Home / A-Level Maths 9709 / Integration / Definite integrals and area
Cambridge International AS & A Level · Topic 1.8
A-Level Maths 9709 / Topic 1.8 / AS & A Level

Integrals that measure.

Put limits on an integral and the constant disappears: what is left is a number, and that number is an area. Learn to evaluate definite integrals, to handle regions below the axis, and to find the area between a curve and a line or between two curves.

Topic 1.8All integration lessons →
The key idea

\(\displaystyle\int_{a}^{b} f(x)\,dx = F(b) - F(a)\), where \(F\) is any integral of \(f\). The area between \(y = f(x)\) and the \(x\)-axis is \(\displaystyle\int_{a}^{b} y\,dx\) when the curve is above the axis; between two graphs, integrate (top \(-\) bottom).

x y −2−112 (−1, 1) (2, 4) y = x² y = x + 2 area = 9/2
Fig. 1: the region between y = x + 2 and y = x², which meet at (−1, 1) and (2, 4), has area 9/2
Section 01

Evaluating definite integrals.

Integrate as usual, write the result in square brackets with the limits, then substitute the top limit and subtract the value at the bottom limit. No \(+ c\) is needed: it would cancel.

Worked example A polynomial
Evaluate \(\displaystyle\int_{1}^{3} (3x^{2} - 2x)\,dx\).
  1. Integrate: \(\left[x^{3} - x^{2}\right]_{1}^{3}\).
  2. Top limit: \(27 - 9 = 18\). Bottom limit: \(1 - 1 = 0\).
  3. Subtract: \(18 - 0 = 18\).
Answer: \(18\)
Worked example A linear bracket
Evaluate \(\displaystyle\int_{0}^{4} \sqrt{2x + 1}\,dx\).
  1. \(\displaystyle\int (2x + 1)^{\frac{1}{2}}\,dx = \dfrac{(2x + 1)^{\frac{3}{2}}}{\frac{3}{2} \times 2} = \tfrac{1}{3}(2x + 1)^{\frac{3}{2}}\).
  2. Top: \(\tfrac{1}{3} \times 9^{\frac{3}{2}} = \tfrac{1}{3} \times 27 = 9\).
  3. Bottom: \(\tfrac{1}{3} \times 1^{\frac{3}{2}} = \tfrac{1}{3}\). The bottom limit is 0, but the bracket is 1 there, not 0.
  4. \(9 - \tfrac{1}{3} = \dfrac{26}{3}\).
Answer: \(\dfrac{26}{3}\)
Examiner trap

A lower limit of 0 does not always give 0: \((2 \times 0 + 1)^{\frac{3}{2}} = 1\). Always substitute properly, and put the bottom-limit value in brackets before subtracting it.

Section 02

Area under a curve.

For a curve above the \(x\)-axis, \(\displaystyle\int_{a}^{b} y\,dx\) is the area between the curve, the axis and the lines \(x = a\), \(x = b\). Below the axis the integral comes out negative. The same idea with \(x\) and \(y\) swapped gives areas against the \(y\)-axis.

Worked example A curve that crosses the axis
Find the total area between \(y = x^{2} - 2x\), the \(x\)-axis and the lines \(x = 0\) and \(x = 3\).
  1. The curve \(y = x(x - 2)\) crosses the axis at \(x = 2\): below it for \(0 < x < 2\), above it for \(2 < x < 3\).
  2. \(\displaystyle\int_{0}^{2} (x^{2} - 2x)\,dx = \left[\tfrac{x^{3}}{3} - x^{2}\right]_{0}^{2} = \tfrac{8}{3} - 4 = -\tfrac{4}{3}\).
  3. \(\displaystyle\int_{2}^{3} (x^{2} - 2x)\,dx = (9 - 9) - \left(\tfrac{8}{3} - 4\right) = \tfrac{4}{3}\).
  4. Total area \(= \tfrac{4}{3} + \tfrac{4}{3}\). (A single integral from 0 to 3 gives 0: the two parts cancel.)
Answer: \(\dfrac{8}{3}\)
Worked example Area against the y-axis
Find the area of the region bounded by \(y = x^{2}\) for \(x \ge 0\), the \(y\)-axis and the line \(y = 4\).
  1. The region lies between the curve and the \(y\)-axis, so integrate with respect to \(y\): \(x = y^{\frac{1}{2}}\).
  2. \(\displaystyle\int_{0}^{4} y^{\frac{1}{2}}\,dy = \left[\tfrac{2}{3}y^{\frac{3}{2}}\right]_{0}^{4} = \tfrac{2}{3} \times 8\).
Answer: \(\dfrac{16}{3}\) (check: with \(\displaystyle\int_{0}^{2} x^{2}\,dx = \tfrac{8}{3}\) it fills the \(2 \times 4\) rectangle, since \(\tfrac{16}{3} + \tfrac{8}{3} = 8\)).
Examiner trap

Sketch first. If the curve crosses the \(x\)-axis between the limits, one integral gives (area above) \(-\) (area below), not the total area.

Stage 1 · Learn

Check your understanding

Four quick checks. Each correct answer earns XP and records this skill on your profile.

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Evaluate \(\displaystyle\int_{0}^{2} 3x^{2}\,dx\).

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Evaluate \(\displaystyle\int_{1}^{2} x\,dx\).

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A region lies wholly below the \(x\)-axis and \(\displaystyle\int_{a}^{b} y\,dx = -6\). What is its area?

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The line \(y = g(x)\) lies above the curve \(y = f(x)\) between their meeting points \(x = a\) and \(x = b\). Which gives the enclosed area?

Section 03

Between a curve and a line.

Find where the graphs meet: those \(x\)-values are the limits. Then integrate (upper graph \(-\) lower graph). This works wherever the region is, even below the \(x\)-axis, because the subtraction removes the part under both graphs.

Worked example A curve and a line
Find the area of the region enclosed by \(y = x^{2}\) and \(y = x + 2\).
  1. Meet where \(x^{2} = x + 2\): \(x^{2} - x - 2 = (x - 2)(x + 1) = 0\), so \(x = -1, 2\).
  2. The line is on top between them (test \(x = 0\): \(2 > 0\)).
  3. \(\displaystyle\int_{-1}^{2} (x + 2 - x^{2})\,dx = \left[\tfrac{x^{2}}{2} + 2x - \tfrac{x^{3}}{3}\right]_{-1}^{2}\).
  4. \(= \left(2 + 4 - \tfrac{8}{3}\right) - \left(\tfrac{1}{2} - 2 + \tfrac{1}{3}\right) = \tfrac{10}{3} + \tfrac{7}{6}\).
Answer: \(\dfrac{9}{2}\) (Fig. 1)
Worked example A region not touching the axis
Find the area of the region between \(y = 9 - x^{2}\) and the line \(y = 5\).
  1. Meet where \(9 - x^{2} = 5\): \(x = \pm 2\).
  2. Curve minus line: \(9 - x^{2} - 5 = 4 - x^{2}\).
  3. \(\displaystyle\int_{-2}^{2} (4 - x^{2})\,dx = \left[4x - \tfrac{x^{3}}{3}\right]_{-2}^{2} = \left(8 - \tfrac{8}{3}\right) - \left(-8 + \tfrac{8}{3}\right)\).
Answer: \(\dfrac{32}{3}\)
Your turn
Find the area enclosed by \(y = x^{2}\) and \(y = 2x\).
Examiner trap

Upper minus lower. If you subtract the wrong way you get the negative of the area; an area is never negative, so take the size, but check your sketch.

Section 04

Improper integrals.

Some integrals have an infinite limit, or a function that is undefined at a limit. Cambridge includes simple cases: evaluate with a letter for the awkward limit and see what happens as it tends to infinity or to zero.

Worked example Two simple improper integrals
Evaluate (a) \(\displaystyle\int_{1}^{\infty} \dfrac{1}{x^{2}}\,dx\) (b) \(\displaystyle\int_{0}^{1} \dfrac{1}{\sqrt{x}}\,dx\).
  1. (a) \(\displaystyle\int_{1}^{N} x^{-2}\,dx = \left[-\tfrac{1}{x}\right]_{1}^{N} = 1 - \dfrac{1}{N}\).
  2. As \(N \to \infty\), \(\dfrac{1}{N} \to 0\), so the integral is \(1\).
  3. (b) \(\displaystyle\int x^{-\frac{1}{2}}\,dx = 2\sqrt{x}\), which is defined at 0 although \(\dfrac{1}{\sqrt{x}}\) is not.
  4. \(\left[2\sqrt{x}\right]_{0}^{1} = 2 - 0 = 2\).
Answer: (a) \(1\)   (b) \(2\)
Examiner trap

An improper integral only has a value if the integrated expression settles to a limit. \(\left[-\tfrac{1}{x}\right]\) at \(\infty\) is 0, but something like \(\left[2\sqrt{x}\right]\) at \(\infty\) grows without limit, so \(\displaystyle\int_{1}^{\infty} \dfrac{1}{\sqrt{x}}\,dx\) has no value.

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Evaluate \(\displaystyle\int_{0}^{4} \sqrt{2x + 1}\,dx\).

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Find the area of the region enclosed by the curve \(y = x^{2}\) and the line \(y = x + 2\).

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Find the total area between the curve \(y = x^{2} - 2x\), the \(x\)-axis and the lines \(x = 0\) and \(x = 3\).

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Evaluate \(\displaystyle\int_{1}^{\infty} \dfrac{6}{x^{3}}\,dx\).

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Definite integrals and areas, measured.

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Six original Cambridge-style questions
Evaluating definite and improper integrals, areas above and below the axis, between a curve and a line, and against the y-axis.