Evaluate _1^∞ (1)/(x^3)\,dx. [2 marks]

CCEA CCEA A2 2021-style practice: Find the Maclaurin series for ln(1 + 2x) up to and including the term in x^3.

Use the standard expansion ln(1 + u) = u - u^22 + u^33 - with u = 2x: ln(1 + 2x) = 2x - (2x)^22 + (2x)^33 - = 2x - 4x^22 + 8x^33 - = 2x - 2x^2 + 83x^3 - Markers reward using the standard ln(1 + u) series, substituting u = 2x, and the three correct terms up to x^3.

CCEA CCEA A2 2018-style practice: Evaluate the improper integral _1^∞ (1)/(x^2)\,dx, or state that it diverges. Then evaluate _0^1 (1)/(sqrt(x))\,dx.

For the first integral, replace the infinite limit by t and take the limit: _1^t x^-2\,dx = [-1x]_1^t = -1t + 1. As t → ∞, -(1)/(t) → 0, so the integral converges to 1. For the second integral, the integrand is unbounded at x = 0, so replace the lower limit by a > 0: _a^1 x^-1/2\,dx = [2x^1/2]_a^1 = 2 - 2sqrt(a). As a → 0^+, 2sqrt(a) → 0, so the integral converges to 2. Markers reward replacing the problem limit with a variable, the antiderivatives, taking the limit, and both values (1 and 2).

Northern IrelandFurther MathsSyllabus dot point

How do you build a Maclaurin series and evaluate improper integrals, arc lengths and surface areas?

Maclaurin series expansions of standard functions, improper integrals with infinite limits or discontinuities, the arc length of a curve, and the area of a surface of revolution.

A CCEA A2 Further Maths answer on Maclaurin series expansions of standard functions, evaluating improper integrals with infinite limits or discontinuities, and finding the arc length of a curve and the area of a surface of revolution.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-16

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

Have a quick question? Jump to the Q&A page

Jump to a section

What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

CCEA wants you to build a Maclaurin series for a function, evaluate improper integrals (with an infinite limit or an integrand that blows up) by taking a limit, and apply the formulae for the arc length of a curve and the area of a surface of revolution.

The answer

Maclaurin series

You either differentiate repeatedly and evaluate at $0$ , or substitute into a standard series.

Improper integrals

Arc length

Surface of revolution

The factor $2\pi y$ is the circumference of the circle traced by each point, and the surd is the slant length element.

An arc-length calculation

Find the length of the curve $y = \tfrac{1}{2}x^2$ from $x = 0$ to $x = 1$ , leaving the answer in terms of an inverse hyperbolic function.

step Differentiate and form the integrand

Here $\dfrac{dy}{dx} = x$ , so $1 + \left(\dfrac{dy}{dx}\right)^2 = 1 + x^2$ and

L = \int_{0}^{1} \sqrt{1 + x^2}\,dx.

step Use the standard result

The standard integral is $\displaystyle\int \sqrt{1 + x^2}\,dx = \tfrac{1}{2}\left[x\sqrt{1 + x^2} + \operatorname{arsinh} x\right]$ .

step Apply the limits

At $x = 1$ : $\tfrac{1}{2}\left[\sqrt{2} + \operatorname{arsinh} 1\right]$ . At $x = 0$ the expression is $0$ . So

L = \tfrac{1}{2}\left(\sqrt{2} + \operatorname{arsinh} 1\right) = \tfrac{1}{2}\left(\sqrt{2} + \ln(1 + \sqrt{2})\right) \approx 1.15.

The inverse hyperbolic functions from the previous topic reappear naturally in arc-length work.

Examples in context

Example 1. Calculators approximating functions. A calculator finds $\sin(0.1)$ from the Maclaurin series $x - \frac{x^3}{6} + \cdots$ ; for small $x$ a couple of terms already give many correct digits. Series are how transcendental functions are actually computed.

Example 2. The surface area of a satellite dish. A parabolic dish is a surface of revolution, and the material needed is its surface area, $\int 2\pi y\sqrt{1 + (y')^2}\,dx$ . The same integral gives the area of any rotationally symmetric shell.

Try this

Q1. Write down the Maclaurin series for $e^x$ up to the $x^2$ term. [1 mark]

Cue. $1 + x + \frac{x^2}{2}$ .

Q2. Evaluate $\displaystyle\int_{1}^{\infty} \frac{1}{x^3}\,dx$ . [2 marks]

Cue. $\left[-\frac{1}{2x^2}\right]_1^{\infty} = 0 + \frac{1}{2} = \frac{1}{2}$ .

Q3. State the arc-length formula for $y = f(x)$ from $a$ to $b$ . [1 mark]

Cue. $\int_a^b \sqrt{1 + (y')^2}\,dx$ .

Exam-style practice questions

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

CCEA A2 20216 marksFind the Maclaurin series for

\ln(1 + 2x)

up to and including the term in

x^3

Show worked answer →

Use the standard expansion $\ln(1 + u) = u - \dfrac{u^2}{2} + \dfrac{u^3}{3} - \cdots$ with $u = 2x$ :

$\ln(1 + 2x) = 2x - \dfrac{(2x)^2}{2} + \dfrac{(2x)^3}{3} - \cdots$

$= 2x - \dfrac{4x^2}{2} + \dfrac{8x^3}{3} - \cdots = 2x - 2x^2 + \dfrac{8}{3}x^3 - \cdots$

Markers reward using the standard $\ln(1 + u)$ series, substituting $u = 2x$ , and the three correct terms up to $x^3$ .

CCEA A2 20186 marksEvaluate the improper integral

\displaystyle\int_{1}^{\infty} \frac{1}{x^2}\,dx

, or state that it diverges. Then evaluate

\displaystyle\int_{0}^{1} \frac{1}{\sqrt{x}}\,dx

Show worked answer →

For the first integral, replace the infinite limit by $t$ and take the limit:

$\int_{1}^{t} x^{-2}\,dx = \left[-\dfrac{1}{x}\right]_{1}^{t} = -\dfrac{1}{t} + 1.$

As $t \to \infty$ , $-\frac{1}{t} \to 0$ , so the integral converges to $1$ .

For the second integral, the integrand is unbounded at $x = 0$ , so replace the lower limit by $a > 0$ :

$\int_{a}^{1} x^{-1/2}\,dx = \left[2x^{1/2}\right]_{a}^{1} = 2 - 2\sqrt{a}.$

As $a \to 0^+$ , $2\sqrt{a} \to 0$ , so the integral converges to $2$ .

Markers reward replacing the problem limit with a variable, the antiderivatives, taking the limit, and both values ( $1$ and $2$ ).

Related dot points

Sources & how we know this

CCEA GCE Further Mathematics specification — CCEA (2018)