ScotlandMathsSyllabus dot point

How do you use integration to measure area and volume and to recover a quantity from its rate of change?

Using integration to find the area enclosed between a curve and a line or between two curves, the area below the x-axis, recovering displacement from velocity, and using a definite integral to evaluate an accumulated quantity in context.

A focused answer to the SQA Higher Mathematics applying integral calculus content, covering the area between a curve and a line or between two curves, the area below the x-axis, recovering displacement from velocity, and evaluating an accumulated quantity with a definite integral.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

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

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What this dot point is asking
Area between curves
Area below the axis
Recovering a quantity
Examples in context
Try this

What this dot point is asking

The SQA wants you to use integration to find the area between a curve and a line or between two curves, handle areas below the x-axis, recover displacement from velocity, and use a definite integral to evaluate an accumulated quantity in a real context.

Area between curves

The key idea is that a definite integral measures signed area between a curve and the x-axis. To find the area trapped between two curves you subtract: the upper boundary minus the lower boundary, integrated between the x-values where the boundaries cross. Finding those crossing points is half the work, so solve the two equations simultaneously before you integrate anything.

Find the area enclosed between

y = 6x - x^2

and the x-axis

Find the area of the region enclosed between the curve $y = 6x - x^2$ and the x-axis.

Step 1: Find the limits by solving for the x-axis crossings

The region is bounded below by the x-axis (where $y = 0$ ), so the limits are the $x$ -values where the curve meets it. Set $y = 0$ :

6x - x^2 = 0 \implies x(6 - x) = 0 \implies x = 0 \text{ or } x = 6

These are the limits of integration. The curve lies above the x-axis between them (check: at $x = 3$ , $y = 18 - 9 = 9 > 0$ ), so the integral gives the area directly.

Step 2: Set up and integrate

\int_0^6 (6x - x^2)\,dx = \left[3x^2 - \dfrac{x^3}{3}\right]_0^6

Step 3: Substitute the limits and subtract

At $x = 6$ : $3(36) - \dfrac{216}{3} = 108 - 72 = 36$ . At $x = 0$ the bracket is $0$ .

36 - 0 = 36

Final answer: The enclosed area is $36$ square units.

Area between a curve and a line

Find the area enclosed between $y = 4 - x^2$ and the line $y = x + 2$ .

Step 1: Find where the curves intersect to get the limits

Set the two expressions equal and solve:

4 - x^2 = x + 2 \implies x^2 + x - 2 = 0 \implies (x + 2)(x - 1) = 0

So the intersection points are at $x = -2$ and $x = 1$ , which become the limits of integration.

Step 2: Decide which function is on top

Between the limits, test a value such as $x = 0$ : the parabola gives $y = 4$ and the line gives $y = 2$ . Since $4 > 2$ , the parabola $y = 4 - x^2$ is the upper boundary. Integrating upper minus lower prevents a negative result:

\text{integrand} = (4 - x^2) - (x + 2) = 2 - x - x^2

Step 3: Integrate and apply the limits

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

At $x = 1$ : $2 - \dfrac{1}{2} - \dfrac{1}{3} = \dfrac{12 - 3 - 2}{6} = \dfrac{7}{6}$ .

At $x = -2$ : $-4 - 2 + \dfrac{8}{3} = \dfrac{-18 + 8}{3} = -\dfrac{10}{3}$ .

\text{Area} = \dfrac{7}{6} - \left(-\dfrac{10}{3}\right) = \dfrac{7}{6} + \dfrac{20}{6} = \dfrac{27}{6} = \dfrac{9}{2} \text{ square units}

Final answer: The enclosed area is $\dfrac{9}{2}$ square units.

Area below the axis

A curve below the x-axis gives a negative integral. Split the interval at each x-axis crossing, integrate each part, and add the magnitudes so areas do not cancel. If a question asks for area it always wants the total positive amount; if it asks for the value of a definite integral, you keep the sign.

Recovering a quantity

Integrating velocity with respect to time gives displacement, and a definite integral of any rate over an interval gives the total change accumulated. Because displacement allows backward motion to cancel forward motion, total distance requires you to split at any time where the velocity is zero and add the magnitudes.

Examples in context

A tank fills at a rate of $r(t) = 8 - 2t$ litres per minute for $0 \le t \le 4$ . The total volume added is $\displaystyle\int_0^4 (8 - 2t)\,dt = \left[8t - t^2\right]_0^4 = 32 - 16 = 16$ litres. The same definite integral that measures area under the rate-time graph here measures accumulated volume, which is the central idea behind every accumulation problem.

Try this

Q1. Find the area between $y = x^2$ and $y = 2x$ from their intersections. [4 marks]

Cue. They meet at $x = 0$ and $x = 2$ ; $\displaystyle\int_0^2 (2x - x^2)\,dx = \left[x^2 - \dfrac{x^3}{3}\right]_0^2 = 4 - \dfrac{8}{3} = \dfrac{4}{3}$ .

Q2. A particle has velocity $v = 3t^2$ m/s. Find the displacement between $t = 0$ and $t = 2$ . [3 marks]

Cue. $\displaystyle\int_0^2 3t^2\,dt = [t^3]_0^2 = 8$ metres.

Q3. Find the area enclosed between $y = x^3$ and $y = x$ in the first quadrant. [4 marks]

Cue. They meet at $x = 0$ and $x = 1$ , with $y = x$ above; $\displaystyle\int_0^1 (x - x^3)\,dx = \left[\dfrac{x^2}{2} - \dfrac{x^4}{4}\right]_0^1 = \dfrac{1}{4}$ square units.

Exam-style practice questions

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

SQA Higher 20186 marksThe curve

y = x^2 - 2x

and the line

y = x

enclose a finite region. Calculate the area of this region.

Show worked answer →

Find the limits by solving for the intersections: $x^2 - 2x = x$ , so $x^2 - 3x = 0$ , giving $x(x - 3) = 0$ and $x = 0$ or $x = 3$ (2 marks).

Decide which is upper: at $x = 1$ the line gives $y = 1$ and the curve gives $1 - 2 = -1$ , so the line $y = x$ is above the curve on $[0, 3]$ (1 mark).

Area $= \displaystyle\int_0^3 \big(x - (x^2 - 2x)\big)\,dx = \int_0^3 (3x - x^2)\,dx$ (1 mark).

Integrate: $\left[\dfrac{3x^2}{2} - \dfrac{x^3}{3}\right]_0^3 = \dfrac{3(9)}{2} - \dfrac{27}{3} = \dfrac{27}{2} - 9 = \dfrac{9}{2}$ square units (2 marks). Markers reward solving for both limits, the correct upper-minus-lower integrand, accurate integration and substitution of limits.

SQA Higher 20225 marksA particle moves in a straight line with velocity

v(t) = 12 - 6t

m/s, where

t

is in seconds. Determine the total distance travelled by the particle between

t = 0

and

t = 4

Show worked answer →

Distance, unlike displacement, must not let motion in opposite directions cancel, so first find where $v = 0$ : $12 - 6t = 0$ gives $t = 2$ (1 mark). The particle moves forward on $[0, 2]$ (where $v > 0$ ) and backward on $[2, 4]$ (where $v < 0$ ).

Forward distance: $\displaystyle\int_0^2 (12 - 6t)\,dt = \left[12t - 3t^2\right]_0^2 = 24 - 12 = 12$ m (1 mark).

Backward part: $\displaystyle\int_2^4 (12 - 6t)\,dt = \left[12t - 3t^2\right]_2^4 = (48 - 48) - (24 - 12) = -12$ , magnitude $12$ m (2 marks).

Total distance $= 12 + 12 = 24$ m (1 mark). Markers reward splitting at $v = 0$ , integrating each part, and adding magnitudes rather than the signed total.

Related dot points

Sources & how we know this

SQA Higher Mathematics Course Specification — SQA (2018)