Integration — Drill Pack (Pure Mathematics

Pure Mathematics — Paper 1 · AQA AS Mathematics (sample) · Drill 3 of 3

Why this topic matters. Integration is examined on every AQA AS Mathematics (sample) Paper 1 and Paper 2, typically carrying 10–18 marks per sitting. Indefinite integration (finding the general function) and definite integration (computing exact areas) both appear consistently. Area under a curve — including areas trapped between two curves — is examined in roughly 4 of every 5 sittings.

The scaffold to use (RICE Framework)

Raise the power by 1: ∫x^n dx ⇒ xⁿ⁺¹/(n+1) + C.
Include the +C for every indefinite integral — never omit it.
Compute limits for definite integrals: [F(x)]ᵃᵇ = F(b) − F(a).
Ensure negative area is handled correctly: area between curve and axis = |∫|; subtract where curve is below axis.

Foundational — recall and single-step application

Q1. Find ∫(3x² + 4x − 2) dx.

Foundational

[2 marks]

M1 Integrate term by term: ∫3x²dx = x³; ∫4x dx = 2x²; ∫(−2) dx = −2x. AO1 — applying power rule integration Common mistake: Forgetting to add +C for an indefinite integral.

A1 ∫(3x² + 4x − 2) dx = x³ + 2x² − 2x + C. AO1 — correct integral with +C

Q2. Evaluate ∫₁²(x² + 1) dx.

Foundational

[3 marks]

M1 F(x) = x³/3 + x. (No +C needed for definite integrals.) AO1 — indefinite integral for definite evaluation Common mistake: Carrying +C through into the definite evaluation.

A1 F(2) = 8/3 + 2 = 14/3. F(1) = 1/3 + 1 = 4/3. AO1 — evaluating at both limits

A1 ∫₁²(x² + 1) dx = 14/3 − 4/3 = 10/3. AO1 — subtracting correctly

Q3. Find ∫(5x^3/2 − 3/x²) dx.

Foundational

[3 marks]

M1 Rewrite: ∫(5x^3/2 − 3x⁻²) dx. Integrate: 5 ⋅ x^5/2/(5/2) − 3 ⋅ x⁻¹/(−1). AO1 — rewriting and integrating Common mistake: Forgetting the negative sign when integrating 3x⁻²: result is +3x⁻¹, not −3x⁻¹.

A1 = 2x^5/2 + 3x⁻¹ + C = 2x^5/2 + 3/x + C. AO1 — correct result with +C

A1 Check by differentiating: d/dx(2x^5/2 + 3x⁻¹) = 5x^3/2 − 3x⁻² ✓. AO2 — differentiation to verify

Q4. A curve has dy/dx = 3x² − 6x and passes through (2, 1). Find the equation of the curve.

Foundational

[4 marks]

M1 Integrate: y = x³ − 3x² + C. AO1 — integrating to find general solution

A1 Substitute (2, 1): 1 = 8 − 12 + C ⇒ C = 5. AO1 — finding the constant Common mistake: Forgetting to find C by substituting the given point.

A1 Equation of curve: y = x³ − 3x² + 5. AO1 — complete equation

A1 Check: y(2) = 8−12+5 = 1 ✓. AO2 — verifying the given point lies on the curve

Q5. Find the area of the region bounded by y = x² + 1, the x-axis, and the lines x = 0 and x = 3.

Foundational

[4 marks]

M1 Area = ∫₀³(x² + 1) dx. F(x) = x³/3 + x. AO1 — setting up the definite integral

A1 F(3) = 9 + 3 = 12. F(0) = 0. AO1 — evaluating at limits

A1 Area = 12 − 0 = 12 (square units). AO1 — correct area

A1 Since y = x²+1 ≥ 1 > 0 on [0,3], the curve is entirely above the x-axis, so the integral gives the area directly. ✓ AO2 — justifying no sign issue

Q6. Evaluate ∫₀⁴(√x + 2) dx.

Foundational

[3 marks]

M1 F(x) = x^3/2/(3/2) + 2x = (2/3)x^3/2 + 2x. AO1 — integrating x^1/2 Common mistake: Integrating √x as x^1/2/(1/2) = 2x^1/2 instead of (2/3)x^3/2.

A1 F(4) = (2/3)(8) + 8 = 16/3 + 8 = 40/3. F(0) = 0. AO1 — evaluating at limits

A1 ∫₀⁴(√x + 2) dx = 40/3. AO1 — final answer

Mid-tier — application and analysis (exam-bulk questions)

Q7. Find the area enclosed between the curve y = x² − 4 and the x-axis.

Mid-tier

[5 marks]

M1 Set y = 0: x² = 4 ⇒ x = ±2. Limits: x = −2 to x = 2. AO1 — finding x-intercepts

M1 ∫₋₂²(x² − 4) dx. F(x) = x³/3 − 4x. AO1 — setting up the integral

A1 F(2) = 8/3 − 8 = −16/3. F(−2) = −8/3 + 8 = 16/3. Integral = −16/3 − 16/3 = −32/3. AO1 — evaluating the definite integral

A1 Area = |−32/3| = 32/3 square units. AO2 — taking absolute value for area below x-axis Common mistake: Giving −32/3 as the area. Area is always positive; take |integral|.

A1 By symmetry about y-axis: Area = 2|∫₀²(x²−4)dx| = 2|(8/3−8)| = 2⋅16/3 = 32/3. ✓ AO2 — symmetry check

Q8. Find the area enclosed between y = 6x − x² and y = x².

Mid-tier

[6 marks]

M1 Set equal: 6x − x² = x² ⇒ 6x = 2x² ⇒ x(x−3) = 0. x = 0 or x = 3. AO1 — finding intersection points Common mistake: Forgetting to find both intersection points, or using the wrong limits.

M1 Top curve: 6x−x² ≥ x² on [0,3] (check at x=1: 5 ≥ 1 ✓). Integrate (top−bottom): ∫₀³(6x−2x²) dx. AO2 — identifying top and bottom curves

M1 F(x) = 3x² − (2/3)x³. F(3) = 27 − 18 = 9. F(0) = 0. AO1 — evaluating the integral

A1 Area = 9 − 0 = 9 square units. AO1 — correct area

A1 Check: the parabola y=x² is below y=6x−x² for 0<x<3 (positive enclosed area). ✓ AO2 — sign verification

A1 Alternatively integrate each curve separately and subtract: more work; the (top−bottom) method is faster. AO2 — noting the efficient method

Q9. Given that ∫₀ᴷ(4x + 1) dx = 12, find the value of k.

Mid-tier

[5 marks]

M1 F(x) = 2x² + x. ∫₀ᴷ(4x+1)dx = [2x²+x]₀ᴷ = 2k² + k − 0 = 2k² + k. AO1 — setting up the integral in terms of k

M1 Set equal to 12: 2k² + k − 12 = 0. AO1 — forming the equation

A1 Solve the quadratic 2k² + k − 12 = 0. It does not factorise over the integers (discriminant 1² + 4·2·12 = 97 is not a perfect square), so use the quadratic formula: k = (−1 ± √97) / 4. AO1 — solving for k

A1 Since k is an upper limit of integration we take the positive root: k = (−1 + √97) / 4 ≈ 2.21 (3 s.f.). AO1 — exact answer

A1 Verify: k ≈ 2.21. 2(4.88)+2.21 ≈ 12.0 ✓. AO2 — numerical verification

Q10. A curve has gradient function dy/dx = 6x² − 10x + 3. The curve passes through (0, 5). Find y in terms of x.

Mid-tier

[4 marks]

M1 y = ∫(6x² − 10x + 3) dx = 2x³ − 5x² + 3x + C. AO1 — integrating the gradient function

A1 At (0, 5): 5 = 0 − 0 + 0 + C ⇒ C = 5. AO1 — finding the constant

A1 y = 2x³ − 5x² + 3x + 5. AO1 — complete equation

A1 Check: dy/dx = 6x²−10x+3 ✓. y(0)=5 ✓. AO2 — verification

Q11. Calculate the area of the region between the curve y = x(4 − x) and the x-axis.

Mid-tier

[5 marks]

M1 y = 4x − x². Roots: x(4−x) = 0 ⇒ x = 0 or x = 4. AO1 — finding x-intercepts

M1 Area = ∫₀⁴(4x − x²) dx. F(x) = 2x² − x³/3. AO1 — setting up the integral

A1 F(4) = 32 − 64/3 = 96/3 − 64/3 = 32/3. F(0) = 0. AO1 — evaluating at limits

A1 Area = 32/3 square units. AO1 — correct area

A1 y = x(4−x) ≥ 0 for 0 ≤ x ≤ 4, so the curve is above the x-axis throughout. Area = integral directly. AO2 — justifying no sign correction needed

Q12. Find ∫(x^1/3 + x⁻³) dx.

Mid-tier

[4 marks]

M1 ∫x^1/3 dx = x^4/3/(4/3) = (3/4)x^4/3. ∫x⁻³ dx = x⁻²/(−2) = −(1/2)x⁻². AO1 — applying power rule to fractional and negative exponents Common mistake: Increasing the exponent without dividing by the new exponent.

A1 ∫(x^1/3 + x⁻³) dx = (3/4)x^4/3 − (1/2)x⁻² + C. AO1 — correct result with +C

A1 Equivalently: (3/4)∛(x⁴) − 1/(2x²) + C. AO2 — writing in alternative notations

A1 Check by differentiating: (3/4)(4/3)x^1/3 + 2/2 ⋅ x⁻³ = x^1/3 + x⁻³ ✓. AO2 — differentiation check

Q13. The region R is bounded by the curve y = 4/x², the x-axis, and the lines x = 1 and x = 2. Find the area of R.

Mid-tier

[4 marks]

M1 4/x² = 4x⁻². ∫4x⁻² dx = 4 ⋅ x⁻¹/(−1) = −4/x. AO1 — integrating 4x⁻²

A1 [−4/x]₁² = (−4/2) − (−4/1) = −2 + 4 = 2. AO1 — applying limits correctly Common mistake: Sign error in [−4/x]: −4/2 = −2, not +2; and −4/1 = −4, not +4.

A1 Area of R = 2 square units. AO1 — correct area

A1 4/x² > 0 for all x ≠ 0, so no sign-correction needed. ✓ AO2 — confirming sign

Q14. Find the x-coordinates of points where y = x³ − 3x intersects y = 2. Hence find the area enclosed between y = x³ − 3x and y = 2 that lies entirely above y = 2.

Mid-tier

[6 marks]

M1 x³ − 3x = 2 ⇒ x³ − 3x − 2 = 0. Test x=−1: −1+3−2=0 ✓. Factor: (x+1)(x²−x−2) = (x+1)(x−2)(x+1) = (x+1)²(x−2). AO1 — finding intersection x-values by factorising

A1 x = −1 (double) and x = 2. AO1 — intersection x-values

M1 On [−1, 2], curve is below y=2 (since x=0: y=0 < 2). Integrate (2 − (x³−3x)) from −1 to 2. AO2 — identifying top minus bottom

A1 ∫₋₁²(2−x³+3x) dx = [2x − x⁴/4 + 3x²/2]₋₁². AO1 — setting up integral

A1 At x=2: 4−4+6=6. At x=−1: −2−1/4+3/2 = −2−0.25+1.5 = −0.75 = −3/4. Area = 6−(−3/4) = 27/4. AO1 — evaluating at both limits

A1 Area = 27/4 (= 6.75) square units. AO2 — exact fraction answer

Q15. Use integration to show that the area under one complete arch of y = sin(x) between x = 0 and x = π equals 2. [Note: ∫sin x dx = −cos x + C.]

Mid-tier

[4 marks]

M1 Area = ∫₀ᶀ sin(x) dx = [−cos(x)]₀ᶀ. AO1 — setting up the definite integral of sin(x)

A1 = (−cos π) − (−cos 0) = (−(−1)) − (−1) = 1 + 1 = 2. ✓ AO1 — correct evaluation Common mistake: cos π = −1 and cos 0 = 1; sign errors here are extremely common.

A1 sin(x) ≥ 0 on [0, π], so the integral gives area directly (no modulus needed). AO2 — justifying sign

A1 This classic result (area = 2) is worth memorising for efficiency. AO2 — highlighting the key result

Q16. The curve C has equation y = x² + 2x − 3. (a) Find where C crosses the x-axis. (b) Find the area between C and the x-axis.

Mid-tier

[6 marks]

M1 (a) x² + 2x − 3 = 0 ⇒ (x+3)(x−1) = 0 ⇒ x = −3 or x = 1. AO1 — finding x-intercepts by factorising

A1 C crosses x-axis at x = −3 and x = 1. AO1 — correct x-values

M1 (b) Between x = −3 and x = 1, the curve is below the x-axis (check at x=0: y=−3 < 0). Area = |∫₋₃¹(x²+2x−3) dx|. AO2 — identifying sign for area calculation

A1 F(x) = x³/3 + x² − 3x. F(1) = 1/3+1−3 = −5/3. F(−3) = −9+9+9 = 9. Integral = −5/3 − 9 = −32/3. AO1 — correct evaluation

A1 Area = |−32/3| = 32/3 square units. AO1 — correct area

A1 Alternative: use symmetry-free approach. Either way, area = 32/3 ✓. AO2 — confirming method

Stretch — discriminators, twist-traps, top-band signals

Q17. Find ∫₁⁴(x − 1/√x) dx.

Stretch

[5 marks]

M1 Rewrite: ∫₁⁴(x − x^−1/2) dx. F(x) = x²/2 − x^1/2/(1/2) = x²/2 − 2√x. AO1 — integrating x^−1/2 correctly Common mistake: Integrating x^−1/2 as −(1/2)x^−3/2 (differentiating instead of integrating).

A1 F(4) = 8 − 4 = 4. F(1) = 1/2 − 2 = −3/2. AO1 — evaluating at limits

A1 Definite integral = 4 − (−3/2) = 4 + 3/2 = 11/2. AO1 — correct subtraction

A1 Check: d/dx(x²/2 − 2x^1/2) = x − x^−1/2 ✓. AO2 — differentiation to verify

A1 Exact answer: 11/2 (= 5.5). Do not convert to decimal unless asked. AO2 — exact form preferred

Q18. The diagram shows the curves y = x² (C₁) and y = √x (C₂). Find the area of the region enclosed between them.

Stretch

[6 marks]

M1 Intersections: x² = √x ⇒ x⁴ = x ⇒ x(x³−1) = 0 ⇒ x=0 or x=1. AO1 — finding intersection points

M1 At x=1/4: √(1/4)=1/2 and (1/4)²=1/16. So √x > x² on (0,1). Integrate (√x−x²) from 0 to 1. AO2 — identifying the top curve

M1 ∫₀₁(x^1/2 − x²) dx = [(2/3)x^3/2 − x³/3]₀₁. AO1 — setting up the integral

A1 At x=1: 2/3 − 1/3 = 1/3. At x=0: 0. AO1 — evaluating the integral

A1 Enclosed area = 1/3 square units. AO1 — correct area Common mistake: Using wrong limits or integrating (x²−√x) giving −1/3; area must be positive.

A1 Note: by symmetry y=x² and y=√x are reflections of each other in y=x; the enclosed area must be symmetric too. AO2 — elegance check

Q19. A function f has f′(x) = 3x² − 12x + 9. The curve y = f(x) has a local maximum at x = 1 with f(1) = 8. Find f(x) and the area between y = f(x) and the line y = f(3) for 1 ≤ x ≤ 3.

Stretch

[8 marks]

M1 f(x) = x³ − 6x² + 9x + C. f(1) = 1−6+9+C = 4+C = 8 ⇒ C = 4. AO1 — integrating and finding C

A1 f(x) = x³ − 6x² + 9x + 4. AO1 — correct f(x)

M1 f(3) = 27−54+27+4 = 4. Line y = 4. Area = ∫₁³(f(x) − 4) dx = ∫₁³(x³−6x²+9x) dx. AO2 — setting up the area integral

A1 F(x) = x⁴/4 − 2x³ + 9x²/2. F(3) = 81/4−54+81/2 = 81/4−216/4+162/4 = 27/4. AO1 — evaluating at x=3

A1 F(1) = 1/4−2+9/2 = 1/4−8/4+18/4 = 11/4. AO1 — evaluating at x=1

A1 Area = 27/4 − 11/4 = 16/4 = 4 square units. AO1 — correct final area

A1 The local max is at x=1 (f′(1)=3−12+9=0 ✓) and f(1)=8 > f(3)=4, so the curve is above y=4 on (1,3). Area is positive. ✓ AO2 — sign and logic verification

A1 Local minimum at x=3: f′(3)=27−36+9=0 ✓. The curve touches y=4 again at the minimum, making the bounded area well-defined. AO2 — confirming the geometry

Q20. The region R is bounded above by y = 4x − x² and below by y = x. (a) Find the area of R. (b) Find the x-coordinate of the vertical line that divides R into two equal areas.

Stretch

[9 marks]

M1 (a) Intersections: 4x−x²=x ⇒ 3x=x² ⇒ x(x−3)=0 ⇒ x=0 or x=3. AO1 — finding intersection points

M1 (a) Area = ∫₀³((4x−x²)−x) dx = ∫₀³(3x−x²) dx = [3x²/2 − x³/3]₀³. AO1 — setting up area integral

A1 (a) F(3)=27/2−9=9/2. Total area = 9/2. AO1 — total area

M1 (b) Area from 0 to p = ∫₀ᴰ(3x−x²) dx = 3p²/2 − p³/3. Set = half of 9/2 = 9/4. AO2 — setting up the equal-area equation

M1 3p²/2 − p³/3 = 9/4. Multiply by 12: 18p² − 4p³ = 27 ⇒ 4p³ − 18p² + 27 = 0. AO2 — forming the cubic equation

A1 This cubic can be tested: p = 3/2 ⇒ 4(27/8)−18(9/4)+27 = 27/2−81/2+27 = (27−81+54)/2 = 0/2 = 0. ✓ AO1 — verifying p = 3/2

A1 The vertical line is at x = 3/2. AO2 — stating the answer clearly

A1 Sense check: p = 3/2 is between 0 and 3 and the parabolic region is not symmetric about x=3/2, so this is plausible. AO2 — plausibility check

A1 Verify area from 0 to 3/2: 3(9/4)/2 − (27/8)/3 = 27/8 − 9/8 = 18/8 = 9/4. ✓ AO2 — numerical verification

Stuck on a question?

Integration — Primer

Pure Mathematics — Paper 1 · AQA AS Mathematics (sample)

What this primer is for. A revision-grade refresher focused on AQA AS integration: what scores marks, not derivation theory.

What this topic is

Integration is the reverse of differentiation. The three key applications examined are: Indefinite Integration (find F(x) + C such that F′(x) = f(x)); Definite Integration (compute [F(x)]ᵃᵇ = F(b) − F(a), a numerical value); and Area Calculations (using the definite integral to find areas bounded by curves, the x-axis, or between two curves).

Why it matters in this paper

Integration appears on virtually every AQA Paper 1 and Paper 2, carrying 10–18 marks per sitting. Area-between-curves questions appear in ~80% of sittings and are a reliable source of 6–8 marks. Mastering the +C discipline and the area sign convention alone recovers several marks that candidates routinely drop.

How to tackle this topic

Match to one of three method types from the question:

Power Rule Integration — ∫ax^n dx = axⁿ⁺¹/(n+1) + C (valid for n ≠ −1). Always rewrite surds and fractions as powers first (e.g. √x = x^1/2; 1/x² = x⁻²). Don’t forget +C for indefinite integrals. Use when: integrating a polynomial or algebraic expression without limits.
Definite Integration (Limits) — compute ∫ᵃᵇ f(x) dx = [F(x)]ᵃᵇ = F(b) − F(a). Upper limit goes first in the bracket; always subtract F(a) from F(b). The +C cancels out (a key reason it’s absent in definite integrals). Use when: limits a and b are given, or when calculating area and the limits are provided.
Area with Axis/Between Curves — if the question asks for area: (1) find x-values where curves meet (set equations equal); (2) determine which curve is on top on the relevant interval; (3) integrate (top − bottom); (4) take |integral| if the result is negative (meaning the curve was below the axis). Use when: area questions involving curves and the x-axis, or two curves.

The key facts you must know

∫x^n dx = xⁿ⁺¹/(n+1) + C for n ≠ −1. ∫x⁻¹ dx = ln|x| + C (A2 only).
Indefinite integral: always write +C. Lose a mark without it.
Definite integral: [F(x)]ᵃᵇ = F(b) − F(a). Upper limit first in the bracket.
Area ≥ 0 always. If ∫ gives a negative number, the curve is below the axis ⇒ area = |integral|.
Area between curves: ∫(top − bottom) dx between intersection points.
∫sin x dx = −cos x + C. ∫cos x dx = sin x + C. (AS level — tested periodically.)
Integration and differentiation are inverses: d/dx[∫f(x)dx] = f(x).

Step-by-step worked example

Question: Find the area enclosed between y = x² and y = x + 2.

Step 1. Find intersections: x² = x + 2 ⇒ x²−x−2=0 ⇒ (x−2)(x+1)=0 ⇒ x=2 or x=−1.
Step 2. On [−1,2], which is on top? At x=0: y=x² gives 0; y=x+2 gives 2. So y=x+2 is on top.
Step 3. Area = ∫₋₁²((x+2)−x²) dx = [x²/2+2x−x³/3]₋₁².
Step 4. At x=2: 2+4−8/3 = 6−8/3 = 10/3. At x=−1: 1/2−2+1/3 = −7/6. Area = 10/3−(−7/6) = 20/6+7/6 = 27/6 = 9/2.

Why this lands top-band: showing which curve is on top before integrating, and presenting the integral correctly as (top−bottom), earns method marks even if arithmetic goes wrong.

Common mistakes (from examiner reports)

Omitting +C in indefinite integrals. — “The absence of the constant of integration was a very common and entirely avoidable error.”
Giving a negative value for an area (curve below x-axis). Area must be positive — take the modulus.
Integrating x^−1/2 as −(1/2)x^−3/2 (differentiating instead of integrating).
Using wrong limits for area between curves (e.g. using x=0 instead of the intersection point).
Forgetting to identify the top curve before integrating (top−bottom), leading to a negative area result being presented as-is.

Quick self-check

Can you state ∫x^n dx without looking? What value of n is excluded?
Write ∫(x^1/2 − 3x⁻²) dx without a calculator.
The area under y=x²−4 from x=−2 to x=2 is negative. What does this mean physically, and how do you correct it?

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Integration High-frequency topic

The scaffold to use (RICE Framework)

Q1. Find ∫(3x² + 4x − 2) dx.

Q2. Evaluate ∫₁²(x² + 1) dx.

Q3. Find ∫(5x3/2 − 3/x²) dx.

Q4. A curve has dy/dx = 3x² − 6x and passes through (2, 1). Find the equation of the curve.

Q5. Find the area of the region bounded by y = x² + 1, the x-axis, and the lines x = 0 and x = 3.

Q6. Evaluate ∫₀⁴(√x + 2) dx.

Q7. Find the area enclosed between the curve y = x² − 4 and the x-axis.

Q8. Find the area enclosed between y = 6x − x² and y = x².

Q9. Given that ∫₀ᴷ(4x + 1) dx = 12, find the value of k.

Q10. A curve has gradient function dy/dx = 6x² − 10x + 3. The curve passes through (0, 5). Find y in terms of x.

Q11. Calculate the area of the region between the curve y = x(4 − x) and the x-axis.

Q12. Find ∫(x1/3 + x−3) dx.

Q13. The region R is bounded by the curve y = 4/x², the x-axis, and the lines x = 1 and x = 2. Find the area of R.

Q14. Find the x-coordinates of points where y = x³ − 3x intersects y = 2. Hence find the area enclosed between y = x³ − 3x and y = 2 that lies entirely above y = 2.

Q15. Use integration to show that the area under one complete arch of y = sin(x) between x = 0 and x = π equals 2. [Note: ∫sin x dx = −cos x + C.]

Q16. The curve C has equation y = x² + 2x − 3. (a) Find where C crosses the x-axis. (b) Find the area between C and the x-axis.

Q17. Find ∫₁⁴(x − 1/√x) dx.

Q18. The diagram shows the curves y = x² (C₁) and y = √x (C₂). Find the area of the region enclosed between them.

Q19. A function f has f′(x) = 3x² − 12x + 9. The curve y = f(x) has a local maximum at x = 1 with f(1) = 8. Find f(x) and the area between y = f(x) and the line y = f(3) for 1 ≤ x ≤ 3.

Q20. The region R is bounded above by y = 4x − x² and below by y = x. (a) Find the area of R. (b) Find the x-coordinate of the vertical line that divides R into two equal areas.

Q3. Find ∫(5x^3/2 − 3/x²) dx.

Q12. Find ∫(x^1/3 + x⁻³) dx.