Calculus

Areas between lines and curves

Areas by integration (14.13) Paper 1 + 2 8 min read

On this page

In this note
Formula bank
Top minus bottom
Regions with a tangent or normal
Ways they test it
From the past papers
Mistakes to avoid
Summary
Practice
Can I do this?
Quiz

In this note

Syllabus 14.13: areas between lines and curves

14.13Find the area between a line and a curve, or two curves, using top − bottom.
14.13Use a triangle or trapezium together with an integral when that is quicker.
14.5, 14.13Handle regions bounded by a tangent or normal: find the line first.

Before you start: Areas under curves; solving simultaneous equations; tangents and normals. This is the last integration page.

Formula bank

Top − bottomLearn it\(\text{Area}=\displaystyle\int_a^b(y_{\text{top}}-y_{\text{bottom}})\,dx\)a and b = where the two graphs meet (or given lines).Use it when region between two graphs.

TriangleLearn it\(\tfrac12\times\text{base}\times\text{height}\)Use it when a straight edge down to the x-axis.

TrapeziumLearn it\(\tfrac12(a+b)h\)the parallel sides are the two vertical heights.Use it when a line above the x-axis between two x values.

Tangent / normalLearn it\(y-y_1=m(x-x_1)\); normal gradient \(=-\tfrac1m\)Use it when region bounded by a tangent or normal.

Top minus bottom

The curve is on top between the meeting points x = 0 and x = 3.

The area under the top graph minus the area under the bottom graph is the area between them, so you can do it in one integral of (top − bottom). This works even if part of the region is below the x-axis.

General methodHow to find the area between two graphs

Find where they meet: set the equations equal and solve. These are the limits.
Decide which graph is on top between them (sketch, or test one x value).
Integrate \((\text{top}-\text{bottom})\) between the limits.
Or: area under one graph ± a triangle or trapezium, when the other boundary is a straight line.

Example Find the area enclosed by \(y=4x-x^{2}\) and \(y=x\).

Meet: 4x − x² = x.

\(x^{2}-3x=0\Rightarrow x=0,\ 3\)
\(\int_0^3(3x-x^{2})\,dx=\big[\tfrac32x^{2}-\tfrac13x^{3}\big]_0^3=\tfrac{27}{2}-9\)

Answer \(\tfrac92\)

Example Find the area enclosed by \(y=x^{2}\) and \(y=8-x^{2}\).

Two curves: meet where 2x² = 8.

\(x=\pm2\)
\(\int_{-2}^{2}(8-2x^{2})\,dx=32-\tfrac{32}{3}\)

Answer \(\tfrac{64}{3}\)

Try it yourself

Area enclosed by \(y=x^{2}\) and \(y=x+2\).

Hint

They meet at x = −1 and x = 2.

Show answer

Answer 4.5

Area between \(y=\cos x\) and \(y=\tfrac12\) for \(-\tfrac{\pi}{3}\le x\le\tfrac{\pi}{3}\). (Type sqrt(3)-pi/3.)

Show answer

Answer \(\sqrt3-\tfrac{\pi}{3}\)

Regions with a tangent or normal

Area under the curve from 0 to 2, minus the triangle under the tangent.

Here the line is not given: you build it. The tangent to \(y=x^{2}\) at \((2,4)\) has gradient 4, so \(y=4x-4\); it meets the x-axis at \(x=1\). The region is the area under the curve from 0 to 2 MINUS the triangle (base 1, height 4): \(\tfrac83-2=\tfrac23\).

General methodHow to answer a tangent/normal area question

Differentiate; find the gradient at the point; for a normal use −1/m.
Write the line's equation; find where it meets the x-axis (or the curve again).
Split the region into 'area under the curve' and a triangle/trapezium under the line.
Add or subtract the pieces as the diagram shows.

Try it yourself

Area enclosed by \(y=\tfrac12x^{2}\) and \(y=2x\). (Fraction ok.)

Show answer

Answer 16/3

Ways they test it

Type 1 Line and curve

"The line and curve intersect at P and Q. Find the shaded area." (8 marks)

Spot it a line crossing a curve twice.

Solve for P and Q.
∫(top − bottom), or trapezium − area under the curve.

For example y = 2x + 5, y = 3x² − 2x + 1

meet at \(-\tfrac23\), 2

Answer \(\tfrac{256}{27}\)

Type 2 Two curves

"Find the area enclosed by the two curves." (7 marks)

Spot it two curves on one diagram.

Meeting points.
∫(top − bottom).

Type 3 Tangent or normal

"…the tangent to the curve at A… Find the area of the shaded region." (9-11 marks)

Spot it 'tangent' or 'normal' in a region question.

Build the line.
Curve area ± triangle.

Type 4 Line from a special point

"…the line AB, where B is the maximum point…" (9-11 marks)

Spot it a line through a root or turning point.

Find the points first (dy/dx = 0, y = 0).
Then the area.

From the past papers

Past paper 1

Curve y = 3x^2 - 2x + 1 and line y = 2x + 5 meet at P and Q. Find the shaded area between them. — 4037/11/M/J/19 Q6Paper 18 marksFull size

How to start: The line is on top between P and Q. Find P and Q by solving, then integrate (line − curve).

✓ Show the solution have a go on paper first

\(3x^{2}-2x+1=2x+5\Rightarrow3x^{2}-4x-4=0\Rightarrow(3x+2)(x-2)=0\), so \(x=-\tfrac23,\ 2\).
\(\int_{-2/3}^{2}\big[(2x+5)-(3x^{2}-2x+1)\big]dx=\int_{-2/3}^{2}(-3x^{2}+4x+4)\,dx\).
\(=\big[-x^{3}+2x^{2}+4x\big]_{-2/3}^{2}=8-\big(-\tfrac{40}{27}\big)\).
Area \(=\tfrac{256}{27}\) (≈ 9.48).

Part	Answer / working	Marks	Guidance
	equate and rearrange to a 3-term quadratic	M1
	\(3x^{2}-4x-4=0\)	A1
	\(x=-\tfrac23\) and \(x=2\)	A1
	subtraction (either way round)	M1
	integration	M1
	\(-x^{3}+2x^{2}+4x\)	A1
	correct use of limits	M1	dep
	\(\tfrac{256}{27}\) or 9.48	A1
Total: 8 marks

Examiner tip Alternative route in the scheme: trapezium under the line minus the area under the curve - same answer, same marks.

Past paper 2

A on y = sqrt(4x+1) has x = 2. Find the area enclosed by the curve, the tangent at A and the x-axis. — 4037/11/M/J/25 Q9Paper 1 · non-calculator10 marksFull size

How to start: Build the tangent at A, find where it meets the x-axis, then: triangle under the tangent MINUS the area under the curve (which starts at \(x=-\tfrac14\)).

✓ Show the solution have a go on paper first

At \(x=2\): \(y=3\). \(\frac{dy}{dx}=\tfrac12(4x+1)^{-1/2}\times4=\dfrac{2}{\sqrt{4x+1}}=\tfrac23\).
Tangent: \(y-3=\tfrac23(x-2)\); at \(y=0\): \(x=-\tfrac52\).
Triangle: \(\tfrac12\times\big(2+\tfrac52\big)\times3=\tfrac{27}{4}\).
Curve meets the axis at \(x=-\tfrac14\). \(\int_{-1/4}^{2}(4x+1)^{1/2}dx=\big[\tfrac16(4x+1)^{3/2}\big]_{-1/4}^{2}=\tfrac{27}{6}=\tfrac92\).
Shaded area \(=\tfrac{27}{4}-\tfrac92=\tfrac94\).

Part	Answer / working	Marks	Guidance
	\(y=3\)	B1
	\(\frac{dy}{dx}=\tfrac12(4x+1)^{-1/2}\times4\)	M1
	gradient \(\tfrac23\)	M1
	tangent equation	M1	dep
	x-intercept \(-\tfrac52\)	A1
	triangle \(\tfrac{27}{4}\)	B1
	curve meets axis at \(-\tfrac14\)	B1
	\(\tfrac16(4x+1)^{3/2}\)	M1
	limits	M1	dep
	\(\tfrac94\)	A1
Total: 10 marks

Examiner tip The curve's area starts where the CURVE meets the axis \(\big(-\tfrac14\big)\), not where the tangent does.

Mistakes to avoid

✗ bottom − top✓ top − bottom (or take the size)
✗ Limits taken from the axis instead of the intersections✓ solve the equations together
✗ Curve area started at the tangent's intercept✓ each piece has its own limits
✗ Normal gradient = m✓ normal gradient \(=-\tfrac1m\)

Summary

Area between graphs \(=\int(\text{top}-\text{bottom})dx\) between the meeting points.
A straight boundary can be a triangle or trapezium instead.
Tangent/normal regions: build the line, then curve area ± triangle.

If you see …

a line and a curvemeeting points, top − bottom.
'tangent' or 'normal'find the line first.

Practice

Practice: easy, medium, hard

Easy

Area enclosed by \(y=4-x^{2}\) and the x-axis. (Fraction ok.)

Show answer

Answer 32/3

Where do \(y=4x-x^{2}\) and \(y=x\) meet? Type the non-zero x.

Show answer

Answer 3

Area enclosed by \(y=4x-x^{2}\) and \(y=x\).

Show answer

Answer 4.5

Medium

Area enclosed by \(y=x^{2}\) and \(y=8-x^{2}\). (Fraction ok.)

Guide me

What is it asking? an area

What do we know? meet at \(\pm2\)

Plan \(\int(8-2x^{2})\)

Show working

\(\tfrac{64}{3}\)

Answer 64/3

The line through \((1,3)\) and \((3,11)\) and the curve \(y=x^{2}+2\): area between them for \(1\le x\le3\). (Fraction ok.)

Guide me

What is it asking? an area

What do we know? line \(y=4x-1\)

Plan \(\int_1^3(-x^{2}+4x-3)\)

Show working

\(\tfrac43\)

Answer 4/3

Region between \(y=x^{2}\), its tangent at \((2,4)\) and the x-axis. (Fraction ok.)

Guide me

What is it asking? an area

What do we know? tangent \(y=4x-4\) meets the axis at 1

Plan \(\tfrac83-2\)

Show working

\(\tfrac23\)

Answer 2/3

Hard

Area between \(y=3x^{2}-2x+1\) and \(y=2x+5\). (Fraction ok.)

Guide me

What is it asking? an area

What do we know? meet at \(-\tfrac23\) and 2

Plan \(\int(-3x^{2}+4x+4)\)

Show working

\(\tfrac{256}{27}\)

Answer 256/27

Part	Answer / working	Marks
	limits	M1A1A1
	top − bottom, integrate	M1M1A1
	limits, answer	M1A1
Total: 8 marks

Area between \(y=\sqrt{4x+1}\), its tangent at \(x=2\) and the x-axis. (Fraction ok.)

Guide me

What is it asking? an area

What do we know? triangle \(\tfrac{27}{4}\)

Plan curve area \(\tfrac92\)

Show working

\(\tfrac94\)

Answer 9/4

Part	Answer / working	Marks
	tangent	B1M1M1M1A1
	triangle, curve	B1B1M1M1
	answer	A1
Total: 10 marks

Area between \(y=\cos x\) and \(y=\tfrac12\) for \(-\tfrac{\pi}{3}\le x\le\tfrac{\pi}{3}\) (3 s.f.).

Guide me

What is it asking? an area

What do we know? \(\int(\cos x-\tfrac12)\)

Plan \(2\sin\tfrac{\pi}{3}-\tfrac{\pi}{3}\)

Show working

\(\sqrt3-\tfrac{\pi}{3}\)

Answer 0.685

Part	Answer / working	Marks
	\(\sin x-\tfrac12x\)	M1A1
	limits	M1
	answer	A1
Total: 4 marks

Can I do this?

I can find the area between a line and a curve, or two curves.
I can use triangles or trapezia with an integral.
I can do regions bounded by a tangent or normal.

Anything unticked: revise that section first. All ticked: take the quiz.

Quiz

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