8 Integration — revision question pack

8 specification points · notes, questions, answers and worked methods

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

8.1 · Know and use the Fundamental Theorem of Calculus.

Explanation

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

Worked example

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

1.Rewrite as $f'(x)=3x^{1/2}-2x^{-2}$ and integrate: $f(x)=2x^{3/2}+2x^{-1}+c$ .
2.Substituting $x=4$ : $2(8)+\frac{2}{4}+c=10$ , so $16.5+c=10$ and $c=-\frac{13}{2}$ .
3.Hence $f(x)=2x^{3/2}+\frac{2}{x}-\frac{13}{2}$ .

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

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

1.
A curve has gradient function $\frac{\mathrm dy}{\mathrm dx}=6x^2-4$ and passes through the point $(1,5)$ . Find the equation of the curve.
(3)
(Total for Question 1 is 3 marks)
2.
The function $H$ is an antiderivative of $q$ . Given that $\displaystyle\int_1^4q(x)\,\mathrm dx=9$ and $H(1)=5$ , find $H(4)$ .
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1.
Given that $F(x)=\dfrac{x^4}{2}-x^3+4x$ is an antiderivative of $2x^3-3x^2+4$ , evaluate $\displaystyle\int_1^3(2x^3-3x^2+4)\,\mathrm dx$ .
(4)
(Total for Question 1 is 4 marks)
2.
A function $F$ satisfies $F'(x)=3x^2+k$ , where $k$ is a constant. Given that $F(0)=2$ and $F(2)=16$ , find $k$ and hence find $F(x)$ .
(4)
(Total for Question 2 is 4 marks)
3.
Let $F$ be an antiderivative of $f$ . Given that $\int_{-2}^{1}f(x)\,\mathrm dx=7$ , $\int_{1}^{5}f(x)\,\mathrm dx=-4$ and $F(5)=9$ , find $F(-2)$ . Hence evaluate $\int_{5}^{-2}f(x)\,\mathrm dx$ .
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
A curve $y=f(x)$ satisfies $f''(x)=6x+2$ . The gradient of the curve at $x=1$ is $3$ , and the curve passes through $(2,4)$ . Find $f(x)$ , and verify your answer by evaluating $\int_1^2 f'(x)\,\mathrm dx$ and comparing it with $f(2)-f(1)$ .
(5)
(Total for Question 1 is 5 marks)
2.
A function $h$ satisfies $h'(x)=|2x-4|$ and $h(1)=7$ . Find $h(5)$ .
(5)
(Total for Question 2 is 5 marks)
3.
A differentiable function $g$ satisfies $g'(x)=(x-1)(x-2)$ for $0\leq x\leq3$ , and $g(0)=0$ . Use the Fundamental Theorem of Calculus to find $g(1)$ , $g(2)$ and $g(3)$ . Hence state the absolute maximum and minimum values of $g$ on this interval.
(6)
(Total for Question 3 is 6 marks)
4.
A function $F$ satisfies $F'(x)=3x^2-12x+9$ and $F(0)=5$ . Find all values of $x$ in $0\leq x\leq3$ for which $F(x)=5$ . Hence find the exact area enclosed by the curve $y=F(x)$ and the line $y=5$ between consecutive points of intersection.
(6)
(Total for Question 4 is 6 marks)
5.
A differentiable function $F$ satisfies $F'(x)=ax+b$ , where $a$ and $b$ are constants. Given that $F(2)-F(0)=6$ , $F(5)-F(2)=3$ and $F(0)=0$ , use the Fundamental Theorem of Calculus to find $a$ and $b$ . Hence find $F(x)$ and the absolute maximum value of $F$ on $0\leq x\leq5$ .
(7)
(Total for Question 5 is 7 marks)

8.2 · 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.

Explanation

For $n\ne-1$ , $\int xⁿ\,\mathrm dx=\frac{x^{n+1}}{n+1}+C$; rewrite roots and reciprocals as powers before applying the rule. Standard forms include $\int e^{kx}\,\mathrm dx=\frac1k e^{kx}+C$ and $\int\frac1x\,\mathrm dx=\ln|x|+C$ on intervals not crossing $0$ .
Divide by the inner coefficient in trigonometric integrals.
Also use $\int\sec^2(kx)\,\mathrm dx=\frac1k\tan(kx)+C$ , $\int\tan x\,\mathrm dx=-\ln|\cos x|+C$ , and identities for powers such as $\sin^2x=\frac12(1-\cos2x)$ .
A common error is to forget the constant of integration or to multiply by $k$ rather than divide by it when integrating a function of $kx$ .
Integrate term by term: the first three terms give $2e^{3x}+4\ln|x|+\frac52\cos(2x)$ .

Worked example

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

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

Answer: $2e^{3x}+4\ln|x|+\frac52\cos(2x)-3\ln|\cos x|+C$

Common mistakes

Don't integrate $e^{kx}$ as $ke^{kx}$ instead of $e^{kx}/k$ .
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

1.
Find $\int(6x^2-4x^{1/2}+3\sec^2(2x))\,\mathrm dx$ .
(4)
(Total for Question 1 is 4 marks)
2.
For $x>0$ , find $\displaystyle\int\dfrac{2x^3-5x+4}{x}\,\mathrm dx$ .
(3)
(Total for Question 2 is 3 marks)

Tier 2 · Standard

1.
For $x>0$ , find $\displaystyle\int\dfrac{(1+\sqrt{x})^2}{x}\,\mathrm dx$ .
(4)
(Total for Question 1 is 4 marks)
2.
Find $\displaystyle\int\left(7e^{-x/2}+4\cos(3x)-5\sec^2(4x)\right)\,\mathrm dx$ .
(4)
(Total for Question 2 is 4 marks)
3.
Evaluate exactly $\displaystyle\int_0^{\pi/6}\tan(2x)\,\mathrm dx$ .
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
Use trigonometric identities to find $\int(4\sin^2x+3\tan^2x)\,\mathrm dx$ .
(6)
(Total for Question 1 is 6 marks)
2.
Find $\displaystyle\int(2\sin x+\cos x)^2\,\mathrm dx$ .
(5)
(Total for Question 2 is 5 marks)
3.
By writing $\sin(3x)\cos x$ in terms of $\sin(3x+x)$ and $\sin(3x-x)$ , using the sine addition and subtraction formulae, find $\displaystyle\int 8\sin(3x)\cos x\,\mathrm dx$ .
(5)
(Total for Question 3 is 5 marks)
4.
Let $f(x)=a\sin(2x)+b\cos(2x)$ , where $a$ and $b$ are constants. Given that $\displaystyle\int_0^{\pi/4}f(x)\,\mathrm dx=3$ and $\displaystyle\int_{\pi/4}^{\pi/2}f(x)\,\mathrm dx=1$ , find $a$ and $b$ . Hence find $\displaystyle\int f(x)\,\mathrm dx$ .
(6)
(Total for Question 4 is 6 marks)
5.
Evaluate exactly $\displaystyle\int_0^{\pi}|1+2\cos x|\,\mathrm dx$ .
(6)
(Total for Question 5 is 6 marks)

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

Explanation

Evaluate $\int_a^b f(x)\,\mathrm dx$ by finding an antiderivative $F$ and calculating $F(b)-F(a)$ . A definite integral is signed area.
Split at roots or intersections if the required geometric area includes regions below the axis or where the upper curve changes. For area between curves, solve their intersection equations first and integrate $y_{\text{upper}}-y_{\text{lower}}$ over each relevant interval.
A common error is to reverse the subtraction or to quote a negative definite integral as a negative geometric area.
Intersections satisfy $4x-x^2=x$ , so $x(3-x)=0$ and $x=0,3$ .
For a parametric curve, its signed area is $\int y\,\dfrac{dx}{dt}\,dt$ over the parameter interval, with limits ordered to match the traced direction.

For a parametric curve, vertical strips give area

\int y(t)\,\dfrac{dx}{dt}\,dt

Worked example

Find the finite area enclosed by the curve $y=4x-x^2$ and the line $y=x$ .

1.Intersections satisfy $4x-x^2=x$ , so $x(3-x)=0$ and $x=0,3$ .
2.The parabola is above the line between them.
3.Thus the area is $\int_0^3(3x-x^2)\,\mathrm dx=[\frac32x^2-\frac13x^3]_0^3=\frac{27}{2}-9=\frac92$ .

Answer: Area $=\frac92$ square units

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.

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

1.
Find the area between the line $y=2x+1$ , the $x$ -axis, and the lines $x=0$ and $x=3$ .
(3)
(Total for Question 1 is 3 marks)
2.
Find the exact area under the curve $y=4e^{-x}$ between $x=0$ and $x=\ln3$ .
(3)
(Total for Question 2 is 3 marks)

Tier 2 · Standard

1.
Find the finite area between the curve $y=x^2-4x+3$ and the $x$ -axis.
(4)
(Total for Question 1 is 4 marks)
2.
Find the exact area between the curves $y=e^x$ and $y=1+x$ , and the lines $x=0$ and $x=1$ .
(4)
(Total for Question 2 is 4 marks)
3.
Find the total area between the curve $y=x(x-1)(x-2)$ and the $x$ -axis for $0\leq x\leq2$ .
(5)
(Total for Question 3 is 5 marks)

Tier 3 · Hard

1.
A curve is given parametrically by $x=t^2$ and $y=t^3$ for $0\leq t\leq2$ . Find the exact area between the curve, the $x$ -axis and the line $x=4$ .
(6)
(Total for Question 1 is 6 marks)
2.
Find the exact total area between the curves $y=\sin x$ and $y=\cos x$ for $0\leq x\leq\dfrac{\pi}{2}$ .
(5)
(Total for Question 2 is 5 marks)
3.
The curve $y=kx-x^2$ , where $k>0$ , and the $x$ -axis enclose a finite region of area $\frac92$ square units. Find $k$ .
(5)
(Total for Question 3 is 5 marks)
4.
The curve $y=4-x^2$ and the $x$ -axis enclose a finite region. The line $x=a$ , where $-2<a<2$ , cuts off an area of $9$ square units between $x=-2$ and $x=a$ . Find $a$ .
(6)
(Total for Question 4 is 6 marks)
5.
A curve is given parametrically by $x=2+t^2$ and $y=t^3-4t$ for $0\leq t\leq3$ . Find the exact area enclosed by the curve and the chord joining its endpoints.
(6)
(Total for Question 5 is 6 marks)

8.4 · Understand and use integration as the limit of a sum.

Explanation

Partition an interval into strips of width $\Delta x$ and form $\sum f(x_r)\Delta x$ ; 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 $\Delta x$ and rewrite the sampled expression in terms of an endpoint such as $x_r=a+r\Delta x$ .
A sum with factor $1/n$ usually samples an interval of length $1$ ; 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 $\lim_{n\to\infty}\frac1n\sum_{r=1}^n\left(1+\frac{3r}{n}\right)^2$ by expressing it as a definite integral.

1.With $x=r/n$ , the limit is $\int_0^1(1+3x)^2\,\mathrm dx$ .
2.Equivalently, with $u=1+3x$ , it is $\frac13\int_1^4u^2\,\mathrm du=\frac19(4^3-1^3)=7$ .

Answer: $7$

Common mistakes

Don't read $i/n$ 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

1.
Express $\int_0^2x^2\,\mathrm dx$ as a limit of a right-endpoint sum and evaluate it.
(4)
(Total for Question 1 is 4 marks)
2.
Evaluate $\displaystyle\lim_{n\to\infty}\frac1n\sum_{r=1}^n\left(2+\frac rn\right)$ by expressing it as a definite integral.
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1.
Express $\displaystyle\int_2^5\dfrac1x\,\mathrm dx$ as the limit of a sum using $n$ equal subintervals and right-hand endpoints, and hence evaluate the limit.
(4)
(Total for Question 1 is 4 marks)
2.
Evaluate exactly $\displaystyle\lim_{n\to\infty}\frac4n\sum_{r=1}^n\dfrac{1}{1+4r/n}$ .
(4)
(Total for Question 2 is 4 marks)
3.
Evaluate $\displaystyle\lim_{n\to\infty}\frac1n\sum_{r=1}^n\left(2+\frac{r}{n}\right)^3$ by expressing it as a definite integral.
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
Evaluate exactly $\lim_{n\to\infty}\frac2n\sum_{r=1}^n\ln\left(1+\frac{2r}{n}\right)$ .
(6)
(Total for Question 1 is 6 marks)
2.
Evaluate exactly $\displaystyle\lim_{n\to\infty}\frac2n\sum_{r=1}^n\sqrt{3+\frac{2r}{n}}$ .
(6)
(Total for Question 2 is 6 marks)
3.
Evaluate exactly $\displaystyle\lim_{n\to\infty}\sum_{r=1}^n\frac{1}{3n+2r}$ by expressing it as a definite integral.
(5)
(Total for Question 3 is 5 marks)
4.
Evaluate exactly $\displaystyle\lim_{n\to\infty}\frac3n\sum_{r=1}^n e^{\,2+3r/n}$ by expressing it as a definite integral.
(5)
(Total for Question 4 is 5 marks)
5.
Evaluate exactly $\displaystyle\lim_{n\to\infty}\frac{2\pi}{3n}\sum_{r=1}^n\sin\left(\frac{\pi}{6}+\frac{2\pi r}{3n}\right)$ by expressing it as a definite integral.
(5)
(Total for Question 5 is 5 marks)

8.5 · 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.

Explanation

Substitution reverses the chain rule: choose $u=g(x)$ so the remaining factor supplies $\mathrm du=g'(x)\,\mathrm dx$ . In particular, recognise $\int\frac{f'(x)}{f(x)}\,\mathrm dx=\ln|f(x)|+C$ .
Integration by parts reverses the product rule: $\int u\,\mathrm dv=uv-\int v\,\mathrm du$ ; choose $u$ 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 $x$ and $u$ in the same transformed integral.
A required standard result is $\int\ln x\,dx=x\ln x-x+c$ , obtained by integration by parts with $u=\ln x$ and $dv=dx$ .

Worked example

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

1.Use integration by parts with $u=x$ and $\mathrm dv=e^{2x}\,\mathrm dx$ .
2.Then $\mathrm du=\mathrm dx$ and $v=\frac12e^{2x}$ .
3.Hence $\int xe^{2x}\,\mathrm dx=\frac x2e^{2x}-\frac12\int e^{2x}\,\mathrm dx=\frac x2e^{2x}-\frac14e^{2x}+C$ .

Answer: $e^{2x}\left(\frac x2-\frac14\right)+C$

Common mistakes

Don't change variable in a definite integral but keep the original $x$ -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

1.
Find $\int\frac{6x}{3x^2+5}\,\mathrm dx$ .
(3)
(Total for Question 1 is 3 marks)
2.
Find $\displaystyle\int8x(2x^2+1)^3\,\mathrm dx$ .
(3)
(Total for Question 2 is 3 marks)

Tier 2 · Standard

1.
Use a substitution to evaluate $\displaystyle\int_0^1\dfrac{2x}{x^2+4}\,\mathrm dx$ .
(4)
(Total for Question 1 is 4 marks)
2.
Evaluate exactly $\displaystyle\int_0^1x\cos(\pi x)\,\mathrm dx$ .
(4)
(Total for Question 2 is 4 marks)
3.
Use integration by parts to evaluate exactly $\displaystyle\int_1^{e^3}\ln x\,\mathrm dx$ .
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
Evaluate exactly $\int_0^1x\ln(1+x^2)\,\mathrm dx$ .
(7)
(Total for Question 1 is 7 marks)
2.
Evaluate exactly $\displaystyle\int_0^{\pi/2}x^2\sin x\,\mathrm dx$ .
(6)
(Total for Question 2 is 6 marks)
3.
Use integration by parts twice to find $\displaystyle\int e^x\cos x\,\mathrm dx$ .
(6)
(Total for Question 3 is 6 marks)
4.
Make the change of variable $u=x^2+1$ to evaluate exactly $\displaystyle\int_0^{\sqrt3}\dfrac{x^3}{x^2+1}\,\mathrm dx$ .
(5)
(Total for Question 4 is 5 marks)
5.
Use integration by parts to evaluate exactly $\displaystyle\int_0^1x^2\ln(1+x)\,\mathrm dx$ .
(6)
(Total for Question 5 is 6 marks)

8.6 · Integrate using partial fractions that are linear in the denominator.

Explanation

Factor the denominator fully into linear factors, then write one partial-fraction term for each factor before solving for its constants. For $\frac{A}{x-a}$ , the integral is $A\ln|x-a|+C$ ; 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 $\int f'(x)/f(x)\,\mathrm dx$ 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 $5x+1=A(x+2)+B(x-1)$ .

Worked example

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

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

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

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

1.
Find $\int\left(\frac3{x-1}+\frac2{x+2}\right)\,\mathrm dx$ .
(3)
(Total for Question 1 is 3 marks)
2.
Given that $\dfrac{1}{(x+1)(x+3)}=\dfrac{1}{2(x+1)}-\dfrac{1}{2(x+3)}$ , integrate $\dfrac{1}{(x+1)(x+3)}$ and combine the logarithms.
(3)
(Total for Question 2 is 3 marks)

Tier 2 · Standard

1.
Express $\dfrac{2x^2+3x-1}{x(x+1)}$ in the form $A+\dfrac{B}{x}+\dfrac{C}{x+1}$ and hence integrate it.
(5)
(Total for Question 1 is 5 marks)
2.
Express $\dfrac{4x-7}{(x-2)(x+3)}$ in partial fractions and hence integrate it.
(4)
(Total for Question 2 is 4 marks)
3.
Express $\dfrac{4x+7}{(x+1)(2x+3)}$ in partial fractions and hence integrate it.
(5)
(Total for Question 3 is 5 marks)

Tier 3 · Hard

1.
Evaluate exactly $\int_0^1\frac{5x+1}{(x+1)(x+2)}\,\mathrm dx$ .
(7)
(Total for Question 1 is 7 marks)
2.
For $x>0$ , a function $F$ satisfies $F'(x)=\dfrac{3}{(x+2)(x+5)}$ and $F(1)=0$ . (a) Find $F(x)$ . (b) Hence solve $F(x)=\ln\left(\dfrac{10}{7}\right)$ .
(6)
(Total for Question 2 is 6 marks)
3.
Evaluate exactly $\displaystyle\int_1^2\dfrac{4x^2+7x+3}{(2x-1)(x+2)}\,\mathrm dx$ .
(6)
(Total for Question 3 is 6 marks)
4.
Given that $\dfrac{9x-8}{(x-4)(3x+2)}=\dfrac{A}{x-4}+\dfrac{B}{3x+2}$ , find $A$ and $B$ . Hence, for $x>5$ , find $\displaystyle\int_5^x\dfrac{9t-8}{(t-4)(3t+2)}\,\mathrm dt$ in the form $\ln f(x)$ . Solve $\displaystyle\int_5^x\dfrac{9t-8}{(t-4)(3t+2)}\,\mathrm dt=\ln\left(\dfrac{80}{17}\right)$ .
(6)
(Total for Question 4 is 6 marks)
5.
Express $\dfrac{2x^2+10x+10}{(x+1)(x+2)(x+3)}$ in partial fractions and hence evaluate exactly $\displaystyle\int_0^1\dfrac{2x^2+10x+10}{(x+1)(x+2)(x+3)}\,\mathrm dx$ .
(6)
(Total for Question 5 is 6 marks)

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

Explanation

For $\frac{\mathrm dy}{\mathrm dx}=g(x)h(y)$, rearrange to place all $y$ -terms with $\mathrm dy$ and all $x$ -terms with $\mathrm dx$ , 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 $y$ without checking whether doing so loses a constant equilibrium solution.

Worked example

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

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

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

Common mistakes

Don't rearrange $\mathrm dy/\mathrm dx=f(x)/g(y)$ as $f(x)\,\mathrm dy=g(y)\,\mathrm dx$ .
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

1.
Solve $\frac{\mathrm dy}{\mathrm dx}=2xy$ , given that $y=3$ when $x=0$ .
(4)
(Total for Question 1 is 4 marks)
2.
Solve $\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{3x^2}{2y}$ , given that $y=2$ when $x=0$ and that $y>0$ .
(3)
(Total for Question 2 is 3 marks)

Tier 2 · Standard

1.
For $\dfrac{\mathrm dy}{\mathrm dx}=3x(y-2)$ , 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)
2.
Solve $\dfrac{\mathrm dy}{\mathrm dx}=(2x+1)e^{-y}$ , given that $y=\ln2$ when $x=0$ .
(4)
(Total for Question 2 is 4 marks)
3.
Solve $\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2x+1}{y^2}$ , given that $y=1$ when $x=0$ and that $y>0$ .
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
Solve $\frac{\mathrm dy}{\mathrm dx}=y(4-y)$ , given that $y=1$ when $x=0$ . State the equilibrium solutions and describe the particular solution's long-term behaviour.
(9)
(Total for Question 1 is 9 marks)
2.
Solve $\dfrac{\mathrm dy}{\mathrm dx}=x(y-1)^2$ , given that $y=0$ when $x=0$ . State the equilibrium solution and the limiting value of the particular solution as $x\to\infty$ .
(6)
(Total for Question 2 is 6 marks)
3.
Solve $\dfrac{\mathrm dy}{\mathrm dx}=(y+2)\sin x$ , given that $y=-1$ when $x=0$ . State the equilibrium solution. Hence find all values of $x$ in $0\leq x\leq2\pi$ for which $y=e-2$ .
(6)
(Total for Question 3 is 6 marks)
4.
Solve $\dfrac{\mathrm dy}{\mathrm dx}=(1+y^2)\cos x$ , given that $y=0$ when $x=0$ . For $0\leq x\leq2\pi$ , find the maximum and minimum values of $y$ and the values of $x$ at which they occur.
(6)
(Total for Question 4 is 6 marks)
5.
Solve $\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{e^x}{y^2+1}$ , given that $y=0$ when $x=0$ , leaving your solution in implicit form. Hence find the value of $x$ for which $y=2$ , and explain why this value is unique.
(6)
(Total for Question 5 is 6 marks)

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

Explanation

Interpret constants and initial values with their units, and examine limiting behaviour to identify equilibrium values or long-term predictions.
In kinematics, $v=\frac{\mathrm ds}{\mathrm dt}$ and $a=\frac{\mathrm dv}{\mathrm dt}$ ; 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 $v=20(1-e^{-0.5t})\,\text{m s}^{-1}$ for $t\geq0$ . Find the limiting velocity and the time when $v=15\,\text{m s}^{-1}$ . State one limitation of the model.

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

Answer: $20\,\text{m s}^{-1}$ ; $t=2\ln4\approx2.77\,\text{s}$ ; 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

1.
A population model has solution $N=500e^{0.12t}$ , where $t$ is measured in years. Interpret the constants $500$ and $0.12$ , and state one limitation of the model.
(4)
(Total for Question 1 is 4 marks)
2.
The temperature $T$ °C of a drink satisfies $\dfrac{\mathrm dT}{\mathrm dt}=-0.25(T-18)$ , where $t$ is measured in minutes. Its solution is $T=18+62e^{-0.25t}$ . State the initial temperature and interpret the limiting value of $T$ .
(3)
(Total for Question 2 is 3 marks)

Tier 2 · Standard

1.
The mass of a substance is modelled by $M=50e^{-0.2t}$ grams, where $t$ is in hours. Find the time at which the mass reaches $5$ grams, giving the time exactly and to $3$ significant figures; hence state when the mass is below $5$ grams. Interpret the limiting value of $M$ and state one limitation of the model.
(5)
(Total for Question 1 is 5 marks)
2.
The mass $M$ grams of dissolved salt in a mixing vessel satisfies $\dfrac{\mathrm dM}{\mathrm dt}=6-0.03M$ , where $t$ is measured in minutes and $M(0)=40$ . The solution is $M=200-160e^{-0.03t}$ . (a) Find the time when $M=120$ , giving your answer exactly and to $3$ significant figures. (b) Explain the significance of $M=200$ in both the differential equation and the model.
(5)
(Total for Question 2 is 5 marks)
3.
A population is modelled by $P=900-100e^{0.06t}$ , where $t$ is the number of years after observations begin. State the initial population. Find the time when the model first predicts no animals, giving your answer exactly and to $3$ significant figures. Hence give the time interval over which the model gives a non-negative population, and state one limitation of the model.
(5)
(Total for Question 3 is 5 marks)

Tier 3 · Hard

1.
A particle has velocity $v=18-24e^{-0.3t}\,\text{m s}^{-1}$ for $t\geq0$ , with displacement $s=0$ at $t=0$ . Find when the particle changes direction, its displacement then, and its limiting velocity. Give one limitation of using this model for arbitrarily large $t$ .
(8)
(Total for Question 1 is 8 marks)
2.
A particle moves in a straight line with velocity $v>0$ . Its motion is modelled by $\dfrac{\mathrm dv}{\mathrm dt}=\dfrac4v$ , where $v=2\,\text{m s}^{-1}$ and displacement $s=0$ when $t=0$ . Find $v$ in terms of $t$ . Hence find the displacement when $t=3$ seconds, giving your answer exactly and to $3$ significant figures. Explain why the model cannot be valid for arbitrarily large $t$ .
(7)
(Total for Question 2 is 7 marks)
3.
The area $A$ hectares of a site that has been restored is modelled by $\dfrac{\mathrm dA}{\mathrm dt}=0.1A\left(1-\dfrac{A}{50}\right)$ , where $t$ is measured in years. A solution of this differential equation is $A=\dfrac{50}{1+4e^{-0.1t}}$ for $t\geq0$ . Interpret the constants $50$ , $4$ and $0.1$ in the model. Find when the restored area first reaches $40$ hectares, giving the time exactly and to $3$ significant figures. Describe the limiting behaviour of both $A$ and $\dfrac{\mathrm dA}{\mathrm dt}$ as $t\to\infty$ , and state one limitation of the model.
(7)
(Total for Question 3 is 7 marks)
4.
The balance, £ $B$ , of an investment account is modelled by $\dfrac{\mathrm dB}{\mathrm dt}=0.04B-1\,200$ , where $t$ is measured in years. The initial balance is £ $40\,000$ , and a solution is $B=30\,000+10\,000e^{0.04t}$ . Interpret the balance £ $30\,000$ using the differential equation, and explain why this equilibrium is unstable. Find when the balance first reaches £ $50\,000$ , giving your answer exactly and to $3$ significant figures. State one limitation of the model.
(7)
(Total for Question 4 is 7 marks)
5.
The mass $M$ kilograms of a pollutant in a treatment tank is modelled by $\displaystyle\frac{\mathrm dM}{\mathrm dt}=\begin{cases}10-0.25M,&0\leq t<4,\\-0.25M,&t\geq4,\end{cases}$ where $t$ is measured in hours and $M(0)=0$ . A solution is $\displaystyle M=\begin{cases}40(1-e^{-0.25t}),&0\leq t<4,\\40(1-e^{-1})e^{-0.25(t-4)},&t\geq4.\end{cases}$ Interpret the terms $10$ and $0.25M$ in the differential equation, and explain what happens at $t=4$ . Find the maximum mass of pollutant and the later time when $M=10$ , giving each answer exactly and to $3$ significant figures. State one limitation of the model.
(7)
(Total for Question 5 is 7 marks)

Answer key

Answers begin on a new printed page so the question pack can be completed without the solutions alongside it.

8.1 · Know and use the Fundamental Theorem of Calculus.

Tier 1 · Easy

Mark scheme for 8.1 Tier 1 · Easy
Question	Scheme	Marks
1	$y=2x^3-4x+7$	3
(3 marks)		3
Notes
Integrating the gradient function gives $y=2x^3-4x+c$ . Substituting $(1,5)$ : $5=2-4+c$ , so $c=7$ . Hence $y=2x^3-4x+7$ .
2	$H(4)=14$	2
(2 marks)		2
Notes
By the Fundamental Theorem of Calculus, $\int_1^4q(x)\,\mathrm dx=H(4)-H(1)$ . Hence $9=H(4)-5$ , giving $H(4)=14$ .

Tier 2 · Standard

Mark scheme for 8.1 Tier 2 · Standard
Question	Scheme	Marks
1	$22$	4
(4 marks)		4
Notes
Use the definite-integral result $\int_a^b f(x)\,dx=F(b)-F(a)$ . Here $F(3)=81/2-27+12=51/2$ and $F(1)=1/2-1+4=7/2$ . Therefore the integral is $51/2-7/2=22$ .
2	$k=3$ $F(x)=x^3+3x+2$	4
(4 marks)		4
Notes
Integrating gives $F(x)=x^3+kx+C$ . The condition $F(0)=2$ gives $C=2$ . Then $F(2)=16$ gives $8+2k+2=16$ , so $k=3$ . Therefore $F(x)=x^3+3x+2$ .
3	$F(-2)=6$ $\displaystyle\int_{5}^{-2}f(x)\,\mathrm dx=-3$	4
(4 marks)		4
Notes
Add the adjacent integrals: $\int_{-2}^{5}f(x)\,\mathrm dx=7-4=3$ . By the Fundamental Theorem of Calculus, this equals $F(5)-F(-2)$ , so $3=9-F(-2)$ and $F(-2)=6$ . Reversing the limits changes the sign, hence $\int_{5}^{-2}f(x)\,\mathrm dx=-3$ .

Tier 3 · Hard

Mark scheme for 8.1 Tier 3 · Hard
Question	Scheme	Marks
1	$f(x)=x^3+x^2-2x-4$ $\int_1^2 f'(x)\,\mathrm dx=8=f(2)-f(1)$	5
(5 marks)		5
Notes
Integrate once: $f'(x)=3x^2+2x+c_1$ ; from $f'(1)=3$ , $3+2+c_1=3$ so $c_1=-2$ . Integrate again: $f(x)=x^3+x^2-2x+c_2$ ; from $f(2)=4$ , $8+4-4+c_2=4$ so $c_2=-4$ . Then $f(2)-f(1)=4-(-4)=8$ , and $\int_1^2(3x^2+2x-2)\,\mathrm dx=\left[x^3+x^2-2x\right]_1^2=8-0=8$ , confirming $\int_a^b f'(x)\,\mathrm dx=f(b)-f(a)$ .
2	$h(5)=17$	5
(5 marks)		5
Notes
The expression inside the modulus changes sign at $x=2$ . Therefore $h(5)-h(1)=\int_1^2(4-2x)\,\mathrm dx+\int_2^5(2x-4)\,\mathrm dx=1+9=10$ . Hence $h(5)=7+10=17$ .
3	$g(1)=\dfrac56$ , $g(2)=\dfrac23$ , $g(3)=\dfrac32$ Absolute maximum $\dfrac32$ at $x=3$ ; absolute minimum $0$ at $x=0$	6
(6 marks)		6
Notes
An antiderivative of $g'(x)=x^2-3x+2$ is $G(x)=x^3/3-3x^2/2+2x$ . Since $g(a)-g(0)=\int_0^a g'(x)\,\mathrm dx=G(a)-G(0)$ and $g(0)=0$ , substitution gives $g(1)=5/6$ , $g(2)=2/3$ and $g(3)=3/2$ . The derivative is positive on $(0,1)$ , negative on $(1,2)$ and positive on $(2,3)$ . Comparing the endpoint and stationary values gives the stated absolute extrema.
4	$F(x)=x^3-6x^2+9x+5$ ; the intersections are at $x=0$ and $x=3$ Area $=\dfrac{27}{4}$ square units	6
(6 marks)		6
Notes
Integrating and using $F(0)=5$ gives $F(x)=x^3-6x^2+9x+5=x(x-3)^2+5$ . Therefore $F(x)=5$ when $x(x-3)^2=0$ , giving $x=0$ and $x=3$ in the stated interval. Since $x(x-3)^2\geq0$ on $[0,3]$ , the curve lies above the line there. Expanding the vertical difference gives the area $\int_0^3(x^3-6x^2+9x)\,\mathrm dx=81/4-54+81/2=27/4$ square units.
5	$a=-\dfrac45$ , $b=\dfrac{19}{5}$ $F(x)=-\dfrac25x^2+\dfrac{19}{5}x$ Absolute maximum $\dfrac{361}{40}$ at $x=\dfrac{19}{4}$	7
(7 marks)		7
Notes
By the Fundamental Theorem of Calculus, $F(2)-F(0)=\int_0^2(ax+b)\,\mathrm dx=2a+2b$ , so $a+b=3$ . Also $F(5)-F(2)=\int_2^5(ax+b)\,\mathrm dx=21a/2+3b=3$ . Solving gives $a=-4/5$ and $b=19/5$ . Integration and $F(0)=0$ then give $F(x)=-2x^2/5+19x/5$ . The only interior stationary point is $x=19/4$ ; the derivative changes from positive to negative there. Comparing with $F(0)=0$ and $F(5)=9$ gives the absolute maximum $F(19/4)=361/40$ .

8.2 · 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.

Tier 1 · Easy

Mark scheme for 8.2 Tier 1 · Easy
Question	Scheme	Marks
1	$2x^3-\frac83x^{3/2}+\frac32\tan(2x)+C$	4
(4 marks)		4
Notes
The power terms give $2x^3-\frac83x^{3/2}$ . Since $\int\sec^2(2x)\,\mathrm dx=\frac12\tan(2x)$ , the final term gives $\frac32\tan(2x)$ . Add $C$ .
2	$\dfrac23x^3-5x+4\ln x+C$	3
(3 marks)		3
Notes
Divide each term by $x$ to obtain $2x^2-5+4/x$ . Integrating term by term gives $\frac23x^3-5x+4\ln x+C$ .

Tier 2 · Standard

Mark scheme for 8.2 Tier 2 · Standard
Question	Scheme	Marks
1	$\ln x+4\sqrt{x}+x+C$	4
(4 marks)		4
Notes
First expand and divide by $x$ : $(1+\sqrt{x})^2/x=(1+2\sqrt{x}+x)/x=x^{-1}+2x^{-1/2}+1$ . Integrating term by term gives $\ln x+4x^{1/2}+x+C$ .
2	$-14e^{-x/2}+\dfrac43\sin(3x)-\dfrac54\tan(4x)+C$	4
(4 marks)		4
Notes
Integrating each term and dividing by its inner coefficient gives $-14e^{-x/2}$ , $\frac43\sin(3x)$ and $-\frac54\tan(4x)$ respectively. Adding the constant gives the stated result.
3	$\dfrac12\ln2$	4
(4 marks)		4
Notes
Since $\int\tan(2x)\,\mathrm dx=-\frac12\ln\|\cos(2x)\|+C$ , the definite integral is $[-\frac12\ln\|\cos(2x)\|]_0^{\pi/6}=-\frac12\ln(1/2)+\frac12\ln1=\frac12\ln2$ .

Tier 3 · Hard

Mark scheme for 8.2 Tier 3 · Hard
Question	Scheme	Marks
1	$3\tan x-x-\sin(2x)+C$	6
(6 marks)		6
Notes
Use $\sin^2x=\frac12(1-\cos2x)$ and $\tan^2x=\sec^2x-1$ . The integrand becomes $2-2\cos2x+3\sec^2x-3=3\sec^2x-1-2\cos2x$ . Integrating gives $3\tan x-x-\sin(2x)+C$ .
2	$\dfrac52x-\dfrac34\sin(2x)-\cos(2x)+C$	5
(5 marks)		5
Notes
Expand and use double-angle identities: $(2\sin x+\cos x)^2=4\sin^2x+4\sin x\cos x+\cos^2x=\frac52-\frac32\cos(2x)+2\sin(2x)$ . Integration gives $\frac52x-\frac34\sin(2x)-\cos(2x)+C$ .
3	$-\cos(4x)-2\cos(2x)+C$	5
(5 marks)		5
Notes
Adding the formulae for $\sin(3x+x)$ and $\sin(3x-x)$ gives $2\sin(3x)\cos x=\sin(4x)+\sin(2x)$ . Hence $8\sin(3x)\cos x=4\sin(4x)+4\sin(2x)$ . Integrating term by term gives $-\cos(4x)-2\cos(2x)+C$ .
4	$a=4$ , $b=2$ $\displaystyle\int f(x)\,\mathrm dx=-2\cos(2x)+\sin(2x)+C$	6
(6 marks)		6
Notes
An antiderivative is $-a\cos(2x)/2+b\sin(2x)/2$ . The first given integral is $(a+b)/2=3$ , while the second is $(a-b)/2=1$ . Hence $a+b=6$ and $a-b=2$ , so $a=4$ and $b=2$ . Substitution into the antiderivative gives $-2\cos(2x)+\sin(2x)+C$ .
5	$\dfrac{\pi}{3}+2\sqrt3$	6
(6 marks)		6
Notes
On $0\leq x\leq\pi$ , $1+2\cos x=0$ only when $x=2\pi/3$ . The expression is non-negative before this point and negative after it. Therefore the integral is $\int_0^{2\pi/3}(1+2\cos x)\,\mathrm dx-\int_{2\pi/3}^{\pi}(1+2\cos x)\,\mathrm dx$ . Using the antiderivative $x+2\sin x$ gives $(2\pi/3+\sqrt3)-(\pi/3-\sqrt3)=\pi/3+2\sqrt3$ .

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

Tier 1 · Easy

Mark scheme for 8.3 Tier 1 · Easy
Question	Scheme	Marks
1	Area $=12$ square units	3
(3 marks)		3
Notes
The line is above the axis on the interval, so the area is $\int_0^3(2x+1)\,\mathrm dx=[x^2+x]_0^3=9+3=12$ .
2	Area $=\dfrac83$ square units	3
(3 marks)		3
Notes
The curve is positive, so the area is $\int_0^{\ln3}4e^{-x}\,\mathrm dx=[-4e^{-x}]_0^{\ln3}=-\frac43+4=\frac83$ square units.

Tier 2 · Standard

Mark scheme for 8.3 Tier 2 · Standard
Question	Scheme	Marks
1	Area $=\dfrac43$ square units	4
(4 marks)		4
Notes
The curve meets the $x$ -axis where $(x-1)(x-3)=0$ , so the limits are $1$ and $3$ . The quadratic is below the axis between these roots, so the area is $\int_1^3-(x^2-4x+3)\,dx=[-x^3/3+2x^2-3x]_1^3=0-(-4/3)=4/3$ square units.
2	Area $=e-\dfrac52$ square units	4
(4 marks)		4
Notes
On $0\leq x\leq1$ , $e^x\geq1+x$ . Therefore the area is $\int_0^1(e^x-1-x)\,\mathrm dx=[e^x-x-\frac12x^2]_0^1=e-\frac52$ square units.
3	Area $=\dfrac12$ square units	5
(5 marks)		5
Notes
The curve is above the axis on $(0,1)$ and below it on $(1,2)$ . An antiderivative of $x(x-1)(x-2)=x^3-3x^2+2x$ is $F(x)=x^4/4-x^3+x^2$ . Hence the total area is $[F(x)]_0^1-[F(x)]_1^2=1/4-(-1/4)=1/2$ square units.

Tier 3 · Hard

Mark scheme for 8.3 Tier 3 · Hard
Question	Scheme	Marks
1	Area $=\frac{64}{5}$ square units	6
(6 marks)		6
Notes
For a parametric curve, area is $\int y\,\mathrm dx=\int_0^2y\frac{\mathrm dx}{\mathrm dt}\,\mathrm dt$ . Here $\frac{\mathrm dx}{\mathrm dt}=2t$ , so the area is $\int_0^2t^3(2t)\,\mathrm dt=2\int_0^2t^4\,\mathrm dt=2[\frac15t^5]_0^2=\frac{64}{5}$ .
2	Area $=2\sqrt2-2$ square units	5
(5 marks)		5
Notes
The curves intersect when $\sin x=\cos x$ , so $x=\pi/4$ . The total area is $\int_0^{\pi/4}(\cos x-\sin x)\,\mathrm dx+\int_{\pi/4}^{\pi/2}(\sin x-\cos x)\,\mathrm dx$ . Each integral equals $\sqrt2-1$ , giving $2\sqrt2-2$ square units.
3	$k=3$	5
(5 marks)		5
Notes
The curve meets the $x$ -axis where $x(k-x)=0$ , so the enclosed region lies between $x=0$ and $x=k$ . Its area is $\int_0^k(kx-x^2)\,\mathrm dx=[kx^2/2-x^3/3]_0^k=k^3/6$ . Thus $k^3/6=9/2$ , so $k^3=27$ . Since $k>0$ , $k=3$ .
4	$a=1$	6
(6 marks)		6
Notes
The curve meets the axis at $x=-2$ and $x=2$ . The stated area condition gives $\int_{-2}^{a}(4-x^2)\,\mathrm dx=9$ , so $4a-a^3/3+16/3=9$ . Hence $a^3-12a+11=0$ , which factors as $(a-1)(a^2+a-11)=0$ . The other roots are $(-1\pm3\sqrt5)/2$ , both outside $(-2,2)$ , so the pinned value is $a=1$ .
5	Area $=\dfrac{423}{10}$ square units	6
(6 marks)		6
Notes
The endpoints are $(2,0)$ and $(11,15)$ , so the chord is $y=5(x-2)/3$ . At parameter $t$ the chord has height $5t^2/3$ . Also $5t^2/3-(t^3-4t)=t(3-t)(t+4/3)\geq0$ for $0\leq t\leq3$ , so the chord is above the curve. Since $\mathrm dx/\mathrm dt=2t$ , the area is $\int_0^3[5t^2/3-(t^3-4t)]2t\,\mathrm dt=\int_0^3(10t^3/3-2t^4+8t^2)\,\mathrm dt=423/10$ square units.

8.4 · Understand and use integration as the limit of a sum.

Tier 1 · Easy

Mark scheme for 8.4 Tier 1 · Easy
Question	Scheme	Marks
1	$\lim_{n\to\infty}\frac8{n^3}\sum_{r=1}^nr^2=\int_0^2x^2\,\mathrm dx=\frac83$	4
(4 marks)		4
Notes
Use $\Delta x=2/n$ and $x_r=2r/n$ . Then $\sum f(x_r)\Delta x=\sum_{r=1}^n(2r/n)^2(2/n)=\frac8{n^3}\sum_{r=1}^nr^2$ . Its limit is $\int_0^2x^2\,\mathrm dx=[x^3/3]_0^2=\frac83$ .
2	$\dfrac52$	2
(2 marks)		2
Notes
Here $\Delta x=1/n$ and the right endpoint is $x_r=r/n$ , so the limit is $\int_0^1(2+x)\,\mathrm dx=[2x+x^2/2]_0^1=5/2$ .

Tier 2 · Standard

Mark scheme for 8.4 Tier 2 · Standard
Question	Scheme	Marks
1	$\displaystyle\lim_{n\to\infty}\frac3n\sum_{r=1}^n\frac{1}{2+3r/n}=\ln\!\left(\dfrac52\right)$	4
(4 marks)		4
Notes
The interval length is $3$ , so each subinterval has width $\Delta x=3/n$ . Using right endpoints $x_r=2+3r/n$ , the Riemann sum is $\frac3n\sum_{r=1}^n\frac{1}{2+3r/n}$ . Its limit is the given integral, which evaluates to $[\ln x]_2^5=\ln5-\ln2=\ln(5/2)$ .
2	$\ln5$	4
(4 marks)		4
Notes
The strip width is $4/n$ and the right endpoints on $[0,4]$ are $x_r=4r/n$ . Hence the limit is $\int_0^4\frac{1}{1+x}\,\mathrm dx=[\ln(1+x)]_0^4=\ln5$ .
3	$\dfrac{65}{4}$	4
(4 marks)		4
Notes
Divide $[0,1]$ into $n$ equal subintervals of width $\Delta x=1/n$ . Their right endpoints are $x_r=0+r\Delta x=r/n$ , so the sum is a right-endpoint Riemann sum for $\int_0^1(2+x)^3\,\mathrm dx$ . Therefore the limit is $[(2+x)^4/4]_0^1=(81-16)/4=65/4$ .

Tier 3 · Hard

Mark scheme for 8.4 Tier 3 · Hard
Question	Scheme	Marks
1	$3\ln3-2$	6
(6 marks)		6
Notes
Here $\Delta x=2/n$ and the right endpoints are $x_r=2r/n$ , so the limit is $\int_0^2\ln(1+x)\,\mathrm dx$ . By substitution followed by integration by parts, an antiderivative is $(1+x)\ln(1+x)-(1+x)$ . Evaluation from $0$ to $2$ gives $(3\ln3-3)-(-1)=3\ln3-2$ .
2	$\dfrac23(5\sqrt5-3\sqrt3)$	6
(6 marks)		6
Notes
Divide $[1,3]$ into $n$ strips of width $\Delta x=2/n$ . The right endpoints are $x_r=1+2r/n$ , and $\sqrt{3+2r/n}=\sqrt{2+x_r}$ . Hence the limit is $\int_1^3\sqrt{x+2}\,\mathrm dx=[\frac23(x+2)^{3/2}]_1^3=\frac23(5\sqrt5-3\sqrt3)$ .
3	$\dfrac12\ln\left(\dfrac53\right)$	5
(5 marks)		5
Notes
Rewrite each term as $1/(3n+2r)=\frac1n\cdot\frac{1}{3+2r/n}$ . The sum is therefore a right-endpoint sum of width $1/n$ on $[0,1]$ , so its limit is $\int_0^1\frac{1}{3+2x}\,\mathrm dx=[\frac12\ln(3+2x)]_0^1=\frac12\ln(5/3)$ .
4	$\displaystyle\int_2^5 e^x\,\mathrm dx=e^5-e^2$	5
(5 marks)		5
Notes
Here $\Delta x=3/n$ and the sample point is the right endpoint $x_r=2+r\Delta x=2+3r/n$ . The given limit is therefore $\int_2^5e^x\,\mathrm dx$ . Evaluating gives $[e^x]_2^5=e^5-e^2$ .
5	$\displaystyle\int_{\pi/6}^{5\pi/6}\sin x\,\mathrm dx=\sqrt3$	5
(5 marks)		5
Notes
The strip width is $\Delta x=2\pi/(3n)$ and the sample point is the right endpoint $x_r=\pi/6+r\Delta x$ . The upper limit is $\pi/6+2\pi/3=5\pi/6$ , so the given limit is $\int_{\pi/6}^{5\pi/6}\sin x\,\mathrm dx$ . Hence its value is $[-\cos x]_{\pi/6}^{5\pi/6}=\sqrt3$ .

8.5 · 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.

Tier 1 · Easy

Mark scheme for 8.5 Tier 1 · Easy
Question	Scheme	Marks
1	$\ln(3x^2+5)+C$	3
(3 marks)		3
Notes
The numerator is the derivative of the denominator. Therefore this has the form $\int f'(x)/f(x)\,\mathrm dx$ , giving $\ln\|3x^2+5\|+C$ . Since $3x^2+5>0$ , this is $\ln(3x^2+5)+C$ .
2	$\dfrac12(2x^2+1)^4+C$	3
(3 marks)		3
Notes
Let $u=2x^2+1$ , so $\mathrm du=4x\,\mathrm dx$ and $8x\,\mathrm dx=2\,\mathrm du$ . Then the integral is $2\int u^3\,\mathrm du=\frac12u^4+C=\frac12(2x^2+1)^4+C$ .

Tier 2 · Standard

Mark scheme for 8.5 Tier 2 · Standard
Question	Scheme	Marks
1	$\ln\!\left(\dfrac54\right)$	4
(4 marks)		4
Notes
Let $u=x^2+4$ , so $du=2x\,dx$ . The limits become $u=4$ when $x=0$ and $u=5$ when $x=1$ . Therefore the integral is $\int_4^5u^{-1}\,du=[\ln u]_4^5=\ln5-\ln4=\ln(5/4)$ .
2	$-\dfrac{2}{\pi^2}$	4
(4 marks)		4
Notes
Use integration by parts with $u=x$ and $\mathrm dv=\cos(\pi x)\,\mathrm dx$ , so $v=\sin(\pi x)/\pi$ . Thus the integral is $[x\sin(\pi x)/\pi]_0^1-\frac1\pi\int_0^1\sin(\pi x)\,\mathrm dx=0-2/\pi^2$ .
3	$2e^3+1$	4
(4 marks)		4
Notes
Choose $u=\ln x$ and $\mathrm dv=\mathrm dx$ . Then $\mathrm du=\mathrm dx/x$ and $v=x$ , so $\int\ln x\,\mathrm dx=x\ln x-x+C$ . Applying the limits gives $[x\ln x-x]_1^{e^3}=2e^3-(-1)=2e^3+1$ .

Tier 3 · Hard

Mark scheme for 8.5 Tier 3 · Hard
Question	Scheme	Marks
1	$\ln2-\frac12$	7
(7 marks)		7
Notes
Let $u=1+x^2$ , so $\mathrm du=2x\,\mathrm dx$ and the limits become $u=1$ to $u=2$ . The integral is $\frac12\int_1^2\ln u\,\mathrm du$ . By parts, $\int\ln u\,\mathrm du=u\ln u-u$ . Therefore the value is $\frac12[u\ln u-u]_1^2=\frac12(2\ln2-2+1)=\ln2-\frac12$ .
2	$\pi-2$	6
(6 marks)		6
Notes
Integrating by parts with $u=x^2$ and $\mathrm dv=\sin x\,\mathrm dx$ gives $[-x^2\cos x]_0^{\pi/2}+2\int_0^{\pi/2}x\cos x\,\mathrm dx$ . A second integration by parts gives $\int x\cos x\,\mathrm dx=x\sin x+\cos x$ . Therefore the value is $2[x\sin x+\cos x]_0^{\pi/2}=\pi-2$ .
3	$\dfrac12e^x(\sin x+\cos x)+C$	6
(6 marks)		6
Notes
Let $I=\int e^x\cos x\,\mathrm dx$ . Integration by parts gives $I=e^x\cos x+\int e^x\sin x\,\mathrm dx$ . Applying integration by parts to the remaining integral gives $\int e^x\sin x\,\mathrm dx=e^x\sin x-I$ . Hence $2I=e^x(\sin x+\cos x)$ , so $I=\frac12e^x(\sin x+\cos x)+C$ .
4	$\dfrac32-\ln2$	5
(5 marks)		5
Notes
Let $u=x^2+1$ , so $\mathrm du=2x\,\mathrm dx$ , $x^2=u-1$ , and the limits become $1$ and $4$ . The integral becomes $\frac12\int_1^4(u-1)/u\,\mathrm du=\frac12\int_1^4(1-1/u)\,\mathrm du$ . Hence its value is $\frac12[u-\ln u]_1^4=3/2-\ln2$ .
5	$\dfrac23\ln2-\dfrac5{18}$	6
(6 marks)		6
Notes
Take $u=\ln(1+x)$ and $\mathrm dv=x^2\,\mathrm dx$ . The integral is $[x^3\ln(1+x)/3]_0^1-\frac13\int_0^1x^3/(1+x)\,\mathrm dx$ . Since $x^3/(1+x)=x^2-x+1-1/(1+x)$ , this becomes $\frac13\ln2-\frac13[x^3/3-x^2/2+x-\ln(1+x)]_0^1=\frac23\ln2-\frac5{18}$ .

8.6 · Integrate using partial fractions that are linear in the denominator.

Tier 1 · Easy

Mark scheme for 8.6 Tier 1 · Easy
Question	Scheme	Marks
1	$3\ln\|x-1\|+2\ln\|x+2\|+C$	3
(3 marks)		3
Notes
Integrate each linear-denominator term directly: $\int3/(x-1)\,\mathrm dx=3\ln\|x-1\|$ and $\int2/(x+2)\,\mathrm dx=2\ln\|x+2\|$ . Add $C$ .
2	$\dfrac12\ln\left\|\dfrac{x+1}{x+3}\right\|+C$	3
(3 marks)		3
Notes
Integrating the supplied terms gives $\frac12\ln\|x+1\|-\frac12\ln\|x+3\|+C$ . Using the logarithm laws, this is $\frac12\ln\|\frac{x+1}{x+3}\|+C$ .

Tier 2 · Standard

Mark scheme for 8.6 Tier 2 · Standard
Question	Scheme	Marks
1	$\dfrac{2x^2+3x-1}{x(x+1)}=2-\dfrac1x+\dfrac{2}{x+1}$ $2x-\ln\|x\|+2\ln\|x+1\|+C$	5
(5 marks)		5
Notes
First divide: $(2x^2+3x-1)/(x^2+x)=2+(x-1)/[x(x+1)]$ . Write $(x-1)/[x(x+1)]=A/x+B/(x+1)$ , so $x-1=(A+B)x+A$ . Hence $A=-1$ and $B=2$ . Integrating $2-1/x+2/(x+1)$ gives $2x-\ln\|x\|+2\ln\|x+1\|+C$ .
2	$\dfrac{4x-7}{(x-2)(x+3)}=\dfrac{1}{5(x-2)}+\dfrac{19}{5(x+3)}$ $\dfrac15\ln\|x-2\|+\dfrac{19}{5}\ln\|x+3\|+C$	4
(4 marks)		4
Notes
Write the fraction as $A/(x-2)+B/(x+3)$ . Then $4x-7=A(x+3)+B(x-2)$ . Substituting $x=2$ gives $A=1/5$ , and substituting $x=-3$ gives $B=19/5$ . Integrating the decomposition gives the stated logarithmic result.
3	$\dfrac{4x+7}{(x+1)(2x+3)}=\dfrac3{x+1}-\dfrac2{2x+3}$ $3\ln\|x+1\|-\ln\|2x+3\|+C$	5
(5 marks)		5
Notes
Write the fraction as $A/(x+1)+B/(2x+3)$ . Then $4x+7=A(2x+3)+B(x+1)$ . Equating coefficients gives $2A+B=4$ and $3A+B=7$ , so $A=3$ and $B=-2$ . Integrating, including the inner coefficient in the second logarithm, gives $3\ln\|x+1\|-\ln\|2x+3\|+C$ .

Tier 3 · Hard

Mark scheme for 8.6 Tier 3 · Hard
Question	Scheme	Marks
1	$9\ln3-13\ln2$	7
(7 marks)		7
Notes
Write $\frac{5x+1}{(x+1)(x+2)}=\frac A{x+1}+\frac B{x+2}$ . Then $5x+1=A(x+2)+B(x+1)$ , giving $A=-4$ and $B=9$ . The integral is $[-4\ln(x+1)+9\ln(x+2)]_0^1=(-4\ln2+9\ln3)-9\ln2=9\ln3-13\ln2$ .
2	$F(x)=\ln\left(\dfrac{2(x+2)}{x+5}\right)$ $x=\dfrac{11}{2}$	6
(6 marks)		6
Notes
$3/[(x+2)(x+5)]=1/(x+2)-1/(x+5)$ , so $F(x)=\ln(x+2)-\ln(x+5)+C$ . From $F(1)=0$ , $C=\ln2$ , giving $F(x)=\ln[2(x+2)/(x+5)]$ . Equating logarithm arguments gives $2(x+2)/(x+5)=10/7$ , so $14x+28=10x+50$ and $x=11/2$ .
3	$2+\dfrac52\ln3-\ln4$	6
(6 marks)		6
Notes
Division and partial fractions give $\frac{4x^2+7x+3}{(2x-1)(x+2)}=2+\frac3{2x-1}-\frac1{x+2}$ . An antiderivative is $2x+\frac32\ln\|2x-1\|-\ln\|x+2\|$ . Evaluation from $1$ to $2$ gives $4+\frac32\ln3-\ln4-(2-\ln3)=2+\frac52\ln3-\ln4$ .
4	$A=2$ , $B=3$ $\displaystyle\ln\left(\dfrac{(x-4)^2(3x+2)}{17}\right)$ $x=6$	6
(6 marks)		6
Notes
Multiplying by $(x-4)(3x+2)$ gives $9x-8=A(3x+2)+B(x-4)$ . Substituting $x=4$ gives $A=2$ , and substituting $x=-2/3$ gives $B=3$ . An antiderivative is $2\ln(t-4)+\ln(3t+2)$ for $t>5$ . Applying the limits gives $\ln[(x-4)^2(3x+2)/17]$ . Equating logarithm arguments gives $(x-4)^2(3x+2)=80$ . The left-hand side is strictly increasing for $x>5$ , and $x=6$ satisfies the equation, so $x=6$ is the unique solution.
5	$\dfrac{2x^2+10x+10}{(x+1)(x+2)(x+3)}=\dfrac1{x+1}+\dfrac2{x+2}-\dfrac1{x+3}$ $3\ln\left(\dfrac32\right)$	6
(6 marks)		6
Notes
Writing the fraction as $A/(x+1)+B/(x+2)+C/(x+3)$ and substituting $x=-1,-2,-3$ gives $A=1$ , $B=2$ and $C=-1$ . An antiderivative is $\ln\|x+1\|+2\ln\|x+2\|-\ln\|x+3\|$ . Evaluation from $0$ to $1$ gives $3\ln3-3\ln2=3\ln(3/2)$ .

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

Tier 1 · Easy

Mark scheme for 8.7 Tier 1 · Easy
Question	Scheme	Marks
1	$y=3e^{x^2}$	4
(4 marks)		4
Notes
Separate variables: $\frac1y\,\mathrm dy=2x\,\mathrm dx$ . Integrating gives $\ln\|y\|=x^2+C$ , so $y=Ae^{x^2}$ . The condition $y(0)=3$ gives $A=3$ , hence $y=3e^{x^2}$ .
2	$y=\sqrt{x^3+4}$	3
(3 marks)		3
Notes
Separate variables: $2y\,\mathrm dy=3x^2\,\mathrm dx$ . Integration gives $y^2=x^3+C$ . The initial condition gives $C=4$ , and the condition $y>0$ selects $y=\sqrt{x^3+4}$ .

Tier 2 · Standard

Mark scheme for 8.7 Tier 2 · Standard
Question	Scheme	Marks
1	$y=2+Ae^{3x^2/2}$ , where $A\ne0$ Equilibrium solution: $y=2$	4
(4 marks)		4
Notes
Before dividing by $y-2$ , note that $y=2$ is a constant equilibrium solution. For $y\ne2$ , separate to obtain $\mathrm dy/(y-2)=3x\,\mathrm dx$ . Integration gives $\ln\|y-2\|=3x^2/2+C$ , so $y-2=Ae^{3x^2/2}$ . Hence the non-equilibrium family is $y=2+Ae^{3x^2/2}$ with $A\ne0$ , alongside the excluded equilibrium solution $y=2$ .
2	$y=\ln(x^2+x+2)$	4
(4 marks)		4
Notes
Separate variables to obtain $e^y\,\mathrm dy=(2x+1)\,\mathrm dx$ . Integration gives $e^y=x^2+x+C$ . Since $y=\ln2$ at $x=0$ , $C=2$ . Taking logarithms gives $y=\ln(x^2+x+2)$ .
3	$y=(3x^2+3x+1)^{1/3}$	4
(4 marks)		4
Notes
Separate variables to obtain $y^2\,\mathrm dy=(2x+1)\,\mathrm dx$ . Integration gives $y^3/3=x^2+x+C$ . The initial condition gives $C=1/3$ , so $y^3=3x^2+3x+1$ . The stated positive branch is $y=(3x^2+3x+1)^{1/3}$ .

Tier 3 · Hard

Mark scheme for 8.7 Tier 3 · Hard
Question	Scheme	Marks
1	$y=\frac4{1+3e^{-4x}}$ Equilibrium solutions $y=0$ and $y=4$ ; the particular solution increases towards $y=4$	9
(9 marks)		9
Notes
The constant solutions lost on division are $y=0$ and $y=4$ . Otherwise separate and integrate to obtain $\frac14(\ln\|y\|-\ln\|4-y\|)=x+C$ , so $\frac{y}{4-y}=Ae^{4x}$ . The initial condition gives $A=1/3$ , hence $y=\frac4{1+3e^{-4x}}$ . It starts at $1$ , has positive gradient while $0<y<4$ , and tends to the upper equilibrium $4$ .
2	$y=\dfrac{x^2}{x^2+2}$ Equilibrium solution $y=1$ ; the particular solution tends to $1$	6
(6 marks)		6
Notes
The equilibrium solution lost by division is $y=1$ . Otherwise $(y-1)^{-2}\,\mathrm dy=x\,\mathrm dx$ , giving $-1/(y-1)=x^2/2+C$ . The initial condition gives $C=1$ , so $1/(1-y)=1+x^2/2$ and hence $y=x^2/(x^2+2)$ . As $x\to\infty$ , this tends to $1$ from below.
3	$y=e^{1-\cos x}-2$ Equilibrium solution $y=-2$ $x=\dfrac{\pi}{2},\dfrac{3\pi}{2}$	6
(6 marks)		6
Notes
The equilibrium solution lost on division is $y=-2$ . Otherwise $\mathrm dy/(y+2)=\sin x\,\mathrm dx$ , so $\ln\|y+2\|=-\cos x+C$ . The initial condition gives $C=1$ and the positive sign, hence $y=e^{1-\cos x}-2$ . Setting this equal to $e-2$ gives $e^{1-\cos x}=e$ , so $\cos x=0$ . The two solutions in the stated interval are $\pi/2$ and $3\pi/2$ .
4	$y=\tan(\sin x)$ Maximum $y=\tan1$ at $x=\dfrac{\pi}{2}$ ; minimum $y=-\tan1$ at $x=\dfrac{3\pi}{2}$	6
(6 marks)		6
Notes
Separating gives $\mathrm dy/(1+y^2)=\cos x\,\mathrm dx$ , so $\arctan y=\sin x+C$ . The initial condition gives $C=0$ . Since $-1\leq\sin x\leq1$ lies wholly inside the principal range $(-\pi/2,\pi/2)$ of $\arctan$ , inversion gives the single continuous branch $y=\tan(\sin x)$ . The tangent function is increasing on this range, so $y$ is largest when $\sin x=1$ , at $x=\pi/2$ , and smallest when $\sin x=-1$ , at $x=3\pi/2$ .
5	$\dfrac{y^3}{3}+y=e^x-1$ $x=\ln\left(\dfrac{17}{3}\right)$ The value is unique because $\mathrm dy/\mathrm dx>0$ for all real $x$	6
(6 marks)		6
Notes
Separating gives $(y^2+1)\,\mathrm dy=e^x\,\mathrm dx$ . Integration gives $y^3/3+y=e^x+C$ , and the initial condition gives $C=-1$ . When $y=2$ , $y^3/3+y=14/3$ , so $e^x=17/3$ and $x=\ln(17/3)$ . Since both $e^x$ and $y^2+1$ are positive, $\mathrm dy/\mathrm dx>0$ for every real $x$ ; the solution is strictly increasing and can take the value $2$ only once.

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

Tier 1 · Easy

Mark scheme for 8.8 Tier 1 · Easy
Question	Scheme	Marks
1	Initial population $500$ ; continuous relative growth rate $0.12$ per year One valid limitation, such as finite resources eventually preventing unlimited exponential growth	4
(4 marks)		4
Notes
At $t=0$ , $N=500$ , so $500$ is the initial population. Also $\frac1N\frac{\mathrm dN}{\mathrm dt}=0.12$ , so $0.12$ is the continuous proportional growth rate per year. The model predicts unbounded growth and therefore ignores a limiting factor such as finite food, space or changing birth and death rates.
2	Initial temperature $80$ °C The temperature tends to the surrounding temperature of $18$ °C	3
(3 marks)		3
Notes
At $t=0$ , $T=18+62=80$ °C. As $t\to\infty$ , $e^{-0.25t}\to0$ , so $T\to18$ °C. In the differential equation, $\mathrm dT/\mathrm dt=0$ when $T=18$ , so this is the equilibrium temperature of the surroundings.

Tier 2 · Standard

Mark scheme for 8.8 Tier 2 · Standard
Question	Scheme	Marks
1	Exact crossing time: $t=5\ln10$ hours Crossing time to $3$ significant figures: $t=11.5$ hours The mass is below $5$ grams for $t>5\ln10$ hours $M$ tends to $0$ grams as $t\to\infty$ . For example, the decay constant may not remain fixed.	5
(5 marks)		5
Notes
At the crossing, $50e^{-0.2t}=5$ , so $e^{-0.2t}=0.1$ and $t=5\ln10=11.512\ldots$ hours, which is $11.5$ hours to $3$ significant figures. Since the model decreases continuously, $M<5$ for $t>5\ln10$ . As $t\to\infty$ , the exponential tends to zero, so the model predicts $M\to0$ . A limitation is that it assumes one constant proportional decay rate indefinitely.
2	$t=\dfrac{100}{3}\ln2$ minutes $=23.1$ minutes to $3$ significant figures $M=200$ g is the equilibrium mass and the limiting mass predicted by the model	5
(5 marks)		5
Notes
Set $120=200-160e^{-0.03t}$ . Then $e^{-0.03t}=1/2$ , so $t=\ln2/0.03=(100/3)\ln2=23.104\ldots$ minutes, which is $23.1$ minutes to $3$ significant figures. In the differential equation, $\mathrm dM/\mathrm dt=0$ when $M=200$ . Also $e^{-0.03t}\to0$ , so the solution approaches $200$ g; it is the model's equilibrium and limiting mass.
3	Initial population $800$ $t=\dfrac{50}{3}\ln9$ years $=36.6$ years to $3$ significant figures $0\leq t\leq\dfrac{50}{3}\ln9$ One valid limitation, such as the rate of population loss not remaining unchanged as the population and conditions change	5
(5 marks)		5
Notes
At $t=0$ , $P=900-100=800$ . Setting $P=0$ gives $e^{0.06t}=9$ , so $t=\ln9/0.06=(50/3)\ln9=36.620\ldots$ years, which is $36.6$ years to $3$ significant figures. The model gives $P\geq0$ only from $t=0$ to this time. It assumes the same exponential loss pattern and unchanged conditions throughout; beyond the zero-population time it becomes physically impossible.

Tier 3 · Hard

Mark scheme for 8.8 Tier 3 · Hard
Question	Scheme	Marks
1	$t=\frac{10}{3}\ln\frac43\approx0.959\,\text{s}$ $s\approx-2.74\,\text{m}$ Limiting velocity $18\,\text{m s}^{-1}$ One valid limitation tied to the assumed force or resistance law	8
(8 marks)		8
Notes
Solve $v=0$ : $18=24e^{-0.3t}$ , so $t=\frac{10}{3}\ln(4/3)\approx0.959\,\text{s}$ . Since $v(0)=-6$ and $v$ tends to $18>0$ , the continuous velocity changes sign there, proving a reversal. Integrating and using $s(0)=0$ gives $s=18t+80e^{-0.3t}-80$ , which is approximately $-2.74\,\text{m}$ at the turn. The limiting velocity is $18\,\text{m s}^{-1}$ . Over long times, the assumed force or resistance relationship and constant conditions may cease to apply.
2	$v=\sqrt{8t+4}\,\text{m s}^{-1}$ $s=\dfrac{14\sqrt7-2}{3}\,\text{m}=11.7\,\text{m}$ to $3$ significant figures The model predicts unbounded speed, so the inverse-speed acceleration law cannot remain valid indefinitely	7
(7 marks)		7
Notes
Separating gives $v\,\mathrm dv=4\,\mathrm dt$ , so $v^2/2=4t+C$ . The initial condition gives $v^2=8t+4$ , and $v>0$ selects $v=\sqrt{8t+4}$ . Since $v=\mathrm ds/\mathrm dt$ , $s(3)=\int_0^3\sqrt{8t+4}\,\mathrm dt=[(8t+4)^{3/2}/12]_0^3=(14\sqrt7-2)/3=11.680\ldots$ , which is $11.7$ m to $3$ significant figures. The formula gives $v\to\infty$ , so the assumed acceleration law is not credible indefinitely.
3	$50$ hectares is the limiting restored area, $4$ encodes the initial area $A(0)=10$ hectares, and $0.1\,\text{year}^{-1}$ is the proportional restoration-rate parameter $t=10\ln16$ years $=27.7$ years to $3$ significant figures $A\to50$ hectares while $\dfrac{\mathrm dA}{\mathrm dt}\to0$ One valid limitation, such as assuming that the available area, funding and restoration conditions remain fixed	7
(7 marks)		7
Notes
The factor $1-A/50$ makes $A=50$ hectares an equilibrium and limiting value. At $t=0$ , $A=50/(1+4)=10$ hectares, so $4$ encodes the initial shortfall-to-restored ratio $(50-10)/10$ . The constant $0.1\,\text{year}^{-1}$ is the model's proportional restoration-rate parameter; when $A$ is small, the relative rate $(1/A)\,\mathrm dA/\mathrm dt$ is approximately $0.1$ per year. Setting $A=40$ gives $1+4e^{-0.1t}=5/4$ , so $e^{-0.1t}=1/16$ and $t=10\ln16=27.7258\ldots$ years, which is $27.7$ years to $3$ significant figures. Since the differential equation gives $\mathrm dA/\mathrm dt>0$ for $0<A<50$ , this is the first such time. As $t\to\infty$ , $A\to50$ and the differential equation then gives $\mathrm dA/\mathrm dt\to0$ . The model assumes a fixed available area and fixed conditions, ignoring changes such as funding, weather or land becoming unsuitable.
4	£ $30\,000$ is an unstable equilibrium balance: balances above it increase and balances below it decrease $t=25\ln2$ years $=17.3$ years to $3$ significant figures One valid limitation, such as assuming a fixed interest rate and a continuous withdrawal of £ $1\,200$ per year indefinitely	7
(7 marks)		7
Notes
At a balance of £ $30\,000$ , the differential equation gives $\mathrm dB/\mathrm dt=0$ . Above £ $30\,000$ the derivative is positive, whereas below £ $30\,000$ it is negative, so balances move away from the equilibrium. Setting the supplied solution equal to £ $50\,000$ gives $e^{0.04t}=2$ , hence $t=25\ln2=17.328\ldots$ years, which is $17.3$ years to $3$ significant figures. The model assumes that both the interest rate and the continuous withdrawal of £ $1\,200$ per year remain fixed, which need not hold over such a long period.
5	$10$ kilograms per hour is the input rate and $0.25M$ kilograms per hour is the removal rate; at $t=4$ the input stops Maximum mass $=40(1-e^{-1})$ kilograms $=25.3$ kilograms to $3$ significant figures $t=4+4\ln\left(4(1-e^{-1})\right)$ hours $=7.71$ hours to $3$ significant figures One valid limitation, such as assuming perfect mixing and a constant proportional removal rate	7
(7 marks)		7
Notes
Before $t=4$ , the first term represents pollutant entering at $10$ kilograms per hour and $0.25M$ is the removal rate. At $t=4$ the input is switched off, leaving removal only. While $0<t<4$ , $M<40$ and the first differential equation gives $\mathrm dM/\mathrm dt>0$ ; after the switch, the second gives $\mathrm dM/\mathrm dt<0$ . The maximum is therefore $M(4)=40(1-e^{-1})=25.284\ldots$ kilograms, or $25.3$ kilograms to $3$ significant figures. On the later branch, $10=40(1-e^{-1})e^{-0.25(t-4)}$ , so $t=4+4\ln(4(1-e^{-1}))=7.710\ldots$ hours, or $7.71$ hours to $3$ significant figures. The model assumes ideal mixing and a fixed removal coefficient, which may not hold in a real tank.