TMUA Integration: Areas & Definite Integrals

Integration is the natural partner to differentiation on the TMUA, and like its partner the calculus is deliberately narrow: you integrate powers of $x$, evaluate definite integrals between limits, and use those integrals to find areas under and between curves. There is no integration by parts, no substitution, and nothing that needs a calculator. What earns marks is reversing the power rule cleanly, substituting limits without arithmetic slips, and understanding that integration and differentiation are two sides of the same coin. This guide sets out exactly what is examined, the techniques that matter, and a real past-paper question to test yourself on.

What integration is examined on the TMUA

The TMUA syllabus keeps calculus to early A-level content, so the integration you need is limited and learnable. You should be completely fluent with:

• Integrating $y = x^n$ for any rational power except $n = -1$, including negative and fractional indices.
• Integrating sums of such terms, including expressions you first rewrite as powers of $x$.
• Evaluating a definite integral $\int_a^b f(x)\, dx$ between two limits.
• Interpreting a definite integral as the area between a curve and the $x$-axis.
• Finding the area enclosed between two curves, or between a curve and a line.
• Recognising that integration reverses differentiation, so the two operations cancel.
• Using the symmetry of a graph to collapse a definite integral without integrating at all. The board's own notes treat this as a technique you are expected to have, and its wording for what it will ask is that any integration you are required to do, if it does not require symmetry or other arguments, will be an integration of sums of powers of $x$ (p.126). The symmetry route is the named exception, so learn it.
• Approximating an area with the trapezium rule, and deciding whether it gives an overestimate or an underestimate. This one is named explicitly in the specification (MM7.5), and the over-or-under judgement is the part that carries the marks: a rule made of straight chords overestimates where the curve is concave up, and underestimates where it is concave down.
• Solving a differential equation of the form $\frac{dy}{dx} = f(x)$ (MM7.6), which is integration with a constant, pinned down by a given point.

Notice what is absent: no integration by parts and no substitution. Two things people assume are absent are not: the specification includes the trapezium rule (MM7.5), together with deciding whether it gives an overestimate or an underestimate, and differential equations of the form $\frac{dy}{dx} = f(x)$ (MM7.6), which is really just integration with a constant. Trigonometric integrals are off the specification as a technique, but that is not the same as saying a trigonometric integral can never appear. The board is explicit that although trigonometric integration is "not on the TMUA/ESAT specification", its symmetry examples are expressions "we could expect you to deduce within the TMUA/ESAT specification" (footnote 69, p.125). The section on symmetry below is where that lives, and it is the highest-value idea in the whole of the board's integration chapter. If you want the full picture of where this sits, our syllabus topics guide lists every assessed area. As with differentiation, the difficulty is never the calculus itself; it is the speed and the indirect phrasing, the recurring theme across the whole test, as we explain in is the TMUA hard.

Integrating powers of x, with worked examples

Everything starts with reversing the power rule. If $y = x^n$ then $\int x^n \, dx = \frac{x^{n+1}}{n+1} + c$, valid for every rational $n$ except $n = -1$. You add one to the power, then divide by the new power, and remember the constant of integration $c$. That single rule, applied term by term, handles the overwhelming majority of TMUA integration.

Take $y = 12x^3 - 10x$. Integrating each term gives $\int \left(12x^3 - 10x\right) dx = 3x^4 - 5x^2 + c$, because $12 \div 4 = 3$ and $10 \div 2 = 5$. Notice that this is exactly the curve you would differentiate to get back $12x^3 - 10x$, which is the whole idea of integration as the reverse of differentiation. We unpack that derivative direction in our differentiation guide.

The marks that separate strong candidates come from terms that do not look like powers of $x$ at first, just as with differentiation. The trick is to rewrite before you integrate. A term like $\frac{1}{x^2}$ becomes $x^{-2}$, so $\int x^{-2} \, dx = \frac{x^{-1}}{-1} + c = -\frac{1}{x} + c$. A surd like $\sqrt{x}$ becomes $x^{1/2}$, so $\int x^{1/2} \, dx = \frac{2}{3}x^{3/2} + c$. The examiners love giving you an expression that looks awkward until you tidy it into powers of $x$, precisely because rewriting fluently is a calculator-free skill. We cover that rewriting habit alongside other shortcuts in our calculator-free techniques guide.

Definite integrals and area under a curve

A definite integral attaches limits to the integral and produces a number rather than a function. To evaluate $\int_a^b f(x)\, dx$ you first integrate to get an antiderivative $F(x)$, then compute $F(b) - F(a)$. The constant of integration cancels in the subtraction, which is why you never write $c$ for a definite integral.

For example, $\int_1^2 \left(3x^2\right) dx = \left[x^3\right]_1^2 = 2^3 - 1^3 = 8 - 1 = 7$. The square-bracket notation records the antiderivative before you substitute the limits, and substituting the top limit minus the bottom limit is the step where careless candidates lose marks under time pressure.

The reason definite integrals matter so much is that they measure area. The integral $\int_a^b f(x)\, dx$ gives the signed area between the curve $y = f(x)$ and the $x$-axis from $x = a$ to $x = b$. Where the curve sits above the axis the area counts as positive, and where it dips below the axis the integral counts that region as negative. This signed behaviour is the single most important thing to understand about area on the TMUA, and it is the source of the most common trap, which we return to below.

Try a real one

Theory only takes you so far. Here is an actual past-paper question. Attempt it fully before revealing the solution, since the indirect phrasing is the real test:

Killing an integral with symmetry

This is the technique almost every TMUA guide misses, and the board teaches it directly. Some definite integrals do not need integrating at all. If the graph has the right symmetry across the interval, the answer falls out in one line.

Two shapes matter. A graph is symmetric about the $y$-axis when reflecting it in the $y$-axis leaves it unchanged, which is the same as saying $f(-x) = f(x)$. Such a function is called even, and $x^2$, $x^4$ and $\cos x$ are the standard examples. A graph is antisymmetric when reflecting it in the $y$-axis and then in the $x$-axis leaves it unchanged, which is the same as $f(-x) = -f(x)$. Such a function is odd, and $x$, $x^3$, $x^5$, $\sin x$ and $\tan x$ are the standard examples.

The two consequences are the whole technique:

• An odd function over an interval symmetric about zero integrates to $0$. Every strip on the right is matched by a strip of the same size on the left with the opposite sign, so the total cancels. In symbols, $\int_{-a}^{a} f(x)\, dx = 0$ when $f$ is odd.
• An even function over an interval symmetric about zero is twice the half-range. The two halves are mirror images with the same sign, so $\int_{-a}^{a} f(x)\, dx = 2\int_0^a f(x)\, dx$ when $f$ is even.

The board also states the general shifted forms: for an even function, $\int_{-a}^{-b} f(x)\, dx = \int_b^a f(x)\, dx$, and for an odd function, $\int_{-b}^{-a} f(x)\, dx = -\int_a^b f(x)\, dx$ (p.123 to p.124).

The board's own examples

On the page immediately after, the board sets these as an exercise and asks you to explain why each is true (p.125). Read each one and say which symmetry does the work:

Notice what these have in common. You never integrate $\sin$, $\cos$ or $\tan$. You look at the picture, decide whether the two halves cancel or double, and write down the answer. That is the reasoning the board says it can ask for, and it is entirely within the specification even though the antiderivatives of the trigonometric functions are not. (The $\tan$ line is the board's own example and is worth a second look: $\tan x$ shoots off to infinity at $\pm\frac{\pi}{2}$, so what is doing the work is the odd symmetry pairing each part of the curve with its mirror image, not an area calculation.)

The polynomial cases are the ones most likely to appear in a Paper 1 disguise. The board's own worked example is $\int_{-3}^{3} x^5\, dx$, which it computes the long way as $\frac{3^6}{6} - \frac{(-3)^6}{6} = 0$ before asking you to see why the answer had to be zero (p.121). Under a four-minute clock, seeing it beats computing it.

The habit to build: before integrating anything with limits, check two things. Is the interval symmetric about zero, or about the midpoint of a periodic hump? And is the integrand odd or even? If both line up, you are done. A quick test for a polynomial: only odd powers of $x$ makes it odd, only even powers (including a constant) makes it even, and a mixture is neither, in which case split it and apply the rule to the odd part.

Area between two curves

A favourite TMUA construction is the area enclosed between two curves, or between a curve and a line. The method is clean once you see it: the enclosed area is the integral of the upper function minus the lower function, taken between the $x$ values where they meet. In symbols, if $y = f(x)$ lies above $y = g(x)$ on the interval, the area is $\int_a^b \left( f(x) - g(x) \right) dx$.

The two limits $a$ and $b$ are almost always the intersection points of the curves, so the first move is usually to solve $f(x) = g(x)$ to find where they cross. Because the test is multiple choice with no calculator, the numbers are chosen to stay clean, so the intersections come out as tidy values and the integral evaluates neatly. If your working is producing ugly fractions you have probably slipped, the same signal that recurs throughout the calculator-free paper.

The beauty of the upper-minus-lower method is that it sidesteps the sign problem entirely. Even if part of the region sits below the $x$-axis, subtracting the lower curve from the upper curve gives a positive height everywhere across the interval, so the integral comes out as a genuine positive area without any special handling. Getting comfortable with reading which curve is on top, perhaps by testing a single $x$ value between the intersections, is the one habit that makes these questions routine. A structured plan that builds that fluency is laid out in our guide to preparing for the TMUA.

Common traps to avoid

Integration marks are usually lost to small, avoidable errors rather than to genuine difficulty. The ones that recur most:

• Forgetting the constant of integration. An indefinite integral must end in $+ c$. It vanishes for a definite integral, but leaving it off an indefinite answer is a careless mark lost.
• Mishandling area below the axis. A region beneath the $x$-axis returns a negative integral, so a single definite integral across a curve that crosses the axis can understate the true area. Split the integral at the crossing point, or use the upper-minus-lower method.
• Forgetting to rewrite first. A term like $\frac{1}{x^2}$ must become $x^{-2}$ before the rule applies, exactly as in differentiation. Trying to integrate it in fraction form is where mistakes start.
• Substituting the limits the wrong way round. A definite integral is top limit minus bottom limit, $F(b) - F(a)$. Reversing them flips the sign of your answer.
• Over-computing. If your arithmetic is getting heavy, re-read the question. Clean numbers are a signal you are on the right path, and ugly ones usually mean a slip.

None of this is hard maths, which is exactly the point. The integration on the TMUA rewards fluency and care, not advanced technique, so the fastest route to these marks is volume on real questions until reversing the power rule and the area method are reflexes.

Frequently asked questions