Anti-differentiation and indefinite integrals

Maths: Analysis and Approaches · Calculus. The idea in plain words, the rules to remember, the mistake to avoid and a worked example, written and checked by Mega Lecture.

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The idea

Integration is the reverse of differentiation. If you differentiate x³ + 5 or x³ − 2 you get 3x² both times, so going backwards you cannot know the constant. This is why every indefinite integral ends with + C, the constant of integration.

A boundary condition, such as a point on the curve or a value of f(x), lets you find C. Substitute the values and solve.

When a function of ax + b is integrated, divide by a. When the integrand is a function of g(x) multiplied by g′(x), or a multiple of it, use substitution with u = g(x), or reverse the chain rule by inspection. You can always check an answer by differentiating it.

Rules to remember

∫axⁿ dx = ax^{n + 1}/(n + 1) + C, for n ≠ −1
∫(1/x) dx = ln|x| + C
∫sin x dx = −cos x + C; ∫cos x dx = sin x + C
∫e^x dx = e^x + C
Linear inside: ∫e^ax dx = (1/a)e^ax + C; ∫cos ax dx = (1/a) sin ax + C; ∫sin ax dx = −(1/a) cos ax + C
∫1/(ax + b) dx = (1/a) ln|ax + b| + C
∫g′(x)/g(x) dx = ln|g(x)| + C; ∫g′(x)e^g(x) dx = e^g(x) + C

Common mistake

Multiplying by the coefficient of x when you should divide. ∫e^3x dx = (1/3)e^3x + C, not 3e^3x + C. Also do not forget + C.

Worked example

A curve has gradient dy/dx = 4 cos 2x and passes through the point (π/4, 5). Find its equation.

y = ∫4 cos 2x dx = 4 × (1/2) sin 2x + C = 2 sin 2x + C
Substitute x = π/4, y = 5: 5 = 2 sin(π/2) + C = 2 × 1 + C
C = 5 − 2 = 3

Answer: y = 2 sin 2x + 3

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Topic 5: Calculus

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