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Topic 14 · 14.10–14.13

Calculus II: integration

Syllabus 14.10–14.13 · Papers 1 and 2 · no calculus formulas are given in the exam

🎯What you need to be able to do

  • Understand integration as the reverse of differentiation, and include the arbitrary constant in indefinite integrals.
  • Integrate sums of powers of \(x\), including \( \dfrac{1}{x} \) and \( \dfrac{1}{ax + b} \).
  • Integrate \( (ax + b)^{n} \) for any rational \(n\), \( \sin(ax + b) \), \( \cos(ax + b) \), \( \sec^{2}(ax + b) \) and \( e^{ax + b} \).
  • Find a curve from its gradient function and a point on it.
  • Evaluate definite integrals, including “hence” questions that use a derivative found earlier.
  • Find plane areas: under a curve, between a line and a curve, between two curves, and areas made of two parts.

📚The mathematics

Reverse differentiation

If \( \dfrac{d}{dx}F(x) = f(x) \), then \( \displaystyle\int f(x)\,dx = F(x) + c \). The constant matters: every curve \( y = F(x) + c \) has the same gradient function, and the syllabus says indefinite integrals must include it. To find which curve, substitute a known point to find \(c\).

The standard integrals

\( \displaystyle\int x^{n}\,dx = \frac{x^{n + 1}}{n + 1} + c \quad (n \ne -1) \)
\( \displaystyle\int \frac{1}{x}\,dx = \ln x + c \)
\( \displaystyle\int (ax + b)^{n}\,dx = \frac{(ax + b)^{n + 1}}{a(n + 1)} + c \)
\( \displaystyle\int \frac{1}{ax + b}\,dx = \frac{1}{a}\ln(ax + b) + c \)
\( \displaystyle\int e^{ax + b}\,dx = \frac{1}{a}e^{ax + b} + c \)
\( \displaystyle\int \sin(ax + b)\,dx = -\frac{1}{a}\cos(ax + b) + c \)
\( \displaystyle\int \cos(ax + b)\,dx = \frac{1}{a}\sin(ax + b) + c \)
\( \displaystyle\int \sec^{2}(ax + b)\,dx = \frac{1}{a}\tan(ax + b) + c \)

Every result on the second row, and the two with \( ax + b \) on the first, follow one rule: integrate as if the bracket were \(x\), then divide by \(a\). Check any answer by differentiating it. Angles are in radians. Products and quotients are not integrated directly at this level: expand or divide first, so \( \dfrac{x^{3} - 4}{x^{2}} = x - 4x^{-2} \).

Definite integrals and “hence” questions

\( \displaystyle\int_{a}^{b} f(x)\,dx = F(b) - F(a) \); the constant cancels, so leave it out. On Paper 1 the limits are chosen to give exact values: \( \tan\tfrac{\pi}{3} = \sqrt{3} \), \( \ln e = 1 \), \( e^{0} = 1 \).

A common 0606 structure is “(a) Differentiate \( F(x) \). (b) Hence find \( \int f(x)\,dx \)”. Part (a) gives \( F'(x) = \) something close to \( f(x) \); rearrange so that the integral you want is isolated, and integrate both sides.

Areas

  • Under a curve (above the axis): \( \int_{a}^{b} y\,dx \).
  • Below the axis the integral is negative. For a region that crosses the axis, integrate each part separately and add the sizes — the “sum of two areas” in the syllabus.
  • Between two graphs: \( \int_{a}^{b} (\text{upper} - \text{lower})\,dx \), with \(a\) and \(b\) where they meet. This works whether or not the region is above the axis.
The cubic y = x(x minus 2)(x minus 3) for 0 to 3. The region above the axis from 0 to 2 has area eight thirds; the region below the axis from 2 to 3 has area five twelfths. A single integral from 0 to 3 gives nine quarters, which is the difference, not the total area thirty-seven twelfths.
\( \int_{0}^{3} \) gives \( \tfrac{8}{3} - \tfrac{5}{12} = \tfrac{9}{4} \). The shaded area is \( \tfrac{8}{3} + \tfrac{5}{12} = \tfrac{37}{12} \).

✏️Worked example 1 (no calculator)

(a) A curve has gradient \( \dfrac{dy}{dx} = 6x^{2} - \dfrac{2}{x^{2}} \) and passes through \( (1, 5) \). Find its equation. [4] (b) Find the exact value of \( \displaystyle\int_{0}^{\pi/6} \left(\sec^{2}2x + 4\cos 2x\right)dx \). [4] (c) Differentiate \( x\ln x \). Hence find the exact value of \( \displaystyle\int_{1}^{e} \ln x\,dx \). [4]

(a) \( y = \displaystyle\int \left(6x^{2} - 2x^{-2}\right)dx = 2x^{3} + 2x^{-1} + c \). At \( (1, 5) \): \( 5 = 2 + 2 + c \), so \( c = 1 \) and \( y = 2x^{3} + \dfrac{2}{x} + 1 \).

(b)

\[ \Big[\tfrac{1}{2}\tan 2x + 2\sin 2x\Big]_{0}^{\pi/6} = \tfrac{1}{2}\tan\tfrac{\pi}{3} + 2\sin\tfrac{\pi}{3} - 0 = \frac{\sqrt{3}}{2} + \sqrt{3} = \frac{3\sqrt{3}}{2} \]

(c) Product rule: \( \dfrac{d}{dx}(x\ln x) = \ln x + 1 \). Integrating both sides, \( x\ln x = \int \ln x\,dx + x \), so \( \int \ln x\,dx = x\ln x - x + c \). Then

\[ \int_{1}^{e} \ln x\,dx = \Big[x\ln x - x\Big]_{1}^{e} = (e - e) - (0 - 1) = 1 \]
Check it. Differentiate each answer. (a) \( \dfrac{d}{dx}\left(2x^{3} + 2x^{-1} + 1\right) = 6x^{2} - 2x^{-2} \) ✓. (c) \( \dfrac{d}{dx}(x\ln x - x) = \ln x + 1 - 1 = \ln x \) ✓.
Forgetting to divide by \(a\). \( \int \sec^{2}2x\,dx = \tfrac{1}{2}\tan 2x \), not \( \tan 2x \); and \( \int \cos 2x\,dx = \tfrac{1}{2}\sin 2x \). Differentiating your answer would show the stray factor of 2 at once.

✏️Worked example 2 (no calculator)

The line \( y = x + 1 \) meets the curve \( y = x^{2} - 4x + 5 \) at \(A\) and \(B\). Find the area of the region enclosed by the line and the curve. [6]
The parabola y = x squared minus 4x plus 5 and the line y = x + 1, meeting at A(1, 2) and B(4, 5). The region between them, under the line and above the curve, is shaded and has area 4.5.
Upper minus lower, integrated between the intersections.

Intersections: \( x^{2} - 4x + 5 = x + 1 \Rightarrow x^{2} - 5x + 4 = 0 \Rightarrow x = 1, 4 \), so \( A(1, 2) \) and \( B(4, 5) \). Between them the line is above the curve (at \( x = 2 \): \( 3 > 1 \)):

\[ \int_{1}^{4} \left[(x + 1) - (x^{2} - 4x + 5)\right]dx = \int_{1}^{4} \left(-x^{2} + 5x - 4\right)dx = \Big[-\tfrac{x^{3}}{3} + \tfrac{5x^{2}}{2} - 4x\Big]_{1}^{4} \]
\[ = \left(-\tfrac{64}{3} + 40 - 16\right) - \left(-\tfrac{1}{3} + \tfrac{5}{2} - 4\right) = \tfrac{8}{3} + \tfrac{11}{6} = \tfrac{9}{2} \]
Check it. Any parabola \( y = x^{2} + \ldots \) cut by a chord between \( x = p \) and \( x = q \) encloses \( \tfrac{(q - p)^{3}}{6} \): here \( \tfrac{27}{6} = 4.5 \) ✓. (A useful check, not a method the mark scheme will accept on its own.)

📝Practise

Written in the style of the current Paper 1 (no calculator) and Paper 2 (calculator) questions.

1. (No calculator.) Find \( \displaystyle\int \left(3x^{2} - \frac{4}{x^{2}} + 2\sqrt{x}\right)dx \). [3]
\( x^{3} + 4x^{-1} + \tfrac{4}{3}x^{\frac{3}{2}} + c \).
2. (No calculator.) Find (a) \( \displaystyle\int (2x - 1)^{5}\,dx \), (b) \( \displaystyle\int \frac{6}{3x + 2}\,dx \), (c) \( \displaystyle\int e^{4x - 1}\,dx \). [5]
(a) \( \dfrac{(2x - 1)^{6}}{12} + c \). (b) \( 2\ln(3x + 2) + c \). (c) \( \tfrac{1}{4}e^{4x - 1} + c \).
3. (No calculator.) Find the exact value of \( \displaystyle\int_{0}^{\pi/4} \sin 2x\,dx \). [2]
\( \Big[-\tfrac{1}{2}\cos 2x\Big]_{0}^{\pi/4} = -\tfrac{1}{2}\cos\tfrac{\pi}{2} + \tfrac{1}{2}\cos 0 = \tfrac{1}{2} \).
4. (No calculator.) Given that \( \displaystyle\int_{1}^{k} (2x - 1)\,dx = 12 \), where \( k > 1 \), find \(k\). [3]
\( \Big[x^{2} - x\Big]_{1}^{k} = k^{2} - k - 0 = 12 \Rightarrow (k - 4)(k + 3) = 0 \), and \( k > 1 \), so \( k = 4 \).
5. (No calculator.) Show that \( \dfrac{d}{dx}\left(xe^{2x}\right) = (1 + 2x)e^{2x} \). Hence find \( \displaystyle\int xe^{2x}\,dx \). [5]
Product rule: \( e^{2x} + 2xe^{2x} = (1 + 2x)e^{2x} \). Integrating: \( xe^{2x} = \int e^{2x}\,dx + 2\int xe^{2x}\,dx = \tfrac{1}{2}e^{2x} + 2\int xe^{2x}\,dx \). So \( \int xe^{2x}\,dx = \tfrac{1}{2}xe^{2x} - \tfrac{1}{4}e^{2x} + c \).
6. (Calculator.) Find the area of the region enclosed by the curve \( y = 6 - x^{2} \) and the line \( y = x \). [5]
\( 6 - x^{2} = x \Rightarrow x^{2} + x - 6 = 0 \Rightarrow x = -3, 2 \). Area \( = \displaystyle\int_{-3}^{2} (6 - x - x^{2})\,dx = \Big[6x - \tfrac{x^{2}}{2} - \tfrac{x^{3}}{3}\Big]_{-3}^{2} = \tfrac{22}{3} - \left(-\tfrac{27}{2}\right) = \tfrac{125}{6} = 20.8 \).
7. (Calculator.) Find the total area between the curve \( y = 2\sin x \) and the \(x\)-axis for \( 0 \le x \le \tfrac{3\pi}{2} \). [4]
Split at \( x = \pi \), where the curve crosses the axis. \( \int_{0}^{\pi} 2\sin x\,dx = \Big[-2\cos x\Big]_{0}^{\pi} = 4 \). \( \int_{\pi}^{3\pi/2} 2\sin x\,dx = \Big[-2\cos x\Big]_{\pi}^{3\pi/2} = 0 - 2 = -2 \), an area of 2 below the axis. Total area \( = 4 + 2 = 6 \). (The single integral from 0 to \( \tfrac{3\pi}{2} \) is 2, which is not the area.)

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

  • Desmos — type an integral with limits to check a Paper 1 answer numerically
  • Khan Academy — definite integrals and the area between curves