| Exam Board | WJEC |
|---|---|
| Module | Further Unit 4 (Further Unit 4) |
| Year | 2023 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration using inverse trig and hyperbolic functions |
| Type | Mean value using inverse trig integral |
| Difficulty | Challenging +1.2 This is a Further Maths question requiring integration of 1/√(x²+4x+3). Part (a) needs completing the square and a standard inverse sinh substitution to find the mean value (integral/width). Part (b) requires volume of revolution where the integrand simplifies nicely to 1/(x²+4x+3) = partial fractions leading to logarithms. While technically demanding with multiple techniques (completing square, hyperbolic/trig substitution, partial fractions), these are standard Further Maths methods applied in a straightforward multi-step context without requiring novel insight. |
| Spec | 4.08d Volumes of revolution: about x and y axes4.08e Mean value of function: using integral |
| Answer | Marks | Guidance |
|---|---|---|
| a) | Mean value \(= \frac{1}{2-0}\int_0^2\frac{1}{\sqrt{x^2+4x+3}}dx\) | M1 |
| \(\frac{1}{2}\int_0^2\frac{1}{\sqrt{(x+2)^2-1}}dx\) | m1, A1 | Completing the square; Must include \(1/(b-a)\) |
| Correct integration of \(\frac{1}{\sqrt{x^2\pm a^2}}\) | ||
| \(= \frac{1}{2}\left[\cosh^{-1}\left(\frac{x+2}{1}\right)\right]_0^2\) | A1 | or \(\frac{1}{2}\ln\left(x + 2 + \sqrt{(x+2)^2-1}\right)\) |
| or \(\frac{1}{2}\ln(x + 2 + \sqrt{x^2+4x+3})\) | ||
| \(= 0.373\) | A1 | cao |
| (5) | No marks awarded for answer only. | |
| b) | METHOD 1: | |
| Volume \(= \pi\int_0^2\left(\frac{1}{\sqrt{x^2+4x+3}}\right)^2dx\) | M1 | Condone omission of \(\pi\) |
| \(= \pi\int_0^2\frac{1}{x^2+4x+3}dx\) | ||
| \(= \pi\int_0^2\frac{1}{(x+2)^2-1}dx\) | A1 | Fully correct; Must include \(\pi\) and limits |
| \(= \frac{\pi}{2}\left[\ln\left | \frac{x+2-1}{x+2+1}\right | \right]_0^2\) |
| A1 | Correct integration | |
| \(= \frac{\pi}{2}\left[\ln\left | \frac{x+1}{x+3}\right | \right]_0^2\) |
| \(= \frac{\pi}{2}\left[\ln\frac{3}{5} - \ln\frac{1}{3}\right]\) | m1 | Correct use of limits |
| A1 | oe Must be exact; Mark final answer | |
| \(= \frac{\pi}{2}\ln\frac{9}{5}\) | ||
| [11] |
a) | Mean value $= \frac{1}{2-0}\int_0^2\frac{1}{\sqrt{x^2+4x+3}}dx$ | M1 | Condone omission of $1/(b-a)$ |
| | | |
| $\frac{1}{2}\int_0^2\frac{1}{\sqrt{(x+2)^2-1}}dx$ | m1, A1 | Completing the square; Must include $1/(b-a)$ |
| | | Correct integration of $\frac{1}{\sqrt{x^2\pm a^2}}$ |
| $= \frac{1}{2}\left[\cosh^{-1}\left(\frac{x+2}{1}\right)\right]_0^2$ | A1 | or $\frac{1}{2}\ln\left(x + 2 + \sqrt{(x+2)^2-1}\right)$ |
| | | or $\frac{1}{2}\ln(x + 2 + \sqrt{x^2+4x+3})$ |
| | | |
| $= 0.373$ | A1 | cao |
| | (5) | No marks awarded for answer only. |
b) | **METHOD 1:** | | |
| Volume $= \pi\int_0^2\left(\frac{1}{\sqrt{x^2+4x+3}}\right)^2dx$ | M1 | Condone omission of $\pi$ |
| | | |
| $= \pi\int_0^2\frac{1}{x^2+4x+3}dx$ | | |
| | | |
| $= \pi\int_0^2\frac{1}{(x+2)^2-1}dx$ | A1 | Fully correct; Must include $\pi$ and limits |
| | | |
| $= \frac{\pi}{2}\left[\ln\left|\frac{x+2-1}{x+2+1}\right|\right]_0^2$ | m1 | Limits not required |
| | A1 | Correct integration |
| $= \frac{\pi}{2}\left[\ln\left|\frac{x+1}{x+3}\right|\right]_0^2$ | | |
| | | |
| $= \frac{\pi}{2}\left[\ln\frac{3}{5} - \ln\frac{1}{3}\right]$ | m1 | Correct use of limits |
| | A1 | oe Must be exact; Mark final answer |
| $= \frac{\pi}{2}\ln\frac{9}{5}$ | | |
| | [11] | |The function $f$ is defined by
$$f(x) = \frac{1}{\sqrt{x^2 + 4x + 3}}.$$
\begin{enumerate}[label=(\alph*)]
\item Find the mean value of the function $f$ for $0 \leqslant x \leqslant 2$, giving your answer correct to three decimal places. [5]
\item The region $R$ is bounded by the curve $y = f(x)$, the $x$-axis and the lines $x = 0$ and $x = 2$. Find the exact value of the volume of the solid generated when $R$ is rotated through four right angles about the $x$-axis. [6]
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 4 2023 Q8 [11]}}