Year 12 Maths Extension 1 (2027) Further applications of calculus

Areas between curves

Q: How do you find the area between two curves?

Integrate top minus bottom, the integral from a to b of f minus g dx, with limits from where the curves intersect.

Q: What techniques appear in Extension 1 areas?

Inverse-trig standard forms, double-angle identities for sin squared and cos squared, and substitution.

Q: How do you get the limits?

Solve f of x equals g of x to find the intersection points, or use the given bounds.

Q: Which integrand gives an inverse trig area?

A root one over the square root of a squared minus x squared gives arcsin; a sum one over a squared plus x squared gives arctan.

20 practice questions 2 video lessons Theory + worked examples

Learn to find areas between curves for NSW Year 12 Mathematics Extension 1. The region enclosed by two curves is found by integrating the difference between the upper and lower functions.

You will learn to locate the points of intersection, set up the definite integral for the bounded region, and evaluate it in both real-life and abstract contexts — an important application of integration in the Extension 1 course.

Practice 20 questions

Practice questions

Every question with a fully worked solution.

Start practising

Watch 2 video(s)

Area Between Two Curves Watch
Calculus 1 Lecture 5.1: Finding Area Between Two Curves Watch

Theory

The area between a top curve $f$ and a bottom curve $g$ is $\int_{a}^{b} (f - g) d x$ , with limits from their intersections. This NSW Year 12 Mathematics Extension 1 topic (NESA outcome ME1-12-05) often needs an inverse-trig form, a double-angle identity, or a substitution.

To find an area, integrate the height of the region. The area under $y = f (x)$ from $a$ to $b$ is $\int_{a}^{b} f (x) d x$ ; the area between a top curve $f$ and a bottom curve $g$ is $\int_{a}^{b} (f (x) - g (x)) d x$ .

In Extension 1 the integrand often needs an inverse-trig standard form, a double-angle identity, or a substitution before it can be integrated.

Limits come from the intersections of the curves, and an enclosed area is positive when you subtract bottom from top.

NESA link. Part of the Year 12 Further applications of calculus focus area, outcome ME1-12-05 ("applies calculus to solve problems involving polynomials, further rates of change, areas and volumes and differential equations") with MAO-WM-01. This is part of the Areas between curves and volumes of solids of revolution focus area.

Area

= \int_{a}^{b} (f - g) d x

, limits at the intersections.

\int_{0}^{3} \frac{d x}{9 + x^{2}} = \frac{π}{12}

— an inverse-trig area.

Area = \int_{a}^{b} (f (x) - g (x)) d x (top - bottom) .

Area = integral a to b of (top - bottom) dx

Common Extension 1 integrands and their tools:

Integrand	Tool
$\frac{1}{\sqrt{a^{2} - x^{2}}}$	$\sin^{- 1} \frac{x}{a}$
$\frac{1}{a^{2} + x^{2}}$	$\frac{1}{a} \tan^{- 1} \frac{x}{a}$
$\sin^{2} x, \cos^{2} x$	double-angle identity
$f (g (x)) g^{'} (x)$	substitution $u = g (x)$

How to find an area

Find the limits — the intersections of the curves (or the given bounds).
Set up top minus bottom: $\int_{a}^{b} (f - g) d x$ .
Choose the technique from the integrand: inverse-trig form, double-angle, or substitution.
Integrate and evaluate for a positive exact area.

Example 1 — Inverse-trig area

Find the area bounded by

y = \frac{1}{9 + x^{2}}

, the

x

-axis,

x = 0

and

x = 3

Solution

$\int_{0}^{3} \frac{d x}{9 + x^{2}}$	$=$	$\frac{1}{3} [\tan^{- 1} \frac{x}{3}]_{0}^{3}$
	$=$	$\frac{1}{3} \cdot \frac{π}{4} = \frac{π}{12}$

area = pi/12

Area $= \frac{π}{12}$ .

Example 2 — Double-angle area

Find the area under

y = \cos^{2} x

from

x = 0

x = \frac{π}{4}

Solution

$\cos^{2} x = \frac{1}{2} (1 + \cos 2 x)$ .

$\int_{0}^{π / 4} \cos^{2} x d x$	$=$	$\frac{1}{2} [x + \frac{1}{2} \sin 2 x]_{0}^{π / 4}$
	$=$	$\frac{π}{8} + \frac{1}{4}$

area = pi/8 + 1/4

Area $= \frac{π}{8} + \frac{1}{4}$ .

Example 3 — Between two curves

Find the area between

y = \cos^{2} x

and

y = \sin^{2} x

from

x = 0

x = \frac{π}{6}

Solution

$\cos^{2} x - \sin^{2} x = \cos 2 x$ .

$\int_{0}^{π / 6} \cos 2 x d x$	$=$	$[\frac{1}{2} \sin 2 x]_{0}^{π / 6}$
	$=$	$\frac{1}{2} \sin \frac{π}{3} = \frac{\sqrt{3}}{4}$

area = sqrt(3)/4

Area $= \frac{\sqrt{3}}{4}$ .

Example 4 — By substitution

Using

u = x^{2} + 4

, find the area under

y = x \sqrt{x^{2} + 4}

from

x = 0

x = \sqrt{5}

Solution

$d u = 2 x d x$ ; $x : 0 \to \sqrt{5}$ gives $u : 4 \to 9$ .

$\int_{0}^{\sqrt{5}} x \sqrt{x^{2} + 4} d x$	$=$	$\frac{1}{2} \int_{4}^{9} \sqrt{u} d u$
	$=$	$\frac{1}{3} (27 - 8) = \frac{19}{3}$

area = 19/3

Area $= \frac{19}{3}$ .

Common pitfalls

Top minus bottom. Between curves, subtract bottom from top with limits from their intersections.

Wrong technique. Read the integrand: root form gives

\sin^{- 1}

, a plain sum gives

\tan^{- 1}

, squared trig needs a double angle.

Sign of area. An enclosed area is positive — integrate top minus bottom, not the reverse.

Missing intersections. Find where the curves meet before setting the limits.

Frequently asked questions

How do you find the area between two curves?

Integrate top minus bottom, $\int_{a}^{b} (f - g) d x$ , with limits from where the curves intersect.

What techniques appear in Extension 1 areas?

Inverse-trig standard forms, double-angle identities for $\sin^{2}$ / $\cos^{2}$ , and substitution.

How do you get the limits?

Solve $f (x) = g (x)$ to find the intersection points, or use the given bounds.

Which integrand gives an inverse trig area?

A root $\frac{1}{\sqrt{a^{2} - x^{2}}}$ gives $\sin^{- 1}$ ; a sum $\frac{1}{a^{2} + x^{2}}$ gives $\tan^{- 1}$ .

Is area always positive?

Yes when you integrate top minus bottom over the enclosed region.

← Previous subtopic

Rate ∝ (Q−P); Newton’s Law of Cooling; Q = P + Ae^{kt}

Next subtopic →

Volumes of solids of revolution

More in Further applications of calculus

Areas between curves

📖 Theory

How to find an area

Common pitfalls

Frequently asked questions

Theory