Regions below the axis, and why the sign has to be discarded before adding

What to assume they know

• The strip construction of the first topic, and both of the chapter's area formulas
• Definite integration of a polynomial and of the two basic trigonometric functions, from the previous chapter
• Solving a simple equation to find where a curve meets the horizontal axis
• The absolute value of a real number, from Class XI
• The graphs of a straight line, a cube, a sine and a cosine over a full turn
• The area of a triangle from its base and its height

What they should be able to do

• Say what a definite integral reports when the curve lies under the horizontal axis, and why that number is not the area
• Restate the chapter's remark in words that survive scrutiny, and identify the clause in the printed version that does not
• Find every place a curve crosses the axis inside the interval of interest, and use those places as the only legitimate cut points
• Split an integral at each crossing, evaluate each piece separately, discard the sign of each and only then add
• Explain why taking the absolute value of the whole integral is not the same procedure and gives a different answer
• Check a computed area against school geometry when the curve is a straight line
• Recognise a case where the signed total over a symmetric interval is nothing at all while the area is not
• Predict, for a multiple-choice item, which printed option corresponds to the sign-blind method
• Say what the chapter's Summary omits about this topic and what a student revising only from it would lose

Where it usually goes wrong

• "Area can be negative." It cannot. The chapter's own remark says the area comes out negative, and that phrasing is where the belief comes from. The integral is what carries a sign; the area is a size. Correct the sentence, once, early.
• "Take the absolute value at the end." That is a different and wrong procedure. Taking the size of the total after the pieces have cancelled is not the same as taking the size of each piece before adding. Both of the chapter's multiple-choice items on Part II p. 298 have the second procedure's answer sitting in the printed options.
• "If the answer comes out negative, just drop the minus sign." That works only when the curve stays on one side throughout — the case of Fig 8.3. The moment there is a crossing inside the interval, dropping the sign at the end gives a number that is too small, sometimes by a lot and sometimes by all of it.
• "Split wherever the formula changes." Split wherever the curve crosses the axis. The two coincide in some problems and not in others: Miscellaneous Exercise question 2 changes formula at a corner where no crossing happens, and question 5's hint changes formula exactly at the crossing. The criterion is the crossing.
• "A symmetric interval means the answer is nothing." Over a symmetric interval the signed integral of an odd function is nothing, and its area is not. Question 5 on Part II p. 298 is exactly this trap, and the first printed option is the trap's answer.
• "The chapter's addition rule has bars on the first piece, so bars go on the first piece." The chapter can write it that way because it has already told you which piece is negative. In an unseen problem you have not been told. Put the bars on every piece.
• "The crossings are the endpoints of the interval." They may be, and usually they are not. In Example 3 the crossing is at minus two thirds while the boundaries are minus one and one; the crossing has to be found by solving, not read off the question.
• "The strip below the axis has negative height." The strip has a height; the curve's value there is negative. Those are different statements. Keeping them apart is what makes the rule intelligible rather than magical.

Questions to check understanding

• Find where a given curve crosses the horizontal axis inside a stated interval
• Compute the area between a curve and the axis over an interval containing one crossing, showing the split
• Explain, in one sentence, why the absolute value goes round each piece rather than round the total
• Choose the area bounded by a cube and two ordinates from four options — the form of Miscellaneous Exercise question 4
• Choose the area bounded by a curve defined through an absolute value from four options — the form of Miscellaneous Exercise question 5
• Confirm a computed area by elementary geometry when the curve is a straight line
• Give a curve and an interval for which the integral is nothing at all and the area is not
• Sketch a graph involving an absolute value and integrate it — the form of Miscellaneous Exercise question 2

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.

• The Remark (§8.2, Part II p. 293). Read closely. It states that when the curve stays under the horizontal axis across the whole interval, the quantity computed comes out negative; that only the size is wanted; and that the fix is to put absolute-value bars round the integral. The clause to handle carefully is the conditional one — the chapter writes as though an area could be negative. An area never is. The integral is.
• Fig 8.3 (Part II p. 293). Read off a 300 dots per inch the printed page: the curve enters from the left already below the axis, descends to a minimum, and rises to cross the axis at a point to the right of the second boundary line, so the shaded region lies wholly on one side. The left boundary is a plain vertical segment with its label to the left of it; the right boundary carries an arrow pointing at it. One narrow strip is drawn in the middle, hanging down from the axis. The figure is deliberately the simple case — one sign throughout — and section 4 should say so, because the next figure is not.
• Fig 8.4 and the sentence under it (Part II pp. 293–294). Read off a 300 dots per inch the printed page: the curve passes through the origin, dips below the axis, turns, rises across the axis at one interior point and climbs to a maximum near the right edge. The left boundary line stands well clear of the origin, inside the part that is below the axis; the right boundary line closes the part that is above it. Both parts are filled, each carries one narrow strip, and the two parts are labelled — the lower one beneath the curve at its lowest point, the upper one above the curve near its peak. There is no shading between the origin and the left boundary. The chapter's sentence declares the lower quantity negative and the upper one positive and then adds the size of the first to the second.
• What the chapter's addition rule hides. Verified reading: the chapter puts bars round the first term only, which is correct here because it has just told you which piece is the negative one. The general rule has to put bars round every piece, because in a problem you have not been told. Say the general form; the chapter's form is a special case of it and will mislead anyone who memorises the shape rather than the reason.
• Example 3 with Fig 8.9 (Part II pp. 296–297). A straight line, the horizontal axis, and two vertical boundaries at minus one and one. Read off a 300 dots per inch the figure: the line runs up from lower left to upper right with arrowheads at both ends; the crossing is marked and labelled with its coordinates; the left vertical carries a point above the axis and a point below it; the right vertical carries a point on the axis and a point on the line. The small piece below the axis is filled in a deeper blue than the large piece above it, and the large piece is itself drawn in two tones, the lighter one to the right of the vertical axis. Verified: the crossing is at minus two thirds; the piece below the axis has size one sixth, the piece above has size twenty-five sixths, and the total is thirteen thirds. Note the extraction defect recorded under Notes before scripting the working.
• The geometry check on Example 3 (not in the book). Verified: the lower piece is a triangle with base one third and height one, so its area is one sixth; the upper piece is a triangle with base five thirds and height five, so its area is twenty-five sixths; and the two agree with the integrals exactly. The chapter offers no check. Give this a section of its own — it is the only place in the chapter where the answer can be confirmed without calculus, and a student who sees the two routes agree stops treating the sign rule as an arbitrary instruction.
• Example 4 with Fig 8.10 (Part II p. 297). A cosine across one full turn. Read off a 300 dots per inch the printed page: the curve starts at height one on the vertical axis, falls through the axis at a quarter turn, reaches its minimum at a half turn, rises through the axis at three quarters, and returns to height one at the full turn, where a vertical segment closes the region. All three pieces are filled in the same shade; the two crossing points and the minimum are lettered; the two quarter-turn positions are labelled below the axis as fractions and the half turn is labelled inside the region below the axis. Verified: the three pieces have sizes one, two and one, so the area is four.
• The reason Example 4 is the sharpest item in the chapter. Verified: the integral taken straight across the full turn, without splitting, is nothing at all — the antiderivative returns to its starting value. So the sign-blind method here does not merely get the wrong number, it reports that the region has no size, while the region is plainly there on the page. Nothing in the chapter points this out. It is the strongest thirty seconds available to this topic.
• Miscellaneous Exercise question 3 (Part II p. 298). The same problem for a sine across one full turn. Verified: two pieces of size two, so the area is four, and again the unsplit integral is nothing at all. The chapter has just worked the cosine; this item is the same lesson with the phase moved, and it is the only unworked item in the chapter that repeats a worked one exactly.
• Miscellaneous Exercise question 4 (Part II p. 298). A cube, the horizontal axis, and boundaries at minus two and one, with four options offered. Verified: the piece below the axis has size four, the piece above has size one quarter, and the area is seventeen quarters, which is the fourth option. The unsplit integral is minus fifteen quarters, and minus fifteen quarters is the second printed option. A student who integrates straight through will find their answer on the page.
• Miscellaneous Exercise question 5 (Part II p. 298). A curve given as the variable times its own absolute value, with boundaries at minus one and one and a printed hint that splits it into two cases. Verified: each piece has size one third, so the area is two thirds, which is the third option. The unsplit integral is nothing at all, and nothing at all is the first printed option. Two items on one page, each with the sign-blind answer sitting in the list. Build section 10 on the pair.
• Miscellaneous Exercise question 2 (Part II p. 298). An absolute-value graph, to be sketched, and its integral over a run of six units. Verified from a 300 dots per inch the printed page — the bars are printed on both the curve and the integrand, and the text layer loses all four of them. Verified: the graph is a V with its corner at minus three, both pieces are triangles with base three and height three, and the integral is nine. It belongs here as the near-miss: the rule changes at the corner, but the curve never goes below the axis, so nothing needs its sign discarded. Use it to sharpen the criterion — you split where the curve crosses, not merely where its formula changes.
• The Summary (Part II p. 298). Two bullets, both about a single curve against an axis, and verified on a 300 dots per inch the printed page: neither mentions the sign, the absolute value, splitting at a crossing, or the below-the-axis case at all. The first bullet states the area as the plain integral with no caveat, which is false for exactly the situations this topic covers.

Figures to have open

• A redraw of Fig 8.3 (Part II p. 293) with the curve entering below the axis, the crossing kept outside the right boundary, both boundary labels, and one hanging strip. The chapter's own diagram; it carries section 4.
• A redraw of Fig 8.4 (Part II p. 294) with the curve through the origin, the left boundary clear of the origin, both parts filled, both parts labelled and one strip in each. The chapter's own diagram; it carries section 5 and is the single most important drawing in this brief.
• A redraw of Fig 8.9 (Part II p. 297) with the line, the crossing labelled with its coordinates, both vertical boundaries and the two triangles. The chapter's own; it carries sections 7 and 8. Draw the two triangles in the same fill, unlike the printed figure — see Notes.
• A redraw of Fig 8.10 (Part II p. 297) with all three pieces, the two crossings and the minimum lettered, and the closing vertical at the full turn. The chapter's own; it carries section 9.
• A slide-through movement for section 1: one region translated downward across the axis while its integral's value is displayed. Not in the book; the chapter draws the two cases as separate figures and never morphs one into the other.
• A four-line rule card for section 12. Not in the book; the chapter states the rule only through two worked instances.

Where this sits in the book

• NCERT Class 12 Mathematics, Chapter 8 "Application of Integrals", the Remark and Fig 8.3, §8.2, Part II p. 293
• Fig 8.4 and the two-part addition rule, Part II pp. 293–294
• Miscellaneous Example 3 with Fig 8.9, Part II pp. 296–297
• Miscellaneous Example 4 with Fig 8.10, Part II p. 297
• Miscellaneous Exercise on Chapter 8, questions 2, 3, 4 and 5, Part II p. 298; Summary, Part II p. 298