Further calculus (Core Pure, section 5) — 9FM0 notes

5.1

Year 1

Derive formulae for and calculate volumes of revolution.

Notes

Evidence from your answers: none yet

Your confidence:

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

Explanation

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

ORIGINAL

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)

Tier 2 · Standard

ORIGINAL

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)

Tier 3 · Hard

ORIGINAL

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)

5.2

Year 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.

Notes

Evidence from your answers: none yet

Your confidence:

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

Explanation

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

ORIGINAL

1.

Evaluate

\displaystyle \int_1^\infty\dfrac{1}{x^2}\,dx

.

(3)

(Total for Question 1 is 3 marks)

Tier 2 · Standard

ORIGINAL

1.

Evaluate

\displaystyle \int_0^4\dfrac{1}{\sqrt{x}}\,dx

.

(3)

(Total for Question 1 is 3 marks)

Tier 3 · Hard

ORIGINAL

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)

5.3

Year 2

Understand and evaluate the mean value of a function.

Notes

Evidence from your answers: none yet

Your confidence:

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

Explanation

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

ORIGINAL

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)

Tier 2 · Standard

ORIGINAL

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)

Tier 3 · Hard

ORIGINAL

1.

Find the exact mean value of

f(x)=xe^{-x}

on

0\leq x\leq2

.

(5)

(Total for Question 1 is 5 marks)

5.4

Year 2

Integrate using partial fractions.

Notes

Evidence from your answers: none yet

Your confidence:

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

Explanation

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

ORIGINAL

1.

Find

\displaystyle \int\dfrac{3x+5}{(x+1)(x+2)}\,dx

.

(4)

(Total for Question 1 is 4 marks)

Tier 2 · Standard

ORIGINAL

1.

Find

\displaystyle \int\dfrac{2x+5}{(x+1)^2}\,dx

.

(4)

(Total for Question 1 is 4 marks)

Tier 3 · Hard

ORIGINAL

1.

Find

\displaystyle \int\dfrac{x^2+1}{x(x^2+4)}\,dx

.

(6)

(Total for Question 1 is 6 marks)

5.5

Year 2

Differentiate inverse trigonometric functions.

Notes

Evidence from your answers: none yet

Your confidence:

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

Explanation

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

ORIGINAL

1.

Differentiate

y=\arcsin(2x)

with respect to

x

.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1.

Differentiate

y=\arctan\left(\dfrac{x-1}{2}\right)

.

(3)

(Total for Question 1 is 3 marks)

Tier 3 · Hard

ORIGINAL

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)

5.6

Year 2

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.

Notes

Evidence from your answers: none yet

Your confidence:

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

Explanation

For $a>0$ , $\int dx/\sqrt{a^2-x^2}=\arcsin(x/a)+C$ 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

ORIGINAL

1.

Find

\displaystyle \int\dfrac{1}{\sqrt{25-x^2}}\,dx

.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1.

Find

\displaystyle \int\dfrac{1}{9-x^2}\,dx

.

(3)

(Total for Question 1 is 3 marks)

Tier 3 · Hard

ORIGINAL

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)

Derive formulae for and calculate volumes of revolution.

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

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

Tier 2 · Standard

Tier 3 · Hard

Understand and evaluate the mean value of a function.

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Integrate using partial fractions.

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Differentiate inverse trigonometric functions.

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

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

Tier 2 · Standard

Tier 3 · Hard

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