Without a sketch you cannot see where the curve is below the axis or where it crosses; a quick sketch prevents most errors.

What are wrong limits?

When the region is bounded by where the curve meets the axis, find those points by solving y = 0.

SEAB Original-style practice: Find the area between the curve y = x^2 - 4 and the x-axis from x = 0 to x = 2.

On 0 x 2 the curve lies below the axis (since x^2 - 4 < 0 there), so the integral is negative. _0^2 (x^2 - 4)\, dx = [x^33 - 4x]_0^2 = (83 - 8) - 0 = -163. Area is the magnitude: 163 square units. What markers reward: recognising the region is below the axis, computing the integral as -163, and taking the positive magnitude for the area.

SingaporeAdditional MathematicsSyllabus dot point

How do we use a definite integral to find the area between a curve and the x-axis?

Find the area of a region bounded by a curve and the x-axis using a definite integral, handling regions below the axis

A focused answer to the N(A)-Level Additional Mathematics outcome on areas. Use a definite integral to find the area between a curve and the x-axis, and handle regions that lie below the axis correctly.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-06

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

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What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

SEAB wants you to find the area of a region between a curve and the $x$ -axis using a definite integral. The integral of a function between two limits gives the signed area: positive where the curve is above the axis and negative where it is below. You need to set up the right integral, evaluate it, and handle below-axis regions by taking the magnitude. A sketch is essential for getting the signs right.

The answer

Area as a definite integral

For a curve that lies above the $x$ -axis between $x = a$ and $x = b$ , the area enclosed by the curve, the axis and the two vertical lines is:

\text{Area} = \int_a^b y\, dx

You integrate the equation of the curve and evaluate between the limits, exactly as for any definite integral.

Signed area and regions below the axis

The definite integral gives a signed result. Where the curve dips below the axis, $y$ is negative, so the integral over that part comes out negative. Area itself is always positive, so for a below-axis region you compute the integral and then take its magnitude (drop the minus sign).

When the curve crosses the axis

If the region spans a point where the curve crosses the axis, the above-axis and below-axis parts would partly cancel if you integrated straight through. To get the total area, split the integral at the crossing point, evaluate each part, and add the magnitudes. This is why a sketch matters: it shows where the curve crosses and which parts are below.

A reliable procedure

Sketch the curve and shade the region.
Find the limits (often where the curve meets the axis, by solving $y = 0$ ).
Integrate and evaluate between the limits.
Take magnitudes for any below-axis parts and add.

Worked example

Find the total area enclosed between the curve $y = x^2 - 1$ and the $x$ -axis from $x = 0$ to $x = 2$ .

Step 1: Sketch and locate the crossing

$y = x^2 - 1$ crosses the axis where $x^2 - 1 = 0$ , that is $x = 1$ (taking the value in range). On $0 \le x < 1$ the curve is below the axis; on $1 < x \le 2$ it is above.

Step 2: Integrate the below-axis part (0 to 1)

\int_0^1 (x^2 - 1)\, dx = \left[\frac{x^3}{3} - x\right]_0^1 = \left(\frac{1}{3} - 1\right) - 0 = -\frac{2}{3}

Magnitude: $\dfrac{2}{3}$ .

Step 3: Integrate the above-axis part (1 to 2)

\int_1^2 (x^2 - 1)\, dx = \left[\frac{x^3}{3} - x\right]_1^2 = \left(\frac{8}{3} - 2\right) - \left(\frac{1}{3} - 1\right) = \frac{2}{3} - \left(-\frac{2}{3}\right) = \frac{4}{3}

Step 4: Add the magnitudes

\text{Total area} = \frac{2}{3} + \frac{4}{3} = 2 \text{ square units}

Examples in context

Example 1. Distance from a velocity-time graph. The area under a velocity-time graph is the distance travelled. When the velocity is negative (motion in the opposite direction), the area lies below the axis, which is exactly the signed-area idea applied in kinematics.

Example 2. Work done by a varying force. In physics, the area under a force-displacement graph gives the work done. Whenever a quantity is the product of two others and one varies, the area under the graph (a definite integral) gives the total, linking this skill to applied problems.

Try this

Q1. Write the integral for the area under $y = x^2$ between $x = 1$ and $x = 4$ . [1 mark]

Cue. $\displaystyle\int_1^4 x^2\, dx$ .

Q2. Find the area under $y = 2x$ from $x = 0$ to $x = 3$ . [2 marks]

Cue. $[x^2]_0^3 = 9$ square units.

Q3. The integral of a curve over a region gives $-5$ . State the area of the region. [1 mark]

Cue. The region is below the axis; the area is the magnitude, $5$ square units.

Exam-style practice questions

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

Original4 marksFind the area of the region bounded by the curve

y = x^2

, the

x

-axis, and the lines

x = 0

and

x = 3

Show worked answer →

The curve is above the axis on this interval, so the area is $\displaystyle\int_0^3 x^2\, dx$ .

$\left[\dfrac{x^3}{3}\right]_0^3 = \dfrac{27}{3} - 0 = 9$ .

The area is $9$ square units.

What markers reward: setting up the definite integral with the correct limits, integrating $x^2$ to $\tfrac{x^3}{3}$ , and evaluating to $9$ .

Original5 marksFind the area between the curve

y = x^2 - 4

and the

x

-axis from

x = 0

x = 2

Show worked answer →

On $0 \le x \le 2$ the curve lies below the axis (since $x^2 - 4 < 0$ there), so the integral is negative.

$\displaystyle\int_0^2 (x^2 - 4)\, dx = \left[\dfrac{x^3}{3} - 4x\right]_0^2 = \left(\dfrac{8}{3} - 8\right) - 0 = -\dfrac{16}{3}$ .

Area is the magnitude: $\dfrac{16}{3}$ square units.

What markers reward: recognising the region is below the axis, computing the integral as $-\tfrac{16}{3}$ , and taking the positive magnitude for the area.