Cambridge A Level Mathematics 9709 · Pure 1

Integration

Integration is differentiation run backwards: given a gradient function, it recovers the original curve; given a curve, it measures the area underneath it. In Cambridge 9709 Pure 1 it is the final topic of the syllabus and one of the heaviest-weighted — and the examiner reports are blunt about where its marks go missing: sign errors and incomplete working, not conceptual failure.

Updated 20 August 2026

The reverse power rule

Raise the power by one, divide by the new power, add a constant:

\int ax^n\,dx = \frac{ax^{n+1}}{n+1} + c \qquad (n \neq -1)

As with differentiation, rewrite roots and fractions as powers before you start: $\sqrt{x} = x^{1/2}$ , $\tfrac{3}{x^2} = 3x^{-2}$ . The negative-power cases are where signs go wrong: $\int 3x^{-2}dx = -3x^{-1} + c$ — raising $-2$ by one gives $-1$ , and dividing by $-1$ flips the sign.

Integrating (ax + b)ⁿ

The chain rule in reverse: raise the power, divide by the new power, and divide by the bracket’s coefficient:

\int (ax+b)^n\,dx = \frac{(ax+b)^{n+1}}{a(n+1)} + c

Example: $\int (3x-2)^5\,dx = \tfrac{(3x-2)^6}{18} + c$ . The divide-by- $a$ is the single most forgotten step in P1 integration — checking by differentiating your answer takes five seconds and catches it every time.

Finding the constant of integration

When a question gives $\tfrac{dy}{dx}$ and a point on the curve, the constant is recoverable — and required:

Integrate, writing $+\,c$ .
Substitute the given point into your expression.
Solve for $c$ and state the full equation of the curve.

“Find the equation of the curve” questions are marked on exactly this chain; an answer with no $c$ (or with $c$ never evaluated) caps the marks immediately.

Definite integrals

A definite integral has limits and produces a number — substitute the top limit, substitute the bottom limit, subtract:

\int_1^3 (2x+1)\,dx = \Big[x^2+x\Big]_1^3 = (9+3) - (1+1) = 10

Write the square-bracket stage explicitly — it carries the method mark — and keep the subtraction in brackets: most sign errors on definite integrals happen when the lower limit’s value is negative and the brackets were skipped.

Areas under and between curves

The area between a curve and the $x$ -axis from $a$ to $b$ is $\int_a^b y\,dx$ . Two refinements matter in P1:

Region below the axis: the integral comes out negative; the area is its magnitude. If the region crosses the axis, split at the root and add the pieces’ absolute values.
Between two curves: $\int_a^b (y_{\text{top}} - y_{\text{bottom}})\,dx$ , with $a$ and $b$ the intersection points — found by solving the curves equal, with working shown.

Worked example

The curve y = 6x − x² and the line y = 2x intersect at the origin and at another point. Find the area enclosed between them.

Step 1 — intersections. $6x - x^2 = 2x \Rightarrow 4x - x^2 = 0 \Rightarrow x(4-x) = 0$ , so $x = 0$ and $x = 4$ .

Step 2 — top minus bottom, integrated between the intersections. On $0 < x < 4$ the curve sits above the line:

\int_0^4 \big(6x - x^2 - 2x\big)dx = \int_0^4 \big(4x - x^2\big)dx = \Big[2x^2 - \tfrac{x^3}{3}\Big]_0^4

Step 3 — evaluate. $\big(32 - \tfrac{64}{3}\big) - 0 = \tfrac{32}{3}$ square units. Leave it exact — $\tfrac{32}{3}$ , not 10.67 — unless the question says otherwise.

The mistakes that lose the marks

1.Sign errors on negative powers
$\int x^{-3}dx = \tfrac{x^{-2}}{-2} + c = -\tfrac{1}{2x^2} + c$ . Raise the power towards zero and let the division carry the sign — then differentiate back to check.
2.Forgetting to divide by the bracket's coefficient
$\int(3x-2)^5 dx$ needs division by both 6 and 3. If the inside of the bracket has a coefficient other than 1, your answer owes a division by it.
3.Dropping + c in indefinite integrals
Every indefinite integral ends $+\,c$ — and in “find the equation of the curve” questions the whole point is evaluating it. Write it the moment you integrate.
4.Treating below-axis area as negative (or ignoring the crossing)
Areas are positive. If the curve crosses the axis inside your limits, one integral straight across cancels area against itself — split at the root, take magnitudes, add.
5.Skipping the bracketed subtraction stage in definite integrals
Write $\big[F(x)\big]_a^b = F(b) - F(a)$ with both values in brackets before simplifying. The stage is a method mark, and the brackets are what stop $-(-2)$ becoming $-2$ .

Common questions

Why can't n = −1 in the reverse power rule?

Raising $-1$ by one gives zero, and dividing by zero is undefined. The integral of $x^{-1}$ is $\ln|x|$ — which belongs to Pure 3, not Pure 1, so a P1 question will never need it.

When do you use integration instead of differentiation?

Differentiate when you’re asked about gradient, tangent, normal, stationary points or rates. Integrate when you’re given a gradient and asked for the curve, or asked for an area. Examiners note candidates doing the opposite operation under time pressure — the question’s nouns tell you which way to go.

Do you need volumes of revolution in Pure 1?

Yes. Rotating a region about the $x$ -axis gives volume $V = \pi\int_a^b y^2\,dx$ (about the $y$ -axis, $\pi\int x^2\,dy$ ). Square the whole expression for $y$ before integrating — expanding that square is usually where the marks sit.

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