Home / IGCSE Add Maths 0606 / Calculus / Definite integrals and areas
Cambridge IGCSE · Topic 14.13
IGCSE Add Maths 0606 / Topic 14 / IGCSE

Put in the limits, take out the area.

An indefinite integral gives a family of functions. Add limits and the answer becomes a single number: \(F(b) - F(a)\). That number measures the area under a curve, as long as you watch for regions below the \(x\)-axis, and by subtracting one curve from another it measures the area trapped between a line and a curve or between two curves.

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The key idea

If \(F'(x) = f(x)\), then \(\displaystyle\int_{a}^{b} f(x)\,dx = \big[F(x)\big]_{a}^{b} = F(b) - F(a)\), with no \(+\,c\). For a curve above the \(x\)-axis this is the area between the curve and the axis; below the axis the integral is negative, so split at the roots and add the sizes. The area between two graphs from \(x = a\) to \(x = b\) is \(\displaystyle\int_{a}^{b} (\text{top} - \text{bottom})\,dx\), where the limits are usually the \(x\)-coordinates of the points of intersection.

x y O 1 2 3 4 5 6 5 10 A(1, 5) B(4, 8) R y = 6x − x² y = x + 4
Fig. 1: the curve y = 6x - x² and the line y = x + 4 meet at A(1, 5) and B(4, 8). The shaded region R between them has area 4.5, found by integrating (curve - line) from x = 1 to x = 4
Section 01

Evaluating a definite integral.

Integrate as usual, write the result in square brackets with the limits on the right, then substitute the top limit and subtract the value at the bottom limit. The constant of integration cancels in the subtraction, so leave it out.

\[\int_{a}^{b} f(x)\,dx = \Big[F(x)\Big]_{a}^{b} = F(b) - F(a)\]

Questions often ask for an exact value. Keep \(\ln\), \(e\), \(\pi\) and surds in the answer, use the laws of logarithms to tidy up, and work in radians for every trigonometric integral (\(\sin\frac{\pi}{3} = \frac{\sqrt{3}}{2}\), \(\ln 1 = 0\), \(e^{0} = 1\)).

Worked example A polynomial
Evaluate \(\displaystyle\int_{1}^{3} (3x^{2} - 4x + 1)\,dx\).
  1. Integrate each term: \(\Big[x^{3} - 2x^{2} + x\Big]_{1}^{3}\).
  2. Top limit: \(27 - 18 + 3 = 12\).
  3. Bottom limit: \(1 - 2 + 1 = 0\).
  4. Subtract: \(12 - 0\).
Answer: \(12\)
Worked example Exact answers with ln and pi
Find the exact values of (a) \(\displaystyle\int_{1}^{5} \frac{4}{2x - 1}\,dx\) (b) \(\displaystyle\int_{0}^{\pi/6} \cos 2x\,dx\).
  1. (a) \(\displaystyle\int \frac{4}{2x - 1}\,dx = 4 \times \frac{1}{2}\ln(2x - 1) = 2\ln(2x - 1)\).
  2. \(\Big[2\ln(2x - 1)\Big]_{1}^{5} = 2\ln 9 - 2\ln 1 = 2\ln 9\).
  3. \(2\ln 9 = \ln 81 = 4\ln 3\).
  4. (b) \(\Big[\tfrac{1}{2}\sin 2x\Big]_{0}^{\pi/6} = \tfrac{1}{2}\sin\tfrac{\pi}{3} - \tfrac{1}{2}\sin 0\).
  5. \(= \tfrac{1}{2} \times \tfrac{\sqrt{3}}{2} - 0\).
Answer: (a) \(4\ln 3\)   (b) \(\dfrac{\sqrt{3}}{4}\)
Your turn
Given that \(\displaystyle\int_{1}^{k} (2x + 1)\,dx = 18\) and \(k > 1\), find \(k\).
Examiner trap

The order is top limit MINUS bottom limit, and the whole bottom value is subtracted. Bracket it: \(\big(27 - 18 + 3\big) - \big(1 - 2 + 1\big)\). Forgetting the bottom limit because 'it is probably zero' loses marks whenever it is not, as with \(e^{0} = 1\) or \(\cos 0 = 1\).

Section 02

Area under a curve, above and below the axis.

For a curve above the \(x\)-axis, the area between the curve, the axis and the lines \(x = a\) and \(x = b\) is \(\displaystyle\int_{a}^{b} y\,dx\). When the curve dips below the axis, \(y\) is negative there, so the integral of that part is negative. An area is never negative.

Worked example A region above the axis
Find the area of the region enclosed by the curve \(y = 4x - x^{2}\) and the \(x\)-axis.
  1. Roots: \(x(4 - x) = 0\), so \(x = 0\) and \(x = 4\). Between them \(y > 0\).
  2. \(\displaystyle\int_{0}^{4} (4x - x^{2})\,dx = \Big[2x^{2} - \tfrac{1}{3}x^{3}\Big]_{0}^{4}\).
  3. \(= \big(32 - \tfrac{64}{3}\big) - 0 = \tfrac{32}{3}\).
Answer: \(\dfrac{32}{3}\) square units
Worked example Part of the region below the axis
The curve \(y = x^{2} - 4x + 3\) crosses the \(x\)-axis at \(x = 1\) and \(x = 3\). Find the total area between the curve, the \(x\)-axis and the lines \(x = 0\) and \(x = 3\).
  1. Let \(F(x) = \tfrac{1}{3}x^{3} - 2x^{2} + 3x\). Then \(F(0) = 0\), \(F(1) = \tfrac{4}{3}\), \(F(3) = 9 - 18 + 9 = 0\).
  2. From 0 to 1 (above the axis): \(F(1) - F(0) = \tfrac{4}{3}\).
  3. From 1 to 3 (below the axis): \(F(3) - F(1) = -\tfrac{4}{3}\), an area of \(\tfrac{4}{3}\).
  4. Total area \(= \tfrac{4}{3} + \tfrac{4}{3}\).
  5. Check: \(\displaystyle\int_{0}^{3} y\,dx = 0\) in one go, because the two equal pieces cancel. That is why you must split.
Answer: \(\dfrac{8}{3}\) square units
Your turn
Find the exact area of the region bounded by \(y = e^{2x}\), the \(x\)-axis and the lines \(x = 0\) and \(x = \ln 3\).
Examiner trap

Never integrate straight across a root. A curve that is above the axis for part of the interval and below it for the rest gives (area above) minus (area below), which can even be zero. Find the roots first, integrate each piece, then add the magnitudes.

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Evaluate \(\displaystyle\int_{1}^{2} 3x^{2}\,dx\).

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Evaluate \(\displaystyle\int_{0}^{\pi/2} \sin x\,dx\).

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\(\displaystyle\int_{-1}^{1} (x^{2} - 1)\,dx = -\tfrac{4}{3}\). What is the area of the region between \(y = x^{2} - 1\) and the \(x\)-axis?

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Find the total area enclosed between \(y = \sin x\) and the \(x\)-axis from \(x = 0\) to \(x = 2\pi\).

Section 03

Between a line and a curve, or two curves.

To find the area trapped between two graphs, find where they meet by solving their equations simultaneously; the \(x\)-coordinates of the intersections are usually the limits. Then integrate (upper graph minus lower graph). Subtracting first means you need only one integral, and it works even when part of the region lies below the \(x\)-axis.

\[\text{Area} = \int_{a}^{b} \big(y_{\text{top}} - y_{\text{bottom}}\big)\,dx\]

For a straight line you can instead find the area under the curve and subtract the area of the trapezium (or triangle) under the line. Some regions are bounded by different graphs on different parts, so the area is a sum of two areas: split where the boundary changes.

Worked example A line and a curve (Fig. 1)
Find the area of the region \(R\) enclosed by the curve \(y = 6x - x^{2}\) and the line \(y = x + 4\).
  1. Intersections: \(6x - x^{2} = x + 4 \Rightarrow x^{2} - 5x + 4 = 0 \Rightarrow (x - 1)(x - 4) = 0\), so \(A(1, 5)\) and \(B(4, 8)\).
  2. Between \(x = 1\) and \(x = 4\) the curve is on top (at \(x = 2\): curve 8, line 6).
  3. \(\displaystyle\int_{1}^{4} \big(6x - x^{2} - (x + 4)\big)\,dx = \int_{1}^{4} (-x^{2} + 5x - 4)\,dx = \Big[-\tfrac{1}{3}x^{3} + \tfrac{5}{2}x^{2} - 4x\Big]_{1}^{4}\).
  4. \(= \big(-\tfrac{64}{3} + 40 - 16\big) - \big(-\tfrac{1}{3} + \tfrac{5}{2} - 4\big) = \tfrac{8}{3} - \big(-\tfrac{11}{6}\big) = \tfrac{9}{2}\).
  5. Check by the trapezium method: \(\displaystyle\int_{1}^{4} (6x - x^{2})\,dx = \Big[3x^{2} - \tfrac{1}{3}x^{3}\Big]_{1}^{4} = 24\); trapezium under the line \(= \tfrac{1}{2}(5 + 8)(3) = 19.5\); \(24 - 19.5 = 4.5\).
Answer: \(\dfrac{9}{2}\) square units
Worked example Two curves
Find the area of the region enclosed by the curves \(y = x^{2}\) and \(y = 4x - x^{2}\).
  1. Intersections: \(x^{2} = 4x - x^{2} \Rightarrow 2x^{2} - 4x = 0 \Rightarrow 2x(x - 2) = 0\), so \(x = 0\) and \(x = 2\).
  2. At \(x = 1\): \(4x - x^{2} = 3\) and \(x^{2} = 1\), so \(y = 4x - x^{2}\) is on top.
  3. \(\displaystyle\int_{0}^{2} \big(4x - x^{2} - x^{2}\big)\,dx = \Big[2x^{2} - \tfrac{2}{3}x^{3}\Big]_{0}^{2} = 8 - \tfrac{16}{3}\).
Answer: \(\dfrac{8}{3}\) square units
Your turn
The region \(S\) is bounded by the curve \(y = x^{2}\), the line \(y = 6 - x\) and the \(x\)-axis, with \(x \ge 0\). The curve and line meet at \((2, 4)\) and the line meets the \(x\)-axis at \((6, 0)\). Find the area of \(S\).
Examiner trap

Integrating (bottom minus top) gives the right size with a minus sign. Decide which graph is on top by testing one \(x\)-value between the limits, and never quote a negative area: if your answer comes out negative, you subtracted the wrong way round.

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Find the exact value of \(\displaystyle\int_{0}^{\ln 2} e^{2x}\,dx\).

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Find the area of the region enclosed by the curve \(y = x^{2}\) and the line \(y = 3x\).

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Given that \(\displaystyle\int_{0}^{k} (2x + 3)\,dx = 10\), where \(k > 0\), find \(k\).

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Limits in, area out.

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Six original Cambridge-style questions
Definite integrals with exact answers, a curve crossing the axis, the area between a line and a curve, two trigonometric curves and finding unknown limits.