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Calculus

Calculus

Areas under curves

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

On this page
  1. In this note
  2. Formula bank
  3. Area above the x-axis
  4. Below the axis, and crossing it
  5. Ways they test it
  6. From the past papers
  7. Mistakes to avoid
  8. Summary
  9. Practice
  10. Can I do this?
  11. Quiz

In this note

Syllabus 14.13: areas between a curve and the x-axis
  • 14.13Find the area between a curve, the x-axis and two lines \(x=a\), \(x=b\).
  • 14.13Deal with a region below the x-axis, and a curve that crosses it (a sum of two areas).
  • 14.13Find an unknown limit when the area is given.

Before you start: Definite integrals; finding where a curve meets the x-axis. Next: Areas between lines and curves.

Formula bank

Area above the axisLearn it\(\text{Area}=\displaystyle\int_a^b y\,dx\)curve above the x-axis between a and b.Use it when 'area under the curve'.
Area below the axisLearn it\(\text{Area}=\left|\displaystyle\int_a^b y\,dx\right|\)the integral comes out negative; the area is its size.Use it when region under the x-axis.
Crossing the axisLearn itsplit at each root; add the SIZES of the piecesUse it when region on both sides of the axis.
Sum of two areasLearn it\(\displaystyle\int_a^c f\,dx+\int_c^b g\,dx\)different curves bound different parts.Use it when the boundary changes at a point.

Area above the x-axis

Area under y = x² + 1 from x = 1 to x = 3123246810xy
Shaded: between the curve, the x-axis, x = 1 and x = 3.

The definite integral adds up thin strips of height y and width dx, so it measures the area between the curve and the x-axis. \(\int_1^3(x^{2}+1)\,dx=\big[\tfrac{x^{3}}{3}+x\big]_1^3=12-\tfrac43=\tfrac{32}{3}\).

General methodHow to find any area under a curve

  1. Sketch it. Find where the curve meets the x-axis if the limits are not given.
  2. Check whether the region is above or below the axis (or both).
  3. Integrate between the limits (split at every root inside the interval).
  4. Area = the SIZE of each piece; add them. Area is never negative.

Try it yourself

1
Area between \(y=6x-x^{2}\) and the x-axis.
Hint
It meets the axis at 0 and 6.
Show answer

Answer 36

2
Area under \(y=\frac{4}{x^{2}}\) from \(x=1\) to \(x=4\).
Show answer

Answer 3

Below the axis, and crossing it

y = x² − 1 from 0 to 2: part below the axis, part above12-1123xy
The pink piece is below the axis: its integral is negative.

Below the x-axis y is negative, so the integral is negative: \(\int_0^4(x^{2}-4x)\,dx=-\tfrac{32}{3}\), area \(\tfrac{32}{3}\). If the curve crosses the axis, one integral over the whole range lets the pieces cancel. For \(y=x^{2}-1\) from 0 to 2: \(\int_0^2=\tfrac23\), but the real area is \(\tfrac23+\tfrac43=2\).

Example Find the total area between \(y=x^{2}-1\), the x-axis and \(x=0\), \(x=2\).

Root at x = 1: split there.

  1. \(\int_0^1(x^{2}-1)dx=-\tfrac23\) → area \(\tfrac23\)
  2. \(\int_1^2(x^{2}-1)dx=\tfrac43\)

Answer 2

Example The area under \(y=\frac3x\) from \(x=1\) to \(x=k\) is 6. Find k.

Integrate, set equal to 6.

  1. \(\big[3\ln x\big]_1^k=3\ln k=6\)
  2. \(\ln k=2\)

Answer \(k=e^{2}\)

Try it yourself

1
Total area between \(y=x^{3}-x\) and the x-axis from \(x=-1\) to \(x=1\).
Show answer

Answer 0.5

Ways they test it

Type 1 Area under a curve

"Find the area enclosed by the curve, the x-axis and the lines x = 1 and x = 3." (4 marks)

Spot it two vertical lines.

  1. Integrate between the lines.

Type 2 Find the limits first

"The curve meets the x-axis at A… find the shaded area." (6 marks)

Spot it a root or a turning point labelled.

  1. Solve y = 0 (or dy/dx = 0).
  2. Then integrate.

Type 3 Two curves meeting the axis

"Find the area of the shaded region." (5-7 marks)

Spot it the top boundary changes at an intersection.

  1. Find the meeting point.
  2. One integral for each part; add.

For example \(y=4x^{2/3}\), \(y=(x-3)^{2}\)

  1. \(\tfrac{12}{5}+\tfrac83\)

Answer \(\tfrac{76}{15}\)

Type 4 Area given

"The area is 6. Find k." (3-5 marks)

Spot it an area value and a letter.

  1. Integrate with the letter as a limit.
  2. Set equal; solve.

From the past papers

Past paper 1

0606/21/M/J/20 Q10Paper 27 marksFull size
The graphs of y = 4x^(2/3) and y = (x-3)^2 meet at B(b, 4); (x-3)^2 meets the x-axis at A(a, 0). Find a, b and the shaded area.

How to start: The top boundary is \(4x^{2/3}\) from O to B, then \((x-3)^{2}\) from B to A. Find a and b, then add two integrals.

✓ Show the solution have a go on paper first
  1. (a) \((x-3)^{2}=0\Rightarrow a=3\). \(4x^{2/3}=4\Rightarrow x=1\), and \((1-3)^{2}=4\) ✓, so \(b=1\).
  2. (b) \(\int_0^1 4x^{2/3}dx=\big[\tfrac{12}{5}x^{5/3}\big]_0^1=\tfrac{12}{5}\).
  3. (b) \(\int_1^3(x-3)^{2}dx=\big[\tfrac{(x-3)^{3}}{3}\big]_1^3=0-\big(-\tfrac83\big)=\tfrac83\).
  4. (b) area \(=\tfrac{12}{5}+\tfrac83=\tfrac{76}{15}\) (≈ 5.07).
PartAnswer / workingMarksGuidance
(a)\(a=3\), \(b=1\)B2B1 for each
(b)splitting into two integrals with their limitsM1
\(\tfrac{12}{5}x^{5/3}\) and \(\tfrac{(x-3)^{3}}{3}\)M2M1 for each
correct use of both sets of limitsM1
\(\tfrac{76}{15}\) or 5.07A1
Total: 7 marks

Examiner tip Integrating \((x-3)^{2}\) as a bracket (no expanding) is fastest; at the lower limit \((1-3)^{3}=-8\).

Past paper 2

0606/22/M/J/21 Q12Paper 28 marksFull size
Non-calculator: curve y = 1/(2x+1) and line 5y = x - 1 meet at C. Find A, B, verify C has x = 2, and find the exact shaded area.

How to start: The shaded region = area under the curve from 0 to 2, MINUS the triangle under the line from B to C.

✓ Show the solution have a go on paper first
  1. (a)(i) \(A(0,1)\) (curve at x = 0); \(B(1,0)\) (line at y = 0).
  2. (a)(ii) At \(x=2\): curve \(y=\tfrac15\), line \(y=\tfrac{2-1}{5}=\tfrac15\) - the same, so C has \(x=2\).
  3. (b) \(\int_0^2\frac{1}{2x+1}dx=\big[\tfrac12\ln(2x+1)\big]_0^2=\tfrac12\ln5\).
  4. (b) triangle BC and the axis: \(\tfrac12\times1\times\tfrac15=0.1\).
  5. (b) area \(=\tfrac12\ln5-0.1\).
PartAnswer / workingMarksGuidance
(a)(i)\(A(0,1)\), \(B(1,0)\)B1
(a)(ii)both y-values shown to be 1/5B2
(b)triangle \(=\tfrac12\times1\times0.2\)B1
\(\tfrac12\ln(2x+1)\)B2B1 for \(k\ln(2x+1)\)
F(2) − F(0) − their 0.1M1
\(\tfrac12\ln5-0.1\)A1
Total: 8 marks

Examiner tip Read the region carefully: it is bounded by the curve, the axes AND the line, so a triangle comes off.

Mistakes to avoid

  • ✗ A negative area✓ take the size of a below-axis integral
  • ✗ One integral across a root✓ split at the rootThe pieces cancel.
  • ✗ Limits from the wrong curve✓ find where EACH boundary starts and stops

Summary

  • Area under a curve \(=\int_a^b y\,dx\).
  • Below the axis: the integral is negative; use its size.
  • Crossing the axis or changing boundary: split and add.

If you see …

  • shaded region on a sketchfind the limits first.
  • part below the axissplit at the root.

Practice

Practice: easy, medium, hard

Easy

1
Area under \(y=x^{2}+1\) from 1 to 3. (Fraction ok.)
Show answer

Answer 32/3

2
Area under \(y=\sin2x\) from 0 to \(\tfrac{\pi}{2}\).
Show answer

Answer 1

3
Area under \(y=\sqrt x\) from 0 to 9.
Show answer

Answer 18

Medium

1
Area between \(y=x^{2}-4x\) and the x-axis. (Fraction ok.)
Guide me

What is it asking? an area

What do we know? roots 0 and 4

Plan integral \(-\tfrac{32}{3}\)

Show working
  1. \(\tfrac{32}{3}\)

Answer 32/3

2
Total area between \(y=x^{2}-1\), the axis, \(x=0\) and \(x=2\).
Guide me

What is it asking? an area

What do we know? split at 1

Plan \(\tfrac23+\tfrac43\)

Show working
  1. 2

Answer 2

3
\(\int_0^k(2x+1)\,dx=20\), \(k>0\). Find k.
Guide me

What is it asking? a limit

What do we know? \(k^{2}+k=20\)

Plan \((k+5)(k-4)=0\)

Show working
  1. 4

Answer 4

Hard

1
Area under \(y=\frac3x\) from 1 to k is 6. Find k. (Type e^2.)
Guide me

What is it asking? a limit

What do we know? \(3\ln k=6\)

Plan \(k=e^{2}\)

Show working
  1. \(e^{2}\)

Answer \(e^{2}\)

PartAnswer / workingMarksGuidance
\(3\ln x\)B1
= 6M1
\(e^{2}\)A1
Total: 3 marks
2
Shaded area under \(4x^{2/3}\) (0 to 1) then \((x-3)^{2}\) (1 to 3). (Fraction ok.)
Guide me

What is it asking? an area

What do we know? \(\tfrac{12}{5}\)

Plan \(+\tfrac83\)

Show working
  1. \(\tfrac{76}{15}\)

Answer 76/15

PartAnswer / workingMarksGuidance
splitM1
two integralsM2
limitsM1
76/15A1
Total: 5 marks
3
Exact area: under \(y=\frac{1}{2x+1}\) from 0 to 2 minus the triangle of area 0.1. (Type ln(5)/2-0.1.)
Guide me

What is it asking? an area

What do we know? \(\tfrac12\ln(2x+1)\)

Plan \(\tfrac12\ln5\)

Show working
  1. \(\tfrac12\ln5-0.1\)

Answer \(\tfrac12\ln5-0.1\)

PartAnswer / workingMarksGuidance
triangle 0.1B1
\(\tfrac12\ln(2x+1)\)B2
limits − 0.1M1
answerA1
Total: 5 marks

Can I do this?

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

Quiz

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