| Exam Board | SPS |
|---|---|
| Module | SPS SM Pure (SPS SM Pure) |
| Year | 2021 |
| Session | May |
| Marks | 14 |
| Topic | Integration by Substitution |
| Type | Show integral transforms via substitution then evaluate (trigonometric/Weierstrass) |
| Difficulty | Challenging +1.3 This is a structured multi-part calculus question requiring second derivatives to find inflection points, setting up an integral with reversed variables (integrating with respect to y), applying a trigonometric substitution, and evaluating the result. While it involves several techniques and the substitution is non-standard, each part is heavily scaffolded with clear instructions ('show that'), making it more methodical than insightful. The techniques are all A-level standard but combined in a moderately sophisticated way. |
| Spec | 1.07p Points of inflection: using second derivative1.08h Integration by substitution |
\begin{enumerate}[label=(\roman*)]
\item Show that the two non-stationary points of inflection on the curve $y = \ln(1 + 4x^2)$ are at $x = \pm\frac{1}{2}$. [6]
\end{enumerate}
\includegraphics{figure_9}
The diagram shows the curve $y = \ln(1 + 4x^2)$. The shaded region is bounded by the curve and a line parallel to the $x$-axis which meets the curve where $x = \frac{1}{2}$ and $x = -\frac{1}{2}$.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Show that the area of the shaded region is given by
$$\int_0^{\ln 2} \sqrt{e^y - 1} \, dy.$$ [3]
\item Show that the substitution $e^y = \sec^2\theta$ transforms the integral in part (ii) to $\int_0^{\frac{\pi}{4}} 2\tan^2\theta \, d\theta$. [2]
\item Hence find the exact area of the shaded region. [3]
\end{enumerate}
\hfill \mbox{\textit{SPS SPS SM Pure 2021 Q9 [14]}}