Lesson 16: Integration (A-level Intro)

Integration is differentiation in reverse — and the tool for totals: areas, distances from speeds, volumes of water accumulated over a day. If differentiation asks 'how fast is it changing?', integration asks 'how much has built up?' This lesson introduces the integral, the constant of integration (and how to pin it down), definite integrals, and areas under curves — the second chapter of calculus and the perfect partner to Lesson 15.

The indefinite integral and the +C

To integrate xⁿ: raise the power by 1 and divide by the new power: ∫xⁿ dx = xⁿ⁺¹/(n + 1) + C. So ∫x² dx = x³/3 + C, ∫x³ dx = x⁴/4 + C, ∫1 dx = x + C. Terms integrate separately: ∫(6x² − 4x) dx = 2x³ − 2x² + C.

The +C (constant of integration) is essential: derivatives of x³/3 + 5 and x³/3 − 2 are both x², so integrating x² recovers a family of curves, not one. Forgetting +C in an indefinite integral is the most common integration error in existence — write it every time. Check any integral by differentiating your answer: d/dx (2x³ − 2x² + C) = 6x² − 4x. ✓

Worked example: integrate 4x³ − 2x + 7: ∫ = x⁴ − x² + 7x + C. (The constant 7 integrates to 7x — constants become terms in x; only the extra +C vanishes on differentiating.) Worked example with rewritten powers: ∫√x dx = ∫x^(1/2) dx = (2/3)x^(3/2) + C, and ∫(3/x²) dx = ∫3x^(−2) dx = −3x^(−1) + C = −3/x + C. Differentiate both to check: d/dx ((2/3)x^(3/2)) = x^(1/2) = √x. ✓ d/dx (−3x^(−1)) = 3x^(−2) = 3/x². ✓ One warning: the power rule fails for x^(−1) — dividing by the new power 0 is impossible. Integrating 1/x gives ln|x|, which you will meet properly at A-level.

Finding C from a point

A point on the curve fixes C. Worked example: a curve has gradient function dy/dx = 4x − 3 and passes through (2, 5). Integrate: y = 2x² − 3x + C. Substitute (2, 5): 5 = 2(4) − 3(2) + C = 8 − 6 + C = 2 + C, so C = 3 and y = 2x² − 3x + 3. Check by differentiating: 4x − 3. ✓ And at x = 2: 8 − 6 + 3 = 5. ✓

Worked example: the gradient of a curve is 3x² and it passes through (1, 4). y = x³ + C; 4 = 1 + C → C = 3; y = x³ + 3. These 'recover the curve' questions test the whole chain — integrate, substitute, solve — and appear wherever calculus is examined.

Real context: a car's acceleration is a = 2t m/s². Velocity is the integral: v = t² + C. If the car starts from rest (v = 0 at t = 0), C = 0, so v = t²; after 3 s, v = 9 m/s. Distance is the integral of velocity: s = t³/3 + C; starting at s = 0 gives s = t³/3, so 9 m after 3 s. Integration chains: acceleration → velocity → distance, each step adding a constant fixed by a condition.

Definite integrals and areas

A definite integral ∫ₐᵇ f(x) dx has limits and no +C (the constants cancel): evaluate the antiderivative at b, subtract the value at a. Worked example: ∫₁⁴ 3x² dx = [x³]₁⁴ = 64 − 1 = 63. The geometric meaning: the area under the curve y = f(x) from a to b (when f ≥ 0). So the area under y = 3x² from 1 to 4 is 63 square units.

Worked example: area under y = x² + 1 from x = 0 to x = 3: ∫₀³ (x² + 1) dx = [x³/3 + x]₀³ = (27/3 + 3) − 0 = 9 + 3 = 12. Sketch it: a parabola shifted up — the area exceeds the 3 × 1 rectangle beneath y = 1 (area 3) plus the curved part (9). Plausible. ✓

Worked example (area between a curve and a line): find the area enclosed by y = x² and y = x + 2. First find the intersections: x² = x + 2 → x² − x − 2 = 0 → (x − 2)(x + 1) = 0 → x = −1 and x = 2. The line is on top between these. Area = ∫₋₁² [(x + 2) − x²] dx = [x²/2 + 2x − x³/3]₋₁² = (2 + 4 − 8/3) − (1/2 − 2 + 1/3) = (10/3) − (−7/6) = 20/6 + 7/6 = 27/6 = 4.5 square units. 'Top minus bottom, between the intersection points' — the recipe never changes.

Negative areas: if a curve dips below the x-axis, its integral there is negative and cancels positive area. For an actual geometric area, split the integral at the roots and take absolute values of each piece. Example: y = x² − 4 from 0 to 3 crosses zero at x = 2; the integral ∫₀³ = [x³/3 − 4x]₀³ = 9 − 12 = −3, but the true shaded area is |∫₀²| + ∫₂³ = 16/3 + 7/3 = 23/3. Examiners set this trap deliberately — sketch first.

Applications — accumulation in the real world

Worked example: water flows into a reservoir in Snowdonia at rate r = 6t + 100 litres per minute (t minutes after midnight). Total water in the first 10 minutes: ∫₀¹⁰ (6t + 100) dt = [3t² + 100t]₀¹⁰ = 300 + 1,000 = 1,300 litres. Differentiation gives rates from totals; integration rebuilds totals from rates — the two directions of the same idea, formalised by the fundamental theorem of calculus.

Worked example: a train's velocity is v = 20 − 2t m/s for the first 10 s of braking. Distance travelled while braking: ∫₀¹⁰ (20 − 2t) dt = [20t − t²]₀¹⁰ = 200 − 100 = 100 m. Check with the graph: a triangle with base 10 and height 20 has area 100. ✓ The integral of a linear function is just the trapezium/triangle area — a useful check.

Estimation check: the trapezium rule (A-level) approximates integrals by strips; even at GCSE level, comparing an integral with a rough rectangle estimate (width × average height) catches order-of-magnitude errors.

Worked example — the area between a curve and the x-axis, carefully

Find the exact area enclosed by y = x² − 2x and the x-axis between x = 0 and x = 3. The curve crosses the axis where x² − 2x = 0 → x = 0 and x = 2; it dips below the axis on [0, 2] and rises above on [2, 3]. Integrate each piece separately and take absolute values. Below: ∫₀² (x² − 2x) dx = [x³/3 − x²]₀² = 8/3 − 4 = −4/3 → area 4/3. Above: ∫₂³ = [x³/3 − x²]₂³ = (27/3 − 9) − (8/3 − 4) = 0 − (−4/3) = 4/3. Total shaded area = 4/3 + 4/3 = 8/3 ≈ 2.67 square units. Had you integrated straight from 0 to 3, the pieces would have cancelled to 0 — the trap from the theory section, caught by sketching first.

Worked example: the area under y = 4 − x² from x = 0 to x = 2 (the curve sits above the axis here, touching it at x = 2): ∫₀² (4 − x²) dx = [4x − x³/3]₀² = 8 − 8/3 = 16/3 ≈ 5.33. Check plausibility: the rectangle 2 × 4 = 8 bounds it above, and the curve sags — 5.33 is sensible. Every integral answer deserves this two-second plausibility audit.

Definite integrals and the trapezium idea

A definite integral is an exact area; the trapezium rule estimates it by slicing the region into strips and treating each as a trapezium. Estimate ∫₀² x² dx using two strips of width 1: heights are f(0) = 0, f(1) = 1, f(2) = 4. Area ≈ ½ × 1 × (0 + 1) + ½ × 1 × (1 + 4) = 0.5 + 2.5 = 3. The exact value is [x³/3]₀² = 8/3 ≈ 2.67. The trapezium rule overestimates here because the curve bends upwards (each strip's top edge lies above the curve) — knowing the direction of the error, not just its size, is the A-level-ready insight, and the estimate improves rapidly as strips narrow.

Common mistakes

• Forgetting +C on indefinite integrals — and wrongly adding it to definite ones (it cancels; a definite answer is a number).
• Power-rule slip: ∫x² dx = x³/3, not x³/2. The new exponent is the divisor. Differentiate to check.
• Integrating 1/x with the power rule. ∫x^(−1) dx is ln|x| (A-level); at this stage, flag it rather than forcing x⁰/0.
• Limits substituted backwards. It is F(b) − F(a), top limit minus bottom limit.
• Negative 'area'. If the curve crosses the axis, split the integral; a negative area answer signals the trap.
• Losing terms: ∫(6x² − 4x) dx needs both terms integrated and a single +C at the end.

Exam Experience

Question 1. Integrate 6x² − 4x with respect to x. [2 marks]

Question 2. Work out ∫₁⁴ 3x² dx. [2 marks]

Question 3. A curve has gradient function dy/dx = 4x − 3 and passes through the point (2, 5). Show that the equation of the curve is y = 2x² − 3x + 3. [3 marks]

Mark scheme.

• Q1: M1 for raising each power by 1 and dividing by the new power. A1 for 2x³ − 2x² + C (the +C is required for the accuracy mark).
• Q2: M1 for the antiderivative [x³] and substituting the limits in the correct order. A1 for 64 − 1 = 63.
• Q3: M1 for integrating: y = 2x² − 3x + C. M1 for substituting (2, 5): 5 = 8 − 6 + C. A1 for C = 3, hence y = 2x² − 3x + 3 as required.