tricks.
- "Reflecting the interval gives you the answer." It gives you a second expression for the same number. The answer comes from what happens when you add the two — usually a collapse to something trivial. Teach the two moves as one move.
- "Even times odd, odd times odd — I will work the parity out on the spot." In Example 31 the parity of a product decides the whole question in one line, and it is the single most reliable mark in this section. Drill it separately.
- "A symmetric interval means the answer is zero." Only for an odd integrand. For an even one it means the answer is twice a half. Both halves of the parity property carry equal weight and students remember only the zero.
- "Bars around the integrand are a symmetry problem." They are a splitting problem. Find where the inside changes sign, cut there, and drop the bars with the right sign in each piece.
- "The reflection point is always the middle of a nice interval." In Example 33 it is the sum of a sixth and a third of a half turn, which is why the example is in the book at all. The property is stated for any pair of limits.
- "If reflecting does not simplify, the property was the wrong choice." In Example 34 reflecting produces something no simpler, and the simplification comes two steps later from a logarithm identity. Persistence is part of the method here.
- "The doubled-range property finishes the job." Exercise 7.10 Q14 shows it handing you a smaller integral that still needs work.
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Worked answers: Exercise 7.1 · Exercise 7.2 · Exercise 7.3 · Exercise 7.4 · Exercise 7.5 · Exercise 7.6 · Exercise 7.7 · Exercise 7.8 · Exercise 7.9 · Exercise 7.10 · Miscellaneous Exercise · this video explains Exercise 7.10 Q1, Exercise 7.10 Q2, Exercise 7.10 Q3, Exercise 7.10 Q4, Exercise 7.10 Q5, Exercise 7.10 Q6, Exercise 7.10 Q7, Exercise 7.10 Q8, Exercise 7.10 Q9, Exercise 7.10 Q10, Exercise 7.10 Q11, Exercise 7.10 Q12, Exercise 7.10 Q13, Exercise 7.10 Q14, Exercise 7.10 Q15, Exercise 7.10 Q16, Exercise 7.10 Q17, Exercise 7.10 Q18, Exercise 7.10 Q19, Exercise 7.10 Q20, Exercise 7.10 Q21, Miscellaneous Exercise Q31, Miscellaneous Exercise Q34, Miscellaneous Exercise Q40
Transcript3,238 words
Eight rules for definite integrals usually arrive together, numbered, in a list. Eight things to hold. They are not eight things, and this video is going to take them apart and count what is actually there. One of them changes nothing a sum could ever notice. One is the fold, and the one after it is the fold again with a limit set to nought. One is right about everything and finishes nothing. What is left is a single move, and beside it a single glance.
The move is worth learning properly, because when it works it does not make an integral easier. It removes it. And the glance is the most reliable line of working in the whole subject, as long as you know what it is quietly assuming. One promise first, because everything after it leans on it. Not a single value in this video comes from an anti derivative. Every integral here is cut into two hundred pieces. The integrand is read once in the middle of each piece, multiplied by the width, and the pieces are added.
That is deliberate. A rule about integrals, checked with the very machinery the rule is there to justify, has not been checked at all. But a sum is only a reading when the width of a piece has stopped mattering. So every integrand here is read twice, once with half as many pieces, and the two answers have to sit on top of each other. Eight ordinary integrands, eight settled readings, and not one of them moves by so much as a thousandth.
From here on, when a rule is called right, it means two sums agreed. When it is called wrong, it means two sums did not. The first rule says you may change the letter carrying the variable. So change it, and write the whole integral out again. All eight integrands come back with the number they had. Nought of them changed. Now look at what that test could possibly have found. Nothing at all.
A sum reads values. A letter is not a value. There is no arrangement of arithmetic that could have noticed the difference, which is why the answer was eight out of eight before anything ran. So the first rule is not a fact about integrals. It is permission to stop worrying about something that was never going to matter. It earns its place on the list for one reason only: you are about to change the letter constantly, and it is polite to be told you may.
The second rule looks like more of the same. It is not. Turn the two limits round -- put the top one at the bottom -- and the number turns over in sign. On all eight integrands. This one a sum can see. Something measurable happened. And here is the control that makes that sentence worth anything. Run the identical test but leave the sign alone -- claim the number is unchanged -- and nought of the eight survive.
That matters more than it looks. If several of these integrals had happened to be nought, they would have passed both versions, and the test would have been agreeing with itself. None of them is nought. Eight out of eight turn over, nought out of eight stay put. The third rule cuts the stretch in two. Integrate from the start to some place in the middle, integrate from there to the end, add.
Three cutting places on each of the eight integrands, twenty-four readings. The two pieces come to the whole in all twenty-four, and miss in nought. The mistake worth showing is not a hard one. It is reading the cut as a place to start again rather than a place to carry on from -- so the second piece runs from the beginning rather than from the cut. Nought of the twenty-four survive that.
One more thing about this rule, and it is the thing lists tend not to say. Where you are allowed to cut is part of the rule. Cut a unit outside either end, where the integrand was never defined, and every one of those sixteen readings comes back as nothing at all. Not a wrong number. Nothing. Which is the right behaviour, and it is worth knowing that the machinery here does it, because a great many of the counts later in this video are counts of nothing coming back.
Now the one that does the work. Take the stretch and fold it back onto itself. Every point at some distance from the left end swaps with the point the same distance from the right end. In symbols: replace the variable by the sum of the two limits, less the variable. The limits stay exactly where they were. The integral is unchanged. All eight, nought wrong. That is worth pausing on, because it is not obvious. You have changed the integrand -- genuinely changed it, into a different function -- and the area under it between the same two numbers is the same.
The picture is the reason. Folding a region left-to-right does not change how much of it there is. The commonest way to write that rule down wrong is to fold by the difference of the two limits instead of by their sum. Run it. Something odd happens. It is right on four of the eight and unreadable on the other four, and never once wrong. The four it is right on are exactly the four whose stretch starts at nought -- where the sum of the limits and their difference are the same number. There are exactly four such rows.
Which tells you what the next rule on the list is. The fifth rule is the fold written for a stretch that starts at nought. And on those stretches it is not merely close to the fourth rule. It returns the same reading, digit for digit. Four the same, nought apart. Ask it anywhere else and it cannot be asked at all: what it wants to fold about runs off the end of the stretch, and all four come back as nothing.
That is what a narrower statement of one fact looks like from the outside. Not a second fact. A convenience. Here is the move the whole list is for, and it is two steps, not one. Folding hands you a second expression for the same number. That is all it hands you. Nothing has been solved. The second step is to add the two expressions together and hope the hard part cancels.
So: add the integrand to its own fold, and read what comes back at twenty-one places along the stretch. If it is the same number everywhere, that number is a constant, and twice your integral is that constant times the length. Nine integrands. On six of them the sum is flat, and on those six the answer the move hands over -- the constant, halved, times the length -- agrees with the summed value every single time.
The other three are run through the move anyway, rather than quietly left out. It returns a perfectly ordinary number on all three, and the number is wrong on all three. That is the honest shape of this technique. It does not fail loudly. It hands you an answer whether or not it was entitled to. Watch what that means. One over one plus a power of the tangent, across a quarter turn, at four different powers: a square root, the tangent itself, its square, its cube.
Those are four genuinely different functions. Read at seven places they give twenty-five different numbers out of twenty-eight, and they agree at exactly one of the seven places, in the middle of the stretch. The fold collapses on all four. And the four integrals hold exactly one value between them. All four are a quarter of a half turn. Look again at what the move actually used. Not the integrand. Only what the integrand and its fold came to when added.
Which is why the power made no difference. Any power at all would have done. One caution before moving on, because it is where this technique is usually lost. The fold is about the middle of the two limits you actually have. Not about the middle of some nice standard interval. One over one plus the square root of the tangent, between a sixth and a third of a half turn. Neither limit is special. But add them: a sixth plus a third is a half, so the two limits sum to a quarter turn -- exactly what the tangent's fold needs.
Add the integrand to its fold and the roots cancel to one. The value is a half turn over twelve. So do not look for a familiar interval. Add the two limits and see what you get. The sixth rule takes a stretch of twice the length, cuts it in half, and folds the far half back onto the near one. It is right on all eight of the doubled integrands tested. Nought wrong.
And it is followed on the list by something that looks like it and is not: double the near half, and read nothing else. That shortcut is right on five and wrong on three. What decides it is a hypothesis -- whether the fold leaves the integrand alone -- and that has to be tested, not read off the shape of the formula. Paired up row by row rather than counted twice: the five it is right on are exactly the five where folding leaves the integrand alone, and the three it is wrong on are exactly the three where it does not. Nought licensed and wrong, nought unlicensed and right.
So the general rule is right about everything and finishes nothing, and its special case finishes things and is right about less. That is what a rule on a list to prove the next rule looks like. Now the glance. On a stretch balanced about nought, read the integrand at the negated input. Either it comes back turned over in sign, or it comes back unchanged, or neither. Eight integrands. Two come back turned over, four come back unchanged, and two are neither -- and those two are the reason this is a measurement rather than a description.
The two turned over integrate to nought. The four unchanged come to twice their own half. The two that are neither get nothing from either half of the rule. Students remember the nought and forget the twice-a-half, and it costs them, because the second half is the one that saves actual work. And the balance is not decoration. Slide one limit by a half so the stretch is no longer balanced, ask exactly the same questions, and the answer is nought noughts, nought twice-a-halfs, all eight neither.
Something quietly connects the glance to the move, and the arithmetic says so plainly. Of twenty-one integrals collected for this video, four have both of the finishing tests fire at once -- the fold's collapse and the turning over. And on all four of those, the constant the fold collapses to is nought. So the parity rule is not a separate technique at all. It is the move, in the case where what you add the integrand to is its own negative, and the constant that comes out is nought.
One idea, wearing two numbers on the list. But here is where the glance can hand you a nought you have not earned, and it is worth seeing once. Take a fifth power of the tangent on a balanced stretch that stops a tenth, then a fiftieth, then a two-hundred-and-fiftieth short of where the tangent runs away to nothing. All three read nought. All three are turned over by the fold. All three are unmoved when the piece width is halved. The reading looks perfect.
Now read just the upper halves on their own. Nought of the three settle. Halve the piece width and the answer moves, every time -- because there is nothing there for it to move towards. The balanced reading was nought because the sample points come in pairs that cancel each other before anything else is looked at. It would have been nought whatever was happening out at the ends. For comparison, a cubed variable on a balanced stretch reads nought the same way, and its upper half settles without any trouble at all.
So the glance is worth its reputation. But its nought is only worth having when the two halves it is cancelling are themselves worth something. Bars around the integrand are not a symmetry problem. They are a cutting problem, and they belong to the third rule. A cubed variable less the variable, in bars, from minus one to two. Do not state where it changes sign. Find out. Walk the stretch in four hundred steps, notice every step whose two ends disagree in sign, and halve down onto each one.
Exactly two places come back, both strictly inside. Cut at both, drop the bars with the right sign in each piece, and the three pieces come to eleven quarters -- which is what the whole thing reads. Now count how many pieces still have the inside changing sign within them. With no cut, one. With only the first of the two cuts, still one. With both, nought. Which is why the rule that cuts a stretch once has to be used twice here. That is allowed. It is worth saying out loud rather than letting it slide past.
And ignoring the bars is not a small error. That reads nine quarters -- a half short of the truth. Everything so far has made the fold look obliging. Here is the one that does not. The logarithm of the sine, across a quarter turn. Fold it. The sine becomes the cosine. Add the two and you have the logarithm of the sine plus the logarithm of the cosine -- which is the logarithm of their product, and that is no simpler than what you started with.
The fold's sum is not flat here. The move does not finish it. If your rule was to abandon a technique the moment it stops simplifying, you would stop. Do not stop. The product of the sine and the cosine is half the sine of the doubled angle. So the sum of the two logarithms is the logarithm of the sine of twice the angle, less the logarithm of two.
That is the step that usually gets a question mark next to it and no answer. There is the answer: a doubled-angle identity, and one logarithm split into two. Substitute to bring the doubled angle back to a quarter turn, use the doubled-stretch rule to fold it in half, and what returns is the integral you started with. The value is minus a half turn over two, times the logarithm of two.
And that value is not asserted here. It is approached: read the sum at one hundred, two hundred and four hundred pieces, and each reading sits closer than the last with the gap halving each time. Aim the same test at a value a hundredth away and the gaps still shrink, but they stop halving. Drifting past a number and closing on it are different things, and only one of them is worth writing down.
So what does a reader actually have to carry? Four tests. Two of them hand back a number and you are finished. Two of them hand back another integral over half the stretch, and you are not. Twenty-one integrals, put through all four. Six have nothing fire at all -- they are ordinary work. Nine are finished outright by a test that returns a number. Five are handed a half integral on which nothing further fires.
And exactly one is handed a half integral on which a test does fire. A fifth power of the cosine across a full turn: the doubled-stretch rule halves it, and then the half needs folding as well. One integral out of twenty-one that takes two of these in sequence. Everything else takes one or none. Every number a test returned is right against the sum -- thirteen out of thirteen. Every half a test returned comes to half of its whole -- six out of six. And the twenty-one integrals hold fourteen different values between them, so they are not twenty-one copies of the same question.
That is the real shape of the work. Not eight rules to try in turn. Four tests, and most integrals answer to one of them or to none. One last thing, because it is the whole technique with the particular function taken out. Suppose a function is unchanged when you fold it about the middle of two limits. Then the variable times that function integrates to the middle of the two limits, times the function alone.
Nine functions were tried, of which exactly six are left alone by their own fold -- tested at twenty-one places each, not read off the shape. On those six, the rule is right six times and wrong nought. Now the three ways to half-remember it. Put the difference of the limits where their sum belongs: right on three of the six, wrong on three -- and the three it survives are exactly the three whose stretch starts at nought, where the sum and the difference are the same number. So getting that one right on a stretch starting at nought proves nothing.
Leave the sum unhalved: nought of six. Keep the right factor and hand it the wrong integral: nought of six. And on the three functions the fold does not leave alone, the rule itself fails on all three. The hypothesis is not decoration. The famous instance of this is the variable times the sine, over one plus a squared cosine, across a half turn. Its fold leaves the awkward part alone, so the rule applies, the lighter integral comes to a quarter turn, and the answer is the square of a half turn, over four.
That example is not a trick. It is this rule at one particular pair of limits. So count what is actually on that list of eight. One is notation. One is a sign. One is a cut, and it is the one the bars need. One is the fold, written twice -- once for any two limits, once for a stretch starting at nought, and those two are the same reading.
One splits a doubled stretch, is right about everything, and finishes nothing; the one after it is its special case with a hypothesis attached, and that one does the finishing. And one is parity, which turns out to be the move again with a constant of nought. What you carry is smaller than eight. Fold the stretch about the middle of its own two limits, add the result to what you started with, and see whether the hard part goes.
If it does, you are finished without integrating anything. If it does not, you have lost one line and learned that this integral wants something else. And on a balanced stretch, before anything else, read the integrand at the negated input. It costs one line, and when it fires you have the answer without integrating anything at all.
Where this fits
Either side of this one
- Substituting inside a definite integral, and why the limits have to move with itClass 12 · Ch 7, Integrals
- Slicing a region into thin strips, and adding the strips with an integralClass 12 · Ch 8, Application of Integrals