Evaluate _1^2 6x\,dx. [2 marks]

A curve has (dy)/(dx) = 2x + 1 and passes through (0, 5). Find y. [3 marks]

CCEA CCEA 2019-style practice: Find the area of the region bounded by the curve y = 4x - x^2 and the x-axis.

The curve meets the x-axis where 4x - x^2 = 0, that is x(4 - x) = 0, so x = 0 and x = 4. The area is _0^4 (4x - x^2)\,dx = [2x^2 - (x^3)/(3)]_0^4. At x = 4: 2(16) - (64)/(3) = 32 - 21.33 = 10.67. At x = 0: 0. So the area is (32)/(3) ≈ 10.7 square units. Markers reward the limits from the roots, the correct integration, substituting both limits, and the final area.

Northern IrelandMathsSyllabus dot point

How does integration reverse differentiation and let us find areas under curves?

Integration as the reverse of differentiation, indefinite integrals with a constant of integration, the definite integral and its evaluation, and finding the area under a curve.

A CCEA A-Level Mathematics answer on integration as the reverse of differentiation, indefinite integrals of powers of x with the constant of integration, evaluating definite integrals, and finding the area between a curve and the x-axis including regions below the axis.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-16

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

CCEA wants you to treat integration as the reverse of differentiation, find indefinite integrals of powers of $x$ (never forgetting the constant of integration), evaluate definite integrals between limits, and use the definite integral to find the area between a curve and the $x$ -axis. This completes the AS calculus and pairs directly with the differentiation dot point.

The answer

Integration as reverse differentiation

Integrating powers of x

To find a particular function from its gradient function, integrate and then use a known point (a boundary condition) to fix the constant $c$ .

The definite integral and area

Areas below the axis

Where a curve dips below the $x$ -axis, the definite integral over that part is negative, because the signed area is below zero. To find the total geometric area, integrate each region separately between the relevant roots, take the modulus of any negative result, and add the magnitudes.

Worked example: a region in two parts

Find a total area split by the axis

Find the total area enclosed between $y = x^3 - x$ and the $x$ -axis.

step Find the roots

x^3 - x = x(x^2 - 1) = x(x - 1)(x + 1) = 0 \Rightarrow x = -1, 0, 1.

The curve is above the axis on

[-1, 0]

and below on

[0, 1]

(by symmetry, an odd function).

step Integrate over the first region

\int_{-1}^{0}(x^3 - x)\,dx = \left[\frac{x^4}{4} - \frac{x^2}{2}\right]_{-1}^{0} = 0 - \left(\frac{1}{4} - \frac{1}{2}\right) = \frac{1}{4}.

step Integrate over the second region and combine

\int_{0}^{1}(x^3 - x)\,dx = \left[\frac{x^4}{4} - \frac{x^2}{2}\right]_{0}^{1} = \frac{1}{4} - \frac{1}{2} = -\frac{1}{4}.

Taking the modulus, each region has area

\frac{1}{4}

, so the total area is

\frac{1}{4} + \frac{1}{4} = \frac{1}{2}

square units.

Examples in context

Example 1. Distance from velocity. If a body's velocity is $v = 3t^2$ , the distance travelled between $t = 0$ and $t = 2$ is $\int_0^2 3t^2\,dt = [t^3]_0^2 = 8\,\text{m}$ . The definite integral converts a rate into a total, the inverse of differentiating displacement.

Example 2. Area between curve and line. To find the area enclosed between a curve and a chord, integrate the difference "upper minus lower" between their intersection points. This is the same definite-integral idea, extended to two boundaries, and recurs throughout applied calculus.

Try this

Q1. Find $\int (4x^3 + 2)\,dx$ . [2 marks]

Cue. $x^4 + 2x + c$ .

Q2. Evaluate $\int_1^2 6x\,dx$ . [2 marks]

Cue. $[3x^2]_1^2 = 12 - 3 = 9$ .

Q3. A curve has $\frac{dy}{dx} = 2x + 1$ and passes through $(0, 5)$ . Find $y$ . [3 marks]

Cue. $y = x^2 + x + c$ ; using $(0, 5)$ gives $c = 5$ , so $y = x^2 + x + 5$ .

Exam-style practice questions

Practice questions written in the style of CCEA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

CCEA 20196 marksFind the area of the region bounded by the curve

y = 4x - x^2

and the

x

-axis.

Show worked answer →

The curve meets the $x$ -axis where $4x - x^2 = 0$ , that is $x(4 - x) = 0$ , so $x = 0$ and $x = 4$ .

The area is

$\int_0^4 (4x - x^2)\,dx = \left[2x^2 - \frac{x^3}{3}\right]_0^4.$

At $x = 4$ : $2(16) - \frac{64}{3} = 32 - 21.33 = 10.67$ . At $x = 0$ : $0$ .

So the area is $\frac{32}{3} \approx 10.7$ square units.

Markers reward the limits from the roots, the correct integration, substituting both limits, and the final area.

CCEA 20215 marksThe curve

y = f(x)

passes through the point

(1, 4)

and has gradient function

\frac{dy}{dx} = 6x^2 - 2x

. Find

f(x)

Show worked answer →

Integrate the gradient function:

$f(x) = \int (6x^2 - 2x)\,dx = 2x^3 - x^2 + c.$

Use the point $(1, 4)$ to find $c$ :

$4 = 2(1)^3 - (1)^2 + c = 2 - 1 + c = 1 + c,$

so $c = 3$ and $f(x) = 2x^3 - x^2 + 3$ .

Markers reward integrating each term, including the constant of integration, substituting the point, and the final expression.

Related dot points

Sources & how we know this

CCEA GCE Mathematics specification — CCEA (2018)