| Exam Board | SPS |
|---|---|
| Module | SPS SM Pure (SPS SM Pure) |
| Year | 2024 |
| Session | February |
| Marks | 10 |
| Topic | Integration by Substitution |
| Type | Show substitution transforms integral, then evaluate |
| Difficulty | Standard +0.8 This is a structured integration problem requiring substitution followed by partial fractions. While the substitution x = u² + 1 is given (removing discovery difficulty), students must correctly handle dx/du, transform limits, apply partial fractions to 6/[u(3+2u)], and simplify logarithms to reach a rational answer. The multi-step nature and algebraic manipulation elevate this above routine integration exercises, but the scaffolding and standard techniques keep it from being exceptionally difficult. |
| Spec | 1.02y Partial fractions: decompose rational functions1.08h Integration by substitution1.08j Integration using partial fractions |
14.
\begin{enumerate}[label=(\alph*)]
\item Use the substitution $x = u ^ { 2 } + 1$ to show that
$$\int _ { 5 } ^ { 10 } \frac { 3 \mathrm {~d} x } { ( x - 1 ) ( 3 + 2 \sqrt { x - 1 } ) } = \int _ { p } ^ { q } \frac { 6 \mathrm {~d} u } { u ( 3 + 2 u ) }$$
where $p$ and $q$ are positive constants to be found.
\item Hence, using algebraic integration, show that
$$\int _ { 5 } ^ { 10 } \frac { 3 \mathrm {~d} x } { ( x - 1 ) ( 3 + 2 \sqrt { x - 1 } ) } = \ln a$$
where $a$ is a rational constant to be found.\\
(6)
Use this page for any additional working.
Use this page for any additional working.
Use this page for any additional working.
Use this page for any additional working.
\end{enumerate}
\hfill \mbox{\textit{SPS SPS SM Pure 2024 Q14 [10]}}