What are area between two curves?

If the curve y = f(x) lies above y = g(x) on [a, b], the enclosed area is

What are wrong order in area between curves?

Use upper minus lower; reversing gives a negative result.

Evaluate _1^3 (2x + 1)\,dx. [2 marks]

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

The curve meets the x-axis where 4 - x^2 = 0, that is x = 2. Area = _-2^2 (4 - x^2)\,dx = [4x - x^33]_-2^2. At x = 2: 8 - 83 = 163. At x = -2: -8 + 83 = -163. Area = 163 - (-163) = 323. Markers reward finding the limits from the intercepts, integrating, and the correct positive area.

SEAB Original-style practice: Find the area enclosed between the curves y = x^2 and y = 2x.

Intersections: x^2 = 2x, so x^2 - 2x = 0, x(x - 2) = 0, giving x = 0 and x = 2. On 0 < x < 2, the line y = 2x is above the parabola y = x^2. Area = _0^2 (2x - x^2)\,dx = [x^2 - x^33]_0^2 = (4 - 83) - 0 = 43. Markers reward the intersection points, integrating (upper minus lower), and the enclosed area 43.

SingaporeMathsSyllabus dot point

How does the definite integral measure area, and how do we handle areas below the axis or between curves?

Evaluate definite integrals, use them to find the area under a curve and between curves, and apply the fundamental theorem of calculus

A focused answer to the H2 Mathematics outcome on definite integrals and area. The fundamental theorem, evaluating definite integrals, signed area below the axis, and area between two curves.

Generated by Claude Opus 4.810 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 evaluate definite integrals, use them to find areas under a curve and between curves, handle regions below the $x$ -axis correctly, and understand the fundamental theorem of calculus that links integration with the antiderivative.

The answer

The fundamental theorem

If $\mathrm{F}$ is an antiderivative of $\mathrm{f}$ (so $\mathrm{F}' = \mathrm{f}$ ), then

\int_a^b \mathrm{f}(x)\,dx = \mathrm{F}(b) - \mathrm{F}(a).

This converts the area-summing definite integral into evaluating an antiderivative at the two limits. No constant of integration is needed because it cancels.

Area under a curve

For $\mathrm{f}(x) \geq 0$ on $[a, b]$ , the area between the curve and the $x$ -axis is $\displaystyle\int_a^b \mathrm{f}(x)\,dx$ .

Regions below the axis

Where $\mathrm{f}(x) < 0$ , the integral gives a negative value (signed area). For a true geometric area you must split at the $x$ -intercepts and take the modulus of each piece, then add. Integrating straight through a sign change underestimates the area because positive and negative parts cancel.

Area between two curves

If the curve $y = \mathrm{f}(x)$ lies above $y = \mathrm{g}(x)$ on $[a, b]$ , the enclosed area is

\int_a^b \big(\mathrm{f}(x) - \mathrm{g}(x)\big)\,dx,

(upper minus lower). Find the limits from the intersection points and decide which curve is on top.

Worked example

Find the total area between the curve $y = x^3 - x$ and the $x$ -axis from $x = -1$ to $x = 1$ .

Step 1: Find where the curve crosses the axis

$x^3 - x = x(x - 1)(x + 1) = 0$ at $x = -1, 0, 1$ . The curve changes sign at $x = 0$ inside the interval.

Step 2: Split the integral at the sign change

On $-1 < x < 0$ the curve is above the axis (positive); on $0 < x < 1$ it is below (negative).

Step 3: Integrate each piece

The antiderivative is $\dfrac{x^4}{4} - \dfrac{x^2}{2}$ .

From $-1$ to $0$ : $\left[0\right] - \left[\dfrac{1}{4} - \dfrac{1}{2}\right] = 0 - \left(-\dfrac{1}{4}\right) = \dfrac{1}{4}$ .

From $0$ to $1$ : $\left[\dfrac{1}{4} - \dfrac{1}{2}\right] - 0 = -\dfrac{1}{4}$ .

Step 4: Take moduli and add

Total area $= \left|\dfrac{1}{4}\right| + \left|-\dfrac{1}{4}\right| = \dfrac{1}{2}$ .

Integrating straight through would have given $0$ , the wrong area.

Examples in context

Example 1. Distance from velocity. Integrating a velocity-time function over an interval gives the displacement; taking moduli of the pieces where velocity is negative gives the total distance travelled, the physical version of signed-versus-geometric area.

Example 2. Consumer surplus. In economics the area between a demand curve and the price line is consumer surplus, computed as a definite integral of upper minus lower, a direct use of area between curves.

Try this

Q1. Evaluate $\displaystyle\int_1^3 (2x + 1)\,dx$ . [2 marks]

Cue. $[x^2 + x]_1^3 = 12 - 2 = 10$ .

Q2. Find the area between $y = x^2$ and the $x$ -axis from $x = 0$ to $x = 3$ . [2 marks]

Cue. $\left[\frac{x^3}{3}\right]_0^3 = 9$ .

Q3. State how to find the area enclosed between two curves. [2 marks]

Cue. Integrate (upper curve minus lower curve) between their intersection points.

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 = 4 - x^2

and the

x

-axis.

Show worked answer →

The curve meets the $x$ -axis where $4 - x^2 = 0$ , that is $x = \pm 2$ .

Area $= \displaystyle\int_{-2}^{2} (4 - x^2)\,dx = \left[4x - \dfrac{x^3}{3}\right]_{-2}^{2}$ .

At $x = 2$ : $8 - \dfrac{8}{3} = \dfrac{16}{3}$ . At $x = -2$ : $-8 + \dfrac{8}{3} = -\dfrac{16}{3}$ .

Area $= \dfrac{16}{3} - \left(-\dfrac{16}{3}\right) = \dfrac{32}{3}$ .

Markers reward finding the limits from the intercepts, integrating, and the correct positive area.

Original5 marksFind the area enclosed between the curves

y = x^2

and

y = 2x

Show worked answer →

Intersections: $x^2 = 2x$ , so $x^2 - 2x = 0$ , $x(x - 2) = 0$ , giving $x = 0$ and $x = 2$ .

On $0 < x < 2$ , the line $y = 2x$ is above the parabola $y = x^2$ .

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

Markers reward the intersection points, integrating (upper minus lower), and the enclosed area $\dfrac{4}{3}$ .