HookThe half-time surge the grid has to add up
At around 21:30 on 4 July 1990, England went out of the World Cup semi-final to West Germany on penalties, and roughly a minute later the National Grid recorded one of the largest electricity surges in its history — about \(2\,800\) megawatts as millions of kettles went on at once. The control engineers at Wokingham had to have that power ready in advance, and the quantity they actually plan around is not the peak power but the total energy delivered during the surge — kilowatt-hours, not kilowatts. Energy is power accumulated over time: if you draw the power on a graph against time, the energy is the area under that curve. Adding up an ever-changing quantity over an interval — that is integration, and the grid does it every night.
Chapter 8 is the reverse of Chapter 7 and, at the same time, the mathematics of area and accumulation. Integration undoes differentiation, so the first job is to run every standard derivative backwards. Then the Fundamental Theorem of Calculus makes the deep claim that the antiderivative and the area-under-a-curve are the same object, which is why a definite integral computes an area. You will find the area under a curve and between two curves, see integration as the limit of a sum of thin rectangles, learn the two reversal techniques — substitution (undoing the chain rule) and by parts (undoing the product rule) — split a fraction with partial fractions before integrating, and finally solve separable differential equations, the equations Chapter 7 taught you to write down.
ModelIntegration as reverse differentiation — and why area is a limit of a sum
To integrate is to answer 'what had this as its derivative?'. Because \(\dfrac{d}{dx}(x^3)=3x^2\), we say \(\displaystyle\int 3x^2\,dx=x^3+C\). The \(+C\), the constant of integration, is unavoidable: differentiation destroys constants, so \(x^3\), \(x^3+7\) and \(x^3-2\) all have the same derivative, and the indefinite integral must record that whole family. Forgetting \(+C\) is the single most-penalised habit in the chapter.
The deeper idea is why this reversal computes an area. Chop the region under \(y=f(x)\) from \(x=a\) to \(x=b\) into \(n\) thin strips of width \(\delta x\), approximate each by a rectangle of height \(f(x_i)\), and add: the area is roughly \(\sum f(x_i)\,\delta x\). As the strips get thinner (\(n\to\infty\), \(\delta x\to 0\)) that sum tends to a limit written \(\displaystyle\int_a^b f(x)\,dx\) — integration as the limit of a sum, which is where the elongated-S symbol comes from. The Fundamental Theorem of Calculus then delivers the miracle: this limiting area equals \(F(b)-F(a)\), where \(F\) is any antiderivative of \(f\). Accumulation and antidifferentiation are the same thing, so you never actually sum infinitely many rectangles — you differentiate in reverse and subtract.
Find \(\displaystyle\int (6x^2+2)\,dx\) and then use it to confirm the Fundamental Theorem on \([0,2]\).
Reverse the power rule term by term: something whose derivative is \(6x^2\) is \(2x^3\); something whose derivative is \(2\) is \(2x\). So \(\displaystyle\int(6x^2+2)\,dx=2x^3+2x+C\). Now the definite version on \([0,2]\): the theorem says the answer is \(F(2)-F(0)\) with \(F(x)=2x^3+2x\) (the \(C\) cancels in the subtraction, which is why definite integrals never need it). \(F(2)=2(8)+4=20\); \(F(0)=0\). So \(\displaystyle\int_0^2(6x^2+2)\,dx=20\). That single number is exactly the limit the thin-rectangle sum would crawl towards — the theorem just lets you skip the crawling.
MechanismThe standard integrals — every derivative, run backwards
The exam list mirrors the differentiation list. For powers, raise the index by one and divide by the new index: \(\displaystyle\int x^n\,dx=\dfrac{x^{n+1}}{n+1}+C\) — valid for every rational \(n\) except \(n=-1\), because that would divide by zero. That excluded case is exactly \(\displaystyle\int \dfrac{1}{x}\,dx=\ln|x|+C\), with the modulus signs because you can only take the log of a positive number and \(x\) might be negative. The rest: \(\displaystyle\int e^{kx}\,dx=\dfrac{1}{k}e^{kx}+C\); \(\displaystyle\int \sin kx\,dx=-\dfrac{1}{k}\cos kx+C\); \(\displaystyle\int \cos kx\,dx=\dfrac{1}{k}\sin kx+C\).
Notice the \(k\) now divides where in differentiation it multiplied — integration and differentiation pull the constant in opposite directions, and swapping the two is a classic slip. As with derivatives, you integrate sums and differences term by term and pull constant multiples through. Two sign traps deserve flagging: the \(\sin\)-into-minus-\(\cos\) result surprises people every year, and the \(\ln|x|\) case is easy to miss when a question hides a \(\dfrac{1}{x}\) as \(x^{-1}\) and tempts you to use the power rule that explicitly forbids it.
Find \(\displaystyle\int\left(6x^2+4e^{2x}-\sin 2x+\dfrac{1}{x}\right)dx\).
Term by term. \(6x^2\to \dfrac{6x^3}{3}=2x^3\). \(4e^{2x}\to 4\cdot\dfrac{1}{2}e^{2x}=2e^{2x}\). \(-\sin 2x\to -\left(-\dfrac{1}{2}\cos 2x\right)=+\dfrac{1}{2}\cos 2x\) — two minus signs make a plus, the year's favourite trap. \(\dfrac{1}{x}\to \ln|x|\), not \(\dfrac{x^0}{0}\). One constant covers the lot:
\[\int\left(6x^2+4e^{2x}-\sin 2x+\tfrac{1}{x}\right)dx=2x^3+2e^{2x}+\tfrac{1}{2}\cos 2x+\ln|x|+C.\]
You can always check an integral by differentiating your answer: \(\dfrac{d}{dx}\big(2e^{2x}\big)=4e^{2x}\) and \(\dfrac{d}{dx}\big(\tfrac12\cos 2x\big)=-\sin 2x\) — back to the integrand, so the work is sound.
MechanismDefinite integrals and area — including the sign trap and area between curves
A definite integral \(\displaystyle\int_a^b f(x)\,dx=F(b)-F(a)\) evaluates to a number, and when the curve sits above the \(x\)-axis on \([a,b]\) that number is the area between the curve and the axis. The catch is that the integral measures signed area: any part of the curve below the axis contributes a negative amount. So to find a genuine geometric area where the curve crosses the axis, you must find the crossing points, integrate each piece separately, and add the magnitudes. Integrating straight through a crossing lets a positive lobe cancel a negative one and reports too little area — a guaranteed lost mark.
For the area between two curves, integrate the difference 'top minus bottom' between their points of intersection: \(\displaystyle\int_a^b\big(y_{\text{top}}-y_{\text{bottom}}\big)\,dx\). Finding the limits is part of the problem — you get \(a\) and \(b\) by solving the two equations simultaneously to locate where the curves meet. Doing the difference this way is cleaner than computing two areas and subtracting, and it automatically handles regions that dip below the axis, because the vertical gap 'top minus bottom' stays positive throughout.
Find the area enclosed between the line \(y=x+2\) and the parabola \(y=x^2\).
First the limits: the curves meet where \(x^2=x+2\), i.e. \(x^2-x-2=0\), i.e. \((x-2)(x+1)=0\), so \(x=-1\) and \(x=2\). Between these the line is on top (test \(x=0\): line gives \(2\), parabola gives \(0\)). So the area is
\[\int_{-1}^{2}\big((x+2)-x^2\big)\,dx=\left[\frac{x^2}{2}+2x-\frac{x^3}{3}\right]_{-1}^{2}.\]
Upper limit \(x=2\): \(\dfrac{4}{2}+4-\dfrac{8}{3}=2+4-2.667=3.333\). Lower limit \(x=-1\): \(\dfrac{1}{2}-2+\dfrac{1}{3}=-1.167\). Subtract: \(3.333-(-1.167)=4.5\). The enclosed area is \(\dfrac{9}{2}\) square units. Because we integrated 'top minus bottom', the answer came out positive in one clean pass — no need to split, and no risk of the sign trap.
MechanismSubstitution and by parts — undoing the chain and product rules
Two integrands resist the standard list, and each is a differentiation rule run backwards. Integration by substitution undoes the chain rule: spot an inner function \(u\) whose derivative also appears (up to a constant), replace \(u\) and \(du=\dfrac{du}{dx}\,dx\), integrate in \(u\), then substitute back. So \(\displaystyle\int 2x(x^2+1)^5\,dx\) with \(u=x^2+1\), \(du=2x\,dx\) becomes \(\displaystyle\int u^5\,du=\dfrac{u^6}{6}+C=\dfrac{(x^2+1)^6}{6}+C\). For a definite integral, either convert the limits to \(u\)-values as you go, or substitute back before evaluating — but never mix \(x\)-limits with a \(u\)-integrand.
Integration by parts undoes the product rule: \(\displaystyle\int u\,\dfrac{dv}{dx}\,dx=uv-\int v\,\dfrac{du}{dx}\,dx\). The skill is choosing which factor is \(u\) (the one that gets simpler when differentiated) and which is \(\dfrac{dv}{dx}\) (the one you can integrate). The mnemonic LIATE — Logs, Inverse trig, Algebra, Trig, Exponentials — ranks what to pick as \(u\): whatever comes first in that list. A polynomial times \(e^{x}\) takes the polynomial as \(u\); a lone \(\ln x\) takes \(u=\ln x\) and \(\dfrac{dv}{dx}=1\).
Find \(\displaystyle\int x\,e^{2x}\,dx\) by parts.
By LIATE the algebra beats the exponential, so let \(u=x\) and \(\dfrac{dv}{dx}=e^{2x}\). Then \(\dfrac{du}{dx}=1\) and \(v=\dfrac{1}{2}e^{2x}\). Apply the formula: \(\displaystyle\int x\,e^{2x}\,dx=x\cdot\dfrac{1}{2}e^{2x}-\int \dfrac{1}{2}e^{2x}\cdot 1\,dx=\dfrac{1}{2}xe^{2x}-\dfrac{1}{2}\int e^{2x}\,dx\). The remaining integral is standard: \(\displaystyle\int e^{2x}\,dx=\dfrac{1}{2}e^{2x}\). So
\[\int x\,e^{2x}\,dx=\tfrac{1}{2}xe^{2x}-\tfrac{1}{4}e^{2x}+C=\tfrac{1}{4}e^{2x}(2x-1)+C.\]
Had you chosen \(u=e^{2x}\) instead, the leftover integral would have got worse, not better — which is the signal you picked \(u\) the wrong way round and should swap.
MechanismIntegration by partial fractions
A single algebraic fraction with a factorising denominator is impossible to integrate as it stands, but splitting it into partial fractions turns it into a sum of \(\dfrac{1}{x}\)-type pieces you can integrate on sight. The A-Level restriction is denominators that are linear in each factor, so \(\dfrac{5}{(x-1)(x+4)}\) decomposes into \(\dfrac{A}{x-1}+\dfrac{B}{x+4}\). Each piece integrates to a natural logarithm, because \(\displaystyle\int\dfrac{1}{x-a}\,dx=\ln|x-a|+C\).
The efficient way to find \(A\) and \(B\) is the cover-up method: multiply through by the common denominator to get an identity in the numerators, then substitute the value of \(x\) that kills one bracket at a time. The two logs that result are usually collected into a single log of a quotient using the log laws, which is the form mark schemes print. Watch the sign inside each modulus and remember the constant of integration survives to the end.
Find \(\displaystyle\int \dfrac{5}{(x-1)(x+4)}\,dx\).
Write \(\dfrac{5}{(x-1)(x+4)}=\dfrac{A}{x-1}+\dfrac{B}{x+4}\). Multiply through: \(5=A(x+4)+B(x-1)\). Substitute \(x=1\) to isolate \(A\): \(5=A(5)+B(0)\Rightarrow A=1\). Substitute \(x=-4\) to isolate \(B\): \(5=A(0)+B(-5)\Rightarrow B=-1\). So the integral splits as \(\displaystyle\int\left(\dfrac{1}{x-1}-\dfrac{1}{x+4}\right)dx\). Each piece is a standard log:
\[\int\!\left(\frac{1}{x-1}-\frac{1}{x+4}\right)dx=\ln|x-1|-\ln|x+4|+C=\ln\left|\frac{x-1}{x+4}\right|+C.\]
The collapse into one log via \(\ln a-\ln b=\ln\frac{a}{b}\) is the tidy final line examiners look for.
CaseSeparable differential equations — and reading the answer in context
Chapter 7 taught you to write \(\dfrac{dP}{dt}=kP\); integration lets you solve it. A first-order equation is separable when you can get all the \(y\)'s on one side and all the \(x\)'s (or \(t\)'s) on the other: \(\dfrac{dy}{dx}=f(x)g(y)\Rightarrow \displaystyle\int\dfrac{1}{g(y)}\,dy=\int f(x)\,dx\). Integrate both sides — a single \(+C\) suffices — and rearrange. That \(C\) is why an initial condition matters: substituting a known point (the value at \(t=0\), say) fixes \(C\) and turns the general family of solutions into the one particular solution the problem is about.
The last spec point, and the most examinable, is interpreting the solution and its limitations. A model such as unbounded exponential growth is only ever locally true: it cannot describe a bacterial colony forever, because food and space run out. Naming that — 'the exponential model predicts the population grows without limit, which is unrealistic in the long run; a logistic model with a ceiling would refine it' — is exactly the evaluative sentence that lifts an answer into the top band, and it links straight back to the kinematics and growth contexts of Chapter 7.
A town of \(8\,000\) people grows so that \(\dfrac{dP}{dt}=0.03P\), with \(t\) in years. Find \(P\) after \(10\) years, and comment.
Separate the variables: \(\displaystyle\int\dfrac{1}{P}\,dP=\int 0.03\,dt\), giving \(\ln P=0.03t+C\). Exponentiate: \(P=e^{0.03t+C}=Ae^{0.03t}\), where \(A=e^{C}\). Apply the initial condition \(P=8\,000\) when \(t=0\): \(8\,000=A e^{0}=A\), so \(A=8\,000\) and the particular solution is \(P=8\,000\,e^{0.03t}\). At \(t=10\): \(P=8\,000\,e^{0.3}=8\,000\times 1.3499\approx 10\,799\) people. Now the interpretation: the model gives smooth \(3\%\)-a-year compound growth, but it is unbounded — extend it to \(t=200\) and it predicts millions, which no town can house. The exponential model is a good local description and a poor long-run one; a realistic refinement caps the growth as the town approaches its housing limit.
VocabularyKey terms the mark scheme pays for
TrapsMisconceptions that cost marks
ExamWhat examiners want
Every indefinite integral must carry \(+C\); every \(\int\frac{1}{x}\)-type result must carry the modulus, \(\ln|x|\). These are cheap marks the examiner is actively looking to dock, so make them automatic. On area questions, sketch the curve first to see whether it dips below the \(x\)-axis on the interval — if it crosses, split at the crossing and add magnitudes; if you are between two curves, integrate 'top minus bottom' and solve simultaneously for the limits before you integrate anything.
For a definite integral, show the antiderivative in square brackets with the limits, then the \(F(b)-F(a)\) subtraction line explicitly — the substitution of limits is a marked step and 'answer only' throws it away. In by-parts and substitution questions, state your choice of \(u\) (and \(v\), or \(u\) and \(du\)) on its own line so the method is visible even if the arithmetic slips. When you solve a separable differential equation, write the separated integral form, keep a single constant, and use the initial condition immediately to find it; then, if the question asks you to comment, add one sentence on the model's limitation — unbounded growth, a linear model outrunning reality — because that evaluative line is where the final marks live and it costs only a sentence.