Areas Under Curves

This lesson covers the use of definite integration to find the area enclosed between a curve and the x-axis. This is one of the most important applications of integration at A-Level and is tested extensively on AQA papers.

The Fundamental Idea

The definite integral of a function f(x) from a to b gives the signed area between the curve y = f(x), the x-axis, and the vertical lines x = a and x = b:

∫(from a to b) f(x) dx = F(b) − F(a)

where F(x) is an antiderivative (indefinite integral) of f(x).

If the curve lies above the x-axis on [a, b], the integral gives a positive value equal to the area.

If the curve lies below the x-axis on [a, b], the integral gives a negative value.

Key Point: The integral calculates signed area (also called net area). If the question asks for the total area (which it usually does), you must handle regions below the x-axis separately and take their absolute value.

Example 1: Area Above the x-axis

Find the area under y = x² + 1 between x = 0 and x = 3.

Since x² + 1 > 0 for all x, the curve is entirely above the x-axis.

Area = ∫₀³ (x² + 1) dx
     = [x³/3 + x]₀³
     = (27/3 + 3) − (0 + 0)
     = 9 + 3
     = 12

The area is 12 square units.

Example 2: Area Below the x-axis

Find the area enclosed between y = x² − 4 and the x-axis.

First, find where the curve crosses the x-axis:

x² − 4 = 0
x = ±2

For −2 < x < 2, the curve is below the x-axis (since x² − 4 < 0).

∫₋₂² (x² − 4) dx = [x³/3 − 4x]₋₂²
= (8/3 − 8) − (−8/3 + 8)
= (8/3 − 8) − (−8/3 + 8)
= 8/3 − 8 + 8/3 − 8
= 16/3 − 16
= −32/3

The integral is negative because the region is below the x-axis. The area is:

Area = |−32/3| = 32/3 square units

Exam Tip: When the integral gives a negative answer and the question asks for "the area", take the absolute value. But if the question says "evaluate the integral", give the signed value.

Example 3: Part Above, Part Below

Find the total area enclosed between y = x³ − x and the x-axis for −1 ≤ x ≤ 1.

First, find the roots: x³ − x = x(x² − 1) = x(x − 1)(x + 1) = 0, so x = −1, 0, 1.

Check signs:

For −1 < x < 0: try x = −0.5: (−0.5)³ − (−0.5) = −0.125 + 0.5 = 0.375 > 0 (above x-axis).
For 0 < x < 1: try x = 0.5: (0.5)³ − 0.5 = 0.125 − 0.5 = −0.375 < 0 (below x-axis).

So we must split the integral:

Area = ∫₋₁⁰ (x³ − x) dx + |∫₀¹ (x³ − x) dx|

First integral:

∫₋₁⁰ (x³ − x) dx = [x⁴/4 − x²/2]₋₁⁰
= (0) − (1/4 − 1/2)
= 0 − (−1/4)
= 1/4

Second integral:

∫₀¹ (x³ − x) dx = [x⁴/4 − x²/2]₀¹
= (1/4 − 1/2) − 0
= −1/4

Total area:

Area = 1/4 + |−1/4| = 1/4 + 1/4 = 1/2

Note: if we had simply computed ∫₋₁¹ (x³ − x) dx, we would get 1/4 + (−1/4) = 0, which is incorrect as the area is clearly not zero.

Example 4: Area Under a Trigonometric Curve

Find the area under y = sin x from x = 0 to x = π.

Area = ∫₀^π sin x dx = [−cos x]₀^π
= (−cos π) − (−cos 0)
= (−(−1)) − (−1)
= 1 + 1
= 2

The area is 2 square units.

Example 5: Area Under an Exponential Curve

Find the area under y = eˣ between x = 0 and x = 2.

Area = ∫₀² eˣ dx = [eˣ]₀²
= e² − e⁰
= e² − 1
≈ 6.389

Area Between a Curve and the y-axis

Sometimes you need the area between a curve and the y-axis. In this case, express x as a function of y and integrate with respect to y:

Area = ∫(from y₁ to y₂) x dy

Example 6

Find the area between x = y², the y-axis, and the lines y = 0 and y = 3.

Area = ∫₀³ y² dy = [y³/3]₀³ = 27/3 = 9

Combining Areas

Example 7

Find the total area enclosed between y = x(x − 2)(x + 1) and the x-axis.

Roots are x = −1, 0, 2.

Sign check:

For −1 < x < 0: try x = −0.5: (−0.5)(−2.5)(0.5) = 0.625 > 0
For 0 < x < 2: try x = 1: (1)(−1)(2) = −2 < 0

Area = ∫₋₁⁰ x(x − 2)(x + 1) dx + |∫₀² x(x − 2)(x + 1) dx|

Expand: x(x − 2)(x + 1) = x(x² − x − 2) = x³ − x² − 2x.

First part:

∫₋₁⁰ (x³ − x² − 2x) dx = [x⁴/4 − x³/3 − x²]₋₁⁰
= 0 − (1/4 + 1/3 − 1)
= 0 − (3/12 + 4/12 − 12/12)
= 0 − (−5/12)
= 5/12

Second part:

∫₀² (x³ − x² − 2x) dx = [x⁴/4 − x³/3 − x²]₀²
= (16/4 − 8/3 − 4) − 0
= (4 − 8/3 − 4)
= −8/3

Total area = 5/12 + 8/3 = 5/12 + 32/12 = 37/12 square units.

Common Mistakes

Not splitting the integral when the curve crosses the x-axis — this leads to cancellation and an incorrect answer.
Forgetting absolute values for regions below the x-axis.
Wrong limits of integration — always check where the curve meets the x-axis.
Arithmetic errors in evaluating definite integrals — show all substitution steps clearly.

Summary

The definite integral ∫ₐᵇ f(x) dx gives the signed area between the curve, the x-axis, and x = a, x = b.
Regions above the x-axis contribute positive area; regions below contribute negative area.
For total area, split the integral at roots and take absolute values of negative parts.
For area between a curve and the y-axis, express x in terms of y and integrate with respect to y.

Exam Tip: When finding areas, always start by sketching the curve (or at least identifying the roots) to determine which regions are above and below the x-axis. Show clearly how you split the integral. A common AQA question gives you a curve that crosses the x-axis and asks for the "total area enclosed" — this is testing whether you know to split and take absolute values.

A-Level Deep Dive: Areas Under Curves

Spec mapping

AQA 7357 specification, Paper 1 — Pure Mathematics, Section H — Integration covers evaluate definite integrals; use a definite integral to find the area under a curve and the area between two curves (refer to the official specification document for exact wording). Areas under curves sit at the heart of integration on AQA 7357. The definite integral $\int_a^b f(x)\,dx$ is defined (informally at A-Level, rigorously at university) as the signed area between $y = f(x)$ and the $x$ -axis on $[a, b]$ . Section H is examined on Paper 1 (Pure) and connects structurally to Section J — Numerical methods (trapezium rule for approximating area), Section O — Mechanics (displacement = $\int v\,dt$ , the area under a velocity-time graph) and Section P — Statistics (probabilities from continuous distributions are areas under PDFs, $P(a \leq X \leq b) = \int_a^b f(x)\,dx$ ). The AQA formula booklet lists standard integrals but not the definition of area or the linearity properties — those must be known. Three core properties underpin every Paper 1 area question: linearity $\int [\alpha f + \beta g]\,dx = \alpha \int f\,dx + \beta \int g\,dx$ , reversal $\int_a^b f = -\int_b^a f$ , and additivity $\int_a^c f = \int_a^b f + \int_b^c f$ for any $b \in [a, c]$ .

Worked example with full mark scheme

Question (8 marks):

The curve $C$ has equation $y = 3x^2 - \dfrac{4}{\sqrt{x}} + 2$ for $x > 0$ . Find the exact area of the region bounded by $C$ , the $x$ -axis and the lines $x = 1$ and $x = 4$ . (8)

Solution with mark scheme:

Step 1 — set up the definite integral.

The required area is

$A = \int_1^4 \left(3x^2 - \dfrac{4}{\sqrt{x}} + 2\right) dx$

M1 — correct setup of a definite integral with the given integrand and limits $1$ and $4$ . Candidates who reverse the limits ( $\int_4^1$ ) or omit the constant term $+2$ lose this mark.

Step 2 — rewrite $\dfrac{4}{\sqrt{x}}$ in index form.

$\dfrac{4}{\sqrt{x}} = 4x^{-1/2}$ .

M1 — converting the surd term to a fractional index ready for the power rule. Leaving the term as $\dfrac{4}{\sqrt{x}}$ blocks the power rule entirely.

Step 3 — integrate term by term.

Apply the power rule $\int x^n\,dx = \dfrac{x^{n+1}}{n+1} + C$ (for $n \neq -1$ ):

$\int 3x^2\,dx = x^3, \qquad \int 4x^{-1/2}\,dx = 4 \cdot \dfrac{x^{1/2}}{1/2} = 8x^{1/2} = 8\sqrt{x}, \qquad \int 2\,dx = 2x$

$\int \left(3x^2 - 4x^{-1/2} + 2\right) dx = x^3 - 8\sqrt{x} + 2x \; (+ C)$

M1 — correct integration of $3x^2$ to $x^3$ .

M1 — correct integration of $-4x^{-1/2}$ to $-8\sqrt{x}$ (the index becomes $1/2$ , and dividing by $1/2$ multiplies the coefficient by $2$ ).

A1 — fully correct antiderivative $x^3 - 8\sqrt{x} + 2x$ . The constant of integration is dropped because we are about to evaluate between limits.

Step 4 — evaluate at the limits.

At $x = 4$ : $4^3 - 8\sqrt{4} + 2 \cdot 4 = 64 - 16 + 8 = 56$ .

At $x = 1$ : $1^3 - 8\sqrt{1} + 2 \cdot 1 = 1 - 8 + 2 = -5$ .

M1 — substituting both limits into the antiderivative.

Step 5 — subtract.

$A = 56 - (-5) = 61$

A1 — exact value $A = 61$ .

A1 — answer presented as an exact integer (square units) matching the question's "exact" instruction.

Total: 8 marks (M5 A3, split as shown).

Specimen question modelled on the AQA 7357 Paper 1 format

Question (6 marks): The curve $y = x^2 - 4x + 3$ crosses the $x$ -axis at $x = 1$ and $x = 3$ .

(a) Sketch the curve, indicating the intercepts. (2)

(b) Find the exact area of the region enclosed between the curve and the $x$ -axis on $1 \leq x \leq 3$ . (4)

Mark scheme decomposition by AO:

(a)

B1 (AO1.1b) — parabola opening upwards with $x$ -intercepts at $1$ and $3$ .
B1 (AO2.5) — $y$ -intercept at $3$ marked, minimum at $x = 2$ (where $y = -1$ ) shown below the $x$ -axis.

(b)

M1 (AO1.1a) — recognising that on $1 \leq x \leq 3$ the curve lies below the $x$ -axis, so $\int_1^3 (x^2 - 4x + 3)\,dx$ will be negative and the area is the absolute value.
M1 (AO1.1b) — correct antiderivative $\dfrac{x^3}{3} - 2x^2 + 3x$ .
A1 (AO1.1b) — evaluating: at $x = 3$ , $9 - 18 + 9 = 0$ ; at $x = 1$ , $\tfrac{1}{3} - 2 + 3 = \tfrac{4}{3}$ . So $\int_1^3 = 0 - \tfrac{4}{3} = -\tfrac{4}{3}$ .
A1 (AO2.5) — area = $\left|-\tfrac{4}{3}\right| = \tfrac{4}{3}$ square units, with a written justification that the negative sign indicates the region lies below the axis.

Total: 6 marks split AO1 = 4, AO2 = 2. The AO2 marks reward the interpretation — knowing that a negative integral signals "below the axis" rather than a calculation slip.

Synoptic links

Connects to:

Section H — Fundamental Theorem of Calculus: the FTC states $\int_a^b f(x)\,dx = F(b) - F(a)$ where $F'(x) = f(x)$ . This is why areas can be computed without summing infinitesimal rectangles by hand. Every definite integral on Paper 1 silently invokes the FTC.
Section H — Area between two curves: if $f(x) \geq g(x)$ on $[a, b]$ , the area between the curves is $\int_a^b [f(x) - g(x)]\,dx$ . Identifying which curve is "upper" requires sketching or solving $f = g$ for the intersection points first — a synoptic skill drawing on Section B (algebra of equations).
Section J — Trapezium rule: when an integrand has no elementary antiderivative (e.g. $\int_0^1 e^{-x^2}\,dx$ ), the trapezium rule approximates the area numerically as $\tfrac{h}{2}[y_0 + 2(y_1 + y_2 + \cdots + y_{n-1}) + y_n]$ . Trapezium-rule overestimation/underestimation is judged by the concavity of the integrand — synoptic with Section G (second derivatives).
Section O — Mechanics (displacement and impulse): for a particle moving along a line, $\text{displacement} = \int v(t)\,dt$ and $\text{impulse} = \int F(t)\,dt$ . The "area under the velocity-time graph" interpretation is exactly this. Negative velocity (motion in the opposite direction) gives signed area below the $t$ -axis, just like the pure-maths interpretation.
Section P — Statistics (continuous distributions): for a continuous random variable $X$ with PDF $f(x)$ , $P(a \leq X \leq b) = \int_a^b f(x)\,dx$ . The total area under any PDF is $1$ . Normalising constants are computed by demanding $\int_{-\infty}^{\infty} f(x)\,dx = 1$ .

Mark-scheme literacy

Areas-under-curves questions on AQA 7357 Paper 1 split AO marks roughly as follows:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	60–70%	Setting up the integral, applying the power rule, evaluating at limits, subtracting correctly
AO2 (reasoning / interpretation)	20–30%	Identifying when to take absolute values (region below axis), splitting the integral where the integrand changes sign, justifying the choice of limits from a sketch
AO3 (problem-solving)	5–15%	Modelling questions where the integrand encodes a physical quantity; multi-stage problems combining curve sketching, intersection, and area

Areas Under Curves

Areas Under Curves

The Fundamental Idea

Example 1: Area Above the x-axis

Example 2: Area Below the x-axis

Example 3: Part Above, Part Below

Example 4: Area Under a Trigonometric Curve

Example 5: Area Under an Exponential Curve

Area Between a Curve and the y-axis

Example 6

Combining Areas

Example 7

Common Mistakes

Summary

A-Level Deep Dive: Areas Under Curves

Spec mapping

Worked example with full mark scheme

Specimen question modelled on the AQA 7357 Paper 1 format

Synoptic links

Mark-scheme literacy

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