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O-Level Additional Mathematics · Secondary 3 and 4

Integration

Integration is the reverse of differentiation. If you know the gradient function dy/dx, integration recovers y, up to a constant. In A-Math you integrate powers of x, functions of the form (ax + b)^n, trigonometric and exponential functions, evaluate definite integrals, and use them to find areas bounded by curves and lines. You also use a result from differentiation to work out an integral that has no direct rule. Integration completes the calculus strand of the course and is the foundation for the much wider integration work in H2 Mathematics.

What the syllabus expects

  • Understand integration as the reverse of differentiation and include the arbitrary constant in indefinite integrals
  • Integrate x^n for rational n (n ≠ -1), and sums and constant multiples of such terms
  • Integrate (ax + b)^n, sin(ax + b), cos(ax + b), sec²(ax + b) and e^(ax + b)
  • Find the equation of a curve given its gradient function and a point on it
  • Evaluate definite integrals
  • Find the area between a curve and the x-axis, including regions below the axis, and between a curve and a straight line
  • Use a given derivative to find a related integral

Indefinite integrals and the constant

If the derivative of F(x) is f(x), then ∫f(x) dx = F(x) + c. The constant c appears because the derivative of any constant is zero, so the integral is only known up to a constant. For powers of x, ∫x^n dx = x^(n+1)/(n + 1) + c for n ≠ -1: raise the power by one and divide by the new power. As with differentiation, rewrite roots and fractions first: ∫(6x² + 4/x³) dx = ∫(6x² + 4x^(-3)) dx = 2x³ - 2x^(-2) + c = 2x³ - 2/x² + c. Expand products before integrating, so ∫(x + 2)(x - 2) dx = ∫(x² - 4) dx = x³/3 - 4x + c. To find a curve from its gradient function, integrate and then use a given point to find c. If dy/dx = 3x² - 2 and the curve passes through (1, 4), then y = x³ - 2x + c and 4 = 1 - 2 + c, so c = 5 and y = x³ - 2x + 5.

Standard integrals

Reversing the chain rule for a linear inside function gives these results, each with + c: ∫(ax + b)^n dx = (ax + b)^(n+1)/(a(n + 1)) for n ≠ -1, ∫sin(ax + b) dx = -1/a cos(ax + b), ∫cos(ax + b) dx = 1/a sin(ax + b), ∫sec²(ax + b) dx = 1/a tan(ax + b), and ∫e^(ax + b) dx = 1/a e^(ax + b). The pattern is the same each time: integrate as if the bracket were just x, then divide by a, the coefficient of x. For example, ∫(2x - 5)⁴ dx = (2x - 5)⁵/10 + c and ∫e^(3x) dx = e^(3x)/3 + c. Check by differentiating your answer. Trigonometric integrals need x in radians. When the integrand is sin²x or cos²x, use a double angle formula first, such as cos²x = ½(1 + cos 2x).

Definite integrals and areas

A definite integral has limits. The integral of f(x) from x = a to x = b equals F(b) - F(a), where F is any integral of f, and the constant cancels so it is left out. For example, the integral of 3x² from x = 1 to x = 2 is [x³] from 1 to 2 = 8 - 1 = 7. If y = f(x) is above the x-axis between x = a and x = b, this integral is the area between the curve and the axis. If the curve is below the axis, the integral is negative and the area is its absolute value. When a region crosses the axis, find where the curve cuts the axis and integrate each part separately, otherwise the positive and negative parts cancel. For example, y = x² - 4x cuts the axis at x = 0 and x = 4. The integral from 0 to 4 is [x³/3 - 2x²] from 0 to 4 = 64/3 - 32 = -32/3, so the area enclosed by the curve and the axis is 32/3 units².

Areas with lines, and integrals from derivatives

For a region between an upper graph y = f(x) and a lower graph y = g(x) from x = a to x = b, the area is the integral of (f(x) - g(x)) from a to b. Find a and b by solving f(x) = g(x). Alternatively, find the area under the upper graph and subtract the area of a triangle or trapezium under the line. Some integrals have no direct rule in A-Math but can be found from a derivative you are told to work out first. For example, differentiating x ln x by the product rule gives ln x + 1. So ∫(ln x + 1) dx = x ln x + c, which rearranges to ∫ln x dx = x ln x - x + c. The word 'hence' is the usual signal to use this method.

Worked example

The line y = x - 1 meets the curve y = 6x - x² - 5 at the points A and B. Find the x-coordinates of A and B, and find the area of the region enclosed by the line and the curve.

  1. At the points of intersection, 6x - x² - 5 = x - 1, so x² - 5x + 4 = 0.
  2. Factorise: (x - 1)(x - 4) = 0, so A and B have x = 1 and x = 4.
  3. Between x = 1 and x = 4 the curve is above the line. For example, at x = 2 the curve gives 3 and the line gives 1.
  4. Area = integral from 1 to 4 of [(6x - x² - 5) - (x - 1)] dx = integral from 1 to 4 of (5x - x² - 4) dx.
  5. Integrate: [5x²/2 - x³/3 - 4x] from 1 to 4.
  6. At x = 4: 40 - 64/3 - 16 = 8/3. At x = 1: 5/2 - 1/3 - 4 = -11/6.
  7. Area = 8/3 - (-11/6) = 16/6 + 11/6 = 27/6 = 9/2.

Answer: x = 1 and x = 4, and the enclosed area is 9/2 = 4.5 units²

Common mistakes

  • Leaving out + c in an indefinite integral. Without it the answer is incomplete, and you cannot find the equation of a curve through a given point.
  • Forgetting to divide by a when integrating a function of (ax + b). ∫cos 2x dx is ½ sin 2x + c, not sin 2x + c.
  • Getting the sign wrong on trigonometric integrals. ∫sin x dx = -cos x + c and ∫cos x dx = sin x + c.
  • Integrating in one go across a point where the curve crosses the x-axis. The parts above and below cancel, so split the integral at the crossing point and add the sizes of the areas.
  • Integrating a product as the product of two integrals, such as treating ∫x(x + 1) dx as ∫x dx × ∫(x + 1) dx. Expand into separate powers of x first.
  • Using ∫x^n dx = x^(n+1)/(n + 1) when n = -1, which would mean dividing by zero. The power rule for integration does not cover 1/x.

Exam tips

  • Differentiate your answer as a check. It takes a few seconds and catches a missing 1/a or a wrong sign.
  • For area questions, sketch the curve and mark where it crosses the x-axis before you integrate. The sketch tells you whether to split the integral.
  • Keep exact values such as 9/2 or π/4 until the end, and round only if the question asks for a decimal.

Practice questions

1. Find ∫(3x² - 4/x² + 2) dx.

Show answer

x³ + 4/x + 2x + c. Write -4/x² as -4x^(-2), which integrates to -4x^(-1)/(-1) = 4x^(-1) = 4/x. The other terms give x³ and 2x.

2. Evaluate (a) the integral of (2x - 1)³ from x = 1 to x = 2, and (b) the integral of cos 3x from x = 0 to x = π/6.

Show answer

(a) 10. The integral is [(2x - 1)⁴/8] from 1 to 2 = (81 - 1)/8 = 10. (b) 1/3. The integral is [1/3 sin 3x] from 0 to π/6 = 1/3 sin(π/2) - 0 = 1/3.

3. Differentiate xe^(2x) with respect to x. Hence find ∫xe^(2x) dx.

Show answer

The derivative is e^(2x) + 2xe^(2x), and ∫xe^(2x) dx = ½xe^(2x) - ¼e^(2x) + c. Integrating the derivative gives ∫(e^(2x) + 2xe^(2x)) dx = xe^(2x), so 2∫xe^(2x) dx = xe^(2x) - ∫e^(2x) dx = xe^(2x) - ½e^(2x). Dividing by 2 and adding the constant gives the result.

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