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Or take a quick quiz: 7 questions on this subtopic in AS Maths (Edexcel), marked as you tap.
A definite integral \(\int_a^b \mathrm{f}(x)\,\mathrm{d}x\) is a number, found by integrating and substituting the limits. Where the curve is above the \(x\)-axis it gives the area between the curve and the axis; where the curve is below the axis it comes out negative. Expect 4 to 6 mark questions with a sketch, often asking for an exact area.
\(\int_a^b \mathrm{f}(x)\,\mathrm{d}x = \left[\mathrm{F}(x)\right]_a^b = \mathrm{F}(b) - \mathrm{F}(a)\), where \(\mathrm{F}\) is any integral of \(\mathrm{f}\). There is no \(+\,c\)
Write the integrated expression in square brackets with the limits, then substitute. Put negative limits in brackets: \((-2)^3 = -8\)
Area under a curve
diagram
If \(y = \mathrm{f}(x) \ge 0\) for \(a \le x \le b\), the area between the curve, the \(x\)-axis and the lines \(x = a\) and \(x = b\) is \(\int_a^b y\,\mathrm{d}x\)
If the region is bounded by the curve and the \(x\)-axis only, the limits are the roots: solve \(\mathrm{f}(x) = 0\)
The limits are the roots: area \(= \int_1^5 (-x^2 + 6x - 5)\,\mathrm{d}x = \dfrac{32}{3}\)
Sketch the curve first, so you can see which parts are above or below the axis.
Regions below the x-axis
diagram
Where the curve is below the axis, \(y \lt 0\), so \(\int y\,\mathrm{d}x\) is negative. The area is the positive value: area \(= \left|\int_a^b y\,\mathrm{d}x\right|\)
If a region is partly above and partly below the axis, split at the root, work out each integral separately and add the sizes of the answers.
\(\int_0^1 y\,\mathrm{d}x = \dfrac{1}{4}\) and \(\int_1^2 y\,\mathrm{d}x = -\dfrac{1}{4}\): the total area is \(\dfrac{1}{2}\), but \(\int_0^2 y\,\mathrm{d}x = 0\)
One integral across a root gives the net (signed) area: the parts cancel, and the result can even be zero when the total area is not.
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 of a region bounded by a curve and the x-axis
5 marksGrade D
Solve \(y = 0\) to find the limits.
Integrate term by term.
Substitute the limits, upper minus lower.
Check the sign: an area is always positive.
Example. The curve \(C\) has equation \(y = -x^2 + 6x - 5\). The finite region \(R\) is bounded by \(C\) and the \(x\)-axis. Find the exact area of \(R\)
Total area\[\begin{aligned}[b]&= \frac{1}{4} + \frac{1}{4} \\ &= \frac{1}{2}\end{aligned}\]
A1
(Note that \(\int_0^2 y\,\mathrm{d}x = 0\): the two parts cancel, which is why you must split.)
Don’t lose marks
Integrating straight across a root and giving the net value as the area.
Giving a negative answer for an area.
Substituting the limits the wrong way round, or dropping the brackets around negative limits.
Using the wrong limits: they must come from the question or from solving \(y = 0\), not from the edges of the diagram.
More tips
Memory tricks
Sketch first: the roots and the shape tell you where the curve dips below the axis.
Use your calculator's definite integral function to check a value (it does not replace working when you are told to show all stages).
An area can never be negative: a negative integral means the region is below the axis.
Exam technique
'Find the exact area' means no decimals: leave fractions, \(\ln\) and \(\pi\)
Show the square-bracket line and the two substituted values: they carry the method marks.
If there is a diagram, check that your limits match it: roots, marked points or the lines given.
In a context, give units: square units, or e.g. m2.
What each grade needs
What you need to be able to do, from the first marks up to the top grade.
Grade E
Evaluate a definite integral of a polynomialIntegrate, substitute the upper limit, then subtract the value at the lower limit.
Grade D
Find the area under a curveFor a curve above the \(x\)-axis, the area between the curve, the axis and the lines \(x = a\) and \(x = b\) is \(\int_a^b y\,\mathrm{d}x\)
Grade C
Find the limits where a curve meets the axisSolve \(y = 0\) to find where a region bounded by the curve and the \(x\)-axis starts and ends.
Grade B
Deal with regions below the x-axisThe integral is negative there, so split the integral at each root and add the sizes of the separate answers.
Sample questions
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy3 marks
In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.
Find the exact value of \(\int_{0}^{4} (6x^{2} + 6x + 2) \,\mathrm{d}x\).[3]
Show the answer and mark scheme
Answer: \(184\)
M1 for \(x^{n} \to x^{n + 1}\) for at least one term
M1 for substituting the limits 4 and 0 into their integrated expression and subtracting
The diagram shows part of the curve \(C\) with equation \(y = 9\sqrt{x} - 3x\), \(x \geqslant 0\). The curve meets the \(x\)-axis at the origin \(O\) and at the point \(A\). The region \(R\), shown shaded, is bounded by \(C\) 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 \(A\).[2]
(b) Use integration to find the exact area of \(R\).[4]
The diagram shows part of the curve \(C\) with equation \(y = x^{3} - 7x^{2} + 10x\). The curve crosses the \(x\)-axis at the origin and at the points where \(x = 2\) and \(x = 5\). The region \(R_{1}\) is bounded by \(C\) and the \(x\)-axis for \(0 \leqslant x \leqslant 2\). The region \(R_{2}\) is bounded by \(C\) and the \(x\)-axis for \(2 \leqslant x \leqslant 5\). 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 exact area of \(R_{1}\).[4]
(b) Find the exact total area of the shaded regions.[4]
Show the answer and mark scheme
(a)Answer: \(\frac{16}{3}\)
M1 for \(x^{n} \to x^{n + 1}\) for at least one term
A1 for \(\frac{1}{4}x^{4} - \frac{7}{3}x^{3} + 5x^{2}\)
M1 for evaluating \(\int_{2}^{5} y \,\mathrm{d}x\) with their integrated expression
A1 for \(-\frac{63}{4}\)
M1 for using the modulus of this value (the region lies below the \(x\)-axis) and adding to \(R_{1}\)
A1 for \(\frac{253}{12}\)
Worked solution: \(\int_{2}^{5} y \,\mathrm{d}x = \left[\frac{1}{4}x^{4} - \frac{7}{3}x^{3} + 5x^{2}\right]_{2}^{5} = \left(-\frac{125}{12}\right) - \left(\frac{16}{3}\right) = -\frac{63}{4}\), which is negative because \(R_{2}\) lies below the \(x\)-axis. So \(R_{2} = \frac{63}{4}\) and the total area is \(\frac{16}{3} + \frac{63}{4} = \frac{253}{12}\).