Integration

This lesson covers Integration as required by the A-Level Mathematics Pure 1 specification. Integration is the reverse process of differentiation. It is used to find areas under curves, recover functions from their derivatives, and solve a variety of mathematical problems. Integration and differentiation together form the core of calculus.

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Integration as the Reverse of Differentiation

If dy/dx = f(x), then y = ∫f(x) dx.

Integration "undoes" differentiation. Since differentiating a constant gives zero, integration always introduces an arbitrary constant C.

The Power Rule for Integration

If f(x) = xⁿ (where n ≠ −1), then:

∫xⁿ dx = xⁿ⁺¹ / (n + 1) + C

Key rule: "Add one to the power, divide by the new power, add C."

Function	Integral
xⁿ (n ≠ −1)	xⁿ⁺¹/(n + 1) + C
axⁿ	axⁿ⁺¹/(n + 1) + C
k (constant)	kx + C
x³	x⁴/4 + C
x⁻²	−x⁻¹ + C = −1/x + C
x^(1/2)	(2/3)x^(3/2) + C

Example: Find ∫(3x² − 4x + 5) dx.

= x³ − 2x² + 5x + C

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Preparing Expressions for Integration

Like differentiation, you often need to rewrite expressions before integrating.

Example: Find ∫(2x³ + 3)/x² dx.

Rewrite: ∫(2x + 3x⁻²) dx.

= x² − 3x⁻¹ + C = x² − 3/x + C

Example: Find ∫(2x⁻³ + 4√x) dx.

Rewrite as ∫(2x⁻³ + 4x^(1/2)) dx.

= 2x⁻²/(−2) + 4x^(3/2)/(3/2) + C
= −1/x² + 8x√x/3 + C

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Finding the Constant of Integration

If you know a point on the curve, you can find the value of C.

Example: A curve has gradient function dy/dx = 6x² − 2 and passes through the point (1, 7). Find the equation of the curve.

y = ∫(6x² − 2) dx = 2x³ − 2x + C

Substituting (1, 7): 7 = 2 − 2 + C → C = 7.

The equation is y = 2x³ − 2x + 7.

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Definite Integration

A definite integral has limits and gives a numerical value:

∫(from a to b) f(x) dx = [F(x)](from a to b) = F(b) − F(a)

where F(x) is the antiderivative of f(x). Note: no constant C is needed for definite integrals.

Example: Evaluate ∫₁³ (2x + 1) dx.

= [x² + x]₁³ = (9 + 3) − (1 + 1) = 12 − 2 = 10

Example: Evaluate ∫₀² (x³ − 3x) dx.

= [x⁴/4 − 3x²/2]₀² = (16/4 − 12/2) − 0 = 4 − 6 = −2

A negative result means the net signed area is below the x-axis.

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Area Under a Curve

The area between a curve y = f(x), the x-axis, and the lines x = a and x = b is:

Area = ∫(from a to b) f(x) dx

Important Notes

1. If the curve is below the x-axis, the integral gives a negative value. Take the absolute value for the area.
2. If the curve crosses the x-axis in the interval, split the integral at the crossing points and add the absolute values.

Example: Find the area enclosed between y = x² − 4 and the x-axis.

The curve crosses the x-axis at x = ±2 (where x² − 4 = 0). Between x = −2 and x = 2, the curve is below the x-axis.

Area = |∫₋₂² (x² − 4) dx| = |[x³/3 − 4x]₋₂²|
     = |(8/3 − 8) − (−8/3 + 8)|
     = |−16/3 − 16/3|
     = 32/3 ≈ 10.67 square units

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Area Between Two Curves

The area between y = f(x) and y = g(x) (where f(x) ≥ g(x)) from x = a to x = b is:

Area = ∫(from a to b) [f(x) − g(x)] dx

Example: Find the area between y = 6x − x² and y = 2x.

Intersection: 6x − x² = 2x → x² − 4x = 0 → x(x − 4) = 0 → x = 0 or x = 4.

Area = ∫₀⁴ [(6x − x²) − 2x] dx = ∫₀⁴ (4x − x²) dx
     = [2x² − x³/3]₀⁴ = (32 − 64/3) − 0
     = 96/3 − 64/3 = 32/3 ≈ 10.67 square units

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The Trapezium Rule

When a function cannot be integrated exactly, the trapezium rule provides an approximation:

∫(from a to b) f(x) dx ≈ h/2 × [y₀ + yₙ + 2(y₁ + y₂ + ... + yₙ₋₁)]

where h = (b − a)/n is the strip width, and y₀, y₁, ..., yₙ are the function values.

Example: Use 4 strips to estimate ∫₀² √(1 + x³) dx.

h = 0.5

x:    0      0.5      1.0      1.5      2.0
y:    1    1.0308   1.4142   2.0917   3.0000

Area ≈ 0.5/2 × [1 + 3 + 2(1.0308 + 1.4142 + 2.0917)]
     = 0.25 × [4 + 2(4.5367)]
     = 0.25 × [4 + 9.0734]
     = 0.25 × 13.0734
     ≈ 3.27

The trapezium rule overestimates for convex curves and underestimates for concave curves.

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Summary

• Integration is the reverse of differentiation.
• Power rule: ∫xⁿ dx = xⁿ⁺¹/(n + 1) + C (n ≠ −1).
• Always include the constant of integration C for indefinite integrals.
• Definite integrals have limits and give a numerical value: F(b) − F(a).
• The area under a curve is found using definite integration.
• When the curve goes below the x-axis, split the integral and take absolute values.
• The area between two curves is ∫[f(x) − g(x)] dx.
• The trapezium rule approximates integrals numerically.

Exam Tip: The most common errors in integration are: forgetting the + C in indefinite integrals, not rewriting expressions before integrating, and not handling negative areas correctly. Always simplify your integrand before integrating, and when finding areas, sketch the curve to identify where it crosses the x-axis.

A-Level Deep Dive: Integration

Spec mapping

AQA 7357 specification, Paper 1 — Pure Mathematics, section H (Integration) covers the Fundamental Theorem of Calculus; integrate $x^n$xn for rational $n \neq -1$n=−1, together with constant multiples, sums and differences; evaluate definite integrals; use definite integration to find the area under a curve and the area between a curve and the $x$x-axis (refer to the official specification document for exact wording). Although H is the explicit home of integration, the topic is examined synoptically across the entire 7357 award. Integration techniques (substitution and integration by parts) live in Pure section H at A2, but their groundwork is laid here. Area between two curves appears in mixed Pure questions, and integration of velocity to recover displacement is examined in Paper 3 — Mechanics (kinematics). The AQA formula booklet lists $\int x^n\,dx$∫xndx for $n \neq -1$n=−1 but expects fluent recall of the FTC and of standard $\int e^{kx}$∫ekx, $\int 1/x$∫1/x, $\int \sin$∫sin, $\int \cos$∫cos results.

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Worked example with full mark scheme

Question (8 marks):

The curve $C$C has equation $y = 3x^2 - 6x$y=3x2−6x.

(a) Evaluate $\displaystyle\int_0^3 (3x^2 - 6x)\,dx$∫03​(3x2−6x)dx, showing all working. (4)

(b) The curve $C$C crosses the $x$x-axis at $x = 0$x=0 and $x = 2$x=2. Explain why your answer to (a) is not equal to the total area enclosed between $C$C and the $x$x-axis from $x = 0$x=0 to $x = 3$x=3, and find that total area exactly. (4)

Solution with mark scheme:

(a) Step 1 — antidifferentiate term by term.

$\int (3x^2 - 6x)\,dx = x^3 - 3x^2 + C$∫(3x2−6x)dx=x3−3x2+C

M1 — correct application of the power rule $\int x^n\,dx = \dfrac{x^{n+1}}{n+1}$∫xndx=n+1xn+1​ to both terms. The constant of integration $C$C is not required for a definite integral but writing it costs nothing and signals understanding.

Step 2 — apply the limits using the FTC.

$\Big[x^3 - 3x^2\Big]_0^3 = (27 - 27) - (0 - 0) = 0$[x3−3x2]03​=(27−27)−(0−0)=0

M1 — substitution of upper and lower limits in the correct order $F(b) - F(a)$F(b)−F(a).

A1 — answer $0$0.

B1 — final mark for clear, unambiguous bracket notation $[\,\cdot\,]_0^3$[⋅]03​ or equivalent and correct arithmetic. Total (4).

(b) Step 1 — interpret the result. The integral evaluates to $0$0 because the signed area below the $x$x-axis on $[0, 2]$[0,2] exactly cancels the signed area above on $[2, 3]$[2,3]. A definite integral measures signed area, not total area.

B1 — explicit statement that the integral gives signed area and that the curve is below the axis on $(0, 2)$(0,2).

Step 2 — split the interval at the root.

$\text{Area} = \left|\int_0^2 (3x^2 - 6x)\,dx\right| + \int_2^3 (3x^2 - 6x)\,dx$Area=​∫02​(3x2−6x)dx​+∫23​(3x2−6x)dx

M1 — correct splitting at the root $x = 2$x=2 with the absolute value applied to the negative piece.

Step 3 — evaluate each piece.

$\int_0^2 (3x^2 - 6x)\,dx = [x^3 - 3x^2]_0^2 = (8 - 12) - 0 = -4$∫02​(3x2−6x)dx=[x3−3x2]02​=(8−12)−0=−4 $\int_2^3 (3x^2 - 6x)\,dx = [x^3 - 3x^2]_2^3 = 0 - (-4) = 4$∫23​(3x2−6x)dx=[x3−3x2]23​=0−(−4)=4

M1 — correct evaluation of both pieces.

A1 — total area $|-4| + 4 = 8$∣−4∣+4=8 square units. Total (4).

Total: 8 marks. The most common loss is reporting $0$0 as "the area" in (a) and never re-examining the question — examiners reward candidates who interpret their answer.

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Specimen question modelled on the AQA 7357 Paper 1 format

Question (6 marks):

The curve $C$C has equation $y = 4 - x^2$y=4−x2. The region $R$R is enclosed between $C$C and the $x$x-axis.

(a) State the $x$x-coordinates of the points where $C$C meets the $x$x-axis. (1)

(b) Find the exact area of $R$R. (5)

Mark scheme decomposition by AO:

(a)

• B1 (AO1.1a) — $x = -2$x=−2 and $x = 2$x=2 from solving $4 - x^2 = 0$4−x2=0.

(b)

• M1 (AO3.1a) — setting up $\displaystyle\int_{-2}^{2}(4 - x^2)\,dx$∫−22​(4−x2)dx as the model for the area, recognising the curve is above the axis on $(-2, 2)$(−2,2).
• M1 (AO1.1b) — antidifferentiating to $4x - \dfrac{x^3}{3}$4x−3x3​.
• A1 (AO1.1b) — substituting upper limit: $4(2) - \dfrac{8}{3} = 8 - \dfrac{8}{3} = \dfrac{16}{3}$4(2)−38​=8−38​=316​.
• A1 (AO1.1b) — substituting lower limit: $4(-2) - \dfrac{-8}{3} = -8 + \dfrac{8}{3} = -\dfrac{16}{3}$4(−2)−3−8​=−8+38​=−316​.
• A1 (AO2.5) — exact area $\dfrac{16}{3} - \left(-\dfrac{16}{3}\right) = \dfrac{32}{3}$316​−(−316​)=332​ square units, presented as a single exact fraction.

Total: 6 marks split AO1 = 4, AO2 = 1, AO3 = 1. AQA Pure integration questions reward (a) symmetry recognition ($y = 4 - x^2$y=4−x2 is even, so $\int_{-2}^{2} = 2\int_0^2$∫−22​=2∫02​), and (b) exact-form discipline — $10.\overline{6}$10.6 or $10.67$10.67 would lose the AO2.5 mark.

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Synoptic links

1. Pure section G — Differentiation: integration is the formal inverse of differentiation. Verifying an indefinite integral by differentiating the answer and checking it returns the integrand is a free self-check that A* candidates use routinely.
2. Pure section H at A2 — Integration techniques: integration by substitution generalises the chain rule in reverse, and integration by parts inverts the product rule. Both rest on the comfortable handling of $\int x^n\,dx$∫xndx taught in Year 1.
3. Pure section F — Coordinate geometry: finding the area enclosed between two curves $y = f(x)$y=f(x) and $y = g(x)$y=g(x) requires solving $f(x) = g(x)$f(x)=g(x) for the intersection points (Pure section B) and then computing $\int [f - g]\,dx$∫[f−g]dx between those limits.
4. Paper 3 — Mechanics (kinematics): for variable acceleration, displacement is recovered from velocity by $s = \int v\,dt$s=∫vdt, and velocity from acceleration by $v = \int a\,dt$v=∫adt. The constants of integration are pinned down by initial conditions at $t = 0$t=0.
5. Pure section E — Exponentials and logarithms: the standard result $\int \dfrac{1}{x}\,dx = \ln|x| + C$∫x1​dx=ln∣x∣+C is the definition used in many treatments of $\ln$ln, and is the reason the power rule has the exclusion $n \neq -1$n=−1.

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Mark-scheme literacy

Definite integration questions on AQA 7357 split AO marks roughly as follows:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	60–70%	Correct antidifferentiation, correct substitution of limits, correct arithmetic
AO2 (reasoning / interpretation)	20–30%	Recognising signed vs total area, presenting exact answers, justifying the splitting of intervals at roots
AO3 (problem-solving / modelling)	10–20%	Setting up the integral from a worded or geometric description, choosing limits from intersection points, modelling kinematics or area-between-curves problems

Examiner-rewarded phrasing: "the curve lies below the $x$x-axis on $(0, 2)$(0,2), so the signed integral is negative — the geometric area is its absolute value"; "by the Fundamental Theorem of Calculus, $\int_a^b f(x)\,dx = F(b) - F(a)$∫ab​f(x)dx=F(b)−F(a) where $F'(x) = f(x)$F′(x)=f(x)"; "the area between the curves is $\int_a^b [f(x) - g(x)]\,dx$∫ab​[f(x)−g(x)]dx where $f \geq g$f≥g on $[a, b]$[a,b]". Phrases that lose marks: "area = $-4$−4" (areas are non-negative); leaving an indefinite integral without $+C$+C when one is asked for; quoting a decimal where the question demands "exact" form.

A specific AQA pattern: questions phrased "find the area enclosed by …" require the candidate to identify the limits themselves from the geometry, not be handed them. The first M1 in such a question is for a correctly set-up integral with correct limits — an integral with the right antiderivative but wrong limits scores 0 on that mark and cascades a lost A1.

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Grade-band model answers

3-mark question

Question: Find $\displaystyle\int_1^2 (6x^2 - 2)\,dx$∫12​(6x2−2)dx.