AQA AQA 2020-style practice: Evaluate the improper integral _0^∞ x e^-x\,dx, showing the limiting process clearly.

Replace the infinite limit with a variable: _0^∞ x e^-x\,dx = _t → ∞ _0^t x e^-x\,dx. Integrate by parts with u = x, (dv)/(dx) = e^-x, so v = -e^-x: int x e^-x\,dx = -x e^-x + int e^-x\,dx = -x e^-x - e^-x = -(x + 1)e^-x. Evaluate between 0 and t: [-(x+1)e^-x]_0^t = -(t+1)e^-t + 1. As t → ∞, (t+1)e^-t → 0 (the exponential dominates), so the integral converges to 1. Markers reward the limit statement, integration by parts, and showing the boundary term vanishes so the integral converges.

EnglandFurther MathsSyllabus dot point

How do you find improper integrals, arc lengths, surface areas and the mean value of a function, and how do you use the Maclaurin series?

Improper integrals, volumes of revolution, mean value of a function, arc length, surface area of revolution, integration using partial fractions and the Maclaurin series of standard functions.

A focused answer to the AQA A-Level Further Mathematics further calculus content, covering improper integrals, volumes of revolution, the mean value of a function, arc length, surface area of revolution, integration with partial fractions and the Maclaurin series of standard functions.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-02

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

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What this dot point is asking
Improper integrals
Volumes of revolution
Mean value of a function
Arc length and surface area
Integration using partial fractions
The Maclaurin series
Common traps

What this dot point is asking

AQA wants you to evaluate improper integrals using limits, find volumes of revolution about both axes, calculate the mean value of a function, find arc lengths and surface areas of revolution, integrate using partial fractions, and write down and use the Maclaurin series of standard functions.

Improper integrals

An integral is improper if a limit is infinite or the integrand is undefined somewhere in the range. You rewrite the problem with a parameter and take a limit.

Evaluate an improper integral

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

Step 1: Replace the infinite limit with a variable

An improper integral with an infinite upper limit cannot be evaluated directly. We replace $\infty$ with a finite variable $t$ and take the limit as $t \to \infty$ ; this makes the limiting process explicit, which AQA requires.

\int_1^{\infty} \frac{1}{x^2}\,dx = \lim_{t \to \infty} \int_1^{t} x^{-2}\,dx

Step 2: Integrate and apply the limits

Integrate $x^{-2}$ using the power rule to get $-x^{-1}$ , then evaluate between 1 and $t$ .

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

Step 3: Take the limit and state convergence

As $t \to \infty$ , the term $\frac{1}{t} \to 0$ , so the expression tends to $1$ . Because the limit exists and is finite, the integral converges.

\int_1^{\infty} \frac{1}{x^2}\,dx = 1

Final answer: The integral converges to $1$ .

Volumes of revolution

Rotating the region under $y = f(x)$ between $x = a$ and $x = b$ through $2\pi$ about the $x$ axis sweeps out a solid of volume $V = \pi\int_a^b y^2\,dx$ , summing thin discs of radius $y$ and thickness $dx$ . About the $y$ axis the volume is $V = \pi\int_c^d x^2\,dy$ , where the limits are now $y$ values. For a region between two curves you subtract the inner volume from the outer.

Mean value of a function

Find the mean value of a function

Find the mean value of $f(x) = x^2$ over the interval $[0, 3]$ .

Step 1: Set up the mean value integral

The mean value formula averages the function over the interval by dividing the area under the curve by the width of the interval. Identify $a = 0$ , $b = 3$ and write the formula with the specific function and limits substituted.

\bar{f} = \frac{1}{b - a}\int_a^b f(x)\,dx = \frac{1}{3 - 0}\int_0^3 x^2\,dx

Step 2: Evaluate the definite integral

Integrate $x^2$ using the power rule and apply the limits. The lower limit contributes zero.

\int_0^3 x^2\,dx = \left[\frac{x^3}{3}\right]_0^3 = \frac{27}{3} - 0 = 9

Step 3: Divide by the interval length to get the mean value

The $\frac{1}{b - a}$ factor scales the integral down to the average height. Geometrically this is the height of the rectangle with base $[0, 3]$ that has the same area as the region under the parabola.

\bar{f} = \frac{1}{3} \times 9 = 3

Final answer: The mean value of $f(x) = x^2$ over $[0, 3]$ is $3$ .

Arc length and surface area

For a curve $y = f(x)$ the arc length from $x = a$ to $x = b$ is $\int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2}\,dx$ . Rotating that arc about the $x$ axis produces a surface of area $\int_a^b 2\pi y \sqrt{1 + \left(\frac{dy}{dx}\right)^2}\,dx$ . For curves given parametrically by $x(t)$ and $y(t)$ , use $\int \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\,dt$ .

Integration using partial fractions

Splitting a rational function into partial fractions turns one hard integral into a sum of standard logarithm and arctangent integrals. A linear denominator factor gives a logarithm, an irreducible quadratic such as $x^2 + a^2$ gives an arctangent, and a repeated linear factor gives a reciprocal-power term. Decompose first, then integrate each piece separately and combine the logarithms at the end.

Integrate using partial fractions

Find $\displaystyle\int \frac{1}{x(x+1)}\,dx$ .

Step 1: Decompose into partial fractions

A product of distinct linear factors in the denominator can be split into separate fractions, each with a simpler denominator. This turns one difficult integral into a sum of standard logarithm integrals.

\frac{1}{x(x+1)} = \frac{1}{x} - \frac{1}{x+1}

Step 2: Integrate each term separately

Each term now has the standard form $\frac{1}{ax + b}$ , which integrates to a natural logarithm.

\int \left(\frac{1}{x} - \frac{1}{x+1}\right)dx = \ln|x| - \ln|x+1| + c

Step 3: Combine the logarithms

Using the logarithm subtraction rule, the two log terms can be written as a single logarithm of a fraction.

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

Final answer: $\displaystyle\int \frac{1}{x(x+1)}\,dx = \ln\left|\dfrac{x}{x+1}\right| + c$ .

The Maclaurin series

Common traps

These techniques recur across a paper: an improper integral may appear as a limit of a volume, an arc length may need a partial-fraction integral, and a Maclaurin series may be integrated term by term to approximate an awkward definite integral. Always state convergence explicitly for any improper integral, keep the squared derivative under the root for arc length, and divide each Maclaurin coefficient by the correct factorial.

Exam-style practice questions

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

AQA 20205 marksEvaluate the improper integral

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

, showing the limiting process clearly.

Show worked answer →

Replace the infinite limit with a variable: $\int_0^{\infty} x e^{-x}\,dx = \lim_{t \to \infty} \int_0^{t} x e^{-x}\,dx$ .

Integrate by parts with $u = x$ , $\frac{dv}{dx} = e^{-x}$ , so $v = -e^{-x}$ :
$\int x e^{-x}\,dx = -x e^{-x} + \int e^{-x}\,dx = -x e^{-x} - e^{-x} = -(x + 1)e^{-x}$ .

Evaluate between $0$ and $t$ : $\left[-(x+1)e^{-x}\right]_0^{t} = -(t+1)e^{-t} + 1$ .

As $t \to \infty$ , $(t+1)e^{-t} \to 0$ (the exponential dominates), so the integral converges to $1$ .

Markers reward the limit statement, integration by parts, and showing the boundary term vanishes so the integral converges.

AQA 20226 marksFind the volume generated when the region under

y = \sqrt{x}

between

x = 0

and

x = 4

is rotated through

2\pi

radians about the

x

axis.

Show worked answer →

The volume of revolution about the $x$ axis is $V = \pi \int_a^b y^2\,dx$ .

Here $y^2 = x$ , so $V = \pi \int_0^4 x\,dx$ .

Integrate: $V = \pi \left[\frac{x^2}{2}\right]_0^4 = \pi\left(\frac{16}{2} - 0\right) = 8\pi$ .

Markers reward the correct formula with $y^2$ , the substitution $y^2 = x$ , correct integration, and the answer $8\pi$ (cubic units).

Related dot points

Sources & how we know this

AQA A-level Further Mathematics (7367) specification — AQA (2017)