Notes · Further MathematicsUK · A-Levels

Further calculus

Further calculus extends integration well beyond the standard A-Level techniques. It covers improper integrals evaluated as limits, the mean value of a function, volumes of revolution about either axis, arc length and the area of a surface of revolution, and reduction formulae that evaluate whole families of integrals by recursion. Integration by parts and partial fractions from A-Level Mathematics are assumed throughout.

4 sections~11 min reading time3 competenciesLevel Standard 1 · Advanced 3

T·0555 / 13

Exam profile

AO1 · Evaluate improper integrals, volumes, arc lengths and reduction formulae accuratelyAO2 · Derive a reduction formula and justify the limit process for an improper integralAO3 · Select and combine integration techniques in extended problems

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basic level

AS Further Mathematics includes volumes of revolution and the mean value of a function.

higher level

The full A-Level adds improper integrals, arc length, surface area of revolution and reduction formulae derived by integration by parts.

Depth

Text

Contents · 4 sections

Further calculus

Improper integrals and the mean value#

AdvancedAQA 7367 §3.5DfE Further Mathematics — Further calculus

An infinite region with finite area

Fig. 1The shaded region under

y=1/x^2

from

x=1

stretches to infinity yet has area exactly

1

An improper integral is one where either a limit of integration is infinite, or the integrand becomes unbounded somewhere on the interval. Such an integral is defined by first integrating over a finite, safe interval and then taking a limit:

\displaystyle\int_a^{\infty} f(x)\,\mathrm{d}x = \lim_{t\to\infty} \int_a^{t} f(x)\,\mathrm{d}x

. If the limit exists and is finite, the integral converges to that value; if the limit is infinite or fails to exist, the integral diverges and has no value.

The classic contrast is between

\int_1^{\infty}\dfrac{1}{x^2}\,\mathrm{d}x

and

\int_1^{\infty}\dfrac{1}{x}\,\mathrm{d}x

. The first gives

\left[-\dfrac{1}{x}\right]_1^{t} = 1 - \dfrac{1}{t} \to 1

, so it converges to

1

— a region of infinite length but finite area. The second gives

[\ln x]_1^{t} = \ln t \to \infty

, so it diverges. The difference is how fast the integrand decays:

1/x^2

decays fast enough for the area to be finite,

1/x

does not.

The same limiting idea handles an integrand that is undefined at an endpoint. For

\int_0^{1}\dfrac{1}{\sqrt{x}}\,\mathrm{d}x

, the integrand blows up at

x=0

, so evaluate

\lim_{t\to 0^+}\int_t^{1} x^{-1/2}\,\mathrm{d}x = \lim_{t\to 0^+}\big[2\sqrt{x}\big]_t^{1} = 2 - 0 = 2

: convergent despite the vertical asymptote. Always identify which endpoint is the problem and take the limit from the correct side; never substitute the singular value directly.

The mean value of a continuous function over

[a,b]

\bar f = \dfrac{1}{b-a}\displaystyle\int_a^{b} f(x)\,\mathrm{d}x

— the constant whose rectangle on

[a,b]

has the same area as the region under the curve. It generalises the ordinary average to a continuously-varying quantity and is a straightforward but frequently examined definite-integral calculation. Together, improper integrals and the mean value sharpen the meaning of the integral as accumulated area.

\int_a^{\infty} f(x)\,\mathrm{d}x = \lim_{t\to\infty} \int_a^{t} f(x)\,\mathrm{d}x

Improper integral (infinite limit)

Converges to the limit when it exists and is finite; otherwise diverges.

\bar f = \frac{1}{b-a}\int_a^{b} f(x)\,\mathrm{d}x

Mean value of a function

The height of the rectangle on $[a,b]$ with the same area as the region under $f$ .

Worked example

A convergent improper integral

Evaluate $\displaystyle\int_0^{\infty} x\,\mathrm{e}^{-x}\,\mathrm{d}x$ .

Write as a limit
$\displaystyle\int_0^{\infty} x\mathrm{e}^{-x}\,\mathrm{d}x = \lim_{t\to\infty} \int_0^{t} x\mathrm{e}^{-x}\,\mathrm{d}x$ .
Integrate by parts
With $u=x$ , $\mathrm{d}v = \mathrm{e}^{-x}\mathrm{d}x$ : $\int x\mathrm{e}^{-x}\,\mathrm{d}x = -x\mathrm{e}^{-x} - \int -\mathrm{e}^{-x}\,\mathrm{d}x = -x\mathrm{e}^{-x} - \mathrm{e}^{-x}$ .
$\int x\mathrm{e}^{-x}\,\mathrm{d}x = -(x+1)\mathrm{e}^{-x}$
Apply the limits
$\big[-(x+1)\mathrm{e}^{-x}\big]_0^{t} = -(t+1)\mathrm{e}^{-t} + 1$ .
Take the limit
As $t\to\infty$ , $(t+1)\mathrm{e}^{-t}\to 0$ (exponential beats the linear term), so the integral converges to $1$ .

Result: $\displaystyle\int_0^{\infty} x\mathrm{e}^{-x}\,\mathrm{d}x = 1$ .

Typical mistakes

Substituting $\infty$ (or the singular endpoint) directly instead of taking a limit, which is meaningless.
Declaring an integral convergent without checking that the limit is actually finite ( $\int_1^\infty 1/x\,\mathrm{d}x$ looks similar to the convergent case but diverges).

Active revision

Determine whether $\displaystyle\int_1^{\infty} \frac{1}{x^3}\,\mathrm{d}x$ converges and, if so, find its value; then find the mean value of $f(x)=x^2$ over $[0,3]$ .

Active recall

Recall the key points — then reveal.

Sources: Further mathematics: AS and A level content (GCE subject content) (Department for Education)

Volumes of revolution#

StandardAQA 7367 §3.5DfE Further Mathematics — Further calculus

A cone as a solid of revolution

Fig. 2Rotating the line

y = \tfrac{r}{h}x

about the

x

-axis sweeps out a cone of volume

\tfrac{1}{3}\pi r^2 h

Rotating the region under a curve about an axis sweeps out a solid whose volume can be found by integration. About the

x

-axis, a thin strip at position

x

becomes a disc of radius

y

and thickness

\mathrm{d}x

, with volume

\pi y^2\,\mathrm{d}x

; summing gives

V = \pi\displaystyle\int_a^{b} y^2\,\mathrm{d}x

. The key is that the radius of each disc is the

y

-coordinate of the curve, so it is

y^2

— not

y

— that is integrated.

Rotation about the

y

-axis is handled symmetrically, with the roles of

x

and

y

swapped:

V = \pi\displaystyle\int_c^{d} x^2\,\mathrm{d}y

, where the curve is expressed with

x

as a function of

y

and the limits are

y

-values. Choosing the right axis and expressing the curve appropriately is often the first decision in a volume question.

For a curve given parametrically,

x = x(t)

y = y(t)

, the same formulae apply with a change of variable:

V = \pi\displaystyle\int y^2\,\dfrac{\mathrm{d}x}{\mathrm{d}t}\,\mathrm{d}t

for rotation about the

x

-axis, with the limits converted to values of the parameter

t

. Keeping track of the limits under the change of variable is where care is needed.

The results reproduce the familiar formulae as special cases and thereby serve as checks. Rotating the line

y = \dfrac{r}{h}x

from

0

h

about the

x

-axis gives a cone:

V = \pi\int_0^h \dfrac{r^2}{h^2}x^2\,\mathrm{d}x = \dfrac{1}{3}\pi r^2 h

, the standard cone volume. Recovering such a known result at the end of a calculation is a reliable way to catch an error, and it connects the integral machinery to elementary mensuration.

V = \pi\int_a^{b} y^2\,\mathrm{d}x \quad (\text{about the } x\text{-axis}), \qquad V = \pi\int_c^{d} x^2\,\mathrm{d}y \quad (\text{about the } y\text{-axis})

Volumes of revolution

Each disc has area $\pi(\text{radius})^2$ ; integrate along the axis of rotation.

Worked example

Volume rotating a square-root curve

The region under $y = \sqrt{x}$ from $x = 0$ to $x = 4$ is rotated fully about the $x$ -axis. Find the exact volume.

Set up the integral
$V = \pi\displaystyle\int_0^{4} y^2\,\mathrm{d}x = \pi\int_0^{4} (\sqrt{x})^2\,\mathrm{d}x = \pi\int_0^{4} x\,\mathrm{d}x$ .
Integrate
$= \pi\left[\dfrac{x^2}{2}\right]_0^{4}$ .
$V = \pi\left[\frac{x^2}{2}\right]_0^{4}$
Evaluate
$= \pi\left(\dfrac{16}{2} - 0\right) = 8\pi$ .

Result: $V = 8\pi$ .

Typical mistakes

Integrating $\pi y$ instead of $\pi y^2$ — forgetting that the disc area is $\pi(\text{radius})^2$ .
Rotating about the wrong axis, or failing to change the limits when the curve is re-expressed or given parametrically.

Active revision

The region bounded by $y = \sqrt{x}$ , the $x$ -axis and $x = 4$ is rotated fully about the $x$ -axis. Find the exact volume of the solid formed.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Further Mathematics 7367 specification (AQA)

Arc length and surface area of revolution#

AdvancedAQA 7367 §3.5DfE Further Mathematics — Further calculus

The arc-length element

Fig. 3A small element of arc

\delta s

is the hypotenuse of a right triangle with legs

\delta x

and

\delta y

The length of a curve is found by adding up tiny straight pieces. A small element of arc has horizontal extent

\mathrm{d}x

and vertical extent

\mathrm{d}y

, so by Pythagoras its length is

\mathrm{d}s = \sqrt{\mathrm{d}x^2 + \mathrm{d}y^2}

. Dividing through by

\mathrm{d}x

gives the Cartesian formula

s = \displaystyle\int_a^{b} \sqrt{1 + \left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)^2}\,\mathrm{d}x

; the integrand is always at least

1

, reflecting that a curve is at least as long as its horizontal projection.

For a curve given parametrically, dividing

\mathrm{d}s

\mathrm{d}t

gives

s = \displaystyle\int_{t_1}^{t_2} \sqrt{\left(\dfrac{\mathrm{d}x}{\mathrm{d}t}\right)^2 + \left(\dfrac{\mathrm{d}y}{\mathrm{d}t}\right)^2}\,\mathrm{d}t

. The two forms are the same element

\mathrm{d}s

measured against different variables, and the parametric form is often the more convenient when the curve is naturally described by a parameter.

The area of the surface obtained by rotating an arc about the

x

-axis multiplies each element of arc by the circumference

2\pi y

of the circle it traces:

S = \displaystyle\int 2\pi y\,\mathrm{d}s = \int_a^{b} 2\pi y\,\sqrt{1 + \left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)^2}\,\mathrm{d}x

. It is essential to keep the arc-length element

\mathrm{d}s

inside the integral — the surface is swept by the arc, not by the horizontal strip, so

\mathrm{d}x

alone would understate the area.

In practice these integrals are often awkward, so questions are set on curves for which

1 + (\mathrm{d}y/\mathrm{d}x)^2

simplifies to a perfect square. A designed example is

y = \dfrac{x^2}{4} - \dfrac{\ln x}{2}

or curves involving

\cosh

, where the surd disappears and the integral becomes elementary. Recognising the algebra that makes the root simplify — and choosing the Cartesian or parametric form accordingly — is the examinable skill.

s = \int_a^{b} \sqrt{1 + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2}\,\mathrm{d}x, \qquad s = \int_{t_1}^{t_2}\sqrt{\left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)^2 + \left(\frac{\mathrm{d}y}{\mathrm{d}t}\right)^2}\,\mathrm{d}t

Arc length (Cartesian and parametric)

Both integrate the same element $\mathrm{d}s = \sqrt{\mathrm{d}x^2 + \mathrm{d}y^2}$ .

S = \int 2\pi y\,\mathrm{d}s = \int_a^{b} 2\pi y\,\sqrt{1 + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2}\,\mathrm{d}x

Surface area of revolution

Each arc element sweeps a circle of circumference $2\pi y$ .

Worked example

Arc length with a perfect-square integrand

Find the length of the arc of $y = \tfrac{2}{3}x^{3/2}$ from $x = 0$ to $x = 3$ .

Differentiate
$\dfrac{\mathrm{d}y}{\mathrm{d}x} = x^{1/2}$ , so $\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)^2 = x$ .
Form the integrand
$1 + \left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)^2 = 1 + x$ , so $s = \displaystyle\int_0^{3} \sqrt{1+x}\,\mathrm{d}x$ .
$s = \int_0^{3} (1+x)^{1/2}\,\mathrm{d}x$
Integrate
$= \left[\dfrac{2}{3}(1+x)^{3/2}\right]_0^{3} = \dfrac{2}{3}\big(4^{3/2} - 1^{3/2}\big)$ .
Evaluate
$= \dfrac{2}{3}(8 - 1) = \dfrac{14}{3}$ .

Result: $s = \dfrac{14}{3}$ .

Typical mistakes

Forgetting the $+1$ under the root, or squaring $\mathrm{d}y/\mathrm{d}x$ incorrectly.
Using $\mathrm{d}x$ in place of the arc element $\mathrm{d}s$ when computing a surface area.

Active revision

Find the arc length of the curve $y = \tfrac{2}{3}x^{3/2}$ from $x = 0$ to $x = 3$ .

Active recall

Recall the key points — then reveal.

Sources: Further mathematics: AS and A level content (GCE subject content) (Department for Education)

Reduction formulae#

AdvancedAQA 7367 §3.5DfE Further Mathematics — Further calculus

Powers of sine decrease in area

Fig. 4On

[0,\pi]

the graphs of

\sin^2 x

\sin^4 x

and

\sin^6 x

enclose progressively less area, so

I_n

decreases with

n

A reduction formula expresses an integral

I_n

depending on a positive integer

n

in terms of an integral of the same form but lower index, such as

I_{n-1}

I_{n-2}

. Applied repeatedly, it grinds a high-index integral down to a base case that can be evaluated directly. This turns an otherwise intractable family of integrals into a simple recursive computation.

The formula is almost always derived by integration by parts, with the choice of parts arranged so that one factor's index drops. For

I_n = \displaystyle\int x^n \mathrm{e}^x\,\mathrm{d}x

, taking

u = x^n

and

\mathrm{d}v = \mathrm{e}^x\mathrm{d}x

gives

I_n = x^n\mathrm{e}^x - n\displaystyle\int x^{n-1}\mathrm{e}^x\,\mathrm{d}x = x^n\mathrm{e}^x - nI_{n-1}

. Each application removes one power of

x

, and the recursion terminates at

I_0 = \int \mathrm{e}^x\,\mathrm{d}x = \mathrm{e}^x

For definite integrals the boundary term is evaluated at once, often vanishing and leaving a clean recursion. The famous case is

I_n = \displaystyle\int_0^{\pi/2} \sin^n x\,\mathrm{d}x

, for which integration by parts yields

I_n = \dfrac{n-1}{n}\,I_{n-2}

(Wallis's reduction), because the boundary term vanishes at both

0

and

\tfrac{\pi}{2}

. The two base cases

I_0 = \tfrac{\pi}{2}

and

I_1 = 1

then generate every value. Because

\sin^n x \leq \sin^{n-2} x

on the interval, the integrals decrease with

n

, which the figure of

\sin^2 x

\sin^4 x

\sin^6 x

makes visible.

The disciplined part of the work is bookkeeping. When deriving the formula, keep the boundary term

[uv]

separate and only discard it after confirming it is zero; when applying the recursion, track carefully whether

n

is even or odd, since the chain of

I_{n-2}

steps lands on a different base case in each parity. A neatly stated reduction formula, followed by an orderly descent to the base case, is exactly what earns full marks.

I_n = \int x^n\mathrm{e}^x\,\mathrm{d}x = x^n\mathrm{e}^x - n\,I_{n-1}

A reduction formula by parts

Integration by parts drops the power of $x$ by one each time; terminate at $I_0 = \mathrm{e}^x$ .

I_n = \int_0^{\pi/2}\sin^n x\,\mathrm{d}x = \frac{n-1}{n}\,I_{n-2}, \qquad I_0 = \tfrac{\pi}{2},\ I_1 = 1

Wallis's reduction

The boundary term vanishes; the recursion halves the index each step.

Worked example

Deriving and using a reduction formula

For $I_n = \displaystyle\int_0^{1} x^n\mathrm{e}^{x}\,\mathrm{d}x$ , show $I_n = \mathrm{e} - nI_{n-1}$ and evaluate $I_2$ .

Integrate by parts
$u = x^n$ , $\mathrm{d}v = \mathrm{e}^x\mathrm{d}x$ : $I_n = \big[x^n\mathrm{e}^x\big]_0^{1} - \displaystyle\int_0^{1} n x^{n-1}\mathrm{e}^x\,\mathrm{d}x$ .
Evaluate the boundary term
$\big[x^n\mathrm{e}^x\big]_0^{1} = 1^n\mathrm{e}^1 - 0 = \mathrm{e}$ , so $I_n = \mathrm{e} - n I_{n-1}$ .
$I_n = \mathrm{e} - n\,I_{n-1}$
Find the base case
$I_0 = \displaystyle\int_0^1 \mathrm{e}^x\,\mathrm{d}x = \mathrm{e} - 1$ ; then $I_1 = \mathrm{e} - 1\cdot I_0 = \mathrm{e} - (\mathrm{e}-1) = 1$ .
Compute I₂
$I_2 = \mathrm{e} - 2 I_1 = \mathrm{e} - 2$ .

Result: $I_n = \mathrm{e} - nI_{n-1}$ , and $I_2 = \mathrm{e} - 2 \approx 0.718$ .

Typical mistakes

Choosing the parts so that the index rises rather than falls, so the recursion never terminates.
Dropping a non-zero boundary term, or landing on the wrong base case by miscounting the steps of the recursion.

Active revision

Given $I_n = \displaystyle\int_0^{1} x^n \mathrm{e}^{x}\,\mathrm{d}x$ , show that $I_n = \mathrm{e} - nI_{n-1}$ and hence evaluate $I_2$ .

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Further Mathematics 7367 specification (AQA)

References & sources

Sources

Department for Education

Further mathematics: AS and A level content (GCE subject content)

AQA

AQA A-level Further Mathematics 7367 specification

Further calculus

Key points

A convergent improper integral

Key points

Volume rotating a square-root curve

Key points

Arc length with a perfect-square integrand

Key points

Deriving and using a reduction formula

Sources