1.8 - Areas between curves | Cambridge A-Level Mathematics

An integral adds signed contributions, whereas a geometrical area is non-negative. You will split a region when its boundary crosses an axis, find intersections before subtracting curves, and choose vertical or horizontal strips to calculate an enclosed area.

Signed area and total area

The integral $\int_0^3(x-1)\,dx$ is $3/2$ . Is that the area between the line and the $x$ -axis? A sketch reveals a small triangle below the axis as well as a larger one above it. Their contributions subtract in the integral but add in the total area.

A thin vertical strip has width $dx$ . Above the axis, its height is $y$ , so its positive area is approximately $y\,dx$ . Below the axis, the positive height is $-y$ , so the area contribution is approximately $-y\,dx$ .

Signed integral and geometrical area

For increasing limits $a<b$ , $\int_a^b f(x)\,dx$ equals area above the $x$ -axis minus area below it. The total area between the curve and the axis adds the magnitudes of all these pieces. Split the interval at every crossing where the sign changes.

If a curve stays below the axis throughout an interval, negate its integral. If it changes sign, do this separately on each piece. Taking the magnitude of one combined integral can lose area through cancellation.

Worked example 1: a region wholly above the axis

Question. Find the area bounded by $y=x-1$ , the $x$ -axis and the lines $x=1$ , $x=3$ .

Plan. Check the sign on the interval, then integrate the positive height.

For $1\le x\le3$ , $x-1\ge0$ . Therefore

A=\int_1^3(x-1)\,dx=\left[\frac{x^2}{2}-x\right]_1^3 =\left(\frac92-3\right)-\left(\frac12-1\right)=2.

Answer. $2$ square units.

Check. The region is a triangle with base $2$ and height $2$ , so its area is $\frac12(2)(2)=2$ .

Move only the left boundary to $x=0$ . The new piece lies below the axis and must be added with its positive area.

Worked example 2: a region crossing the axis

Question. Find the total area between $y=x-1$ and the $x$ -axis, from $x=0$ to $x=3$ .

Plan. Split at $x=1$ , where the line crosses the axis, and reverse the sign of the negative piece.

\begin{aligned} A&=-\int_0^1(x-1)\,dx+\int_1^3(x-1)\,dx\\ &=-\left(\frac12-1\right)+2\\ &=\frac52. \end{aligned}

Answer. $5/2$ square units. The signed integral is only $2-1/2=3/2$ .

Check. The two triangles have areas $\frac12(1)(1)=1/2$ and $2$ . Their sum is $5/2$ , as required.

The picture keeps the two positive areas visible even though the integrals have opposite signs.

[DIAGRAM: asset_name: Cancellation is not total area; asset_slug: signed-versus-total-area; caption: The two triangles add to 5 over 2 square units, while their signed contributions give 3 over 2; description: Graph of y equals x minus 1 over x equals zero to three. The triangle below the axis from zero to one is shaded clay and labelled area one half and integral minus one half. The triangle above from one to three is shaded blue and labelled area 2 and integral plus 2. The crossing at 1 and vertical boundary at 3 are marked.]
The two triangles add to 5 over 2 square units, while their signed contributions give 3 over 2

A horizontal boundary works in the same way: between $y=f(x)$ and $y=c$ , use the vertical distance $f(x)-c$ where the curve is above the line, or $c-f(x)$ where it is below. Split if their order changes.

Watch out

In example 2, $\left|\int_0^3(x-1)\,dx\right|=3/2$ , which is still wrong for the total area. Split before taking positive magnitudes.

Locate the crossing before calculating this area.

A line and a curve

Between two graphs, a vertical strip stretches from the lower boundary to the upper boundary. Its height is their difference. The intercepts with an axis are usually irrelevant: the limits come from the intersections of the two boundaries.

First solve the two equations simultaneously. Then test an input between successive intersections, or inspect a reliable sketch, to decide which graph is above the other.

Area from vertical strips

If $u(x)\ge l(x)$ throughout $a\le x\le b$ , then the area between them is

A=\int_a^b[u(x)-l(x)]\,dx.

Split into intervals if the upper and lower boundaries exchange places.

Not stated in the formula booklet: learn upper minus lower.

Subtract the complete lower expression, with brackets. This gives a positive strip height even when both graphs lie below the axis.

Worked example 3: a line through the origin

Question. Find the area enclosed by $y=2x$ and $y=x^2$ .

Plan. Find their intersections, identify the upper graph, and integrate the difference.

x^2=2x\quad\Longrightarrow\quad x(x-2)=0,

so the limits are $0$ and $2$ . At $x=1$ , the line has height $2$ and the curve height $1$ , so the line is above the curve between these intersections.

A=\int_0^2(2x-x^2)\,dx =\left[x^2-\frac{x^3}{3}\right]_0^2 =4-\frac83=\frac43.

Answer. $4/3$ square units.

Check. The strip height $2x-x^2=1-(x-1)^2$ is non-negative between $0$ and $2$ and no greater than $1$ . The answer is positive and less than the enclosing rectangle's area $2$ .

Move only the line upwards by $3$ . Both intersection limits now change, including one to the left of the origin.

Worked example 4: a shifted line

Question. Find the area enclosed by $y=2x+3$ and $y=x^2$ .

Plan. Recalculate the intersections rather than reusing the earlier limits.

x^2=2x+3\quad\Longrightarrow\quad(x-3)(x+1)=0.

The limits are $-1$ and $3$ . At $x=0$ , the line is above the curve. Hence

\begin{aligned} A&=\int_{-1}^{3}(2x+3-x^2)\,dx\\ &=\left[x^2+3x-\frac{x^3}{3}\right]_{-1}^{3}\\ &=9-\left(-\frac53\right)=\frac{32}{3}. \end{aligned}

Answer. $32/3$ square units.

Check. The height is $4-(x-1)^2$ , positive between $-1$ and $3$ , with maximum $4$ . An area of about $10.7$ fits inside the width- $4$ , height- $4$ rectangle of area $16$ .

Continue this lesson.

You’re previewing NovaLearn without an account. Create a free account to read the rest.

Create free account

Last Accessed

Pure Mathematics 1

Advanced pure mathematics

Pure Mathematics 3 extensions

Pure Mathematics 2 only

Mechanics

Probability and Statistics 1

Probability and Statistics 2

1.8 - Areas between curves

Signed area and total area

A line and a curve

Continue this lesson.

Learn faster with Nova