| Exam Board | SPS |
|---|---|
| Module | SPS FM Pure (SPS FM Pure) |
| Year | 2024 |
| Session | June |
| Marks | 7 |
| Topic | Parametric integration |
| Type | Parametric area under curve |
| Difficulty | Standard +0.3 This is a straightforward parametric area question requiring standard techniques: finding the domain from ln(t+2), then computing area using the formula ∫y(dx/dt)dt with appropriate limits. The integration is simple (1/(t+1) × 1/(t+2)) and can be done with partial fractions or substitution. While it requires multiple steps, each is routine for Further Maths students and the answer is given to verify the working. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.08a Fundamental theorem of calculus: integration as reverse of differentiation1.08e Area between curve and x-axis: using definite integrals |
13.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{ace492d8-1dd0-401e-af74-505ca19d5e9c-28_583_917_155_676}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
Figure 4 shows a sketch of the curve $C$ with parametric equations
$$x = \ln ( t + 2 ) , y = \frac { 1 } { t + 1 } , \quad t > - \frac { 2 } { 3 }$$
\begin{enumerate}[label=(\alph*)]
\item State the domain of values of $x$ for the curve $C$.
The finite region $R$, shown shaded in Figure 4, is bounded by the curve $C$, the line with equation $x = \ln 2$, the $x$-axis and the line with equation $x = \ln 4$
\item Use calculus to show that the area of $R$ is $\ln \left( \frac { 3 } { 2 } \right)$.\\[0pt]
\end{enumerate}
\hfill \mbox{\textit{SPS SPS FM Pure 2024 Q13 [7]}}