WJEC Further Unit 4 2024 June — Question 7 12 marks

Exam BoardWJEC
ModuleFurther Unit 4 (Further Unit 4)
Year2024
SessionJune
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIntegration using inverse trig and hyperbolic functions
TypeMean value using inverse trig integral
DifficultyChallenging +1.8 This Further Maths question requires completing the square to recognize inverse trig integrals (arcsin), computing a volume of revolution with a rational integrand, and evaluating an improper integral with exponential decay. While the techniques are standard for Further Maths (arcsin integration, substitution for the improper integral), the multi-step nature, algebraic manipulation required, and combination of advanced calculus topics place it well above average difficulty.
Spec4.08c Improper integrals: infinite limits or discontinuous integrands4.08d Volumes of revolution: about x and y axes4.08e Mean value of function: using integral

  1. A curve C is defined by the equation \(y = \frac{1}{\sqrt{16-6x-x^2}}\) for \(-3 \leq x \leq 1\).
    1. Find the mean value of \(y = \frac{1}{\sqrt{16-6x-x^2}}\) between \(x = -3\) and \(x = 1\). [4]
    2. The region \(R\) is bounded by the curve C, the \(x\)-axis and the lines \(x = -3\) and \(x = 1\). Find the volume of the solid generated when \(R\) is rotated through four right-angles about the \(x\)-axis. [5]
  2. Evaluate the improper integral $$\int_1^{\infty} \frac{8e^{-2x}}{4e^{-2x} - 5} \mathrm{d}x,$$ giving your answer correct to three decimal places. [3]

Question 7:
AnswerMarks
712
Question 7:
7 | 12
\begin{enumerate}[label=(\alph*)]
\item A curve C is defined by the equation $y = \frac{1}{\sqrt{16-6x-x^2}}$ for $-3 \leq x \leq 1$.

\begin{enumerate}[label=(\roman*)]
\item Find the mean value of $y = \frac{1}{\sqrt{16-6x-x^2}}$ between $x = -3$ and $x = 1$. [4]

\item The region $R$ is bounded by the curve C, the $x$-axis and the lines $x = -3$ and $x = 1$. Find the volume of the solid generated when $R$ is rotated through four right-angles about the $x$-axis. [5]
\end{enumerate}

\item Evaluate the improper integral
$$\int_1^{\infty} \frac{8e^{-2x}}{4e^{-2x} - 5} \mathrm{d}x,$$
giving your answer correct to three decimal places. [3]
\end{enumerate}

\hfill \mbox{\textit{WJEC Further Unit 4 2024 Q7 [12]}}