OCR A Level Mathematics A: Integration basics: fundamental theorem, integrals and areas (H240) – Revision Notes

These notes condense OCR A Level Mathematics A (H240) content references 1.08a to 1.08f from the OCR specification (Version 3.1, August 2026). The 1.08a, 1.08b, 1.08d and 1.08e statements are Stage 1, common to AS and A Level. Integrating e^kx, 1/x, sin kx and cos kx (1.08c) and the area between two curves (1.08f) are A Level only. The study guide has the full explanations. Test yourself with the practice set, tick topics off on the checklist, see the course hub and try the free diagnostics. Reverse chain rule, parts, substitution, partial fractions, the trapezium rule and differential equations are not covered here.

Key results

Ref	Result
1.08a	∫ f(x) dx = F(x) + c when f(x) = d/dx F(x)
1.08b	xⁿ integrates to xⁿ⁺¹/(n + 1) + c, for n ≠ -1
1.08c	∫ e^kx dx = (1/k)e^kx + c; ∫ 1/x dx = ln|x| + c
1.08c	∫ sin kx dx = -(1/k)cos kx + c; ∫ cos kx dx = (1/k)sin kx + c
1.08c	cos²A = (1 + cos 2A)/2; sin²A = (1 - cos 2A)/2
1.08d	∫ₐᵇ f(x) dx = F(b) - F(a)
1.08e	Area under a curve above the axis = ∫ₐᵇ y dx
1.08f	Area between curves = ∫ₐᵇ (upper - lower) dx

1.08a Reverse of differentiation

• Integrating undoes differentiating. Check any indefinite integral by differentiating your answer.
• Indefinite: no limits, answer is a function plus c. Definite: limits, answer is a number.
• “Hence” questions: differentiate the given function, rearrange so the target integrand appears, then integrate both sides.

Worked reminder. d/dx (x² cos x) = 2x cos x - x² sin x. So ∫ (x² sin x - 2x cos x) dx = -x² cos x + c, with no new integration needed.

1.08b Powers of x (Stage 1)

Method in steps

1. Rewrite each term as a multiple of a power of x (roots are fractions, 1/xᵐ is x⁻ᵐ).
2. Expand brackets and split fractions over a single term in the denominator.
3. Raise each index by 1, then divide by the index you now have.
4. Add c once. For a curve through a given point, substitute to find c.

Worked reminders.

• ∫ (6√x + 2/x³) dx = ∫ (6x^(1/2) + 2x⁻³) dx = 4x^(3/2) - 1/x² + c.
• ∫ (2x - 3)(x + 1) dx: expand to 2x² - x - 3, then integrate to (2/3)x³ - x²/2 - 3x + c.
• A curve with dy/dx = 2x - 3 through (2, 5): y = x² - 3x + c, so c = 7.

1.08c Standard integrals (A Level only)

• The factor 1/k appears for e^kx and cos kx. For sin kx it is -1/k.
• ln|x| needs the modulus because x can be negative.
• These cover kx only. Do not extend them to other inner functions without the method from another topic.
• The specification says integrals of arcsin, arccos and arctan will be given if required.

Method for sin² and cos²

1. Replace sin²(ax) by (1 - cos 2ax)/2 or cos²(ax) by (1 + cos 2ax)/2.
2. Integrate term by term.

Worked reminders.

• ∫ 6 cos² 2x dx = ∫ 3(1 + cos 4x) dx = 3x + (3/4) sin 4x + c.
• Divide before integrating: (e^(2x) + 4)/eˣ = eˣ + 4e^(-x), so the integral is eˣ - 4e^(-x) + c.

1.08d Definite integrals (Stage 1)

Method in steps

1. Integrate (no c).
2. Put the upper limit in, then the lower, and subtract.
3. Simplify exactly: leave π, e and ln in the answer unless told to give decimals.

Worked reminders.

• ∫₀^(ln 4) 6e^(-x) dx = [-6e^(-x)]₀^(ln 4) = -6/4 + 6 = 9/2.

• ∫(-4)^(-1) (2/x) dx = [2 ln|x|](-4)^(-1) = 2 ln 1 - 2 ln 4 = -4 ln 2. The value is negative because 1/x is negative on this interval.

• Swapping limits changes the sign. Splitting at a point: ∫ₐᶜ = ∫ₐᵇ + ∫ᵇᶜ.

• Show the antiderivative and the substitution of limits. One of the specification’s examples of detailed reasoning is a definite integral, and it says a bare calculator answer would not be awarded any marks there.

1.08e Area under a curve (Stage 1)

Method in steps

1. Sketch or sign-test the curve between the limits.
2. Find any x-intercepts inside the interval.
3. Integrate on each piece. A piece below the axis gives a negative value.
4. Add the positive sizes.

A Level only extensions (Stage 2 wording)

• Against the y-axis: express x using y, then area = ∫ x dy between the two y-values.
• Parametric: area = ∫ y (dx/dt) dt, with limits converted to t.

Worked reminders.

• y = x² - 4x between x = 0 and x = 4 is below the axis: ∫₀⁴ (x² - 4x) dx = -32/3, so the area is 32/3.
• x = 2t, y = 3t² for t from 1 to 2: ∫₁² 3t² × 2 dt = [2t³]₁² = 14.

1.08f Area between curves (A Level only)

Method in steps

1. Set f(x) = g(x) and solve; the roots are the limits.
2. Pick an x-value between the roots to see which curve is higher.
3. Integrate (upper - lower).
4. If the curves cross inside the interval, split there and reverse the subtraction.

Worked reminder. The curve y = 8 - x² and the line y = 2x meet where x² + 2x - 8 = 0, so x = -4 or 2. At x = 0 the curve (8) is above the line (0). Area = ∫(-4)^2 (8 - 2x - x²) dx = [8x - x² - x³/3](-4)^2 = 28/3 + 80/3 = 36.

Exam technique in short

• Read the verb. “Find” an indefinite integral needs +c. “Evaluate” and “find the exact value” need limits substituted and a simplified exact answer.
• In “hence” questions the earlier part is the intended route, so use it and show where it feeds in.
• When a question says to show detailed reasoning, write every line: the antiderivative, the limits substituted, and the simplification.
• Sketch the curve before any area question. Two minutes on signs and intersections saves a lost mark.
• Check an indefinite integral by differentiating it, and an area by asking whether the size looks sensible against the graph.
• For an area between curves, write the subtraction as one bracket, (upper - lower), before you expand. It keeps the signs under control.

Must-know distinctions

Idea	Contrast
Indefinite vs definite	A function with +c vs a number with limits
Net integral vs area	Signed total vs positive size of each piece added
∫ 1/x dx vs ∫ xⁿ dx	ln|x| vs the power rule (n = -1 excluded)
∫ e^kx vs d/dx e^kx	Divide by k vs multiply by k
Stage 1 vs Stage 2	Curve and axis vs curve and curve (A Level only)

Quick self-test

1. Find ∫ (4x³ - 6x⁻⁴) dx.
2. Find ∫ (9e^(3x) + 5/x) dx.
3. Find ∫ (cos 4x - 2 sin(x/2)) dx.
4. Evaluate ∫₁² (12x² - 5) dx.
5. Evaluate ∫₁^(e³) (2/x) dx.
6. Evaluate ∫₀^(ln 2) 10e^(-5x) dx.
7. Evaluate ∫₀^(π/4) 8 cos 2x dx.
8. Evaluate ∫₀^(π/2) cos²x dx.
9. Given that k > 1 and ∫₁^k 3x² dx = 63, find k.
10. A curve has dy/dx = 2x - 3 and passes through (2, 5). Find its equation.
11. Find the area between y = x² - 9 and the x-axis.
12. Find the area between y = x² and y = 2x + 3.
13. Find the area enclosed between y = sin x and the x-axis for 0 ≤ x ≤ 2π.

Answers

1. x⁴ + 2x⁻³ + c, which is x⁴ + 2/x³ + c
2. 3e^(3x) + 5 ln|x| + c
3. (1/4) sin 4x + 4 cos(x/2) + c
4. [4x³ - 5x]₁² = (32 - 10) - (4 - 5) = 23
5. [2 ln x]₁^(e³) = 6
6. [-2e^(-5x)]₀^(ln 2) = -2/32 + 2 = 31/16
7. [4 sin 2x]₀^(π/4) = 4
8. ∫ (1 + cos 2x)/2 dx = [x/2 + (sin 2x)/4]₀^(π/2) = π/4
9. k³ - 1 = 63, so k³ = 64 and k = 4
10. y = x² - 3x + c; 4 - 6 + c = 5, so c = 7 and y = x² - 3x + 7
11. The curve is below the axis between x = -3 and 3: area = 36
12. The curves meet at x = -1 and 3: area = ∫_(-1)^3 (2x + 3 - x²) dx = 32/3
13. The net integral is 0. Area = 2 × 2 = 4

Where marks are usually lost

• Leaving out +c on an indefinite integral, or carrying +c into a definite one.
• Multiplying instead of dividing by k when integrating e^kx, sin kx and cos kx.
• Dropping the minus sign in ∫ sin kx dx.
• Writing ln x instead of ln|x| when the interval or question allows negative x.
• Not expanding a bracket or splitting a fraction before integrating.
• Quoting a net integral as an area when the curve crosses the axis.
• Using the wrong upper curve, or missing a crossing inside the limits.
• Giving a decimal where the question wants an exact answer.
• Substituting x-limits into an integral that is already in terms of t, or the reverse.
• Working out ln of a negative number instead of using ln|x| and then simplifying.
• Writing down only a calculator value for a definite integral that asks for detailed reasoning.

Official syllabus

These notes follow content references 1.08a to 1.08f of the OCR A Level in Mathematics A (H240) specification, Version 3.1 (August 2026), for first assessment in 2018, published by OCR.