| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2011 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Substitution |
| Type | Algebraic manipulation before substitution |
| Difficulty | Standard +0.3 This is a standard C4 integration question with two straightforward methods. Part (i) requires routine substitution with clear guidance, and part (ii) uses difference of squares rationalisation—both are textbook techniques. The algebraic manipulation is mechanical rather than requiring insight, making this slightly easier than average. |
| Spec | 1.02b Surds: manipulation and rationalising denominators1.08h Integration by substitution |
| Answer | Marks | Guidance |
|---|---|---|
| Attempt to connect \(dx\) and \(du\) | M1 | Including \(\frac{du}{dt} = \) or \(du = ...dx\) ; not \(dx = du\) |
| \(5 - x = 4 - u^2\) | B1 | perhaps in conjunction with next line |
| Show \(\int \frac{4-u^2}{2+u}2u\,du\) reduced to \(\int 4u - 2u^2\,du\) AG | A1 | In a fully satisfactory & acceptable manner |
| Clear explanation of why limits change | B1 | e.g. when \(x = 2, u = 1\) and when \(x = 5, u = 2\) |
| \(\frac{4}{3}\) | B1 5 | not dependent on any of first 4 marks |
| (ii)(a) \(5 - x\) | \*B1 1 | Accept \(4 - x - 1 = 5 - x\) (this is not AG) |
| (ii)(b) Show reduction to \(2 - \sqrt{x-1}\) | dep\*B1 | |
| \(\int \sqrt{x-1}\,dx = \frac{2}{3}(x-1)^{\frac{3}{2}}\) | B1 | Indep of other marks, seen anywhere in (b) |
| \((10 - \frac{2}{3} \cdot 8) - (4 - \frac{2}{3}) = \frac{4}{3}\) or \(4\frac{2}{3} - 3\frac{1}{3} = \frac{4}{3}\) | B1 3 | Working must be shown |
**(i)**
Attempt to connect $dx$ and $du$ | M1 | Including $\frac{du}{dt} = $ or $du = ...dx$ ; not $dx = du$
$5 - x = 4 - u^2$ | B1 | perhaps in conjunction with next line
Show $\int \frac{4-u^2}{2+u}2u\,du$ reduced to $\int 4u - 2u^2\,du$ AG | A1 | In a fully satisfactory & acceptable manner
Clear explanation of why limits change | B1 | e.g. when $x = 2, u = 1$ and when $x = 5, u = 2$
$\frac{4}{3}$ | B1 5 | not dependent on any of first 4 marks
**(ii)(a)** $5 - x$ | \*B1 1 | Accept $4 - x - 1 = 5 - x$ (this is not AG)
**(ii)(b)** Show reduction to $2 - \sqrt{x-1}$ | dep\*B1 |
$\int \sqrt{x-1}\,dx = \frac{2}{3}(x-1)^{\frac{3}{2}}$ | B1 | Indep of other marks, seen anywhere in (b)
$(10 - \frac{2}{3} \cdot 8) - (4 - \frac{2}{3}) = \frac{4}{3}$ or $4\frac{2}{3} - 3\frac{1}{3} = \frac{4}{3}$ | B1 3 | Working must be shown
---5 In this question, $I$ denotes the definite integral $\int _ { 2 } ^ { 5 } \frac { 5 - x } { 2 + \sqrt { x - 1 } } \mathrm {~d} x$. The value of $I$ is to be found using two different methods.
\begin{enumerate}[label=(\roman*)]
\item Show that the substitution $u = \sqrt { x - 1 }$ transforms $I$ to $\int _ { 1 } ^ { 2 } \left( 4 u - 2 u ^ { 2 } \right) \mathrm { d } u$ and hence find the exact value of $I$.
\item (a) Simplify $( 2 + \sqrt { x - 1 } ) ( 2 - \sqrt { x - 1 } )$.\\
(b) By first multiplying the numerator and denominator of $\frac { 5 - x } { 2 + \sqrt { x - 1 } }$ by $2 - \sqrt { x - 1 }$, find the exact value of $I$.
\end{enumerate}
\hfill \mbox{\textit{OCR C4 2011 Q5 [9]}}