Integration

• Differentiate your answer to check it. Sketch before any area or volume: a region below the axis gives a negative integral, and a hole in a solid is subtracted after squaring, never before.
• Add one to the power, divide by the new power (and by a), add c, differentiate to check. Integrate to N and say what tends to 0. Sketch first; split at the axis; a washer is π∫(y12 − y22) dx.

• 1.8.1 I can understand integration as the reverse process of differentiation, and integrate (ax + b)ⁿ (for any rational n except −1), together with constant multiples, sums and differences
• 1.8.2 I can solve problems involving the evaluation of a constant of integration
• 1.8.3 I can evaluate definite integrals
• 1.8.4 I can use definite integration to find (a) the area of a region bounded by a curve and lines parallel to the axes, or between a curve and a line or between two curves (b) a volume of revolution about one of the axes

Accuracy for this chapter. Leave integrals and areas exact as fractions (\(\tfrac{32}{3}\), \(\tfrac{13}{3}\), \(\tfrac{45}{4}\)) and volumes exact in \(\pi\) (\(\tfrac{704\pi}{5}\)). Give a decimal only when the question asks, to 3 significant figures, from the exact value. Show the integral with its limits substituted before the number: a value read off a calculator's integration key, with no working, earns nothing, and a calculator that integrates symbolically is not permitted.

• “The region between \(y = 9 - x^2\) and \(y = 5\), rotated about the \(x\)-axis, has volume \(\pi\displaystyle\int_{-2}^{2} \left((9 - x^2) - 5\right)^2 dx\).” Correct A washer is \(\pi\displaystyle\int (y_1^2 - y_2^2)\,dx\), never \(\pi\displaystyle\int (y_1 - y_2)^2\,dx\). Each slice is a disc of radius \(y_1\) with a hole of radius \(y_2\), both measured from the axis: \(\pi\displaystyle\int_{-2}^{2} \left((9 - x^2)^2 - 5^2\right)dx = \tfrac{704\pi}{5}\). Squaring the gap gives \(\tfrac{512\pi}{15}\), the volume of a different, smaller solid.
• “The area between \(y = x^2 - 4x\) and the \(x\)-axis from \(x = 0\) to \(x = 5\) is \(\displaystyle\int_0^5 (x^2 - 4x)\,dx = -\tfrac{25}{3}\).” Correct A region below the \(x\)-axis gives a negative integral: split at the intercept and take the modulus. \(\displaystyle\int_0^4 = -\tfrac{32}{3}\) (area \(\tfrac{32}{3}\)) and \(\displaystyle\int_4^5 = \tfrac{7}{3}\), so the area is \(13\). One integral across the intercept lets the two parts cancel.
• “\(\displaystyle\int 3x^2\,dx = x^3\).” Correct \(x^3 + c\). Every indefinite integral carries its constant; without it, a curve cannot be found from its gradient.
• “\(\displaystyle\int (2x + 3)^{-2}\,dx = -(2x + 3)^{-1} + c\).” Correct Divide by the new power and by \(a = 2\): \(-\tfrac{1}{2}(2x + 3)^{-1} + c\). Differentiating the wrong answer gives \(2(2x + 3)^{-2}\), twice the integrand, which is how you catch it.
• “\(\displaystyle\int (x^2 + 1)^2\,dx = \dfrac{(x^2 + 1)^3}{3 \times 2x} + c\).” Correct The bracket rule is for a linear bracket \(ax + b\) only. Expand first: \(\displaystyle\int (x^4 + 2x^2 + 1)\,dx = \tfrac{1}{5}x^5 + \tfrac{2}{3}x^3 + x + c\).
• “\(\displaystyle\int_1^\infty x^{-2}\,dx = \left[-x^{-1}\right]_1^\infty = -\tfrac{1}{\infty} + 1 = 1\).” Correct \(\infty\) is not a number to substitute. Integrate to \(N\): \(1 - \tfrac{1}{N}\), then say that \(\tfrac{1}{N} \to 0\) as \(N \to \infty\), so the value is 1.
• “Add one to the power and divide by the new power: it works for every power of \(x\).” Correct It works for every rational \(n\) except \(n = -1\), where the new power would be 0 and the division is impossible. No integral in this chapter has \(n = -1\); that case is chapter 13 (Paper 3).
• Wrong: “\(\displaystyle\int 3x^2\,dx = x^3\).” Repair An indefinite integral has a constant: \(x^3 + c\). Without it, a 1.8.2 problem cannot even be started.
• Wrong: “\(\displaystyle\int x^{-2}\,dx = \dfrac{x^{-3}}{-3} + c\).” Repair Integration raises the power: \(\dfrac{x^{-1}}{-1} + c = -\dfrac{1}{x} + c\). Differentiating the wrong answer gives \(x^{-4}\), which is the check that catches it.
• Wrong: “\(\displaystyle\int \dfrac{6}{x^3}\,dx = \dfrac{6}{x^4/4} + c\).” Repair Rewrite as \(6x^{-3}\) first; the integral is \(\dfrac{6x^{-2}}{-2} + c = -\dfrac{3}{x^2} + c\). A denominator is never integrated on its own.
• Wrong: “\(\displaystyle\int \dfrac{x^2 + 1}{x^2}\,dx = \dfrac{\tfrac{1}{3}x^3 + x}{-1/x}\).” Repair There is no quotient rule for integrals. Split: \(\displaystyle\int \left(1 + x^{-2}\right)dx = x - \dfrac{1}{x} + c\).
• Wrong: “\(\displaystyle\int (2x + 3)^{-2}\,dx = -(2x + 3)^{-1} + c\).” Repair Divide by \(a\) as well as by the new power: \(-\dfrac{(2x + 3)^{-1}}{2} + c\).
• Wrong: “\(\displaystyle\int (x^2 + 1)^2\,dx = \dfrac{(x^2 + 1)^3}{3 \times 2x} + c\).” Repair The bracket rule is for a linear bracket only. Expand: \(\displaystyle\int \left(x^4 + 2x^2 + 1\right)dx = \tfrac{1}{5}x^5 + \tfrac{2}{3}x^3 + x + c\).
• Wrong: “\(y = x^2 + x + c\) passes through \((1, -2)\), so \(c = -2\).” Repair Substitute both coordinates: \(-2 = 1 + 1 + c\), so \(c = -4\).
• Wrong: “\(\displaystyle\int_1^3 (3x^2 - 2x)\,dx = 27 - 9 = 18 - 0\).” Repair The lower limit is 1, not 0: subtract \((1 - 1)\), and write that bracket even when it happens to be zero. The answer 18 is right here only because \(1^3 - 1^2\) happens to be 0; with most integrands the skipped bracket costs the answer.
• Wrong: “\(\displaystyle\int_1^\infty x^{-2}\,dx = \Big[-\tfrac{1}{x}\Big]_1^\infty = -\tfrac{1}{\infty} + 1\).” Repair \(\infty\) is not a number to substitute. Integrate to \(N\), get \(1 - \tfrac{1}{N}\), and say \(\tfrac{1}{N} \to 0\) as \(N \to \infty\).
• Wrong: “\(\displaystyle\int_1^\infty x^{-1/2}\,dx = \Big[2\sqrt{x}\Big]_1^\infty = 2\).” Repair \(2\sqrt{N} - 2\) grows without limit as \(N \to \infty\), so this integral has no finite value. Check the limit before writing a value.
• Wrong: “The area between \(y = x^2 - 4x\) and the \(x\)-axis from 0 to 5 is \(-\tfrac{25}{3}\).” Repair An area is positive, and a region crossing the axis must be split: \(\tfrac{32}{3} + \tfrac{7}{3} = 13\).
• Wrong: “The area between \(y = x^2\) and \(y = 2x + 3\) is \(\displaystyle\int_{-1}^{3} (x^2 - 2x - 3)\,dx = -\tfrac{32}{3}\).” Repair Upper minus lower: the line is on top, so integrate \(2x + 3 - x^2\) and get \(\tfrac{32}{3}\). A negative answer is the signal that the subtraction was the wrong way round.
• Wrong: “\(V = \pi\displaystyle\int_{-2}^{2} \left((9 - x^2) - 5\right)^2 dx\).” Repair The slice is a washer, not a disc: \(V = \pi\displaystyle\int_{-2}^{2} \left((9 - x^2)^2 - 5^2\right)dx = \tfrac{704\pi}{5}\).
• Wrong: “The volume about the \(y\)-axis of the region bounded by \(y = x^3\), the \(y\)-axis and \(y = 8\) is \(\pi\displaystyle\int_0^8 y^2\,dy\).” Repair About the \(y\)-axis the radius is \(x\): write \(x^2 = y^{2/3}\) and integrate \(\pi\displaystyle\int_0^8 y^{2/3}\,dy = \tfrac{96\pi}{5}\).

• Read the command word before you decide how much to write. This syllabus uses eleven: calculate, describe, determine, evaluate, explain, identify, justify, show (that), sketch, state and verify. Show that and verify give you the answer and mark the route to it, so every step must be visible and the argument must run forwards from what is given, never backwards from the result. Sketch means a simple freehand drawing showing the key features, taking care over proportions; it is not a plot. Determine means establish with certainty; justify means support a case with evidence or argument. Find, solve, express and hence are ordinary question wording; hence means the previous part is the intended route.
• Interleave with the chapters that use this one. Chapter 13 (Paper 3) extends every method here to new functions: before starting it, redo the bracket drill. Chapter 16 solves differential equations by integrating: redo worked example 4. Chapter 19 (Mechanics) integrates velocity to find displacement: redo the constant finder, because the constant there is the starting position. Chapter 30 (Statistics 2) uses improper integrals for probability density functions: redo worked example 5. Recalling a method inside a new problem is worth more than another pass over this chapter on its own: Paper 1 is assumed knowledge for every other paper, and an individual examination question may involve ideas and methods from more than one section of that paper’s content, so nothing here is ever finished with.

• Cambridge International AS & A Level Mathematics 9709 has six components, and a candidate takes two of them for the AS Level and four for the A Level. This chapter is Pure Mathematics 1 content, examined in Paper 1. Paper 1 (Pure Mathematics 1) is compulsory for both the AS Level and the A Level: it is 60% of the AS Level and 30% of the A Level, and its content is assumed knowledge for every other paper. Every paper is a written examination of compulsory structured questions, answered on the question paper, with MF19 (the list of formulae and statistical tables) supplied. Examinations are available in the June and November series, and in March in India.
• Across the whole qualification the assessment objectives are weighted AO1 55% (knowledge and understanding: concepts, terminology, notation and accurate manipulative technique) and AO2 45% (application and communication: choosing the procedure, combining techniques to solve problems, and presenting the work clearly and logically) at AS Level, and AO1 52%, AO2 48% at A Level. AS candidates are graded a–e; A Level candidates A*–E.
• The four outcomes combine naturally, and Paper 1's structured parts let one question use several: a gradient function and a point, leading to the equation of a curve, which can lead on to its stationary points; a “show that” two graphs meet at stated points, followed by the area between them; an area or volume from a sketch, where the sketch decides the limits and any split; and an integral with an infinite limit or an unbounded integrand, evaluated by a limit. The method line — the integral with its limits substituted — is the reasoning, and it must be visible.
• \(\displaystyle\int x^n\,dx = \dfrac{x^{n+1}}{n + 1}\) \((n \ne -1)\) is printed, without its \(+\,c\). Everything else is to be known: the \((ax + b)^n\) form, \(F(b) - F(a)\), the limiting processes, \(\int y\,dx\), \(\int x\,dy\), upper minus lower, \(\pi\int y^2\,dx\), \(\pi\int x^2\,dy\) and the washer. The other rows of the integration table are Paper 3. See the MF19 card.
• Leave areas as fractions and volumes as multiples of \(\pi\) unless a decimal is asked for, and then give it to 3 significant figures from the exact value. Write \(+\,c\) on every indefinite integral and never on a definite one. Show both brackets of every substitution line, the \(N\) or \(\varepsilon\) of every improper integral, and the modulus of every negative part of an area. A calculator value for a definite integral, with no working, is an unsupported answer.
• Read the command word before you decide how much to write. This syllabus uses eleven: calculate, describe, determine, evaluate, explain, identify, justify, show (that), sketch, state and verify. Show that and verify give you the answer and mark the route to it, so every step must be visible and the argument must run forwards from what is given, never backwards from the result. Sketch means a simple freehand drawing showing the key features, taking care over proportions; it is not a plot. Determine means establish with certainty; justify means support a case with evidence or argument. Find, solve, express and hence are ordinary question wording; hence means the previous part is the intended route.