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

What to assume they know

• Integration by substitution, from the first topic of module m02, including the conversion of the differential
• The two-step evaluation of a definite integral, from the first topic of this module
• The six quadratic-denominator formulas and completing the square, from the third topic of module m02
• The exponential shape from the last topic of module m02
• Reading a substitution as a map that sends each old value to a new one
• The principal-value range of the inverse sine, from Chapter 2 of Part I

What they should be able to do

• State the chapter's four-step procedure and identify which step the limits enter at
• State the shortcut the chapter's Note offers, and say which of the four steps it removes
• Convert a pair of limits through a substitution, and check the conversion by evaluating the substitution at each end
• Evaluate one definite integral by both routes and confirm the answers agree
• Explain what goes wrong when a substitution is made and the limits are left unchanged
• Read a printed hint as a statement about which substitution was intended
• Recognise an item whose simplification needs a restriction, and check whether the stated limits supply it
• Choose a substitution for a definite integral from the shape of the integrand

Where it usually goes wrong

• "Substitution in a definite integral is a different technique." It is the same technique with one extra decision: whether to bring the original letter back or to move the limits instead. Both are in the section and both are correct.
• "The limits go with the integral sign, so they are unaffected by a substitution." They are numbers attached to the variable, and changing the variable changes them. Leaving them alone while renaming the variable is the single commonest error in this section and it produces a plausible wrong number rather than an obvious one.
• "The new limits are found by substituting the old limits into the anti derivative." They are found by substituting the old limits into the substitution. Say which expression the numbers go into; the two are different and the confusion is easy.
• "You must resubstitute; changing the limits is a shortcut for the confident." The chapter offers both and prefers neither. What it does say is that skipping step three requires the limits to be changed, and that is not optional.
• "If the new limits come out in the wrong order, something has gone wrong." They can come out in either order, and Exercise 7.9 Q9 is a case where the upper end maps below the lower. Nothing is wrong; the next topic makes the sign behaviour a numbered property.
• "A printed hint is a courtesy." It is a statement about which substitution the item was built for. Exercise 7.9 Q4's hint clears a root that a more obvious substitution would leave behind.
• "An inverse ratio can always be rewritten by the standard identity." Only on the range where the identity holds. Exercise 7.9 Q3 is safe because its limits confine the variable; the same integrand without limits is not.
• "Every item in Exercise 7.9 is a substitution." Two of them complete the square and land on a §7.4 formula, one is the exponential shape, and one is the first fundamental theorem in disguise. Four of the ten are not what the heading advertises.

Questions to check understanding

• Evaluate a definite integral by substituting and then resubstituting — the four-step route
• Evaluate the same integral by converting the limits instead, and confirm the two agree
• Convert a stated pair of limits through a stated substitution
• Evaluate a definite integral whose hint names the substitution — the form of Exercise 7.9 Q4
• Recognise an item in the exercise that is not a substitution at all
• Decide whether a standard inverse-ratio rewriting is admissible on the stated interval — the form of Exercise 7.9 Q3
• Choose the correct value from four options — the form of Exercise 7.9 Q9

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.

• §7.9's four steps (Part II p. 271). Numbered 1 to 4 and read off the page image. Step 1 drops the limits and substitutes. Step 2 integrates in the new letter and says explicitly not to write the constant. Step 3 puts the original letter back. Step 4 evaluates at the two limits and subtracts. The limits are absent for three of the four steps and reappear only at the end, which is exactly the point of the section and the reason the shortcut in the Note is worth having.
• The boxed Note (Part II p. 272). It offers a faster route: after steps 1 and 2, skip step 3 entirely, keep the integral in the new letter, and change the limits to match. One paragraph.
• Example 26, worked both ways (Part II p. 272). Five times a fourth power times the square root of one more than a fifth power, from minus one to one. Route one, the four steps: substitute for one more than the fifth power, integrate, put the original letter back, evaluate between minus one and one. Route two, marked as the alternative: the same substitution, but the limits are converted first — at the lower end the new letter is zero and at the upper end it is two — and the integral is evaluated in the new letter without ever returning. Verified by working added here: the anti derivative in the new letter is two thirds of its three-halves power; both routes give four times the square root of two, divided by three. Read the exponents and the radical off the page image. This is the only place in the chapter where one integral is evaluated twice by two routes to show they agree, and it is section 6.
• Example 27 (Part II pp. 272–273). An inverse tangent over one plus a square, from zero to one, done by the shortcut route only. The substitution is the inverse tangent itself. The new limits are zero and an eighth of a full turn, and the chapter states them before integrating. Verified: the integral in the new letter is half its square, and the value is the square of a half-turn divided by thirty-two. This is section 7: the upper limit changes into something that does not look like the number it came from, which is where students lose confidence in the method.
• Exercise 7.9, questions 1 to 8 (Part II p. 273). Grouped by substitution — an added classification, not the chapter's: Q1 substitutes for a quadratic denominator; Q2 and Q5 substitute for a trigonometric ratio after a factor is peeled off; Q4 carries a printed hint naming its substitution; Q6 and Q7 complete the square and land on a §7.4 formula rather than on a substitution at all; and Q8 is the exponential shape from the last topic of module m02 with a doubled exponent, so the anti derivative is spotted rather than substituted for. Q3 is the one that needs care and is treated in section 10.
• Exercise 7.9, question 4's printed hint (Part II p. 273). The item is a variable times the square root of two more than it, and the hint sets the whole bracket equal to the square of a new letter. Verified as the right reading: setting the bracket to a plain new letter also works but leaves a half power to integrate, whereas setting it to a square clears the root entirely. The hint is therefore a statement about which of two workable substitutions is shorter, and section 9 should say so — the chapter prints five hints in this chapter and never explains any of them.
• Exercise 7.9, question 3 (Part II p. 273). An inverse sine applied to twice the variable over one more than its square, from zero to one. Every solver will rewrite the integrand as twice an inverse tangent. Verified: that rewriting needs the variable's modulus not to exceed one — and here it does not, because the limits are zero and one. So the item is safe as printed. Contrast it with Exercise 7.6 Q22, which is the same integrand with no limits and therefore no such guarantee; that item is recorded in g12-maths-ch07-m02-t05.md. The pair is the best available illustration of what attaching limits can settle.
• Exercise 7.9, question 9 (Part II p. 273). A multiple-choice item: a cube root of a difference of the variable and its cube, over a fourth power, from a third to one. Verified by working added here: pulling a cube out of the bracket and substituting for the reciprocal of the square less one turns it into a plain power integral; the new limits are eight and zero, and the value is six, which is the first option. The chapter prints no answers.
• Exercise 7.9, question 10 (Part II p. 273). A multiple-choice item that has nothing to do with substitution: it differentiates an integral whose upper end is the variable, which is Theorem 1 of §7.8.2. It is claimed by g12-maths-ch07-m03-t02.md and is not treated here. Worth knowing it is there, because a student working straight down the exercise will meet it without warning.

Figures to have open

• A two-column parallel working of Example 26 for section 6, the two routes running down the page and meeting at one value. Both routes are the chapter's own, on Part II p. 272; the parallel layout is added here.
• A limit-mapping diagram for section 5: two number lines, one in the old letter and one in the new, with the two limits joined by arrows through the substitution. Not in the book; the chapter states the two new numbers in words and draws nothing.
• A grouping table of Exercise 7.9 for section 11. The items are the chapter's own, on Part II p. 273; the grouping is added here. Build it with the repo's DataTable component.
• A deliberately wrong worked example for section 8, with the limits left behind and the resulting value set beside the correct one. Not in the book, and it must be labelled as wrong throughout — the chapter prints no such counter-example anywhere.
• 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.

Where this sits in the book

• NCERT Class 12 Mathematics, Chapter 7 "Integrals", §7.9 Evaluation of Definite Integrals by Substitution, the four steps, Part II p. 271
• The boxed Note offering the shorter route, Part II p. 272
• Example 26, worked by both routes, Part II p. 272
• Example 27, Part II pp. 272–273
• Exercise 7.9, questions 1 to 9, with the printed hint on question 4, Part II p. 273