Lesson 6: Integration

Integration is differentiation in reverse — and far more. It computes areas under curves, total distance from speed, the volume of solids, and the solutions of differential equations. Where differentiation is mechanical, integration is an art: recognising which technique fits requires pattern-matching that only practice builds. This lesson covers the A-level core: indefinite and definite integrals, integration by substitution and by parts, areas between curves, and the trapezium rule for numerical integration.

Antidifferentiation — reversing the chain

Integration reverses differentiation: ∫ xⁿ dx = xⁿ⁺¹/(n + 1) + c for n ≠ −1. The constant c appears because differentiation destroys constants — the family of antiderivatives differs by vertical shifts. Never omit +c in indefinite integrals; examiners deduct for it. Example: ∫ (3x² − 4√x + 2/x) dx = x³ − 4 · (2/3)x^{3/2} + 2 ln|x| + c = x³ − (8/3)x^{3/2} + 2 ln x + c (for x > 0).

The integral of 1/x is ln|x| — the exception to the power rule (n = −1 would divide by zero). The modulus matters: ∫ (from −2 to −1) 1/x dx = [ln|x|] = ln 1 − ln 2 = −ln 2. Forgetting the modulus gives ln(−1), which is undefined in real analysis. Standard integrals to memorise: ∫ eˣ dx = eˣ, ∫ sin x dx = −cos x, ∫ cos x dx = sin x, ∫ sec²x dx = tan x, ∫ 1/(1 + x²) dx = arctan x.

Definite integrals evaluate the antiderivative at bounds: ∫ (from a to b) f(x) dx = F(b) − F(a). Example: ∫ (from 1 to 4) (2x + 3/x) dx = [x² + 3 ln x] = (16 + 3 ln 4) − (1 + 0) = 15 + 3 ln 4 ≈ 19.16. The +c cancels — definite integrals give numbers. Geometrically this is the signed area under the curve: area below the axis counts negative.

Integration by substitution

Substitution reverses the chain rule: for ∫ f(g(x)) g'(x) dx, set u = g(x), du = g'(x) dx. Example: ∫ 2x(x² + 1)⁵ dx. Let u = x² + 1, du = 2x dx: the integral becomes ∫ u⁵ du = u⁶/6 + c = (x² + 1)⁶/6 + c. The 2x was the clue — it is (up to a constant) the derivative of the inside. Train yourself to spot 'function and its derivative' pairs.

Definite integrals with substitution: change the limits too. ∫ (from 0 to 2) x√(x² + 1) dx: u = x² + 1, du = 2x dx, so x dx = du/2; when x = 0, u = 1; x = 2, u = 5. Integral = (1/2) ∫ (from 1 to 5) √u du = (1/2) · (2/3)[u^{3/2}] = (1/3)(5√5 − 1) ≈ 3.39. Changing limits avoids back-substitution — cleaner and less error-prone. Alternatively keep x-limits and back-substitute; either is accepted.

Trigonometric substitutions handle roots: for ∫ √(1 − x²) dx, set x = sin θ (then √(1 − x²) = cos θ, dx = cos θ dθ): the integral becomes ∫ cos²θ dθ = ∫ (1 + cos 2θ)/2 dθ = θ/2 + (sin 2θ)/4 + c = (1/2) arcsin x + (x√(1 − x²))/2 + c. This is the area of a circular segment — geometry emerging from substitution. Standard substitutions: √(a² − x²) → x = a sin θ; √(a² + x²) → x = a tan θ; √(x² − a²) → x = a sec θ.

Integration by parts

Integration by parts reverses the product rule: ∫ u dv = uv − ∫ v du. The art is choosing u (to differentiate) and dv (to integrate). The LIATE priority — Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential — picks u: differentiate the higher-priority type. Example: ∫ x eˣ dx: u = x (algebraic), dv = eˣ dx: = x eˣ − ∫ eˣ dx = x eˣ − eˣ + c = eˣ(x − 1) + c.

Worked example: ∫ x² sin x dx needs parts twice. First: u = x², dv = sin x dx → −x² cos x + ∫ 2x cos x dx. Second: u = 2x, dv = cos x dx → 2x sin x − ∫ 2 sin x dx = 2x sin x + 2 cos x. Total: −x² cos x + 2x sin x + 2 cos x + c. Tabular integration (differentiate x² → 2x → 2 → 0 down one column, integrate sin x repeatedly down another, alternate signs) does this in one pass — learn it for repeated parts.

Worked example (parts with a logarithm). ∫ x ln x dx. LIATE says u = ln x (logarithmic beats algebraic), dv = x dx, so v = x²/2 and du = dx/x. Then ∫ x ln x dx = (x²/2) ln x − ∫ (x²/2)(1/x) dx = (x²/2) ln x − ∫ x/2 dx = (x²/2) ln x − x²/4 + c. Check by differentiating: (x²/2)(1/x) + x ln x − x/2 = x ln x ✓.

Cyclic integrals: ∫ eˣ sin x dx. Parts twice returns to the start: let I = ∫ eˣ sin x dx = eˣ sin x − ∫ eˣ cos x dx = eˣ sin x − [eˣ cos x + ∫ eˣ sin x dx] = eˣ(sin x − cos x) − I. So 2I = eˣ(sin x − cos x), I = eˣ(sin x − cos x)/2 + c. Solving for I algebraically — a trick worth memorising. Also: ∫ ln x dx = x ln x − x + c (write as ∫ 1 · ln x dx, u = ln x).

Areas, volumes, and the trapezium rule

The definite integral gives signed area; for total area, split at roots. Area between y = x² − 4 and the x-axis from −3 to 3: the curve crosses at ±2. The antiderivative of x² − 4 is x³/3 − 4x. From −3 to −2: (−8/3 + 8) − (−9 + 12) = 16/3 − 3 = 7/3. From −2 to 2: (8/3 − 8) − (−8/3 + 8) = −32/3, absolute value 32/3. From 2 to 3: (9 − 12) − (8/3 − 8) = 7/3. Total = 7/3 + 32/3 + 7/3 = 46/3 ≈ 15.33. Sketch first — the sketch tells you where to split.

Area between two curves: ∫ (top − bottom) dx between intersection x-values. For y = x² and y = 2x + 3: intersections at x² = 2x + 3 → x = −1, 3. Area = ∫ (from −1 to 3) (2x + 3 − x²) dx = [x² + 3x − x³/3] = (9 + 9 − 9) − (1 − 3 + 1/3) = 9 − (−5/3) = 32/3. Volumes of revolution: rotating y = f(x) about the x-axis gives V = π ∫ y² dx. Rotating y = √x from 0 to 4: V = π ∫ (from 0 to 4) x dx = π[x²/2] = 8π.

The trapezium rule approximates ∫ (from a to b) f(x) dx ≈ (h/2)[f₀ + fₙ + 2(f₁ + ... + fₙ₋₁)] with h = (b − a)/n. Example: ∫ (from 0 to 1) e^{x²} dx (no elementary antiderivative!) with n = 4, h = 0.25: values e⁰, e^{0.0625}, e^{0.25}, e^{0.5625}, e¹ ≈ 1, 1.0645, 1.2840, 1.7551, 2.7183. Approximation = 0.125 × [1 + 2.7183 + 2(1.0645 + 1.2840 + 1.7551)] = 0.125 × [3.7183 + 8.2072] ≈ 1.4907 (true ≈ 1.4627 — overestimate, since e^{x²} is convex). More strips → better accuracy; Simpson's rule (Further Maths) converges faster.

Differential equations preview and exam technique

Integration solves the simplest differential equations: dy/dx = f(x) ⟹ y = ∫ f(x) dx. With dy/dx = 3x² − 2 and y(0) = 5: y = x³ − 2x + c, and 5 = 0 − 0 + c gives c = 5, so y = x³ − 2x + 5. The initial condition fixes the constant — the '+c' is not decoration but the degree of freedom of the general solution. This is the seed of Lesson 11.

Exam technique: (1) for indefinite integrals, always +c; (2) for definite integrals, write the [F(x)] line before substituting — it earns method marks; (3) for area problems, sketch and split at roots — signed vs total area is the classic trap; (4) choose substitution vs parts by pattern: composition → substitution, product of different types → parts; (5) in trapezium rule, tabulate x and f(x) values carefully — one table error ruins the answer; (6) give exact answers (in terms of π, ln 2) unless decimals are requested.

Check by differentiating: the answer to ∫ (2x + 1)³ dx = (2x + 1)⁴/8 + c differentiates (chain rule) to 4(2x + 1)³ · 2/8 = (2x + 1)³ ✓. Differentiation-checking takes seconds and catches most errors. Make it a habit on every integration question.

Common mistakes

• Dropping the +c. An indefinite integral without +c is an incomplete answer and loses the accuracy mark — differentiation destroys constants, so integration must restore the family.
• Applying the power rule to 1/x. ∫ x⁻¹ dx is ln|x| + c, not x⁰/0. And the modulus matters on negative intervals.
• Substituting without changing the limits. In a definite integral, either convert the limits to u-values or back-substitute to x before evaluating. Mixing x-limits with a u-antiderivative is a guaranteed error.
• Reporting signed area as area. If part of the curve dips below the axis, split at the roots and take absolute values of each piece. A negative 'area' answer is always wrong.
• Trapezium rule slips. The end ordinates appear once, the middle ordinates twice, and the whole bracket is multiplied by h/2. One doubling error propagates through the entire estimate.

Exam Experience

Questions in the style of the A-level pure paper. Attempt them before reading the mark scheme.

1. Find ∫ (3x² − 4/x) dx. [2]

2. Using the substitution u = x² + 1, show that ∫ (from 0 to 1) x(x² + 1)³ dx = 15/8. [4]

3. Show that ∫ (from 1 to e) ln x dx = 1. [3]

Mark scheme.

• 1. ∫ 3x² dx = x³ and ∫ 4/x dx = 4 ln|x| (M1) for both antiderivatives (note ln, not the power rule, for 1/x). = x³ − 4 ln|x| + c (A1) — the +c is required for the accuracy mark.
• 2. du = 2x dx, so x dx = du/2; limits: x = 0 → u = 1, x = 1 → u = 2 (M1) for the substitution and converted limits. Integral = (1/2) ∫ (from 1 to 2) u³ du = (1/2)[u⁴/4] (M1). = (1/8)(16 − 1) = 15/8 (A1) — target reached exactly.
• 3. Write as ∫ 1 · ln x dx and use parts with u = ln x, dv = dx: ∫ ln x dx = x ln x − ∫ x · (1/x) dx (M1). = x ln x − x (A1). Evaluate from 1 to e: (e · 1 − e) − (0 − 1) = 0 + 1 = 1 (A1).