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

What to assume they know

• The indefinite integral as a family of anti derivatives, and the Note that an equals sign between two of them compares families, from module m01
• Every technique of module m02, since a definite integral is evaluated by first finding an anti derivative
• Continuity on a closed interval, from Chapter 5 of Part I
• The domain of a square root, and what makes an expression undefined on part of an interval
• Reading a bracketed expression evaluated between two values
• Partial fractions, from the fourth topic of module m02

What they should be able to do

• Explain what the two attached numbers do to an indefinite integral, in terms of families and single values
• Name the two attached numbers as the chapter names them
• State both routes the chapter's definition offers, and say which one this edition actually teaches
• Carry out the chapter's two-step recipe on a stated definite integral
• Explain why the constant of integration may be dropped, and demonstrate the cancellation rather than asserting it
• State the condition the fourth Remark imposes on the integrand, and apply it to reject a stated integral
• Evaluate a definite integral whose integrand needs a substitution, a decomposition or a rewrite before the limits can be applied
• Say what a student may and may not claim about the second route the definition mentions

Where it usually goes wrong

• "A definite integral is an indefinite integral with numbers stuck on." It is a different kind of object: the indefinite integral is a family of functions, the definite integral is one number. The two attached numbers do not decorate the family, they collapse it.
• "You have to remember to add the constant and then cancel it." You may add it and watch it cancel — once, to see why — and thereafter you may leave it out. The chapter says so and shows the cancellation on the same page.
• "The definite integral is defined as a limit of a sum, so I should be able to compute one that way." The definition on Part II p. 267 names that route and this edition supplies nothing to carry it out with — no section, no example, no exercise. See section 4 and the note below. A student should know the phrase exists and should not be told it is a method available to them from this book.
• "Any integrand can be integrated between any two numbers." Remark (iv) rejects one outright, and the reason is that the integrand does not exist on part of the interval. Ask the domain question first; the chapter puts it fourth, which is late.
• "The two-step recipe is the hard part." The two steps are two lines. Every part of Example 25 except the first spends its length on the anti derivative, and three of the four need a module m02 technique to get one.
• "Once you have the anti derivative you can stop." You have found a family; the question asked for a number. Students routinely stop one step early on items where the anti derivative is itself hard-won.
• "The order of the two attached numbers is a convention that does not matter." It matters, and the next topic but one makes it a numbered property with a sign attached. Here just say the lower one goes below and the upper one above, and that swapping them is not free.
• "A definite integral is an area, so it cannot be negative." The chapter does not say a definite integral is an area in this section — that identification arrives in §7.8 and is stated for a positive integrand. Nothing in §7.7 forbids a negative value, and several items in Exercise 7.8 produce one.

Questions to check understanding

• Evaluate a definite integral whose anti derivative is a standard form — the form of Exercise 7.8 Q1 to Q6
• Evaluate a definite integral requiring a substitution, a decomposition or a trigonometric rewrite first — the form of Example 25 (ii) to (iv)
• Justify omitting the constant of integration in a definite evaluation
• Decide whether a stated definite integral may be written down at all, given its integrand and its interval
• Name the two attached numbers correctly and write an integral to a stated specification
• Choose the correct value from four options — the form of Exercise 7.8 Q21 and Q22
• State both constructions the chapter's definition names, and say which one this book carries out

Examples worth working on the board

Values marked verified are worked out here from the chapter's own printed data; neither answers file was opened, and this chapter prints no answers to its exercises.

• §7.7 in full (Part II p. 267). It is four sentences long and the whole of this topic's framing sits in it. It says the definite integral has one value; it gives the notation and names the number below and the number above; and it then offers two constructions — building the object as a limiting sum, or, when an anti derivative exists on the closed interval, taking the difference of that anti derivative's values at the two ends. Read off the page image. Only the second construction is carried out anywhere in this chapter. See section 4 and the note below.
• The two-step recipe (Part II p. 268). Find the indefinite integral, then evaluate the difference between its values at the two ends. The chapter prints the second step with the bracket-and-limits notation the rest of the module uses. Two lines, and every exercise in §7.8 is these two lines.
• The demonstration that the constant drops out (Part II p. 268, inside step (i)). The chapter carries an arbitrary constant through the evaluation and shows the two copies of it cancel. Verified: the constant appears once at the upper end and once at the lower end with a minus sign in front, so it contributes nothing. Show the cancellation, do not assert it — this is the same courtesy §7.6's Remark (ii) extends to a different constant, and both times the chapter earns the shortcut rather than declaring it.
• Remarks (i) to (iii) (Part II p. 268). (i) restates the evaluation in words. (ii) says the theorem makes evaluation easier by avoiding the other construction — see the note below about this sentence. (iii) says the hard part is finding an anti derivative at all, which is a fair summary of why module m02 exists and takes six topics.
• Remark (iv) (Part II p. 268). The condition: the integrand must be well defined and continuous throughout the closed interval. The chapter then names an integral that fails it — the variable times a half power of one less than its square, taken from minus two to three — and calls writing it down erroneous, because the integrand does not exist on part of that interval. Verified: one less than the square is negative strictly between minus one and one, so its half power is not a real number there, and the interval from minus two to three contains that whole stretch. Read the exponent off the page image; the text layer sets it as a detached fraction. This is the only place in the chapter where an integral is declared inadmissible, and it deserves its own section.
• Example 25 (Part II pp. 269–270), four parts, each needing a different first move. (i) A square, from two to three, straight off the power rule. Verified: the value is nineteen thirds. (ii) A square root over the square of a bracket containing a three-halves power, from four to nine. The anti derivative is found by substitution first, and the chapter is explicit that this is a separate step done before the limits are used. Verified by working added here: substituting for the bracket gives two thirds of its reciprocal, and evaluating between the ends gives two thirds of the difference of a third and one twenty-second, which is nineteen ninety-ninths. Both the radical and the three-halves exponent were read off the page image. (iii) The variable over a product of two linear factors, from one to two. The anti derivative needs a partial-fraction decomposition first. Verified: the decomposition has coefficients minus one and two, and the value is the logarithm of thirty-two over twenty-seven. (iv) A cube of a sine of twice the variable times a cosine of twice it, from zero to a quarter turn. The anti derivative needs a substitution. Verified: the anti derivative is an eighth of the fourth power of the sine of twice the variable, and the value is one eighth. Use all four; the point of the group is that the definite part is always the same two lines and the work is always somewhere else.
• Exercise 7.8 (Part II pp. 270–271), twenty-two items, of which the last two are multiple choice. Grouped by what has to happen before the limits go in — an added classification, not the chapter's: Q1, Q2, Q3, Q6, Q9, Q10 and Q17 land on a module m01 row directly; Q4, Q5, Q7, Q8 and Q12 need a trigonometric rewrite or one of §7.3.1's four derived integrals; Q11, Q13, Q14, Q16, Q19 and Q20 need a substitution or a §7.4 formula; Q15 and Q18 need a substitution and a trigonometric rewrite respectively; and Q16 needs a decomposition. Q7 and Q8 are the two that should be checked against Remark (iv) before anything else — both integrate a trigonometric ratio over an interval, and a student should be in the habit of asking whether the integrand survives the whole interval before reaching for an anti derivative.
• Exercise 7.8, questions 21 and 22 (Part II p. 271). Two multiple-choice items. Verified by working added here: Q21 integrates the reciprocal of one plus a square from one to the square root of three, giving the difference of two standard inverse-tangent values, which is a twelfth of a half-turn — the fourth option. Read Q21's upper limit off the page image: it carries a radical that the text layer drops, and without the radical the item has no matching option. Q22 integrates the reciprocal of four plus nine times a square from zero to two thirds; the constants come out as one sixth and the inverse tangent evaluates to an eighth of a half-turn, so the value is one twenty-fourth of a half-turn — the third option. The chapter prints no answers.
• The Summary (Part II p. 291). It restates the second fundamental theorem and then gives the collective name for the two attached numbers, repeating the individual names. Read off the page image. It carries no mention of the other construction.

Figures to have open

• A collapsing-stack movement for section 1: the family of anti derivatives drawn as translated copies, then reduced to a single number once the two limits are attached. Not in the book; the chapter draws nothing in §7.7 and its only figure belongs to the next topic.
• A number line with the interval from minus two to three drawn and the stretch between minus one and one shaded out, for section 8. Not in the book; the chapter states the failing stretch in words on Part II p. 268 and draws nothing.
• Two grouping tables — the four parts of Example 25 by first move, and the twenty-two items of Exercise 7.8 by first move — for sections 9 and 10. The content is the chapter's own, from Part II pp. 269–271; the groupings are added here. Build both with the repo's DataTable component.
• The chapter's one figure, Fig 7.1 on Part II p. 267, is not used by this topic. It belongs to g12-maths-ch07-m03-t02.md, which is built around it.

Where this sits in the book

• NCERT Class 12 Mathematics, Chapter 7 "Integrals", §7.7 Definite Integral, Part II p. 267
• Remarks (i) to (iv) and the two-step evaluation recipe, Part II p. 268
• Example 25, four parts, Part II pp. 269–270
• Exercise 7.8, questions 1 to 22, Part II pp. 270–271
• Summary, the second fundamental theorem bullet and the naming of the limits, Part II p. 291
• §7.1 Introduction, the two problems and the two kinds of integral, Part II p. 226