CP-5 Further calculus — revision question pack

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

Checked against Edexcel 9FM0 section CP-5. Review basis: the qualification registry sourced from the Pearson Edexcel Level 3 Advanced GCE in Further Mathematics (9FM0) specification; registry verification recorded 17 July 2026.

How this checking works

CP-5.1 · Derive formulae for and calculate volumes of revolution.

Explanation

Rotating a thin strip perpendicular to an axis produces a disc or washer. About the $x$ -axis, summing cross-sectional areas gives $V=\pi\int y^2\,dx$ ; about the $y$ -axis, $V=\pi\int x^2\,dy$ .
For a region between curves, subtract the squared inner radius from the squared outer radius before integrating.
These formulae follow by taking the limit of a sum of thin cylindrical volumes.
Cartesian or parametric equations may be used: under a parametrisation, include the appropriate derivative such as $dx/dt$ .
A sketch should identify the rotated region, axis, limits and radii before the integral is formed, and the exact answer needs cubic units.

A strip rotated about the x-axis forms a thin disc of radius y and width dx.

Worked example

The region under $y=3-x$ from $x=0$ to $x=3$ is rotated about the $x$ -axis. Find the exact volume.

1.The disc radius is $y=3-x$ .
2. $V=\pi\int_0^3(3-x)^2\,dx$ .
3.Integrating gives $\pi[-(3-x)^3/3]_0^3$ .

Answer: $V=9\pi$ cubic units.

Common mistakes

Don't fall into the trap of integrating the radius instead of the cross-sectional area $\pi r^2$ .
Don't fall into the trap of subtracting the outer squared radius from the inner squared radius in a washer.
Don't fall into the trap of using $dx$ for rotation about the $y$ -axis without first rewriting the radius consistently.

Exam tip

A volume-of-revolution setup should show the axis, outer and inner radii, limits and factor $\pi$ before evaluation.

Tier 1 · Easy

1.
The region bounded by $y=x^2+1$ , the $x$ -axis, the $y$ -axis and the line $x=1$ is rotated through $2\pi$ radians about the $x$ -axis. Determine the exact volume generated.
(3)
(Total for Question 1 is 3 marks)
2.
The region bounded by the line $y=2x+1$ , the $x$ -axis, the $y$ -axis and the line $x=3$ is rotated through $2\pi$ radians about the $x$ -axis. Find the exact volume of the solid formed.
(4)
(Total for Question 2 is 4 marks)

Tier 2 · Standard

1.
The region bounded by $x=y^2+1$ , the $y$ -axis, $y=0$ and $y=2$ is rotated through $2\pi$ radians about the $y$ -axis. Find the exact volume.
(4)
(Total for Question 1 is 4 marks)
2.
The region enclosed by $y=x+2$ and $y=x^2$ is rotated through $2\pi$ radians about the $x$ -axis. Find the coordinates of the points of intersection and hence the exact volume.
(5)
(Total for Question 2 is 5 marks)
3.
A curve has parametric equations $x=t^2$ , $y=t(2-t)$ , where $0\le t\le2$ . The finite region between the curve and the $x$ -axis is rotated through $2\pi$ radians about the $x$ -axis. Find the exact volume generated.
(5)
(Total for Question 3 is 5 marks)

Tier 3 · Hard

1.
The finite region between $y=\sqrt{x}$ and $y=x^2$ is rotated through $2\pi$ radians about the $y$ -axis. Determine the exact volume of the solid formed.
(5)
(Total for Question 1 is 5 marks)
2.
The region between $y=3-|x|$ and $y=x^2+1$ is rotated through $2\pi$ radians about the $x$ -axis. Show that the curves meet only where $x=\pm1$ . Hence find the exact volume of the solid formed.
(7)
(Total for Question 2 is 7 marks)
3.
The curve $C$ has parametric equations $x=a+t(1-t^2)$ , $y=t$ , where $0\le t\le1$ and $a>0$ . The region between $C$ and the line $x=a$ is rotated through $2\pi$ radians about the $y$ -axis. The volume generated is $\dfrac{218\pi}{105}$ . Determine $a$ .
(6)
(Total for Question 3 is 6 marks)
4.
The region enclosed by the ellipse $\dfrac{x^2}{9}+\dfrac{(y-4)^2}{4}=1$ is rotated through $2\pi$ radians about the $x$ -axis. Find the exact volume of the solid formed.
(7)
(Total for Question 4 is 7 marks)
5.
The finite region bounded by $x=y^2$ and $x=3$ is rotated through $2\pi$ radians about the $y$ -axis. Find the exact volume of the solid formed.
(8)
(Total for Question 5 is 8 marks)

CP-5.2 · Evaluate improper integrals where either the integrand is undefined at a value in the range of integration or the range of integration extends to infinity.

Explanation

An improper integral is defined through a limit. Replace an infinite endpoint by a finite variable and then let that variable tend to infinity; replace a singular endpoint by a variable approaching from within the interval.
If the integrand is undefined at an interior point, split the integral there and evaluate both one-sided limits independently. The integral converges only when every required limit is finite.
An antiderivative is evaluated before the limit is taken, and direct substitution of an undefined endpoint is invalid.
Divergent pieces cannot be combined through formal cancellation to manufacture a finite value.
A complete solution states whether the integral converges and, if so, gives its limiting value.

Worked example

Determine whether $\int_1^\infty x^{-3}\,dx$ converges and evaluate it.

1.Write $\lim_{b\to\infty}\int_1^b x^{-3}\,dx$ .
2.An antiderivative is $-1/(2x^2)$ .
3.The expression is $\lim_{b\to\infty}(1/2-1/(2b^2))$ .

Answer: The improper integral converges to $1/2$ .

Common mistakes

Don't fall into the trap of substituting $\infty$ into an antiderivative as though it were a number.
Don't fall into the trap of failing to split an integral at a singular point inside the interval.
Don't fall into the trap of calling the whole integral convergent when only one of two one-sided limits is finite.

Exam tip

For 'evaluate an improper integral', display the defining limit and finish with an explicit convergence statement.

Tier 1 · Easy

1.
Evaluate $\displaystyle \int_1^\infty\dfrac{1}{x^2}\,dx$ .
(3)
(Total for Question 1 is 3 marks)

Tier 2 · Standard

1.
Evaluate $\displaystyle \int_0^4\dfrac{1}{\sqrt{x}}\,dx$ .
(3)
(Total for Question 1 is 3 marks)
2.
Show that $\displaystyle\int_1^\infty\frac{\ln x}{x^2}\,dx$ converges and find its exact value.
(4)
(Total for Question 2 is 4 marks)
3.
(a) Explain why $\displaystyle\int_0^3\frac{dx}{\sqrt[3]{(x-2)^2}}$ is an improper integral. (1) (b) Show that the integral converges and find its exact value. (4)
(5)
(Total for Question 3 is 5 marks)
4.
Use the substitution $u=1-x$ to show that $\displaystyle\int_0^1 x(1-x)^{-1/2}\,dx$ converges and find its exact value.
(5)
(Total for Question 4 is 5 marks)

Tier 3 · Hard

1.
Evaluate $\displaystyle \int_0^\infty\dfrac{1}{\sqrt{x}(1+x)}\,dx$ , showing explicitly how both improper endpoints are handled.
(6)
(Total for Question 1 is 6 marks)
2.
(a) Explain why $\displaystyle\int_0^2\frac{dx}{x-1}$ is an improper integral. (1) (b) Show that the integral is divergent. (3) (c) A student writes $\left[\ln|x-1|\right]_0^2=\ln1-\ln1=0$ and concludes that the integral equals $0$ . Explain why this is incorrect. (2)
(6)
(Total for Question 2 is 6 marks)
3.
For $b>0$ , find $J(b)=\displaystyle\int_{-b}^{b}\frac{e^x}{(1+e^x)^2}\,dx$ in the form $\dfrac{e^b+p}{e^b+q}$ , with integer constants $p$ and $q$ to be found. Then treat the two infinite endpoints separately to show that the corresponding improper integral converges and find its exact value.
(6)
(Total for Question 3 is 6 marks)
4.
Show that the improper integral $\displaystyle\int_0^\infty e^{-x}\sin x\,dx$ converges and find its exact value.
(7)
(Total for Question 4 is 7 marks)
5.
Determine whether the improper integral $\displaystyle\int_0^1\left(\frac{\ln x}{\sqrt{x}}+\frac1{\sqrt{1-x}}\right)\,dx$ converges. If it does, find its exact value.
(8)
(Total for Question 5 is 8 marks)

CP-5.3 · Understand and evaluate the mean value of a function.

Explanation

The mean value of an integrable function $f$ on $[a,b]$ is $\dfrac1{b-a}\int_a^bf(x)\,dx$ . Geometrically, it is the signed height of a rectangle of width $b-a$ whose signed area equals the definite integral.
The function may take negative values, so the mean can also be negative.
Evaluate the definite integral first and divide by the full interval length, not by an endpoint.
Averaging $f(a)$ and $f(b)$ is valid for a linear function but not in general.
In applications, include the quantity's units: integration multiplies output units by input units, and division by the interval length restores the original output units.

The mean-value rectangle has the same signed area as the region under the curve on the interval.

Worked example

Find the mean value of $f(x)=2x+1$ on $1\le x\le4$ .

1.The interval length is $4-1=3$ .
2. $\int_1^4(2x+1)\,dx=[x^2+x]_1^4=18$ .
3.Divide the integral by the interval length.

Answer: The mean value is $18/3=6$ .

Common mistakes

Don't fall into the trap of dividing the definite integral by $b$ instead of by $b-a$ .
Don't fall into the trap of averaging only the endpoint values for a non-linear function.
Don't fall into the trap of replacing signed area by total geometric area when the function crosses the axis.

Exam tip

Write the factor $1/(b-a)$ before integrating so the interval-length division is not lost.

Tier 1 · Easy

1.
Find the mean value of $f(x)=x^2$ on the interval $0\leq x\leq3$ .
(3)
(Total for Question 1 is 3 marks)
2.
The mean value of $f(x)=kx+1$ on $0\le x\le4$ is $7$ . Determine $k$ .
(3)
(Total for Question 2 is 3 marks)

Tier 2 · Standard

1.
Determine the exact mean value of $f(x)=\dfrac{1}{x+1}$ on $0\leq x\leq e-1$ .
(4)
(Total for Question 1 is 4 marks)
2.
Find the exact mean value of $f(x)=x^2+\ln(x+1)$ on $[1,3]$ .
(5)
(Total for Question 2 is 5 marks)
3.
Let $f(x)=\sin x+2\sin(2x)$ . Find the exact mean value of $f$ on each of the intervals $[0,\pi/2]$ and $[0,\pi]$ . State which mean is greater and find the exact difference.
(5)
(Total for Question 3 is 5 marks)

Tier 3 · Hard

1.
Find the exact mean value of $f(x)=xe^{-x}$ on $0\leq x\leq2$ .
(5)
(Total for Question 1 is 5 marks)
2.
A positive constant $a$ is chosen so that the mean value of $x^2$ on $[0,a]$ equals the mean value of $x$ on $[a,a+2]$ . Determine the exact value of $a$ .
(5)
(Total for Question 2 is 5 marks)
3.
The function $f$ is defined on $[0,4]$ by $f(x)=x+a$ for $0\le x\le2$ and $f(x)=(x-4)^2$ for $2<x\le4$ . Given that the mean value of $f$ on $[0,4]$ is $13/6$ , determine $a$ . Hence find the exact mean value of $f$ on $[1,3]$ .
(6)
(Total for Question 3 is 6 marks)
4.
During a four-hour test, the electrical power output of a generator is $P(t)=6+\dfrac t2+2\sin\left(\dfrac{\pi t}{4}\right)$ kW for $0\le t\le4$ , where $t$ is measured in hours. Find the exact mean power output during the test, in kW. Hence find the exact energy generated during the test, in kWh.
(6)
(Total for Question 4 is 6 marks)
5.
A positive constant $a$ is such that the mean value of $\ln x$ on $[a,ea]$ is $2$ . Determine $a$ . Hence find the exact value of $\displaystyle\int_a^{ea}(\ln x)^2\,dx$ .
(8)
(Total for Question 5 is 8 marks)

CP-5.4 · Integrate using partial fractions.

Explanation

Make a rational function proper by polynomial division whenever the numerator degree is at least the denominator degree. A distinct linear factor contributes a constant numerator; a repeated linear factor needs a term for every power.
An irreducible quadratic factor such as $ax^2+c$ requires a linear numerator $Bx+C$ .
After finding the constants, integrate term by term.
Linear denominators give logarithms with the chain-rule factor; a numerator proportional to the derivative of a quadratic also gives a logarithm, while the remaining constant-over-quadratic term gives an inverse tangent after scaling.
The decomposition should be checked by recombination, and logarithmic arguments from linear factors require absolute values.

Worked example

Find $\int\dfrac{x+3}{x(x^2+4)}\,dx$ .

1.Write $\dfrac{x+3}{x(x^2+4)}=\dfrac A x+\dfrac{Bx+C}{x^2+4}$ .
2.Comparison gives $A=3/4$ , $B=-3/4$ and $C=1$ .
3.Integrate the logarithmic terms and use $\int dx/(x^2+4)=\tfrac12\arctan(x/2)$ .

Answer: $\dfrac34\ln|x|-\dfrac38\ln(x^2+4)+\dfrac12\arctan(x/2)+C$ .

Common mistakes

Don't fall into the trap of omitting one of the terms required by a repeated linear factor.
Don't fall into the trap of using a constant numerator over an irreducible quadratic when a linear numerator is required.
Don't fall into the trap of forgetting the derivative factor when integrating a logarithmic partial fraction.

Exam tip

Write the complete decomposition before solving coefficients; missing a term makes every later constant incorrect.

Tier 1 · Easy

1.
Find $\displaystyle \int\dfrac{3x+5}{(x+1)(x+2)}\,dx$ .
(4)
(Total for Question 1 is 4 marks)

Tier 2 · Standard

1.
Find $\displaystyle \int\dfrac{2x+5}{(x+1)^2}\,dx$ .
(4)
(Total for Question 1 is 4 marks)
2.
Evaluate $\displaystyle\int_1^2\frac{x^3+x^2+2x+1}{x(x+1)}\,dx$ . Give your answer in the form $m+\ln k$ , where $m$ and $k$ are constants to be found.
(5)
(Total for Question 2 is 5 marks)
3.
Given that $\dfrac{mx+7}{(2x-1)(x+3)}\equiv\dfrac D{2x-1}+\dfrac D{x+3}$ , (a) find the value of $D$ and the value of $m$ . (2) (b) Hence find $\displaystyle\int\frac{mx+7}{(2x-1)(x+3)}\,dx$ . (3)
(5)
(Total for Question 3 is 5 marks)
4.
Find the exact value of $\displaystyle\int_2^4\frac{x^2+7x+1}{(x-1)^2(x+2)}\,dx$ .
(6)
(Total for Question 4 is 6 marks)

Tier 3 · Hard

1.
Find $\displaystyle \int\dfrac{x^2+1}{x(x^2+4)}\,dx$ .
(6)
(Total for Question 1 is 6 marks)
2.
(a) Express $\dfrac{4x^3-x^2-x+8}{(x-1)^2(x^2+4)}$ in partial fractions. (4) (b) Hence show that $\displaystyle\int\frac{4x^3-x^2-x+8}{(x-1)^2(x^2+4)}\,dx=A\ln|x-1|+\frac{B}{x-1}+C\ln(x^2+4)+D\arctan(x/2)+c$ . Here $c$ is arbitrary; find the constants $A$ , $B$ , $C$ and $D$ . (4)
(8)
(Total for Question 2 is 8 marks)
3.
The constants $m$ and $n$ are such that an antiderivative of $\dfrac{x^2+mx+n}{(x-2)(x^2+25)}$ contains no arctangent term and the coefficient of $\ln|x-2|$ is $3$ . Determine the value of $m$ and the value of $n$ , and find the antiderivative.
(8)
(Total for Question 3 is 8 marks)
4.
Express $\dfrac{x^3+8x^2+19x+18}{(x+1)(x+2)(x+3)(x+4)}$ in partial fractions. Hence find the exact value of $\displaystyle\int_0^1\frac{x^3+8x^2+19x+18}{(x+1)(x+2)(x+3)(x+4)}\,dx$ .
(8)
(Total for Question 4 is 8 marks)
5.
Find the exact value of $\displaystyle\int_1^3\frac{x^4-x^3+33x^2-33x+6}{(x+1)(x^2+36)}\,dx$ .
(8)
(Total for Question 5 is 8 marks)

CP-5.5 · Differentiate inverse trigonometric functions.

Explanation

The standard derivatives are $d(\arcsin x)/dx=1/\sqrt{1-x^2}$ , $d(\arccos x)/dx=-1/\sqrt{1-x^2}$ and $d(\arctan x)/dx=1/(1+x^2)$ .
For an argument $u(x)$ , multiply by $u'(x)$ through the chain rule and replace every occurrence of the inner variable consistently.
Products and quotients involving inverse trigonometric functions still require their usual differentiation rules.
Domain restrictions matter when square roots or inverse functions are present.
The notation $\arcsin x$ means the inverse sine function, not the reciprocal of sine; similarly, the negative sign in the derivative of arccosine must be retained.

Worked example

Differentiate $y=\arctan(3x^2)$ .

1.Set $u=3x^2$ , so $u'=6x$ .
2.Use $d(\arctan u)/dx=u'/(1+u^2)$ .
3.Substitute $u=3x^2$ and simplify.

Answer: $\dfrac{dy}{dx}=\dfrac{6x}{1+9x^4}$ .

Common mistakes

Don't fall into the trap of omitting the derivative of the inner function in a composite inverse-trigonometric expression.
Don't fall into the trap of losing the negative sign when differentiating $\arccos x$ .
Don't fall into the trap of interpreting inverse-function notation as a reciprocal trigonometric function.

Exam tip

Identify the inner argument first, write its derivative, and substitute both into the appropriate standard inverse-trigonometric derivative.

Tier 1 · Easy

1.
Differentiate $y=\arcsin(2x)$ with respect to $x$ .
(2)
(Total for Question 1 is 2 marks)
2.
Find $dy/dx$ for $y=\arccos((x+1)/3)$ and state the values of $x$ for which $dy/dx$ is defined.
(3)
(Total for Question 2 is 3 marks)

Tier 2 · Standard

1.
Differentiate $y=\arctan\left(\dfrac{x-1}{2}\right)$ .
(3)
(Total for Question 1 is 3 marks)
2.
Given $y=x^2\arcsin(x/3)$ , find $dy/dx$ and hence the exact value of $dy/dx$ at $x=3/2$ .
(4)
(Total for Question 2 is 4 marks)
3.
For $-3<x<3$ , the curve $y=\arctan(2x)+k\arccos(x/3)$ passes through $(0,\pi/3)$ , where $k$ is a real constant. Determine $k$ and hence find the exact gradient of the curve at $x=1$ .
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
Given $y=x\arccos x-\sqrt{1-x^2}$ for $-1<x<1$ , show that $\dfrac{dy}{dx}=\arccos x$ .
(5)
(Total for Question 1 is 5 marks)
2.
Let $f(x)=\arcsin(x/2)+\arccos((x-1)/2)$ . Find the domain of $f$ . Find the exact coordinates of the stationary point of $f$ and determine its nature.
(7)
(Total for Question 2 is 7 marks)
3.
The function $f$ is defined for all real $x$ by $f(x)=\arcsin\left(\dfrac{x}{1+x^2}\right)$ . Find $f'(x)$ . Hence find the exact coordinates of all stationary points and determine the nature of each one.
(7)
(Total for Question 3 is 7 marks)
4.
The curve has equation $y=x\arctan x$ . Find the exact equation of the tangent to the curve at $x=\sqrt3$ , giving your answer in the form $y=ax+b$ .
(7)
(Total for Question 4 is 7 marks)
5.
The function $f$ is defined on $[0,1]$ by $f(x)=\arccos(1-2x^2)$ . (a) Show that $f'(x)=\dfrac2{\sqrt{1-x^2}}$ for $0<x<1$ . (3) (b) Hence show that $f(x)=2\arcsin x$ throughout $[0,1]$ . (2) (c) Hence solve the equation $f(x)=\dfrac{2\pi}{3}$ . (2)
(7)
(Total for Question 5 is 7 marks)

CP-5.6 · Integrate functions of the form (a^2 - x^2)^(-1/2) and (a^2 - x^2)^(-1) and be able to choose trigonometric substitutions to integrate associated functions.

Explanation

For $a>0$ , $\int dx/\sqrt{a^2-x^2}=\arcsin(x/a)+C$ and $\int dx/(a^2−x²)=\dfrac₁{2a}\ln\left|\dfrac{a+x}{a-x}\right|+C$. A square root containing $a^2-x^2$ suggests $x=a\sin\theta$ , because the Pythagorean identity changes the root to $a\cos\theta$ on a suitable interval and $dx=a\cos\theta\,d\theta$ .
Associated integrals may then require a trigonometric identity before integration.
The rational form $1/(a^2-x^2)$ can instead be resolved into linear partial fractions.
After integrating in $\theta$ , use a right triangle or identities to return completely to $x$ , including the constant of integration.
Differentiating the result checks the scale factors and signs.

A right triangle for the substitution x equals a sine theta.

Worked example

Use $x=3\sin\theta$ to evaluate $\int\sqrt{9-x^2}\,dx$ .

1. $dx=3\cos\theta\,d\theta$ and $\sqrt{9-x^2}=3\cos\theta$ .
2.The integral becomes $9\int\cos^2\theta\,d\theta$ .
3.Using $\cos^2\theta=(1+\cos2\theta)/2$ gives $9\theta/2+9\sin2\theta/4+C$ .
4.Return to $x$ using $\theta=\arcsin(x/3)$ and $\sin2\theta=2x\sqrt{9-x^2}/9$ .

Answer: $\dfrac92\arcsin(x/3)+\dfrac x2\sqrt{9-x^2}+C$ .

Common mistakes

Don't fall into the trap of replacing the square root by $a\cos\theta$ but omitting $dx=a\cos\theta\,d\theta$ .
Don't fall into the trap of leaving the antiderivative in terms of $\theta$ instead of converting back to $x$ .
Don't fall into the trap of using $x=a\tan\theta$ for the form $a^2-x^2$ and creating a less suitable identity.

Exam tip

State the substitution, transformed differential and transformed square root on separate lines before simplifying the integral.

Tier 1 · Easy

1.
Find $\displaystyle \int\dfrac{1}{\sqrt{25-x^2}}\,dx$ .
(2)
(Total for Question 1 is 2 marks)

Tier 2 · Standard

1.
Find $\displaystyle \int\dfrac{1}{9-x^2}\,dx$ .
(3)
(Total for Question 1 is 3 marks)
2.
Find the exact value of $\displaystyle\int_0^1\frac{x+15}{25-x^2}\,dx$ , giving your answer in the form $\ln m$ , where $m$ is a rational number to be found.
(4)
(Total for Question 2 is 4 marks)
3.
Use a suitable trigonometric substitution to find the exact value of $\displaystyle\int_1^{\sqrt2}\frac{dx}{x^2\sqrt{4-x^2}}$ .
(5)
(Total for Question 3 is 5 marks)
4.
Use the substitution $x=2\sin\theta$ to find the exact value of $\displaystyle\int_0^{\sqrt2}(4-x^2)^{-3/2}\,dx$ .
(5)
(Total for Question 4 is 5 marks)

Tier 3 · Hard

1.
Using the substitution $x=4\sin\theta$ , find $\displaystyle \int\dfrac{x^2}{\sqrt{16-x^2}}\,dx$ .
(6)
(Total for Question 1 is 6 marks)
2.
Use a suitable trigonometric substitution to find the exact value of $\displaystyle\int_0^{3/2}x^2\sqrt{9-x^2}\,dx$ .
(6)
(Total for Question 2 is 6 marks)
3.
For $0<x<2$ , use the substitution $x=2\sin\theta$ to find $\displaystyle\int\frac{\sqrt{4-x^2}}{x^2}\,dx$ .
(6)
(Total for Question 3 is 6 marks)
4.
For $-6<x<6$ , use the substitution $x=6\sin\theta$ to find $\displaystyle\int(36-x^2)^{3/2}\,dx$ .
(7)
(Total for Question 4 is 7 marks)
5.
Use the substitution $x=5\sin\theta$ to find the exact value of $\displaystyle\int_0^{5/2}\frac{x^4}{\sqrt{25-x^2}}\,dx$ .
(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.

CP-5.1 · Derive formulae for and calculate volumes of revolution.

Tier 1 · Easy

Mark scheme for CP-5.1 Tier 1 · Easy
Question	Scheme	Marks
1	$V=\pi\displaystyle\int_0^1(x^2+1)^2\,dx$ $V=\pi[x^5/5+2x^3/3+x]_0^1$ $V=\dfrac{28\pi}{15}$	3
(3 marks)		3
Notes
Using discs, $V=\pi\int_0^1y^2\,dx=\pi\int_0^1(x^2+1)^2\,dx$ . Hence $V=\pi[x^5/5+2x^3/3+x]_0^1=28\pi/15$ .
2	The vertical boundaries give the limits $x=0$ and $x=3$ $V=\pi\displaystyle\int_0^3(2x+1)^2\,dx$ $V=\pi\left[\dfrac{(2x+1)^3}{6}\right]_0^3$ $V=57\pi$	4
(4 marks)		4
Notes
The $y$ -axis and the line $x=3$ give the limits $0$ and $3$ . Disc areas give $V=\pi\int_0^3(2x+1)^2\,dx$ . Since $\int(2x+1)^2\,dx=(2x+1)^3/6$ , evaluation gives $V=\pi(343-1)/6=57\pi$ .

Tier 2 · Standard

Mark scheme for CP-5.1 Tier 2 · Standard
Question	Scheme	Marks
1	$V=\dfrac{206\pi}{15}$	4
(4 marks)		4
Notes
A cross-section perpendicular to the $y$ -axis is a disc of radius $x=y^2+1$ . Therefore $V=\pi\int_0^2(y^2+1)^2\,dy=\pi\int_0^2(y^4+2y^2+1)\,dy$ . Evaluation gives $\pi(32/5+16/3+2)=206\pi/15$ .
2	$x^2=x+2$ gives $x=-1$ and $x=2$ The intersections are $(-1,1)$ and $(2,4)$ $V=\pi\displaystyle\int_{-1}^{2}[(x+2)^2-x^4]\,dx$ $V=\pi[x^3/3+2x^2+4x-x^5/5]_{-1}^{2}$ $V=\dfrac{72\pi}{5}$	5
(5 marks)		5
Notes
Solving $x^2=x+2$ gives $x=-1,2$ . On this interval $x+2$ is the outer radius and $x^2$ the inner radius. Thus $V=\pi\int_{-1}^{2}[(x+2)^2-x^4]\,dx$ . Applying the limits to $x^3/3+2x^2+4x-x^5/5$ gives $V=72\pi/5$ .
3	$dx/dt=2t$ and the parameter limits are $0$ and $2$ $V=\pi\displaystyle\int_0^2[t(2-t)]^2(2t)\,dt$ $V=2\pi\displaystyle\int_0^2(4t^3-4t^4+t^5)\,dt$ $V=2\pi[t^4-4t^5/5+t^6/6]_0^2$ $V=\dfrac{32\pi}{15}$	5
(5 marks)		5
Notes
For rotation about the $x$ -axis, $V=\pi\int y^2\,dx$ . Since $dx/dt=2t$ , this becomes $\pi\int_0^2[t(2-t)]^2(2t)dt$ . Expanding and integrating gives $2\pi[t^4-4t^5/5+t^6/6]_0^2=32\pi/15$ .

Tier 3 · Hard

Mark scheme for CP-5.1 Tier 3 · Hard
Question	Scheme	Marks
1	$V=\dfrac{3\pi}{10}$	5
(5 marks)		5
Notes
The curves meet at $(0,0)$ and $(1,1)$ . Rewriting with $x$ in terms of $y$ : the outer boundary is $x=\sqrt{y}$ (from $y=x^2$ ) and the inner is $x=y^2$ (from $y=\sqrt{x}$ ). Then $V=\pi\int_0^1\left((\sqrt{y})^2-(y^2)^2\right)dy=\pi\int_0^1(y-y^4)\,dy=\pi\left(\tfrac12-\tfrac15\right)=3\pi/10$ .
2	For $x\ge0$ , $x^2+1=3-x$ gives $x=1$ as its only non-negative root For $x\le0$ , $x^2+1=3+x$ gives $x=-1$ as its only non-positive root The intersections are $(-1,2)$ and $(1,2)$ only $V=\pi\displaystyle\int_{-1}^{0}[(3+x)^2-(x^2+1)^2]\,dx+\pi\int_0^1[(3-x)^2-(x^2+1)^2]\,dx$ By symmetry, $V=2\pi\displaystyle\int_0^1[(3-x)^2-(x^2+1)^2]\,dx$ $V=2\pi[8x-3x^2-x^3/3-x^5/5]_0^1$ $V=\dfrac{134\pi}{15}$	7
(7 marks)		7
Notes
For $x\ge0$ , $x^2+1=3-x$ gives the relevant root $x=1$ ; for $x\le0$ , $x^2+1=3+x$ gives $x=-1$ . The outer radius is $3+x$ on $[-1,0]$ and $3-x$ on $[0,1]$ , while the inner radius is $x^2+1$ . Hence the split washer integral reduces by symmetry to $2\pi\int_0^1[(3-x)^2-(x^2+1)^2]\,dx=2\pi[8x-3x^2-x^3/3-x^5/5]_0^1=134\pi/15$ .
3	$t(1-t^2)\ge0$ on $0\le t\le1$ , so the outer radius is $a+t(1-t^2)$ , the inner radius is $a$ and $dy/dt=1$ $V=\pi\displaystyle\int_0^1\{[a+t(1-t^2)]^2-a^2\}\,dt$ $V=\pi\displaystyle\int_0^1[2at(1-t^2)+t^2(1-t^2)^2]\,dt$ $\displaystyle\int_0^1t(1-t^2)\,dt=\frac14$ and $\displaystyle\int_0^1t^2(1-t^2)^2\,dt=\frac8{105}$ $V=\pi\left(\dfrac a2+\dfrac8{105}\right)$ $\dfrac a2+\dfrac8{105}=\dfrac{218}{105}$ , so $a=4$	6
(6 marks)		6
Notes
On the stated parameter interval, $t(1-t^2)\ge0$ , so washers about the $y$ -axis have outer radius $a+t(1-t^2)$ and inner radius $a$ . Since $dy/dt=1$ , $V=\pi\int_0^1([a+t(1-t^2)]^2-a^2)dt$ . The two required integrals are $1/4$ and $8/105$ , giving $V=\pi(a/2+8/105)$ . Equating this to $218\pi/105$ gives $a=4$ .
4	$y=4\pm2\sqrt{1-x^2/9}$ The limits are $x=-3$ and $x=3$ The outer radius is $4+2\sqrt{1-x^2/9}$ and the inner radius is $4-2\sqrt{1-x^2/9}$ $\displaystyle V=\pi\int_{-3}^{3}\left[(4+2\sqrt{1-x^2/9})^2-(4-2\sqrt{1-x^2/9})^2\right]dx$ The difference of the squared radii is $32\sqrt{1-x^2/9}$ $\displaystyle\int_{-3}^{3}\sqrt{1-x^2/9}\,dx=\frac{3\pi}{2}$ $V=32\pi(3\pi/2)=48\pi^2$ cubic units	7
(7 marks)		7
Notes
At each $x$ in $[-3,3]$ , the outer and inner radii are $4\pm2\sqrt{1-x^2/9}$ . Their squared difference is $32\sqrt{1-x^2/9}$ . With $x=3\sin\theta$ , the remaining integral from $-3$ to $3$ is $3\pi/2$ . Therefore the volume is $32\pi(3\pi/2)=48\pi^2$ cubic units.
5	$y^2=3$ gives the intersections $y=-\sqrt3$ and $y=\sqrt3$ For $-\sqrt3\le y\le\sqrt3$ , the outer radius is $3$ The inner radius is $y^2$ $\displaystyle V=\pi\int_{-\sqrt3}^{\sqrt3}(9-y^4)\,dy$ The integrand is even, so $V=2\pi\displaystyle\int_0^{\sqrt3}(9-y^4)\,dy$ $V=2\pi[9y-y^5/5]_0^{\sqrt3}$ $V=2\pi(9\sqrt3-9\sqrt3/5)$ $V=\dfrac{72\sqrt3\pi}{5}$ cubic units	8
(8 marks)		8
Notes
The curves meet at $y=\pm\sqrt3$ . Washers about the $y$ -axis have outer radius $3$ and inner radius $y^2$ . Thus $V=\pi\int_{-\sqrt3}^{\sqrt3}(9-y^4)dy=2\pi[9y-y^5/5]_0^{\sqrt3}=72\sqrt3\pi/5$ .

CP-5.2 · Evaluate improper integrals where either the integrand is undefined at a value in the range of integration or the range of integration extends to infinity.

Tier 1 · Easy

Mark scheme for CP-5.2 Tier 1 · Easy
Question	Scheme	Marks
1	The integral converges to $1$ .	3
(3 marks)		3
Notes
Write the integral as $\lim_{b\to\infty}\int_1^b x^{-2}\,dx$ . This is $\lim_{b\to\infty}[-x^{-1}]_1^b=\lim_{b\to\infty}(1-1/b)=1$ .

Tier 2 · Standard

Mark scheme for CP-5.2 Tier 2 · Standard
Question	Scheme	Marks
1	The integral converges to $4$ .	3
(3 marks)		3
Notes
The integrand is undefined at $x=0$ , so use $\lim_{a\to0^+}\int_a^4x^{-1/2}\,dx$ . This equals $\lim_{a\to0^+}[2\sqrt{x}]_a^4=\lim_{a\to0^+}(4-2\sqrt{a})=4$ .
2	$\displaystyle\int_1^\infty\frac{\ln x}{x^2}\,dx=\lim_{b\to\infty}\int_1^b\frac{\ln x}{x^2}\,dx$ Integration by parts gives $\displaystyle\int_1^b\frac{\ln x}{x^2}\,dx=\left[-\frac{\ln x}{x}-\frac1x\right]_1^b=1-\frac{\ln b+1}{b}$ $\displaystyle\lim_{b\to\infty}\frac{\ln b}{b}=0$ and $\displaystyle\lim_{b\to\infty}\frac1b=0$ Therefore the improper integral converges to $1$	4
(4 marks)		4
Notes
Write the integral as $\lim_{b\to\infty}\int_1^b(\ln x)x^{-2}dx$ . Integration by parts with $u=\ln x$ and $dv=x^{-2}dx$ gives $[-(\ln x)/x-1/x]_1^b=1-(\ln b+1)/b$ . Since $(\ln b)/b\to0$ and $1/b\to0$ , the improper integral converges to $1$ .
3	The integrand is undefined at $x=2$ , which lies inside the interval of integration Split the integral at $x=2$ into separate one-sided limits $\displaystyle\lim_{a\to2^-}\int_0^a(2-x)^{-2/3}\,dx=3\sqrt[3]{2}$ $\displaystyle\lim_{b\to2^+}\int_b^3(x-2)^{-2/3}\,dx=3$ Both one-sided limits are finite, so the integral converges to $3(1+\sqrt[3]{2})$	5
(5 marks)		5
Notes
The integrand is undefined at the interior point $x=2$ , so the integral is improper and must be split there. On the left, $\lim_{a\to2^-}\int_0^a(2-x)^{-2/3}dx=\lim_{a\to2^-}[3\sqrt[3]{2}-3\sqrt[3]{2-a}]=3\sqrt[3]{2}$ . On the right, $\lim_{b\to2^+}\int_b^3(x-2)^{-2/3}dx=\lim_{b\to2^+}[3-3\sqrt[3]{b-2}]=3$ . Both limits are finite, so the integral converges to $3(1+\sqrt[3]{2})$ .
4	The integrand is unbounded at the upper endpoint, so $\displaystyle I=\lim_{b\to1^-}\int_0^b x(1-x)^{-1/2}\,dx$ With $u=1-x$ , $dx=-du$ , and the limits $x=0,b$ become $u=1,1-b$ $\displaystyle I=\lim_{b\to1^-}\int_{1-b}^{1}(1-u)u^{-1/2}\,du$ $\displaystyle\int(1-u)u^{-1/2}\,du=2u^{1/2}-\frac23u^{3/2}$ $\displaystyle I=\lim_{b\to1^-}\left[2u^{1/2}-\frac23u^{3/2}\right]_{1-b}^{1}=2-\frac23=\frac43$ , so the finite one-sided limit proves convergence	5
(5 marks)		5
Notes
The integrand is unbounded at the upper endpoint $x=1$ , so write the integral as $\lim_{b\to1^-}\int_0^b x(1-x)^{-1/2}dx$ . With $u=1-x$ , this becomes $\lim_{b\to1^-}\int_{1-b}^1(1-u)u^{-1/2}du$ . An antiderivative is $2u^{1/2}-(2/3)u^{3/2}$ . Its value at $u=1$ is $4/3$ , while its value at $u=1-b$ tends to zero, so the one-sided limit is finite and the improper integral converges to $4/3$ .

Tier 3 · Hard

Mark scheme for CP-5.2 Tier 3 · Hard
Question	Scheme	Marks
1	The integral converges to $\pi$ .	6
(6 marks)		6
Notes
For $0<a<b$ , set $x=t^2$ , so $dx=2t\,dt$ and $\sqrt{x}=t$ . Then $\int_a^b dx/(\sqrt{x}(1+x))=2\int_{\sqrt a}^{\sqrt b}dt/(1+t^2)=2[\arctan t]_{\sqrt a}^{\sqrt b}$ . Taking the limits separately gives $2(\pi/2-0)=\pi$ as $a\to0^+$ and $b\to\infty$ .
2	The integrand is undefined at $x=1$ , which lies inside the interval of integration Split the integral into separate limits approaching $x=1$ from the left and right $\displaystyle\lim_{a\to1^-}\int_0^a\frac{dx}{x-1}=-\infty$ $\displaystyle\lim_{b\to1^+}\int_b^2\frac{dx}{x-1}=+\infty$ , so the integral is divergent The student has applied the antiderivative across the point $x=1$ where the integrand is undefined Each one-sided limit must exist and be finite for the integral to converge; here neither is finite, so no value can be assigned	6
(6 marks)		6
Notes
The integrand is undefined at the interior point $x=1$ , so the integral is improper and must be split there. The left contribution is $\lim_{a\to1^-}[\ln\|x-1\|]_0^a=\lim_{a\to1^-}\ln(1-a)=-\infty$ . The right contribution is $\lim_{b\to1^+}[\ln\|x-1\|]_b^2=\lim_{b\to1^+}-\ln(b-1)=+\infty$ . Since neither one-sided integral is finite, the improper integral diverges. The student's working is invalid because the antiderivative is applied across the undefined point: convergence requires each one-sided limit to exist and be finite, and here neither does, so the integral cannot be assigned any value.
3	$\displaystyle\int\frac{e^x}{(1+e^x)^2}\,dx=\frac{e^x}{1+e^x}+C$ $\displaystyle J(b)=\frac{e^b}{1+e^b}-\frac{e^{-b}}{1+e^{-b}}$ $\displaystyle J(b)=\frac{e^b-1}{e^b+1}$ , so $p=-1$ and $q=1$ $\displaystyle\lim_{a\to-\infty}\int_a^0\frac{e^x}{(1+e^x)^2}\,dx=\frac12-0=\frac12$ $\displaystyle\lim_{b\to\infty}\int_0^b\frac{e^x}{(1+e^x)^2}\,dx=1-\frac12=\frac12$ Both halves have finite limits, so the improper integral converges to $1/2+1/2=1$	6
(6 marks)		6
Notes
The antiderivative is $e^x/(1+e^x)$ , so evaluation at $-b$ and $b$ gives $J(b)=(e^b-1)/(e^b+1)=\tanh(b/2)$ . For the improper integral, split at zero. The antiderivative tends to $0$ at the left endpoint and to $1$ at the right endpoint, so the separate contributions are $1/2$ and $1/2$ . Both exist and are finite, giving a total of $1$ .
4	Let $I(b)=\displaystyle\int_0^b e^{-x}\sin x\,dx$ Integration by parts gives $I(b)=-e^{-b}\sin b+\displaystyle\int_0^b e^{-x}\cos x\,dx$ $\displaystyle\int_0^b e^{-x}\cos x\,dx=1-e^{-b}\cos b-I(b)$ $2I(b)=1-e^{-b}(\sin b+\cos b)$ $\|e^{-b}(\sin b+\cos b)\|\le\sqrt2e^{-b}$ $\displaystyle\lim_{b\to\infty}e^{-b}(\sin b+\cos b)=0$ Therefore the improper integral converges to $1/2$	7
(7 marks)		7
Notes
For a finite upper limit $b$ , integrate first with the sine factor and then with the cosine factor. This gives $2I(b)=1-e^{-b}(\sin b+\cos b)$ . The remaining term has absolute value at most $\sqrt2e^{-b}$ and tends to zero, so the defining limit exists and the improper integral converges to $1/2$ .
5	At $x=0$ , define $A=\displaystyle\lim_{a\to0^+}\int_a^1\frac{\ln x}{\sqrt{x}}\,dx$ $\displaystyle\int\frac{\ln x}{\sqrt{x}}\,dx=2\sqrt{x}\ln x-4\sqrt{x}+C$ $\displaystyle\lim_{a\to0^+}\sqrt a\ln a=0$ $A=-4$ At $x=1$ , define $B=\displaystyle\lim_{b\to1^-}\int_0^b\frac{dx}{\sqrt{1-x}}$ $\displaystyle\int\frac{dx}{\sqrt{1-x}}=-2\sqrt{1-x}+C$ $B=2$ Both endpoint contributions are finite, so the given integral converges to $A+B=-2$	8
(8 marks)		8
Notes
Separate the two terms and use a one-sided limit at the endpoint where each is undefined. The antiderivatives are $2\sqrt{x}\ln x-4\sqrt{x}$ and $-2\sqrt{1-x}$ . Since $\sqrt{x}\ln x\to0$ as $x\to0^+$ , the two contributions converge to $-4$ and $2$ . Their sum is therefore $-2$ .

CP-5.3 · Understand and evaluate the mean value of a function.

Tier 1 · Easy

Mark scheme for CP-5.3 Tier 1 · Easy
Question	Scheme	Marks
1	Mean value $=3$	3
(3 marks)		3
Notes
The interval length is $3$ . Hence the mean is $\frac13\int_0^3x^2\,dx=\frac13[x^3/3]_0^3=\frac13(9)=3$ .
2	$\dfrac14\displaystyle\int_0^4(kx+1)\,dx=7$ $2k+1=7$ $k=3$	3
(3 marks)		3
Notes
The prescribed mean gives $\frac14\int_0^4(kx+1)dx=7$ . The left side is $\frac14[8k+4]=2k+1$ . Hence $2k+1=7$ and $k=3$ .

Tier 2 · Standard

Mark scheme for CP-5.3 Tier 2 · Standard
Question	Scheme	Marks
1	Mean value $=\dfrac{1}{e-1}$	4
(4 marks)		4
Notes
The interval length is $e-1$ . Also $\int_0^{e-1}dx/(x+1)=[\ln(x+1)]_0^{e-1}=\ln e-\ln1=1$ . Dividing by the interval length gives $1/(e-1)$ .
2	$\displaystyle\operatorname{mean}(f)=\frac12\int_1^3[x^2+\ln(x+1)]\,dx$ Using integration by parts, $\displaystyle\int\ln(x+1)\,dx=(x+1)\ln(x+1)-(x+1)$ $\displaystyle\operatorname{mean}(f)=\frac12\left[\frac{x^3}{3}+(x+1)\ln(x+1)-(x+1)\right]_1^3$ $\displaystyle\int_1^3[x^2+\ln(x+1)]\,dx=\frac{20}{3}+6\ln2$ $\displaystyle\operatorname{mean}(f)=\frac{10}{3}+3\ln2$	5
(5 marks)		5
Notes
The interval length is $2$ , so the mean is $\frac12\int_1^3[x^2+\ln(x+1)]dx$ . Integration by parts gives $\int\ln(x+1)dx=(x+1)\ln(x+1)-(x+1)$ . Hence the definite integral is $[x^3/3+(x+1)\ln(x+1)-(x+1)]_1^3=20/3+6\ln2$ , and the exact mean is $10/3+3\ln2$ .
3	$\displaystyle\operatorname{mean}_{[0,\pi/2]}(f)=\frac2\pi\int_0^{\pi/2}[\sin x+2\sin(2x)]\,dx$ $\displaystyle\int_0^{\pi/2}\sin x\,dx=1$ and $\displaystyle\int_0^{\pi/2}2\sin(2x)\,dx=2$ , so the first mean is $6/\pi$ $\displaystyle\int_0^\pi\sin x\,dx=2$ and $\displaystyle\int_0^\pi2\sin(2x)\,dx=0$ $\displaystyle\operatorname{mean}_{[0,\pi]}(f)=\frac1\pi\int_0^\pi[\sin x+2\sin(2x)]\,dx=\frac2\pi$ The mean on $[0,\pi/2]$ is greater by $\dfrac4\pi$	5
(5 marks)		5
Notes
On $[0,\pi/2]$ , the two terms integrate to $1$ and $2$ , so division by the interval length gives $6/\pi$ . On $[0,\pi]$ , they integrate to $2$ and $0$ , giving a mean of $2/\pi$ . The first is therefore greater by $4/\pi$ .

Tier 3 · Hard

Mark scheme for CP-5.3 Tier 3 · Hard
Question	Scheme	Marks
1	Mean value $=\dfrac{1-3e^{-2}}{2}$	5
(5 marks)		5
Notes
Integration by parts gives $\int xe^{-x}\,dx=-(x+1)e^{-x}+C$ . Therefore $\int_0^2xe^{-x}\,dx=[-(x+1)e^{-x}]_0^2=1-3e^{-2}$ . The interval length is $2$ , so the mean is $(1-3e^{-2})/2$ .
2	The mean value of $x^2$ on $[0,a]$ is $\dfrac1a\displaystyle\int_0^a x^2\,dx=\dfrac{a^2}{3}$ The mean value of $x$ on $[a,a+2]$ is $\dfrac12\displaystyle\int_a^{a+2}x\,dx=a+1$ $\dfrac{a^2}{3}=a+1$ , so $a^2-3a-3=0$ $a=\dfrac{3\pm\sqrt{21}}{2}$ $a=\dfrac{3+\sqrt{21}}{2}$ since $a>0$	5
(5 marks)		5
Notes
The first mean is $a^{-1}\int_0^a x^2dx=a^2/3$ . The second is $\frac12\int_a^{a+2}x\,dx=a+1$ . Equating gives $a^2/3=a+1$ , or $a^2-3a-3=0$ . Thus $a=(3\pm\sqrt{21})/2$ , and positivity selects $(3+\sqrt{21})/2$ .
3	$\displaystyle\operatorname{mean}_{[0,4]}(f)=\frac14\left(\int_0^2(x+a)\,dx+\int_2^4(x-4)^2\,dx\right)$ $\displaystyle\int_0^2(x+a)\,dx=2+2a$ $\displaystyle\int_2^4(x-4)^2\,dx=8/3$ $\dfrac14(2+2a+8/3)=13/6$ , so $a=2$ $\displaystyle\int_1^2(x+2)\,dx=7/2$ and $\displaystyle\int_2^3(x-4)^2\,dx=7/3$ $\displaystyle\operatorname{mean}_{[1,3]}(f)=\frac12\left(\frac72+\frac73\right)=\frac{35}{12}$	6
(6 marks)		6
Notes
Split the first mean at $x=2$ . The two integrals are $2+2a$ and $8/3$ , so $(2+2a+8/3)/4=13/6$ and $a=2$ . For $[1,3]$ , the two pieces contribute $7/2$ and $7/3$ . Dividing their sum by the interval length $2$ gives the mean $35/12$ .
4	$\displaystyle\overline P=\frac14\int_0^4\left[6+\frac t2+2\sin\left(\frac{\pi t}{4}\right)\right]dt$ $\displaystyle\int_0^4 6\,dt=24$ and $\displaystyle\int_0^4\frac t2\,dt=4$ $\displaystyle\int_0^4 2\sin\left(\frac{\pi t}{4}\right)dt=\frac{16}{\pi}$ $\displaystyle\overline P=\frac14\left(28+\frac{16}{\pi}\right)=7+\frac4\pi\text{ kW}$ Energy generated equals mean power multiplied by time $\displaystyle E=4\left(7+\frac4\pi\right)=28+\frac{16}{\pi}\text{ kWh}$	6
(6 marks)		6
Notes
Divide the integral of the power output by the four-hour interval length. The constant, linear and sine terms contribute $24$ , $4$ and $16/\pi$ kWh respectively, so the exact mean power is $7+4/\pi$ kW. Multiplying this mean by four hours gives $28+16/\pi$ kWh.
5	$\displaystyle\operatorname{mean}=\frac1{a(e-1)}\int_a^{ea}\ln x\,dx$ $\displaystyle\int\ln x\,dx=x\ln x-x$ $\displaystyle\int_a^{ea}\ln x\,dx=a(e-1)\ln a+a$ The mean value is $\ln a+1/(e-1)$ $\ln a+1/(e-1)=2$ , so $a=\exp(2-1/(e-1))$ $\displaystyle\int(\ln x)^2\,dx=x[(\ln x)^2-2\ln x+2]+C$ Writing $L=\ln a=2-1/(e-1)$ , the second integral is $a[(e-1)L^2+2L+e-2]$ $\displaystyle\int_a^{ea}(\ln x)^2\,dx=\exp\left(2-\frac1{e-1}\right)\left(5e-6-\frac1{e-1}\right)$	8
(8 marks)		8
Notes
The interval length is $a(e-1)$ . Evaluating $[x\ln x-x]_a^{ea}$ gives the mean $\ln a+1/(e-1)$ , so $a=\exp(2-1/(e-1))$ . Using the elementary antiderivative $x[(\ln x)^2-2\ln x+2]$ and simplifying with $L=\ln a$ gives $a[(e-1)L^2+2L+e-2]=a[5e-6-1/(e-1)]$ .

CP-5.4 · Integrate using partial fractions.

Tier 1 · Easy

Mark scheme for CP-5.4 Tier 1 · Easy
Question	Scheme	Marks
1	$2\ln\|x+1\|+\ln\|x+2\|+C$	4
(4 marks)		4
Notes
Write $(3x+5)/((x+1)(x+2))=A/(x+1)+B/(x+2)$ . Then $3x+5=A(x+2)+B(x+1)$ , giving $A=2$ and $B=1$ . Integrating $2/(x+1)+1/(x+2)$ gives $2\ln\|x+1\|+\ln\|x+2\|+C$ .

Tier 2 · Standard

Mark scheme for CP-5.4 Tier 2 · Standard
Question	Scheme	Marks
1	$2\ln\|x+1\|-\dfrac{3}{x+1}+C$	4
(4 marks)		4
Notes
Resolve $(2x+5)/(x+1)^2=A/(x+1)+B/(x+1)^2$ . Since $2x+5=A(x+1)+B$ , comparison gives $A=2$ and $B=3$ . Thus the integral is $2\ln\|x+1\|+3\int(x+1)^{-2}\,dx=2\ln\|x+1\|-3/(x+1)+C$ .
2	$\dfrac{x^3+x^2+2x+1}{x(x+1)}=x+\dfrac{2x+1}{x(x+1)}$ $\dfrac{2x+1}{x(x+1)}=\dfrac1x+\dfrac1{x+1}$ $\displaystyle\int\left(x+\frac1x+\frac1{x+1}\right)dx=\frac{x^2}{2}+\ln x+\ln(x+1)$ on $[1,2]$ $\displaystyle\left[\frac{x^2}{2}+\ln x+\ln(x+1)\right]_1^2=\frac32+\ln2+\ln(3/2)$ $\displaystyle\int_1^2\frac{x^3+x^2+2x+1}{x(x+1)}\,dx=\frac32+\ln3$ , so $m=3/2$ and $k=3$	5
(5 marks)		5
Notes
Polynomial division gives $(x^3+x^2+2x+1)/[x(x+1)]=x+(2x+1)/[x(x+1)]$ . Then $(2x+1)/[x(x+1)]=1/x+1/(x+1)$ . An antiderivative on $[1,2]$ is $x^2/2+\ln x+\ln(x+1)$ . Applying the limits gives $3/2+\ln2+\ln(3/2)=3/2+\ln3$ , so $m=3/2$ and $k=3$ .
3	$mx+7=D(x+3)+D(2x-1)=D(3x+2)$ $2D=7$ and $m=3D$ , so $D=7/2$ and $m=21/2$ $\displaystyle\int\frac{D}{2x-1}\,dx=\frac D2\ln\|2x-1\|=\frac74\ln\|2x-1\|$ $\displaystyle\int\frac{D}{x+3}\,dx=D\ln\|x+3\|=\frac72\ln\|x+3\|$ $\dfrac74\ln\|2x-1\|+\dfrac72\ln\|x+3\|+C$	5
(5 marks)		5
Notes
If both coefficients equal $D$ , recombination gives $mx+7=D(x+3)+D(2x-1)=D(3x+2)$ . Hence $2D=7$ , so $D=7/2$ , and $m=3D=21/2$ . Termwise integration gives $\frac74\ln\|2x-1\|+\frac72\ln\|x+3\|+C$ .
4	$\displaystyle\frac{x^2+7x+1}{(x-1)^2(x+2)}=\frac A{x-1}+\frac B{(x-1)^2}+\frac C{x+2}$ $x^2+7x+1=A(x-1)(x+2)+B(x+2)+C(x-1)^2$ $A=2$ , $B=3$ and $C=-1$ $\displaystyle\int\left(\frac2{x-1}+\frac3{(x-1)^2}-\frac1{x+2}\right)dx=2\ln(x-1)-\frac3{x-1}-\ln(x+2)$ on $[2,4]$ $\displaystyle\left[2\ln(x-1)-\frac3{x-1}-\ln(x+2)\right]_2^4$ $2\ln3-1-\ln6+3+\ln4=2+\ln6$	6
(6 marks)		6
Notes
Use $A/(x-1)+B/(x-1)^2+C/(x+2)$ . Clearing denominators and comparing coefficients gives $A=2$ , $B=3$ and $C=-1$ . On $[2,4]$ an antiderivative is $2\ln(x-1)-3/(x-1)-\ln(x+2)$ . Applying the limits gives $2\ln3-1-\ln6+3+\ln4=2+\ln6$ .

Tier 3 · Hard

Mark scheme for CP-5.4 Tier 3 · Hard
Question	Scheme	Marks
1	$\dfrac14\ln\|x\|+\dfrac38\ln(x^2+4)+C$	6
(6 marks)		6
Notes
Use $(x^2+1)/(x(x^2+4))=A/x+(Bx+C)/(x^2+4)$ . Multiplying through gives $x^2+1=A(x^2+4)+x(Bx+C)$ , so $A=1/4$ , $C=0$ and $B=3/4$ . Therefore the integral is $\frac14\int dx/x+\frac34\int x\,dx/(x^2+4)=\frac14\ln\|x\|+\frac38\ln(x^2+4)+C$ .
2	$\dfrac{4x^3-x^2-x+8}{(x-1)^2(x^2+4)}=\dfrac P{x-1}+\dfrac Q{(x-1)^2}+\dfrac{Rx+S}{x^2+4}$ $4x^3-x^2-x+8=P(x-1)(x^2+4)+Q(x^2+4)+(Rx+S)(x-1)^2$ $P=1$ and $Q=2$ $R=3$ and $S=4$ $\displaystyle\int\dfrac{dx}{x-1}=\ln\|x-1\|$ , so $A=1$ $\displaystyle\int\dfrac{2\,dx}{(x-1)^2}=-\dfrac2{x-1}$ , so $B=-2$ $\displaystyle\int\dfrac{3x\,dx}{x^2+4}=\dfrac32\ln(x^2+4)$ and $\displaystyle\int\dfrac{4\,dx}{x^2+4}=2\arctan(x/2)$ , so $C=3/2$ and $D=2$ $\ln\|x-1\|-\dfrac2{x-1}+\dfrac32\ln(x^2+4)+2\arctan(x/2)+c$	8
(8 marks)		8
Notes
Use $P/(x-1)+Q/(x-1)^2+(Rx+S)/(x^2+4)$ . Coefficient comparison gives $P=1$ , $Q=2$ , $R=3$ and $S=4$ . Thus integrate $1/(x-1)+2/(x-1)^2+(3x+4)/(x^2+4)$ . The four contributions are $\ln\|x-1\|$ , $-2/(x-1)$ , $(3/2)\ln(x^2+4)$ and $2\arctan(x/2)$ , so $A=1$ , $B=-2$ , $C=3/2$ and $D=2$ , with arbitrary constant $c$ .
3	$\displaystyle\frac{x^2+mx+n}{(x-2)(x^2+25)}=\frac A{x-2}+\frac{Bx+C}{x^2+25}$ $x^2+mx+n=A(x^2+25)+(Bx+C)(x-2)$ No arctangent term requires $C=0$ The coefficient of $\ln\|x-2\|$ gives $A=3$ $A+B=1$ , so $B=-2$ $m=-2B=4$ $n=25A=75$ $\displaystyle\int\frac{x^2+4x+75}{(x-2)(x^2+25)}\,dx=3\ln\|x-2\|-\ln(x^2+25)+c$	8
(8 marks)		8
Notes
Decompose the fraction as $A/(x-2)+(Bx+C)/(x^2+25)$ . An arctangent would arise from the constant numerator $C$ , so $C=0$ , while the stated logarithm coefficient gives $A=3$ . Comparing the $x^2$ , $x$ and constant coefficients then gives $B=-2$ , $m=4$ and $n=75$ . Thus the integrand is $3/(x-2)-2x/(x^2+25)$ , whose antiderivative is $3\ln\|x-2\|-\ln(x^2+25)+c$ .
4	$\displaystyle\frac{x^3+8x^2+19x+18}{(x+1)(x+2)(x+3)(x+4)}=\frac A{x+1}+\frac B{x+2}+\frac C{x+3}+\frac D{x+4}$ $A=1$ $B=-2$ $C=3$ $D=-1$ An antiderivative is $\ln(x+1)-2\ln(x+2)+3\ln(x+3)-\ln(x+4)$ on $[0,1]$ The definite integral is $\ln2-2\ln(3/2)+3\ln(4/3)-\ln(5/4)$ $\displaystyle\int_0^1\frac{x^3+8x^2+19x+18}{(x+1)(x+2)(x+3)(x+4)}\,dx=\ln\left(\frac{2048}{1215}\right)$	8
(8 marks)		8
Notes
Use one constant numerator over each distinct linear factor. Substitution at the four roots of the denominator gives $A=1$ , $B=-2$ , $C=3$ and $D=-1$ . Integrating term by term and applying the limits gives $\ln2-2\ln(3/2)+3\ln(4/3)-\ln(5/4)=\ln(2048/1215)$ .
5	Polynomial division gives quotient $x-2$ $\displaystyle\frac{x^4-x^3+33x^2-33x+6}{(x+1)(x^2+36)}=x-2+\frac A{x+1}+\frac{Bx+C}{x^2+36}$ Clearing denominators and comparing coefficients gives $A=2$ $B=-3$ and $C=6$ $\displaystyle\int(x-2)\,dx=\frac{x^2}{2}-2x$ $\displaystyle\int\left(\frac2{x+1}-\frac{3x}{x^2+36}\right)dx=2\ln(x+1)-\frac32\ln(x^2+36)$ $\displaystyle\int\frac6{x^2+36}\,dx=\arctan(x/6)$ The value is $\displaystyle2\ln2+\frac32\ln\left(\frac{37}{45}\right)+\arctan\left(\frac12\right)-\arctan\left(\frac16\right)$	8
(8 marks)		8
Notes
After division, use $x-2+A/(x+1)+(Bx+C)/(x^2+36)$ . Recombination gives $A=2$ , $B=-3$ and $C=6$ . An antiderivative is $x^2/2-2x+2\ln(x+1)-(3/2)\ln(x^2+36)+\arctan(x/6)$ . Applying $1$ and $3$ gives the stated exact value.

CP-5.5 · Differentiate inverse trigonometric functions.

Tier 1 · Easy

Mark scheme for CP-5.5 Tier 1 · Easy
Question	Scheme	Marks
1	$\dfrac{dy}{dx}=\dfrac{2}{\sqrt{1-4x^2}}$	2
(2 marks)		2
Notes
Let $u=2x$ , so $u'=2$ . The chain rule gives $dy/dx=u'/\sqrt{1-u^2}=2/\sqrt{1-4x^2}$ .
2	$u=(x+1)/3$ and $u'=1/3$ $\dfrac{dy}{dx}=-\dfrac1{\sqrt{9-(x+1)^2}}$ $-4<x<2$	3
(3 marks)		3
Notes
With $u=(x+1)/3$ , $u'=1/3$ , so $dy/dx=-(1/3)/\sqrt{1-(x+1)^2/9}=-1/\sqrt{9-(x+1)^2}$ . This derivative is defined when $\|(x+1)/3\|<1$ , which gives $-4<x<2$ .

Tier 2 · Standard

Mark scheme for CP-5.5 Tier 2 · Standard
Question	Scheme	Marks
1	$\dfrac{dy}{dx}=\dfrac{2}{(x-1)^2+4}$	3
(3 marks)		3
Notes
Take $u=(x-1)/2$ , so $u'=1/2$ . Then $dy/dx=(1/2)/(1+u^2)=(1/2)/(1+(x-1)^2/4)=2/((x-1)^2+4)$ .
2	By the product rule, $\dfrac{dy}{dx}=2x\arcsin(x/3)+x^2\dfrac{d}{dx}[\arcsin(x/3)]$ $\displaystyle\frac{dy}{dx}=2x\arcsin(x/3)+\frac{x^2}{\sqrt{9-x^2}}$ At $x=3/2$ , $\arcsin(1/2)=\pi/6$ and $\sqrt{9-9/4}=3\sqrt3/2$ $\displaystyle\left.\frac{dy}{dx}\right\|_{x=3/2}=3(\pi/6)+\frac{9/4}{3\sqrt3/2}=\frac\pi2+\frac{\sqrt3}{2}$	4
(4 marks)		4
Notes
The product and chain rules give $dy/dx=2x\arcsin(x/3)+x^2/\sqrt{9-x^2}$ . At $x=3/2$ , $\arcsin(1/2)=\pi/6$ and $\sqrt{9-9/4}=3\sqrt3/2$ . Hence the derivative is $3(\pi/6)+(9/4)/(3\sqrt3/2)=\pi/2+\sqrt3/2$ .
3	$y(0)=k\arccos0=k\pi/2=\pi/3$ $k=2/3$ $\displaystyle\frac{dy}{dx}=\frac2{1+4x^2}-\frac{k}{\sqrt{9-x^2}}$ $\displaystyle\left.\frac{dy}{dx}\right\|_{x=1}=\frac25-\frac{2/3}{2\sqrt2}=\frac25-\frac{\sqrt2}{6}$	4
(4 marks)		4
Notes
The stated point gives $k\arccos0=k\pi/2=\pi/3$ , so $k=2/3$ . Differentiating gives $dy/dx=2/(1+4x^2)-k/\sqrt{9-x^2}$ . At $x=1$ , this is $2/5-(2/3)/(2\sqrt2)=2/5-\sqrt2/6$ .

Tier 3 · Hard

Mark scheme for CP-5.5 Tier 3 · Hard
Question	Scheme	Marks
1	$\dfrac{dy}{dx}=\arccos x$	5
(5 marks)		5
Notes
The product rule gives $d(x\arccos x)/dx=\arccos x-x/\sqrt{1-x^2}$ . Also $d(-\sqrt{1-x^2})/dx=x/\sqrt{1-x^2}$ . The two rational terms cancel, leaving $dy/dx=\arccos x$ .
2	$-2\le x\le2$ from $\arcsin(x/2)$ $-1\le x\le3$ from $\arccos((x-1)/2)$ Domain $[-1,2]$ $f'(x)=1/\sqrt{4-x^2}-1/\sqrt{4-(x-1)^2}$ $f'(x)=0$ gives $x=1/2$ $f(1/2)=\pi/2+2\arcsin(1/4)$ The stationary point $(1/2,\,\pi/2+2\arcsin(1/4))$ is a minimum	7
(7 marks)		7
Notes
The two inverse-function domains intersect as $[-2,2]\cap[-1,3]=[-1,2]$ . In the interior, $f'(x)=1/\sqrt{4-x^2}-1/\sqrt{4-(x-1)^2}$ . Equality of the positive denominators gives $x^2=(x-1)^2$ , hence $x=1/2$ . For $x<1/2$ , $x^2<(x-1)^2$ , so the first reciprocal is smaller and $f'(x)<0$ ; for $x>1/2$ , the sign reverses. Therefore the stationary point is a minimum.
3	$u=x/(1+x^2)$ gives $u'=(1-x^2)/(1+x^2)^2$ $\displaystyle f'(x)=\frac{1-x^2}{(1+x^2)^2\sqrt{1-x^2/(1+x^2)^2}}$ $\displaystyle f'(x)=\frac{1-x^2}{(1+x^2)\sqrt{x^4+x^2+1}}$ $f'(x)=0$ at $x=-1$ and $x=1$ The stationary points are $(-1,-\pi/6)$ and $(1,\pi/6)$ $f'$ changes from negative to positive at $x=-1$ , so $(-1,-\pi/6)$ is a minimum $f'$ changes from positive to negative at $x=1$ , so $(1,\pi/6)$ is a maximum	7
(7 marks)		7
Notes
For $u=x/(1+x^2)$ , the quotient rule gives $u'=(1-x^2)/(1+x^2)^2$ . Applying the derivative of $\arcsin u$ and simplifying gives $f'(x)=(1-x^2)/[(1+x^2)\sqrt{x^4+x^2+1}]$ . Its denominator is positive, so stationary points occur at $x=\pm1$ . Their values are $\arcsin(\pm1/2)=\pm\pi/6$ . The numerator is negative outside $[-1,1]$ and positive inside, giving a minimum at $x=-1$ and a maximum at $x=1$ .
4	At $x=\sqrt3$ , $y=\sqrt3\arctan(\sqrt3)=\dfrac{\pi\sqrt3}{3}$ $\displaystyle\frac{dy}{dx}=\arctan x+\frac{x}{1+x^2}$ $\displaystyle a=\left.\frac{dy}{dx}\right\|_{x=\sqrt3}=\frac\pi3+\frac{\sqrt3}{4}$ $\displaystyle y-\frac{\pi\sqrt3}{3}=\left(\frac\pi3+\frac{\sqrt3}{4}\right)(x-\sqrt3)$ $\displaystyle b=\frac{\pi\sqrt3}{3}-\sqrt3\left(\frac\pi3+\frac{\sqrt3}{4}\right)$ $b=-3/4$ $\displaystyle y=\left(\frac\pi3+\frac{\sqrt3}{4}\right)x-\frac34$	7
(7 marks)		7
Notes
The product rule gives $dy/dx=\arctan x+x/(1+x^2)$ . At $x=\sqrt3$ , the point is $(\sqrt3,\pi\sqrt3/3)$ and the gradient is $\pi/3+\sqrt3/4$ . Substitution into the point-gradient form and simplification give $y=(\pi/3+\sqrt3/4)x-3/4$ .
5	For $u=1-2x^2$ , $u'=-4x$ $1-u^2=1-(1-2x^2)^2=4x^2(1-x^2)$ $\displaystyle f'(x)=\frac{4x}{2x\sqrt{1-x^2}}=\frac2{\sqrt{1-x^2}}$ for $0<x<1$ $\displaystyle\frac{d}{dx}[2\arcsin x]=\frac2{\sqrt{1-x^2}}$ $f(0)=0=2\arcsin0$ , so continuity gives $f(x)=2\arcsin x$ on $[0,1]$ $f(x)=2\pi/3$ gives $\arcsin x=\pi/3$ $x=\sqrt3/2$	7
(7 marks)		7
Notes
Apply the chain rule to $\arccos(1-2x^2)$ and use $1-(1-2x^2)^2=4x^2(1-x^2)$ . Since $x>0$ , this simplifies to $2/\sqrt{1-x^2}$ , the derivative of $2\arcsin x$ . Both functions equal zero at $x=0$ , so they agree on the interval. The final equation then gives $x=\sin(\pi/3)=\sqrt3/2$ .

CP-5.6 · Integrate functions of the form (a^2 - x^2)^(-1/2) and (a^2 - x^2)^(-1) and be able to choose trigonometric substitutions to integrate associated functions.

Tier 1 · Easy

Mark scheme for CP-5.6 Tier 1 · Easy
Question	Scheme	Marks
1	$\arcsin\left(\dfrac{x}{5}\right)+C$	2
(2 marks)		2
Notes
This matches the standard form with $a=5$ , so the integral is $\arcsin(x/5)+C$ .

Tier 2 · Standard

Mark scheme for CP-5.6 Tier 2 · Standard
Question	Scheme	Marks
1	$\dfrac{1}{6}\ln\left\|\dfrac{3+x}{3-x}\right\|+C$	3
(3 marks)		3
Notes
Since $9-x^2=(3-x)(3+x)$ , partial fractions give $\dfrac{1}{9-x^2}=\dfrac{1}{6}\left(\dfrac{1}{3+x}+\dfrac{1}{3-x}\right)$ . Integrating term by term gives $\dfrac{1}{6}\left(\ln\|3+x\|-\ln\|3-x\|\right)+C=\dfrac{1}{6}\ln\left\|\dfrac{3+x}{3-x}\right\|+C$ .
2	$\dfrac{x+15}{25-x^2}=\dfrac A{5-x}+\dfrac B{5+x}$ $A=2$ and $B=1$ $\displaystyle\int_0^1\frac{x+15}{25-x^2}\,dx=[-2\ln(5-x)+\ln(5+x)]_0^1$ $2\ln(5/4)+\ln(6/5)=\ln(15/8)$ , so $m=15/8$	4
(4 marks)		4
Notes
Factor $25-x^2=(5-x)(5+x)$ and resolve $(x+15)/(25-x^2)=2/(5-x)+1/(5+x)$ . Therefore the integral is $[-2\ln(5-x)+\ln(5+x)]_0^1=2\ln(5/4)+\ln(6/5)=\ln(15/8)$ , so $m=15/8$ .
3	Use $x=2\sin\theta$ $dx=2\cos\theta\,d\theta$ and $\sqrt{4-x^2}=2\cos\theta$ $x=1,\sqrt2$ gives $\theta=\pi/6,\pi/4$ $\dfrac14\displaystyle\int_{\pi/6}^{\pi/4}\operatorname{cosec}^2\theta\,d\theta$ $\dfrac{\sqrt3-1}{4}$	5
(5 marks)		5
Notes
Choose $x=2\sin\theta$ because it changes $\sqrt{4-x^2}$ to $2\cos\theta$ , positive on the transformed interval. The limits are $\theta=\pi/6$ and $\pi/4$ , and the integral becomes $\frac14\int_{\pi/6}^{\pi/4}\operatorname{cosec}^2\theta\,d\theta$ . Thus its value is $[-\frac14\cot\theta]_{\pi/6}^{\pi/4}=(\sqrt3-1)/4$ .
4	Set $x=2\sin\theta$ $x=0$ and $x=\sqrt2$ give $\theta=0$ and $\theta=\pi/4$ $dx=2\cos\theta\,d\theta$ and $(4-x^2)^{3/2}=8\cos^3\theta$ $\displaystyle\int_0^{\sqrt2}(4-x^2)^{-3/2}\,dx=\frac14\int_0^{\pi/4}\sec^2\theta\,d\theta=\left[\frac14\tan\theta\right]_0^{\pi/4}$ $\tan\theta=x/\sqrt{4-x^2}$ , so the value is $\displaystyle\left[\frac{x}{4\sqrt{4-x^2}}\right]_0^{\sqrt2}=\frac14$	5
(5 marks)		5
Notes
With $x=2\sin\theta$ , the limits become $0$ and $\pi/4$ , while $dx=2\cos\theta d\theta$ and $(4-x^2)^{3/2}=8\cos^3\theta$ . The integral is therefore $(1/4)\int_0^{\pi/4}\sec^2\theta d\theta=(1/4)[\tan\theta]_0^{\pi/4}=1/4$ . Back-substitution gives the equivalent antiderivative $x/[4\sqrt{4-x^2}]$ .

Tier 3 · Hard

Mark scheme for CP-5.6 Tier 3 · Hard
Question	Scheme	Marks
1	$8\arcsin\left(\dfrac{x}{4}\right)-\dfrac{x}{2}\sqrt{16-x^2}+C$	6
(6 marks)		6
Notes
With $x=4\sin\theta$ , $dx=4\cos\theta\,d\theta$ and $\sqrt{16-x^2}=4\cos\theta$ . The integral becomes $16\int\sin^2\theta\,d\theta=8\theta-4\sin2\theta+C$ . Now $\theta=\arcsin(x/4)$ and $\sin2\theta=2(x/4)(\sqrt{16-x^2}/4)=x\sqrt{16-x^2}/8$ . Substitution gives $8\arcsin(x/4)-(x/2)\sqrt{16-x^2}+C$ .
2	Use $x=3\sin\theta$ $x=0,3/2$ gives $\theta=0,\pi/6$ $dx=3\cos\theta\,d\theta$ and $\sqrt{9-x^2}=3\cos\theta$ $81\displaystyle\int_0^{\pi/6}\sin^2\theta\cos^2\theta\,d\theta$ $\sin^2\theta\cos^2\theta=(1-\cos4\theta)/8$ $\dfrac{27\pi}{16}-\dfrac{81\sqrt3}{64}$	6
(6 marks)		6
Notes
Take $x=3\sin\theta$ . Then $dx=3\cos\theta\,d\theta$ , the square root is $3\cos\theta$ , and the limits are $0$ and $\pi/6$ . The integral becomes $81\int_0^{\pi/6}\sin^2\theta\cos^2\theta\,d\theta$ . Using $\sin^2\theta\cos^2\theta=(1-\cos4\theta)/8$ gives $81[\theta/8-\sin4\theta/32]_0^{\pi/6}=27\pi/16-81\sqrt3/64$ .
3	Set $x=2\sin\theta$ , where $0<\theta<\pi/2$ $dx=2\cos\theta\,d\theta$ and $\sqrt{4-x^2}=2\cos\theta$ $\displaystyle\int\frac{\sqrt{4-x^2}}{x^2}\,dx=\int\cot^2\theta\,d\theta$ $\cot^2\theta=\operatorname{cosec}^2\theta-1$ $\displaystyle\int(\operatorname{cosec}^2\theta-1)\,d\theta=-\cot\theta-\theta+C$ $\cot\theta=\sqrt{4-x^2}/x$ and $\theta=\arcsin(x/2)$ , giving $-\dfrac{\sqrt{4-x^2}}x-\arcsin(x/2)+C$	6
(6 marks)		6
Notes
For $0<x<2$ , set $x=2\sin\theta$ with $0<\theta<\pi/2$ . Then $dx=2\cos\theta d\theta$ and the square root is $2\cos\theta$ , so the integrand becomes $\cot^2\theta$ . Since $\cot^2\theta=\operatorname{cosec}^2\theta-1$ , integration gives $-\cot\theta-\theta+C$ . Back-substitution yields $-\sqrt{4-x^2}/x-\arcsin(x/2)+C$ .
4	$x=6\sin\theta$ , $dx=6\cos\theta\,d\theta$ and $\sqrt{36-x^2}=6\cos\theta$ $\displaystyle\int(36-x^2)^{3/2}dx=1296\int\cos^4\theta\,d\theta$ $\displaystyle\cos^4\theta=\frac{3+4\cos2\theta+\cos4\theta}{8}$ $\displaystyle1296\int\cos^4\theta\,d\theta=486\theta+324\sin2\theta+\frac{81}{2}\sin4\theta+C$ $\theta=\arcsin(x/6)$ and $\sin2\theta=x\sqrt{36-x^2}/18$ $\displaystyle\sin4\theta=x\sqrt{36-x^2}(18-x^2)/162$ $\displaystyle\int(36-x^2)^{3/2}dx=486\arcsin(x/6)+\frac{x(90-x^2)}4\sqrt{36-x^2}+C$	7
(7 marks)		7
Notes
The stated substitution changes the integral to $1296\int\cos^4\theta d\theta$ . Use $\cos^4\theta=(3+4\cos2\theta+\cos4\theta)/8$ and integrate. Back-substitution with $\sin2\theta=x\sqrt{36-x^2}/18$ and $\sin4\theta=x\sqrt{36-x^2}(18-x^2)/162$ gives the stated antiderivative.
5	$x=5\sin\theta$ gives the limits $0$ and $\pi/6$ $dx=5\cos\theta\,d\theta$ and $\sqrt{25-x^2}=5\cos\theta$ $\displaystyle\int_0^{5/2}\frac{x^4}{\sqrt{25-x^2}}dx=625\int_0^{\pi/6}\sin^4\theta\,d\theta$ $\displaystyle\sin^4\theta=\frac{3-4\cos2\theta+\cos4\theta}{8}$ $\displaystyle625\int_0^{\pi/6}\sin^4\theta\,d\theta=625\left[\frac{3\theta}{8}-\frac{\sin2\theta}{4}+\frac{\sin4\theta}{32}\right]_0^{\pi/6}$ $\sin(\pi/3)=\sin(2\pi/3)=\sqrt3/2$ $\displaystyle\int_0^{5/2}\frac{x^4}{\sqrt{25-x^2}}dx=\frac{625\pi}{16}-\frac{4375\sqrt3}{64}$	7
(7 marks)		7
Notes
The single substitution changes the integral to $625\int_0^{\pi/6}\sin^4\theta d\theta$ . Use $\sin^4\theta=(3-4\cos2\theta+\cos4\theta)/8$ and apply the transformed limits. Since both required sine values are $\sqrt3/2$ , the exact value is $625\pi/16-4375\sqrt3/64$ .