Further calculus.

HookThe horn you can fill but never paint

ModelImproper integrals — taming infinity with a limit

An integral is improper in two ways: a limit of integration is infinite, like \(\int_1^\infty\frac{1}{x^2}\,dx\), or the integrand is undefined somewhere in the range, like \(\int_0^1\frac{1}{\sqrt{x}}\,dx\) at \(x=0\). Either way the definition of integration has been breached, and the repair is the same ritual: replace the offending endpoint with a parameter, integrate normally, then take the limit. \(\int_1^\infty\frac{1}{x^2}\,dx=\lim_{t\to\infty}\int_1^t\frac{1}{x^2}\,dx=\lim_{t\to\infty}\left(1-\frac{1}{t}\right)=1\). If the limit exists the integral converges to it; if not, it diverges, and 'diverges' is a complete and mark-worthy answer.

The boundary between the two is thinner than intuition suggests. \(\int_1^\infty\frac{1}{x^2}\,dx\) converges, but \(\int_1^\infty\frac{1}{x}\,dx=\lim_{t\to\infty}\ln t\) diverges — even though \(\frac{1}{x}\) also shrinks to zero. An integrand tending to zero is nowhere near enough; it has to shrink fast enough, and that pair of integrals is precisely why the horn holds finite paint behind an infinite wall.

When exponentials and logarithms enter, AQA hands you two standard limits and expects them quoted by name: for any \(k>0\), \(\lim_{x\to\infty}x^k e^{-x}=0\) and \(\lim_{x\to 0^+}x^k\ln x=0\). The slogan: exponentials beat powers, and powers beat logarithms. \(e^{-x}\) drags any polynomial to zero eventually; \(x^k\) crushes the slow divergence of \(\ln x\) near the origin. These two facts are exactly what evaluates the boundary terms that integration by parts leaves behind in improper integrals.

ModelVolumes of revolution and the mean value of a function

Rotate the region under \(y=f(x)\), between \(x=a\) and \(x=b\), about the \(x\)-axis, and slice the solid into discs perpendicular to the axis: each slice is (almost) a cylinder of radius \(y\) and thickness \(\delta x\), volume \(\pi y^2\,\delta x\). Summing and taking the limit gives \(V=\pi\int_a^b y^2\,dx\). About the \(y\)-axis, the roles swap: \(V=\pi\int x^2\,dy\), with limits in \(y\) and \(x^2\) expressed in terms of \(y\). The two mechanical traps: square the whole expression for \(y\) — cross terms included — and match everything (integrand, variable, limits) to the axis of rotation. Rotating between two curves subtracts volumes: \(\pi\int(y_{\text{outer}}^2-y_{\text{inner}}^2)\,dx\), not \(\pi\int(y_{\text{outer}}-y_{\text{inner}})^2\,dx\).

The mean value of a function answers a different question: what constant height would produce the same area over the interval? Define \(\bar{f}=\frac{1}{b-a}\int_a^b f(x)\,dx\) — the area under the curve, shared out evenly across the width. For example, the mean value of \(\sin x\) over \([0,\pi]\) is \(\frac{1}{\pi}\int_0^\pi\sin x\,dx=\frac{2}{\pi}\approx 0.64\): a sine arch, flattened to a rectangle, stands about \(0.64\) high. Physically this is the average of a continuously varying quantity — mean current, mean velocity — and it is emphatically not the average of the endpoint values, because the function spends more 'time' at some heights than others.

MechanismPartial fractions grow up — quadratic factors in the denominator

At A-level Maths, partial fractions handled linear factors. Further Maths adds the irreducible quadratic factor \(ax^2+c\), and the decomposition rule changes with it: a factor \(ax^2+c\) in the denominator demands a numerator of the form \(Bx+C\), a full linear term — not a bare constant. So \(\frac{p(x)}{x(x^2+1)}\) splits as \(\frac{A}{x}+\frac{Bx+C}{x^2+1}\). Find the constants the usual way: multiply out, substitute convenient values, compare coefficients.

The payoff is that each resulting piece is directly integrable, and the quadratic piece splits along its numerator. The \(Bx\) part is a disguised logarithm — \(\int\frac{x}{x^2+c}\,dx=\frac{1}{2}\ln(x^2+c)+k\), since the numerator is half the derivative of the denominator. The constant part is a disguised inverse tangent — \(\int\frac{1}{x^2+c}\,dx\) is an arctan, using the standard form from the next block. One decomposition can therefore feed logarithms and inverse trig into the same answer line, which is exactly the look of a typical exam solution: \(2\ln|x|+\arctan x+c\).

Watch the top-heaviness check first, always: if the numerator's degree is greater than or equal to the denominator's, divide out before decomposing. And keep the modulus signs on logarithms of linear factors — \(\ln|x|\), not \(\ln x\) — unless the given domain makes the argument positive; \(\ln(x^2+c)\) needs no modulus because its argument cannot be negative.

MechanismInverse trig — differentiate once, integrate forever

The derivatives of the inverse trigonometric functions come from one move: implicit differentiation. If \(y=\arcsin x\) then \(x=\sin y\), so \(\frac{dx}{dy}=\cos y=\sqrt{1-\sin^2 y}=\sqrt{1-x^2}\) — the positive root, because \(\cos y\ge 0\) on the principal range \(-\frac{\pi}{2}\le y\le\frac{\pi}{2}\). Hence \(\frac{d}{dx}\arcsin x=\frac{1}{\sqrt{1-x^2}}\). The same manoeuvre gives \(\frac{d}{dx}\arccos x=-\frac{1}{\sqrt{1-x^2}}\) and \(\frac{d}{dx}\arctan x=\frac{1}{1+x^2}\), the last via \(\sec^2 y=1+\tan^2 y\).

Read backwards, these are integration results, and AQA states them with a general constant: \(\int\frac{dx}{\sqrt{a^2-x^2}}=\arcsin\frac{x}{a}+c\) and \(\int\frac{dx}{a^2+x^2}=\frac{1}{a}\arctan\frac{x}{a}+c\) — note the extra \(\frac{1}{a}\) in the arctan result, the most-dropped factor in the section. Recognition is the skill: a square root of (constant minus \(x^2\)) says arcsine; (constant plus \(x^2\)) downstairs with no root says arctangent.

When the integral is dressed up — \(\sqrt{9-4x^2}\), say — the substitutions that undress it are trigonometric, chosen to kill the square root by identity: \(x=a\sin\theta\) turns \(a^2-x^2\) into \(a^2\cos^2\theta\); \(x=a\tan\theta\) turns \(a^2+x^2\) into \(a^2\sec^2\theta\). Match the pattern, choose the substitution that makes the identity fire, and change the limits as you go for definite integrals.

MechanismArc length and surface area — the horn's punchline

Zoom in on any smooth curve and a tiny piece of it is almost straight: Pythagoras on the small triangle with legs \(\delta x\) and \(\delta y\) gives \(\delta s\approx\sqrt{(\delta x)^2+(\delta y)^2}\). Divide through and integrate: arc length is \(s=\int_a^b\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx\) for a Cartesian curve, or \(s=\int\sqrt{\dot{x}^2+\dot{y}^2}\,dt\) for a parametric one — the same triangle, sliced by the parameter instead. Spin the curve about the \(x\)-axis and each element of arc sweeps a thin band of radius \(y\) and width \(\delta s\): surface area of revolution is \(S=2\pi\int y\,ds\), with \(ds\) expanded in whichever form matches the curve. The discipline of the topic is bookkeeping: compute the derivative, square it, add, simplify under the root until something integrable appears — questions are engineered so that \(1+\left(\frac{dy}{dx}\right)^2\) collapses to a perfect square or a linear expression.

Now Torricelli's paradox resolves into two integrals. Volume: \(\pi\int_1^\infty x^{-2}\,dx=\pi\) — converges, as the first block showed. Surface: \(S=2\pi\int_1^\infty\frac{1}{x}\sqrt{1+x^{-4}}\,dx\). The root factor is always at least \(1\), so \(S\ge 2\pi\int_1^\infty\frac{1}{x}\,dx\), which diverges. Finite volume, infinite surface, zero contradiction: volume shrinks like \(\frac{1}{x^2}\) (fast enough), surface like \(\frac{1}{x}\) (not). The 'paradox' was only ever a bet that the two rates were the same — and improper integration is the tool that settles the bet.

CaseReduction formulae — solving a whole family at once

Some integrals come in indexed families: \(I_n=\int_0^{\pi/2}\sin^n x\,dx\), \(\int x^n e^x\,dx\), \(\int\tan^n x\,dx\). Attacking \(n=6\) directly is misery; the professional move is a reduction formula — an equation expressing \(I_n\) in terms of \(I_{n-1}\) or \(I_{n-2}\), derived once, then applied repeatedly until the index hits a trivial base case like \(I_0\) or \(I_1\). It is recursion for integrals, and the derivation is almost always integration by parts: split off one or two factors, integrate them, and manoeuvre the leftover integral — which may be a rearranged copy of \(I_n\) itself — onto the left-hand side.

That last manoeuvre is the step students miss. In the \(\sin^n\) derivation below, parts produces \((n-1)I_{n-2}-(n-1)I_n\) on the right: the unknown reappears. Collect it — \(nI_n=(n-1)I_{n-2}\) — and the formula falls out. Expect exactly this structure whenever the identity \(\cos^2x=1-\sin^2x\) (or \(\sec^2x=1+\tan^2x\)) reintroduces the original power.

Exam questions stage it as derive-then-apply: part (a) 'show that \(nI_n=(n-1)I_{n-2}\)', part (b) 'hence find \(I_4\)' or an improper relative. The word hence is binding — use the formula, walk the index down in steps of two, and evaluate the base case explicitly. This block also closes the section's loop: reduction formulae routinely feed the improper integrals of the first block, with the standard limits killing the boundary terms at each rung.

VocabularyKey terms the mark scheme pays for

TrapsMisconceptions that cost marks

ExamWhat examiners want