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The area between two graphs is \(\int_a^b (y_{\text{top}} - y_{\text{bottom}})\,\mathrm{d}x\) between their intersection points, and a line, tangent or normal can form part of the boundary. These are usually 6 to 10 mark questions with a diagram, combining intersections, tangents or normals and integration.
Key facts
Area between graphs \(= \int_a^b (y_{\text{top}} - y_{\text{bottom}})\,\mathrm{d}x\)
Limits \(a\) and \(b\): the \(x\)-coordinates of the intersection points
Tangent: \(y - y_1 = m(x - x_1)\) with \(m = \dfrac{\mathrm{d}y}{\mathrm{d}x}\) at the point
Normal: gradient \(-\dfrac{1}{m}\)
Notes
Area between two graphs
diagram
Find the intersection points by solving the equations simultaneously: their \(x\)-coordinates are the limits.
Area \(= \int_a^b (y_{\text{top}} - y_{\text{bottom}})\,\mathrm{d}x\). This works wherever the region is, even below the \(x\)-axis, as long as the top graph stays above the bottom one for the whole interval.
Area \(= \int_{-1}^{4} \big((x + 7) - (x^2 - 2x + 3)\big)\,\mathrm{d}x\): top minus bottom, between the intersections.
If the graphs cross inside the interval, split at the crossing point and swap top and bottom.
Subtract before integrating: one integral of the difference is quicker than two separate integrals.
Combining shapes
When a straight line is involved, it can be quicker to find the area under the curve and add or subtract a triangle (\(\dfrac{1}{2}bh\)) or trapezium.
For a tangent or normal: differentiate to find the gradient at the point, use \(y - y_1 = m(x - x_1)\), and find where the line meets the axis. Label every key point on a sketch before deciding which areas to add or subtract.
How to answer each type of question
M = method mark, A = answer mark, B = a mark on its own, dM = depends on the earlier method mark.
Find the area between a curve and a line
5 marksGrade C
Solve the equations simultaneously to find the \(x\)-coordinates where they meet.
Decide which graph is on top between these points (use the sketch or test a value).
Integrate (top − bottom) between the limits.
Example. The curve \(C\) has equation \(y = x^2 - 2x + 3\) and the line \(l\) has equation \(y = x + 7\). The finite region \(R\) is bounded by \(C\) and \(l\). Find the exact area of \(R\)
Find the area of a region bounded by a curve, a tangent and an axis
7 marksGrade B
Differentiate to find the gradient at the point and write down the tangent (or normal).
Find where the line and the curve meet the \(x\)-axis, and mark every point on a sketch.
Split the region into an area under the line (often a triangle) and an area under the curve.
Add or subtract as the sketch shows.
Example. The curve \(C\) has equation \(y = 4 - x^2\). The line \(l\) is the tangent to \(C\) at the point \(P(1, 3)\). The finite region \(R\) is bounded by \(C\), \(l\) and the positive \(x\)-axis. Find the exact area of \(R\).
Show the model answerHide the model answer
\(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -2x = -2\) at \(x = 1\)
M1
\(l\):\(y - 3 = -2(x - 1)\), so \(y = 5 - 2x\)
A1
\(l\) meets the \(x\)-axis at \(x = \dfrac{5}{2}\) and \(C\) meets it at \(x = 2\)
B1
Area under \(l\) from \(x = 1\) to \(x = \dfrac{5}{2}\) is a triangle: \(\dfrac{1}{2} \times \dfrac{3}{2} \times 3 = \dfrac{9}{4}\)
M1
Area under \(C\) from \(x = 1\) to \(x = 2\): \(\left[4x - \dfrac{x^3}{3}\right]_1^2 = \dfrac{16}{3} - \dfrac{11}{3} = \dfrac{5}{3}\)
M1 A1
Area of\[\begin{aligned}[b]&R \\ &= \frac{9}{4} - \frac{5}{3} \\ &= \frac{7}{12}\end{aligned}\]
A1
Don’t lose marks
Integrating (bottom − top), which gives a negative answer, or subtracting only the first term of the bottom function (use brackets).
Using the whole area under the curve when part of the region's boundary is a tangent or other line.
Taking the limits from where the curve meets the axis when the region is bounded by the line instead.
More tips
Memory tricks
'Top minus bottom' works above or below the axis, so you never need to worry about negative areas.
Subtract the functions first, then integrate once.
Label every point (intersections, axis crossings) on your sketch before choosing which areas to add or subtract.
Exam technique
Many questions start 'Show that the curves meet at…': show the algebra even if you can see the answer.
Write the integral you are going to evaluate, with its limits, before integrating: it earns the method mark.
Check your area against the diagram: a rough estimate (compare with a triangle or rectangle) catches big errors.
What each grade needs
What you need to be able to do, from the first marks up to the top grade.
Grade D
Find where two graphs intersectSet the two equations equal and solve to get the limits of the region.
Grade C
Find the area between a curve and a lineIntegrate (top − bottom), or subtract the area of a triangle or trapezium from the area under the curve.
Grade B
Find the area between two curvesIntegrate the difference of the two functions between the intersection points.
Grade B
Handle regions bounded by tangents or normalsFind the line's equation and where it meets the axis, then add or subtract areas.
Sample questions
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy7 marks
The diagram shows part of the curve \(C\) with equation \(y = 5x - x^{2}\) and the straight line \(l\) with equation \(y = 2x\). The line \(l\) meets \(C\) at the origin and at the point \(A\). The region \(R\), shown shaded, is bounded by \(C\) and \(l\). In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.
[object Object]
(a) Find the coordinates of \(A\).[3]
(b) Use integration to find the exact area of \(R\).[4]
Show the answer and mark scheme
(a)Answer: \(A(3, 6)\)
M1 for equating \(5x - x^{2} = 2x\)
A1 for \(x = 3\)
A1 for \(A(3, 6)\)
Worked solution: \(5x - x^{2} = 2x \Rightarrow x^{2} - 3x = 0 \Rightarrow x(x - 3) = 0\), so at \(A\), \(x = 3\) and \(y = 6\).
(b)Answer: \(\frac{9}{2}\)
M1 for \(\int_{0}^{3} \left((5x - x^{2}) - 2x\right) \,\mathrm{d}x\) (curve minus line, or area under curve minus area of triangle)
M1 for \(x^{n} \to x^{n + 1}\) for at least one term
A1 for \(\left[\frac{3}{2}x^{2} - \frac{1}{3}x^{3}\right]_{0}^{3}\)
A1 for \(\frac{9}{2}\)
Worked solution: Area \(= \int_{0}^{3} (3x - x^{2}) \,\mathrm{d}x = \left[\frac{3}{2}x^{2} - \frac{1}{3}x^{3}\right]_{0}^{3}\) \(= \left(\frac{9}{2}\right) - \left(0\right) = \frac{9}{2}\). (Equivalently, area under \(C\) minus the area of the triangle below \(OA\).)
Question 2Medium9 marks
The diagram shows part of the curve \(C\) with equation \(y = -x^{2} + 8x - 5\) and the line \(l\) with equation \(y = x + 1\). The line meets \(C\) at the points \(A\) and \(B\). The region \(R\), shown shaded, is bounded by \(C\) and \(l\). In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.
[object Object]
(a) Find the coordinates of \(A\) and \(B\).[4]
(b) Find the exact area of \(R\).[5]
Show the answer and mark scheme
(a)Answer: \(A(1, 2)\), \(B(6, 7)\)
M1 for equating the equations of \(C\) and \(l\)
A1 for a correct 3TQ, e.g. \(x^{2} - 7x + 6 = 0\)
M1 for solving their 3TQ
A1 for \(A(1, 2)\) and \(B(6, 7)\)
Worked solution: \(-x^{2} + 8x - 5 = x + 1 \Rightarrow x^{2} - 7x + 6 = 0 \Rightarrow (x - 1)(x - 6) = 0\). So \(x = 1\) or \(x = 6\), giving \(A(1, 2)\) and \(B(6, 7)\).
(b)Answer: \(\frac{125}{6}\)
M1 for an integral of the form \(\int (\text{upper} - \text{lower}) \,\mathrm{d}x\) with their limits
A1 for the integrand \(-x^{2} + 7x - 6\)
M1 for \(x^{n} \to x^{n + 1}\)
M1 for substituting their limits and subtracting
A1 for \(\frac{125}{6}\)
Worked solution: The curve is above the line for \(1 \lt x \lt 6\), so area \(= \int_{1}^{6} (-x^{2} + 7x - 6) \,\mathrm{d}x = \left[-\frac{1}{3}x^{3} + \frac{7}{2}x^{2} - 6x\right]_{1}^{6}\) \(= \left(18\right) - \left(-\frac{17}{6}\right) = \frac{125}{6}\).
Question 3Hard8 marks
The diagram shows part of the curve \(C\) with equation \(y = 2x^{2}\), \(x \geqslant 0\), and the line \(l\) with equation \(y = 4 - 2x\). The line \(l\) meets \(C\) at the point \(P\) and meets the \(x\)-axis at the point \(Q\). The region \(R\), shown shaded, is bounded by \(C\), \(l\) and the \(x\)-axis. In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.
[object Object]
(a) Find the coordinates of \(P\).[3]
(b) Find the exact area of \(R\).[5]
Show the answer and mark scheme
(a)Answer: \(P(1, 2)\)
M1 for equating: \(2x^{2} = 4 - 2x\)
M1 for solving the 3TQ \(2x^{2} + 2x - 4 = 0\)
A1 for \(P(1, 2)\), rejecting \(x = -2\)
Worked solution: \(2x^{2} + 2x - 4 = 0 \Rightarrow 2(x - 1)(x + 2) = 0\). As \(x \geqslant 0\), \(x = 1\), so \(P(1, 2)\).
(b)Answer: \(\frac{5}{3}\)
M1 for \(\int_{0}^{1} 2x^{2} \,\mathrm{d}x\) (the area under \(C\) from \(O\) to \(P\))
A1 for \(\frac{2}{3}\)
M1 for finding \(Q(2, 0)\) and the area of the triangle below \(PQ\) (or \(\int_{1}^{2} (4 - 2x) \,\mathrm{d}x\))
A1 for \(1\)
A1 for the total area \(\frac{5}{3}\)
Worked solution: Area under \(C\): \(\int_{0}^{1} 2x^{2} \,\mathrm{d}x = \left[\frac{2}{3}x^{3}\right]_{0}^{1} = \frac{2}{3}\). \(l\) meets the \(x\)-axis at \(Q(2, 0)\), so the triangle below \(PQ\) has area \(\frac{1}{2} \times 1 \times 2 = 1\). Area of \(R = \frac{2}{3} + 1 = \frac{5}{3}\).