Find the volume generated when y = sqrt(x) for 0 x 4 is rotated about the x-axis. [3 marks]

Evaluate _1^∞(1)/(x^3)\,dx. [3 marks]

Pearson Edexcel Edexcel 2018-style practice: Evaluate _0^∞ x e^-x\,dx, or show that it diverges.

Set up the improper integral as a limit and use integration by parts. _0^∞ x e^-x\,dx = _t → ∞_0^t x e^-x\,dx (M1 for the limit set-up). By parts with u = x, (dv)/(dx) = e^-x: int x e^-x\,dx = -xe^-x - int -e^-x\,dx = -xe^-x - e^-x (M1 A1). Evaluating from 0 to t: [-xe^-x - e^-x]_0^t = (-te^-t - e^-t) - (0 - 1) = 1 - te^-t - e^-t (A1). As t → ∞, both te^-t → 0 and e^-t → 0, so the integral converges to 1 (A1).

Pearson Edexcel Edexcel 2023-style practice: Find int (1)/(sqrt(9 + x^2))\,dx and find the mean value of f(x) = x^2 over the interval [0, 3].

Recognise the standard inverse-hyperbolic integral, then apply the mean value formula. The standard result is int (1)/(sqrt(a^2 + x^2))\,dx = arsinh(x)/(a) + c. With a = 3: int (1)/(sqrt(9 + x^2))\,dx = arsinh(x)/(3) + c (M1 A1). Mean value: f = (1)/(b-a)_a^b f\,dx = (1)/(3)_0^3 x^2\,dx (M1). = (1)/(3)[(x^3)/(3)]_0^3 = (1)/(3) · 9 = 3 (A1 A1).

EnglandFurther MathsSyllabus dot point

How do you extend integration to improper integrals, volumes, arc lengths and mean values?

Improper integrals, volumes of revolution, the mean value of a function, integration using partial fractions, and the derivation of standard inverse trig and hyperbolic integrals.

A focused answer to the Edexcel A-Level Further Mathematics further calculus content, covering improper integrals evaluated as limits, volumes of revolution about both axes, the mean value of a function, integration by partial fractions, and the standard inverse trig and inverse hyperbolic integral results.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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Quick answer

An improper integral has an infinite limit or an unbounded integrand; evaluate it as a limit and state whether it converges. The volume of revolution about the $x$ -axis is $V = \pi\int_a^b y^2\,dx$ and about the $y$ -axis is $V = \pi\int_c^d x^2\,dy$ . The mean value of $f$ over $[a, b]$ is $\frac{1}{b-a}\int_a^b f(x)\,dx$ . Partial fractions split a rational function into simple pieces integrating to logarithms and arctangents, and standard results include $\int \frac{1}{a^2 + x^2}\,dx = \frac{1}{a}\arctan\frac{x}{a} + c$ and $\int \frac{1}{\sqrt{a^2 - x^2}}\,dx = \arcsin\frac{x}{a} + c$ .

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What this dot point is asking
Improper integrals
Volumes of revolution and mean value
Standard integrals and partial fractions
Examples in context
Try this

What this dot point is asking

Edexcel wants you to evaluate improper integrals as limits and state convergence, compute volumes of revolution about both axes, find the mean value of a function over an interval, integrate rational functions via partial fractions, and recognise the standard integrals that produce inverse trigonometric and inverse hyperbolic functions. These results are heavily used in differential equations and arc-length work, so fluency pays off across the whole paper.

Improper integrals

An integral is improper if one of its limits is infinite, or if the integrand becomes unbounded somewhere in the range (for example a vertical asymptote). You cannot simply substitute infinity, so you replace the offending limit with a variable, integrate, and take the limit. If the limit is finite the integral converges to that value; if it is infinite or undefined the integral diverges. State your conclusion explicitly, because examiners award a mark for it.

Evaluate an improper integral with an infinite limit

Evaluate $\displaystyle\int_1^{\infty} \frac{1}{x^2}\,dx$ .

Step 1: Replace the infinite limit with a variable

We cannot directly substitute infinity, so we introduce a parameter $t$ and agree to let $t \to \infty$ at the end. This converts the improper integral into an ordinary definite integral:

\int_1^{\infty} x^{-2}\,dx = \lim_{t \to \infty}\int_1^{t} x^{-2}\,dx.

Step 2: Integrate and evaluate the definite integral

Using the power rule $\int x^{-2}\,dx = -x^{-1}$ :

\lim_{t \to \infty}\left[-\frac{1}{x}\right]_1^{t} = \lim_{t \to \infty}\left(-\frac{1}{t} + 1\right).

Step 3: Take the limit and state convergence

As $t \to \infty$ , $\dfrac{1}{t} \to 0$ , so the expression tends to $1$ . The integral converges to $1$ .

Final answer: $\displaystyle\int_1^{\infty} \frac{1}{x^2}\,dx = 1$ .

Volumes of revolution and mean value

Rotating a curve about an axis sweeps out a solid whose cross-sections perpendicular to the axis are discs. Summing the disc volumes gives an integral.

Standard integrals and partial fractions

Once a rational function is split into partial fractions, each piece integrates to a logarithm or an arctangent. The formula booklet supplies the inverse trigonometric and inverse hyperbolic results, but you must recognise which one applies from the shape of the denominator: $a^2 + x^2$ gives arctan, $\sqrt{a^2 - x^2}$ gives arcsin, $\sqrt{x^2 + a^2}$ gives arsinh, and $\sqrt{x^2 - a^2}$ gives arcosh.

Integrate using partial fractions

Find $\displaystyle\int \frac{1}{x^2 - 1}\,dx$ .

Step 1: Factor the denominator and set up partial fractions

The denominator $x^2 - 1 = (x-1)(x+1)$ splits into two distinct linear factors, so the integrand decomposes as:

\frac{1}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1}.

Step 2: Find the constants $A$ and $B$

Multiply both sides by $(x-1)(x+1)$ :

1 = A(x+1) + B(x-1).

Setting $x = 1$ eliminates $B$ : $1 = 2A$ , so $A = \dfrac{1}{2}$ . Setting $x = -1$ eliminates $A$ : $1 = -2B$ , so $B = -\dfrac{1}{2}$ .

Step 3: Integrate each term

Each partial fraction integrates to a logarithm:

\int \frac{1}{x^2 - 1}\,dx = \frac{1}{2}\int\frac{1}{x-1}\,dx - \frac{1}{2}\int\frac{1}{x+1}\,dx = \frac{1}{2}\ln|x-1| - \frac{1}{2}\ln|x+1| + c.

Combining the logarithms:

= \frac{1}{2}\ln\left|\frac{x-1}{x+1}\right| + c.

Final answer: $\dfrac{1}{2}\ln\left|\dfrac{x-1}{x+1}\right| + c$ .

Examples in context

Further calculus tools recur throughout the course. The inverse-hyperbolic standard integrals connect directly to the hyperbolic-functions dot point, where $\operatorname{arsinh} x = \ln(x + \sqrt{x^2 + 1})$ gives the logarithmic form of the answer. Volumes of revolution combine with parametric and polar work, and the disc method extends to regions between two curves. Improper integrals appear when computing the total probability or expectation of a continuous distribution in Further Statistics, and the mean value of a function is the deterministic analogue of an expected value. Partial fractions are the standard preliminary to integrating rational right-hand sides in differential equations.

Try this

Q1. Find the volume generated when $y = \sqrt{x}$ for $0 \le x \le 4$ is rotated about the $x$ -axis. [3 marks]

Cue. $V = \pi\int_0^4 x\,dx = \pi\left[\frac{x^2}{2}\right]_0^4 = 8\pi$ .

Q2. Find the mean value of $f(x) = x^2$ on $[0, 3]$ . [3 marks]

Cue. $\bar{f} = \frac{1}{3}\int_0^3 x^2\,dx = \frac{1}{3}\cdot 9 = 3$ .

Q3. Evaluate $\displaystyle\int_1^{\infty}\frac{1}{x^3}\,dx$ . [3 marks]

Cue. $\lim_{t\to\infty}\left[-\frac{1}{2x^2}\right]_1^{t} = \frac{1}{2}$ ; converges.

Exam-style practice questions

Practice questions written in the style of Pearson Edexcel exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Edexcel 20185 marksEvaluate

\int_0^{\infty} x e^{-x}\,dx

, or show that it diverges.

Show worked answer →

Set up the improper integral as a limit and use integration by parts.

$\int_0^{\infty} x e^{-x}\,dx = \lim_{t \to \infty}\int_0^{t} x e^{-x}\,dx$ (M1 for the limit set-up).

By parts with $u = x$ , $\frac{dv}{dx} = e^{-x}$ : $\int x e^{-x}\,dx = -xe^{-x} - \int -e^{-x}\,dx = -xe^{-x} - e^{-x}$ (M1 A1).

Evaluating from $0$ to $t$ : $\big[-xe^{-x} - e^{-x}\big]_0^{t} = (-te^{-t} - e^{-t}) - (0 - 1) = 1 - te^{-t} - e^{-t}$ (A1).

As $t \to \infty$ , both $te^{-t} \to 0$ and $e^{-t} \to 0$ , so the integral converges to $1$ (A1).

Edexcel 20216 marksThe region

R

is bounded by

y = \frac{1}{\sqrt{x}}

, the

x

-axis, and the lines

x = 1

and

x = 4

. Find the exact volume generated when

R

is rotated through

2\pi

radians about the

x

-axis.

Show worked answer →

Use the volume of revolution formula about the $x$ -axis.

$V = \pi\int_1^4 y^2\,dx = \pi\int_1^4 \frac{1}{x}\,dx$ (M1 A1 for $y^2 = \frac{1}{x}$ ).

$= \pi\big[\ln x\big]_1^4$ (M1 A1).

$= \pi(\ln 4 - \ln 1) = \pi\ln 4 = 2\pi\ln 2$ (A1 A1 for exact simplified form).

Edexcel 20235 marksFind

\int \frac{1}{\sqrt{9 + x^2}}\,dx

and find the mean value of

f(x) = x^2

over the interval

[0, 3]

Show worked answer →

Recognise the standard inverse-hyperbolic integral, then apply the mean value formula.

The standard result is $\int \frac{1}{\sqrt{a^2 + x^2}}\,dx = \operatorname{arsinh}\frac{x}{a} + c$ . With $a = 3$ : $\int \frac{1}{\sqrt{9 + x^2}}\,dx = \operatorname{arsinh}\frac{x}{3} + c$ (M1 A1).

Mean value: $\bar{f} = \frac{1}{b-a}\int_a^b f\,dx = \frac{1}{3}\int_0^3 x^2\,dx$ (M1).

$= \frac{1}{3}\big[\frac{x^3}{3}\big]_0^3 = \frac{1}{3} \cdot 9 = 3$ (A1 A1).

Related dot points

Sources & how we know this

Pearson Edexcel A-Level Further Mathematics (9FM0) specification — Pearson Edexcel (2017)