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Edexcel A-level Maths revision notes

Integration

Section 8
Years 1 and 2
Years 1 and 2: this holds AS subject content and content the exam board adds beyond it for the full A-level.
8 specification points

Notes and three levels of exam-style practice for each registered specification point in this section.

Checked against Edexcel 9MA0 section 8

Checked against Edexcel 9MA0 section 8. Review basis: the qualification registry sourced from the Pearson Edexcel Level 3 Advanced GCE in Mathematics (9MA0) specification; registry verification recorded 11 July 2026.

How this checking works

In the exam: Formulae booklet provided · calculator allowed in every paper

Year 1
Year 1: this is the AS subject content the exam board publishes, which is what most schools teach in Year 12.

Know and use the Fundamental Theorem of Calculus.

Notes
Worked answers & exam appearances →
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Integration is the reverse process of differentiation: if dydx=f(x)\frac{\mathrm dy}{\mathrm dx}=f(x), then y=∫f(x) dxy=\int f(x)\,\mathrm dx recovers the family of curves with that gradient function.
  • Every indefinite integral needs a constant of integration: ∫f(x) dx=F(x)+c\int f(x)\,\mathrm dx=F(x)+c.
  • One known point on the curve fixes the value of cc.
  • A definite integral is evaluated from any antiderivative: ∫abf(x) dx=F(b)−F(a)\int_a^b f(x)\,\mathrm dx=F(b)-F(a); the constant of integration cancels in the subtraction.
  • A common error is to omit +c+c from an indefinite integral, or to substitute the known point into f(x)f(x) instead of the integrated function when finding cc.
Worked example

The function ff satisfies f′(x)=3x−2x2f'(x)=3\sqrt{x}-\dfrac{2}{x^2} for x>0x>0, and f(4)=10f(4)=10. Find f(x)f(x).

  1. 1.Rewrite as f′(x)=3x1/2−2x−2f'(x)=3x^{1/2}-2x^{-2} and integrate: f(x)=2x3/2+2x−1+cf(x)=2x^{3/2}+2x^{-1}+c.
  2. 2.Substituting x=4x=4: 2(8)+24+c=102(8)+\frac{2}{4}+c=10, so 16.5+c=1016.5+c=10 and c=−132c=-\frac{13}{2}.
  3. 3.Hence f(x)=2x3/2+2x−132f(x)=2x^{3/2}+\frac{2}{x}-\frac{13}{2}.

Answer: f(x)=2x3/2+2x−132f(x)=2x^{3/2}+\dfrac{2}{x}-\dfrac{13}{2}

Common mistakes

  • Don't use the supplied boundary value itself as the integration constant without substituting into the antiderivative.
  • Don't integrate the derivative but omit the arbitrary constant before using the given function value.

Exam tip

When recovering a function from its derivative, include the integration constant and determine it from the boundary condition.

Tier 1 · Easy

ORIGINAL

1.

A curve has gradient function dydx=6x2−4\frac{\mathrm dy}{\mathrm dx}=6x^2-4 and passes through the point (1,5)(1,5). Find the equation of the curve.

(3)

(Total for Question 1 is 3 marks)

Tier 2 · Standard

ORIGINAL

1.

Given that F(x)=x42−x3+4xF(x)=\dfrac{x^4}{2}-x^3+4x is an antiderivative of 2x3−3x2+42x^3-3x^2+4, evaluate ∫13(2x3−3x2+4) dx\displaystyle\int_1^3(2x^3-3x^2+4)\,\mathrm dx.

(4)

(Total for Question 1 is 4 marks)

Tier 3 · Hard

ORIGINAL

1.

A curve y=f(x)y=f(x) satisfies f′′(x)=6x+2f''(x)=6x+2. The gradient of the curve at x=1x=1 is 33, and the curve passes through (2,4)(2,4). Find f(x)f(x), and verify your answer by evaluating ∫12f′(x) dx\int_1^2 f'(x)\,\mathrm dx and comparing it with f(2)−f(1)f(2)-f(1).

(5)

(Total for Question 1 is 5 marks)

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Answer conventions

Follow the wording on the question and its mark scheme. awrt means an appropriately rounded value is accepted; an exact answer must stay as a fraction, surd, logarithm or multiple of π when required, and a rounded decimal may be disallowed. Include requested units and forms. A cso tag protects that accuracy mark, while earlier method marks follow the question-specific dependencies.

Year 1
Year 1: this point starts in the AS subject content, which is what most schools teach in Year 12. The full A-level adds in Year 2: Integrating e^kx, 1/x, sin kx and cos kx.Year 2 adds: Integrating e^kx, 1/x, sin kx and cos kx.

Integrate xⁿ (excluding n = −1) and related sums, differences and constant multiples; integrate e^(kx), 1/x, sin kx, cos kx and related sums, differences and constant multiples.

Notes
Worked answers & exam appearances →
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • For n≠−1n\ne-1, $\int xn\,\mathrm dx=\frac{x^{n+1}}{n+1}+C$; rewrite roots and reciprocals as powers before applying the rule. Standard forms include ∫ekx dx=1kekx+C\int e^{kx}\,\mathrm dx=\frac1k e^{kx}+C and ∫1x dx=ln⁡∣x∣+C\int\frac1x\,\mathrm dx=\ln|x|+C on intervals not crossing 00.
  • Divide by the inner coefficient in trigonometric integrals.
  • Also use ∫sec⁡2(kx) dx=1ktan⁡(kx)+C\int\sec^2(kx)\,\mathrm dx=\frac1k\tan(kx)+C, ∫tan⁡x dx=−ln⁡∣cos⁡x∣+C\int\tan x\,\mathrm dx=-\ln|\cos x|+C, and identities for powers such as sin⁡2x=12(1−cos⁡2x)\sin^2x=\frac12(1-\cos2x).
  • A common error is to forget the constant of integration or to multiply by kk rather than divide by it when integrating a function of kxkx.
  • Integrate term by term: the first three terms give 2e3x+4ln⁡∣x∣+52cos⁡(2x)2e^{3x}+4\ln|x|+\frac52\cos(2x).
Worked example

Find ∫(6e3x+4x−5sin⁡(2x)+3tan⁡x) dx\int\left(6e^{3x}+\frac4x-5\sin(2x)+3\tan x\right)\,\mathrm dx.

  1. 1.Integrate term by term: the first three terms give 2e3x+4ln⁡∣x∣+52cos⁡(2x)2e^{3x}+4\ln|x|+\frac52\cos(2x).
  2. 2.Since ∫tan⁡x dx=−ln⁡∣cos⁡x∣\int\tan x\,\mathrm dx=-\ln|\cos x|, the final term gives −3ln⁡∣cos⁡x∣-3\ln|\cos x|.
  3. 3.Add CC.

Answer: 2e3x+4ln⁡∣x∣+52cos⁡(2x)−3ln⁡∣cos⁡x∣+C2e^{3x}+4\ln|x|+\frac52\cos(2x)-3\ln|\cos x|+C

Common mistakes

  • Don't integrate ekxe^{kx} as kekxke^{kx} instead of ekx/ke^{kx}/k.
  • Don't treat the integral of 1/x as a power-rule case or miss reciprocal chain factors.

Exam tip

Match each term to its standard antiderivative and include the integration constant after combining the results.

Tier 1 · Easy

ORIGINAL
Year 2 part
Integrating sec squared kx

1.

Find ∫(6x2−4x1/2+3sec⁡2(2x)) dx\int(6x^2-4x^{1/2}+3\sec^2(2x))\,\mathrm dx.

(4)

(Total for Question 1 is 4 marks)

Tier 2 · Standard

ORIGINAL
Year 2 part
Integrating 1/x

1.

For x>0x>0, find ∫(1+x)2x dx\displaystyle\int\dfrac{(1+\sqrt{x})^2}{x}\,\mathrm dx.

(4)

(Total for Question 1 is 4 marks)

Tier 3 · Hard

ORIGINAL
Year 2 part
Trigonometric identities; integrating trigonometric functions

1.

Use trigonometric identities to find ∫(4sin⁡2x+3tan⁡2x) dx\int(4\sin^2x+3\tan^2x)\,\mathrm dx.

(6)

(Total for Question 1 is 6 marks)

Year 1
Year 1: this point starts in the AS subject content, which is what most schools teach in Year 12. The full A-level adds in Year 2: The area between two curves.Year 2 adds: The area between two curves.

Evaluate definite integrals; use a definite integral to find the area under a curve and the area between two curves.

Notes
Worked answers & exam appearances →
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Definition

Specification 8.3: evaluate definite integrals; use a definite integral to find the area under a curve and the area between two curves.

Explanation

  • ∫abf(x) dx=F(b)−F(a)\int_a^b f(x)\,dx=F(b)-F(a) where FF is an antiderivative; write the antiderivative in square brackets with the limits, substitute both, and subtract in the right order, using e0=1e^0=1 and ln⁡1=0\ln1=0 where they appear.
  • The integral is a signed area, so a region below the xx-axis integrates to a negative number and its area is the modulus; split at the roots when a region crosses the axis.
  • For the area between a curve and a line or a second curve, find the intersections first, then integrate yupper−ylowery_{\text{upper}}-y_{\text{lower}} between them, or subtract the area of a triangle or trapezium under a straight line from the area under the curve.
  • Exact answers keep surds, π\pi, ee and logarithms; “show all stages of working” means calculator-only answers score nothing.
  • A parametric curve has area ∫y dxdt dt\int y\,\dfrac{dx}{dt}\,dt with the limits in tt.
For a parametric curve, vertical strips give area ∫y(t) dxdt dt\int y(t)\,\dfrac{dx}{dt}\,dt.
Worked example

1.

The curve CC has equation y=4x−x2y=4x-x^2 and the line ll has equation y=xy=x. The finite region RR is bounded by CC and ll. Use algebraic integration to find the exact area of RR.

(5)

(Total for Question 1 is 5 marks)

Working

  1. 1.Intersections: 4x−x2=x4x-x^2=x, so x(3−x)=0x(3-x)=0 and x=0x=0 or x=3x=3; between them the curve is above the line.
  2. 2.Area =∫03[(4x−x2)−x]dx=∫03(3x−x2) dx=\displaystyle\int_0^3\left[(4x-x^2)-x\right]dx=\int_0^3(3x-x^2)\,dx.
  3. 3.=[3x22−x33]03=(272−9)−0=92=\left[\dfrac{3x^2}{2}-\dfrac{x^3}{3}\right]_0^3=\left(\dfrac{27}{2}-9\right)-0=\dfrac92.

Answer: 92\dfrac{9}{2}

Mark scheme

Mark scheme for question 1
QuestionSchemeMarks
1.Sets 4x−x2=x4x-x^2=x and solves to find the intersectionsM1
x=0x=0 and x=3x=3A1
Integrates the difference of the two functions, xn→xn+1x^n\to x^{n+1}M1
3x22−x33\dfrac{3x^2}{2}-\dfrac{x^3}{3} or 2x2−x332x^2-\dfrac{x^3}{3} and 92\dfrac92A1
Applies the limits 00 and 33 correctly to obtain 92\dfrac92A1
Notes
M1: or area under the curve minus the triangle 12×3×3\frac12\times3\times3
A1: cso

(5 marks)

Common mistakes

  • Don't use one pair of limits after the upper and lower curves swap, causing part of an enclosed area to cancel.
  • Don't integrate one curve from the axis instead of subtracting lower curve from upper curve over the intersection limits.

Mark-scheme wording to use

  • “y = 8x, 12x + y = 48 ⇒ y = 19.2 (or x = 2.4) M1 Triangle area is (1/2) × 4 × "19.2" (= 38.4 or 192/5) or ∫_0^"2.4" 8x dx + ∫_"2.4"^4 "(48 − 12x)" dx dM1 … M1: Attempts to find either the x or y coordinate of the intersection of line l_1 and line l_2 … dM1: Correct method for the area of the triangle.” — Pearson 9MA0, June 2024 Paper 1, Q10(c)
  • “M1: Substitutes the limits 1 and 0 into an expression of the form ±αx²e^(−3x) ± βxe^(−3x) ± γe^(−3x), α, β, γ ≠ 0 and subtracts the right way round. … Must see evidence of the use of both limits and subtraction and use of e⁰ = 1. … A1: Correct answer of 16/27 − (136/27)e⁻³ but allow equivalent exact fractions and condone 16/27 − 136/(27e³).” — Pearson 9MA0, June 2024 Paper 2, Q11
  • “M1: Substitute both limits, subtract and set the result equal to ln(7). A1: Reach a=ln(7/2) and state k=7/2 or an exact equivalent” — Pearson 9MA0, Sample Paper 1, Q4

What the examiner wants

The simplest correct strategy, kept exact. The June 2024 Paper 1 report says most found the area of the triangle and subtracted the area under the curve, but a significant proportion used Pythagoras and trigonometry for the triangle, which led to errors or premature rounding, and the Paper 2 report notes sign errors when integration by parts was applied twice.

Exam tip

For an enclosed area, solve for every intersection, identify upper minus lower, and check that the final area is positive.

Tier 1 · Easy

ORIGINAL

1.

Find the area between the line y=2x+1y=2x+1, the xx-axis, and the lines x=0x=0 and x=3x=3.

(3)

(Total for Question 1 is 3 marks)

Tier 2 · Standard

ORIGINAL

1.

Find the finite area between the curve y=x2−4x+3y=x^2-4x+3 and the xx-axis.

(4)

(Total for Question 1 is 4 marks)

Tier 3 · Hard

ORIGINAL
Year 2 part
Areas under parametric curves

1.

A curve is given parametrically by x=t2x=t^2 and y=t3y=t^3 for 0≤t≤20\leq t\leq2. Find the exact area between the curve, the xx-axis and the line x=4x=4.

(6)

(Total for Question 1 is 6 marks)

Year 2
Year 2: this is content the exam board adds beyond the AS subject content, for the full A-level.

Understand and use integration as the limit of a sum.

Notes
Worked answers & exam appearances →
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Partition an interval into strips of width Δx\Delta x and form ∑f(xr)Δx\sum f(x_r)\Delta x; its limit as the maximum strip width tends to zero is the definite integral.
  • To recognise a limit, identify the factor playing the role of Δx\Delta x and rewrite the sampled expression in terms of an endpoint such as xr=a+rΔxx_r=a+r\Delta x.
  • A sum with factor 1/n1/n usually samples an interval of length 11; a different interval width introduces a corresponding scale factor.
  • A common error is to identify the integrand but omit the width factor, which changes the value of the limiting integral.
Worked example

Evaluate lim⁡n→∞1n∑r=1n(1+3rn)2\lim_{n\to\infty}\frac1n\sum_{r=1}^n\left(1+\frac{3r}{n}\right)^2 by expressing it as a definite integral.

  1. 1.With x=r/nx=r/n, the limit is ∫01(1+3x)2 dx\int_0^1(1+3x)^2\,\mathrm dx.
  2. 2.Equivalently, with u=1+3xu=1+3x, it is 13∫14u2 du=19(43−13)=7\frac13\int_1^4u^2\,\mathrm du=\frac19(4^3-1^3)=7.

Answer: 77

Common mistakes

  • Don't read i/ni/n as the strip width rather than as the sample-point position within the interval.
  • Don't match a Riemann sum to the wrong interval by overlooking the index increment and endpoint expression.

Exam tip

Rewrite the summand as a function of the sample point and identify both the strip width and integration limits.

Tier 1 · Easy

ORIGINAL

1.

Express ∫02x2 dx\int_0^2x^2\,\mathrm dx as a limit of a right-endpoint sum and evaluate it.

(4)

(Total for Question 1 is 4 marks)

Tier 2 · Standard

ORIGINAL

1.

Express ∫251x dx\displaystyle\int_2^5\dfrac1x\,\mathrm dx as the limit of a sum using nn equal subintervals and right-hand endpoints, and hence evaluate the limit.

(4)

(Total for Question 1 is 4 marks)

Tier 3 · Hard

ORIGINAL

1.

Evaluate exactly lim⁡n→∞2n∑r=1nln⁡(1+2rn)\lim_{n\to\infty}\frac2n\sum_{r=1}^n\ln\left(1+\frac{2r}{n}\right).

(6)

(Total for Question 1 is 6 marks)

Year 2
Year 2: this is content the exam board adds beyond the AS subject content, for the full A-level.

Carry out simple cases of integration by substitution and integration by parts; understand these methods as the inverse processes of the chain and product rules respectively.

Notes
Worked answers & exam appearances →
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Substitution reverses the chain rule: choose u=g(x)u=g(x) so the remaining factor supplies du=g′(x) dx\mathrm du=g'(x)\,\mathrm dx. In particular, recognise ∫f′(x)f(x) dx=ln⁡∣f(x)∣+C\int\frac{f'(x)}{f(x)}\,\mathrm dx=\ln|f(x)|+C.
  • Integration by parts reverses the product rule: ∫u dv=uv−∫v du\int u\,\mathrm dv=uv-\int v\,\mathrm du; choose uu so that differentiating it simplifies the integral.
  • For a definite integral, either change the limits to the new variable or return fully to the original variable before applying the old limits.
  • A common error is to omit the minus sign in integration by parts or to mix xx and uu in the same transformed integral.
  • A required standard result is ∫ln⁡x dx=xln⁡x−x+c\int\ln x\,dx=x\ln x-x+c, obtained by integration by parts with u=ln⁡xu=\ln x and dv=dxdv=dx.
Worked example

Find ∫xe2x dx\int xe^{2x}\,\mathrm dx.

  1. 1.Use integration by parts with u=xu=x and dv=e2x dx\mathrm dv=e^{2x}\,\mathrm dx.
  2. 2.Then du=dx\mathrm du=\mathrm dx and v=12e2xv=\frac12e^{2x}.
  3. 3.Hence ∫xe2x dx=x2e2x−12∫e2x dx=x2e2x−14e2x+C\int xe^{2x}\,\mathrm dx=\frac x2e^{2x}-\frac12\int e^{2x}\,\mathrm dx=\frac x2e^{2x}-\frac14e^{2x}+C.

Answer: e2x(x2−14)+Ce^{2x}\left(\frac x2-\frac14\right)+C

Common mistakes

  • Don't change variable in a definite integral but keep the original xx-limits.
  • Don't choose integration by parts but reverse the selected u and dv, creating a harder integral.

Exam tip

For a product of an algebraic and exponential term, choose the algebraic factor as u and retain the boundary-free constant.

Tier 1 · Easy

ORIGINAL

1.

Find ∫6x3x2+5 dx\int\frac{6x}{3x^2+5}\,\mathrm dx.

(3)

(Total for Question 1 is 3 marks)

Tier 2 · Standard

ORIGINAL

1.

Use a substitution to evaluate ∫012xx2+4 dx\displaystyle\int_0^1\dfrac{2x}{x^2+4}\,\mathrm dx.

(4)

(Total for Question 1 is 4 marks)

Tier 3 · Hard

ORIGINAL

1.

Evaluate exactly ∫01xln⁡(1+x2) dx\int_0^1x\ln(1+x^2)\,\mathrm dx.

(7)

(Total for Question 1 is 7 marks)

Year 2
Year 2: this is content the exam board adds beyond the AS subject content, for the full A-level.

Integrate using partial fractions that are linear in the denominator.

Notes
Worked answers & exam appearances →
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Factor the denominator fully into linear factors, then write one partial-fraction term for each factor before solving for its constants. For Ax−a\frac{A}{x-a}, the integral is Aln⁡∣x−a∣+CA\ln|x-a|+C; retain absolute values unless the domain makes the sign known.
  • Determine constants by equating coefficients or substituting convenient values that make all but one term vanish.
  • A common error is to integrate the unfactorised denominator as though ∫f′(x)/f(x) dx\int f'(x)/f(x)\,\mathrm dx applied when the numerator is not its derivative.
  • Set $\frac{5x+1}{(x-1)(x+2)}=\frac A{x-1}+\frac B{x+2}$.
  • Then 5x+1=A(x+2)+B(x−1)5x+1=A(x+2)+B(x-1).
Worked example

Express 5x+1(x−1)(x+2)\frac{5x+1}{(x-1)(x+2)} in partial fractions and hence integrate it.

  1. 1.Set 5x+1(x−1)(x+2)=Ax−1+Bx+2\frac{5x+1}{(x-1)(x+2)}=\frac A{x-1}+\frac B{x+2}.
  2. 2.Then 5x+1=A(x+2)+B(x−1)5x+1=A(x+2)+B(x-1).
  3. 3.Substituting x=1x=1 gives A=2A=2, and x=−2x=-2 gives B=3B=3.
  4. 4.Integrating the decomposition gives 2ln⁡∣x−1∣+3ln⁡∣x+2∣+C2\ln|x-1|+3\ln|x+2|+C.

Answer: 5x+1(x−1)(x+2)=2x−1+3x+2\frac{5x+1}{(x-1)(x+2)}=\frac2{x-1}+\frac3{x+2}; 2ln⁡∣x−1∣+3ln⁡∣x+2∣+C2\ln|x-1|+3\ln|x+2|+C

Common mistakes

  • Don't drop the absolute-value signs after integrating a linear denominator to a logarithm.
  • Don't split the fraction without solving the coefficient identity, so the logarithmic coefficients are incorrect.

Exam tip

Find partial-fraction constants before integrating and preserve absolute-value signs in logarithmic antiderivatives.

Tier 1 · Easy

ORIGINAL

1.

Find ∫(3x−1+2x+2) dx\int\left(\frac3{x-1}+\frac2{x+2}\right)\,\mathrm dx.

(3)

(Total for Question 1 is 3 marks)

Tier 2 · Standard

ORIGINAL

1.

Express 2x2+3x−1x(x+1)\dfrac{2x^2+3x-1}{x(x+1)} in the form A+Bx+Cx+1A+\dfrac{B}{x}+\dfrac{C}{x+1} and hence integrate it.

(5)

(Total for Question 1 is 5 marks)

Tier 3 · Hard

ORIGINAL

1.

Evaluate exactly ∫015x+1(x+1)(x+2) dx\int_0^1\frac{5x+1}{(x+1)(x+2)}\,\mathrm dx.

(7)

(Total for Question 1 is 7 marks)

Year 2
Year 2: this is content the exam board adds beyond the AS subject content, for the full A-level.

Evaluate the analytical solution of simple first order differential equations with separable variables, including finding particular solutions.

Notes
Worked answers & exam appearances →
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • For $\frac{\mathrm dy}{\mathrm dx}=g(x)h(y)$, rearrange to place all yy-terms with dy\mathrm dy and all xx-terms with dx\mathrm dx, then integrate both sides.
  • Include a constant after integration and use the initial condition to determine it.
  • A family of solutions can be sketched from its equilibrium curves, initial values, gradients and long-term behaviour.
  • When logarithms arise, exponentiate carefully and use the stated domain or initial condition to choose any required sign or branch.
  • A common error is to divide by a factor involving yy without checking whether doing so loses a constant equilibrium solution.
Worked example

Solve dydx=x+1y\frac{\mathrm dy}{\mathrm dx}=\frac{x+1}{y}, given that y=2y=2 when x=0x=0 and y>0y>0.

  1. 1.Separate: y dy=(x+1) dxy\,\mathrm dy=(x+1)\,\mathrm dx.
  2. 2.Integration gives 12y2=12x2+x+C\frac12y^2=\frac12x^2+x+C, so y2=x2+2x+C1y^2=x^2+2x+C_1.
  3. 3.The condition y(0)=2y(0)=2 gives C1=4C_1=4.
  4. 4.Since y>0y>0, y=x2+2x+4y=\sqrt{x^2+2x+4}.

Answer: y=x2+2x+4y=\sqrt{x^2+2x+4}

Common mistakes

  • Don't rearrange dy/dx=f(x)/g(y)\mathrm dy/\mathrm dx=f(x)/g(y) as f(x) dy=g(y) dxf(x)\,\mathrm dy=g(y)\,\mathrm dx.
  • Don't separate variables but lose the integration constant or select a branch inconsistent with the initial condition.

Exam tip

After separation and integration, use the initial condition before choosing the branch required by the stated domain.

Tier 1 · Easy

ORIGINAL

1.

Solve dydx=2xy\frac{\mathrm dy}{\mathrm dx}=2xy, given that y=3y=3 when x=0x=0.

(4)

(Total for Question 1 is 4 marks)

Tier 2 · Standard

ORIGINAL

1.

For dydx=3x(y−2)\dfrac{\mathrm dy}{\mathrm dx}=3x(y-2), find the general non-equilibrium solution and state the equilibrium solution excluded when the variables are separated.

(4)

(Total for Question 1 is 4 marks)

Tier 3 · Hard

ORIGINAL

1.

Solve dydx=y(4−y)\frac{\mathrm dy}{\mathrm dx}=y(4-y), given that y=1y=1 when x=0x=0. State the equilibrium solutions and describe the particular solution's long-term behaviour.

(9)

(Total for Question 1 is 9 marks)

Year 2
Year 2: this is content the exam board adds beyond the AS subject content, for the full A-level.

Interpret the solution of a differential equation in the context of solving a problem, including identifying limitations of the solution; includes links to kinematics.

Notes
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Explanation

  • Interpret constants and initial values with their units, and examine limiting behaviour to identify equilibrium values or long-term predictions.
  • In kinematics, v=dsdtv=\frac{\mathrm ds}{\mathrm dt} and a=dvdta=\frac{\mathrm dv}{\mathrm dt}; the sign of velocity gives direction, while a change of sign marks a reversal of motion.
  • Check whether the mathematical solution remains meaningful on the time interval and within the physical range assumed by the model.
  • A common error is to state only that a model is 'unrealistic'; name a specific assumption, such as constant environmental conditions or neglect of resistance, and explain its effect.
Worked example

The velocity of a particle is modelled by v=20(1−e−0.5t) m s−1v=20(1-e^{-0.5t})\,\text{m s}^{-1} for t≥0t\geq0. Find the limiting velocity and the time when v=15 m s−1v=15\,\text{m s}^{-1}. State one limitation of the model.

  1. 1.As t→∞t\to\infty, e−0.5t→0e^{-0.5t}\to0, so the limiting velocity is 20 m s−120\,\text{m s}^{-1}.
  2. 2.For v=15v=15, 15=20(1−e−0.5t)15=20(1-e^{-0.5t}), so e−0.5t=1/4e^{-0.5t}=1/4 and t=2ln⁡4≈2.77 st=2\ln4\approx2.77\,\text{s}.
  3. 3.A real resistance law or driving force may change with conditions, so the same constants need not remain valid indefinitely.

Answer: 20 m s−120\,\text{m s}^{-1}; t=2ln⁡4≈2.77 st=2\ln4\approx2.77\,\text{s}; One valid limitation, such as a constant resistance law being assumed

Common mistakes

  • Don't extend a population solution beyond the time at which the model predicts an impossible negative value.
  • Don't report a limiting value algebraically and fail to interpret its units or challenge the long-term model.

Exam tip

For a contextual differential model, give numerical conclusions with units and state a limitation tied to its assumptions.

Tier 1 · Easy

ORIGINAL

1.

A population model has solution N=500e0.12tN=500e^{0.12t}, where tt is measured in years. Interpret the constants 500500 and 0.120.12, and state one limitation of the model.

(4)

(Total for Question 1 is 4 marks)

Tier 2 · Standard

ORIGINAL

1.

The mass of a substance is modelled by M=50e−0.2tM=50e^{-0.2t} grams, where tt is in hours. Find the time at which the mass reaches 55 grams, giving the time exactly and to 33 significant figures; hence state when the mass is below 55 grams. Interpret the limiting value of MM and state one limitation of the model.

(5)

(Total for Question 1 is 5 marks)

Tier 3 · Hard

ORIGINAL

1.

A particle has velocity v=18−24e−0.3t m s−1v=18-24e^{-0.3t}\,\text{m s}^{-1} for t≥0t\geq0, with displacement s=0s=0 at t=0t=0. Find when the particle changes direction, its displacement then, and its limiting velocity. Give one limitation of using this model for arbitrarily large tt.

(8)

(Total for Question 1 is 8 marks)

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