| Exam Board | WJEC |
|---|---|
| Module | Further Unit 4 (Further Unit 4) |
| Session | Specimen |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration using inverse trig and hyperbolic functions |
| Type | Improper integral to infinity with inverse trig |
| Difficulty | Standard +0.8 This is a Further Maths question on improper integrals requiring standard techniques: part (a) is routine integration with limits to infinity, while part (b) requires the substitution u = ln x and recognizing the resulting integral diverges. Both parts test understanding of convergence but use well-established methods without requiring novel insight. |
| Spec | 4.08c Improper integrals: infinite limits or discontinuous integrands |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int_0^{\infty} \frac{dx}{(1+x)^3} = -\frac{1}{4}\left[\frac{1}{(1+x)^4}\right]_0^{\infty} = -\frac{1}{4}(0-1) = \frac{1}{4}\) | M1, A1, A1 | AO1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(du = \frac{dx}{x}\); \([2,\infty) \to [\ln 2, \infty)\) | B1, M1 | AO1 |
| \(\int_{\ln 2}^{\infty} \frac{du}{u}\) | A1 | AO1 |
| \(= [\ln u]_{\ln 2}^{\infty} \to \infty\) because \(\ln u \to \infty\) | A1 | AO1 |
## Part (a)
$\int_0^{\infty} \frac{dx}{(1+x)^3} = -\frac{1}{4}\left[\frac{1}{(1+x)^4}\right]_0^{\infty} = -\frac{1}{4}(0-1) = \frac{1}{4}$ | M1, A1, A1 | AO1
## Part (b)
$du = \frac{dx}{x}$; $[2,\infty) \to [\ln 2, \infty)$ | B1, M1 | AO1
$\int_{\ln 2}^{\infty} \frac{du}{u}$ | A1 | AO1
$= [\ln u]_{\ln 2}^{\infty} \to \infty$ because $\ln u \to \infty$ | A1 | AO1
**Total: [7]**
---\begin{enumerate}[label=(\alph*)]
\item Evaluate the integral
$$\int_0^{\infty} \frac{dx}{(1+x)^5}.$$ [3]
\item By putting $u = \ln x$, determine whether or not the following integral has a finite value.
$$\int_2^{\infty} \frac{dx}{x \ln x}.$$ [4]
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 4 Q1 [7]}}