OCR FP2 2008 June — Question 9 12 marks

Exam BoardOCR
ModuleFP2 (Further Pure Mathematics 2)
Year2008
SessionJune
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIntegration by Parts
TypeImproper integral with parts
DifficultyChallenging +1.3 This is a multi-part FP2 question requiring integration by parts (standard technique), understanding of Riemann sums vs integrals (conceptual but well-scaffolded), and numerical calculation. Part (i) is routine integration by parts. Parts (ii)(a-b) require geometric reasoning about upper/lower sums but the diagram and structure guide students clearly. Part (ii)(c) combines results algebraically. While this is Further Maths content and requires careful multi-step work, the scaffolding is strong and each component uses standard techniques without requiring novel insight.
Spec1.08g Integration as limit of sum: Riemann sums1.08i Integration by parts4.08d Volumes of revolution: about x and y axes

9
  1. Prove that \(\int _ { 0 } ^ { N } \ln ( 1 + x ) \mathrm { d } x = ( N + 1 ) \ln ( N + 1 ) - N\), where \(N\) is a positive constant.
  2. \includegraphics[max width=\textwidth, alt={}, center]{63a316f6-1c18-4224-930f-0b58112c9f71-4_616_1261_406_482} The diagram shows the curve \(y = \ln ( 1 + x )\), for \(0 \leqslant x \leqslant 70\), together with a set of rectangles of unit width.
    1. By considering the areas of these rectangles, explain why $$\ln 2 + \ln 3 + \ln 4 + \ldots + \ln 70 < \int _ { 0 } ^ { 70 } \ln ( 1 + x ) d x$$
    2. By considering the areas of another set of rectangles, show that $$\ln 2 + \ln 3 + \ln 4 + \ldots + \ln 70 > \int _ { 0 } ^ { 69 } \ln ( 1 + x ) d x$$
    3. Hence find bounds between which \(\ln ( 70 ! )\) lies. Give the answers correct to 1 decimal place.

AnswerMarks
(i) Attempt use of quotient rule*M1
Obtain \(\frac{75 - 15x^2}{(x^2 + 5)^2}\)A1
Equate attempt at first derivative to zero and rearrange to solvable formM1
Obtain \(x = \sqrt{5}\) or \(2.24\)A1
Recognise range as values less than y-coord of st ptM1
Obtain \(0 \leq y \leq \frac{3}{5}\sqrt{5}\)A1
(ii) State \(\sqrt{5}\)B1√
(iii) Equate attempt at first derivative to \(-1\) and attempt simplification*M1
Obtain \(x^4 - 5x^2 + 100 = 0\)A1
Attempt evaluation of discriminant or equivM1
Obtain \(-375\) or equiv and conclude appropriatelyA1
4724 Core Mathematics 4
| | |
|---|---|
| (i) Attempt use of quotient rule | *M1 |
| Obtain $\frac{75 - 15x^2}{(x^2 + 5)^2}$ | A1 |
| Equate attempt at first derivative to zero and rearrange to solvable form | M1 |
| Obtain $x = \sqrt{5}$ or $2.24$ | A1 |
| Recognise range as values less than y-coord of st pt | M1 |
| Obtain $0 \leq y \leq \frac{3}{5}\sqrt{5}$ | A1 |
| (ii) State $\sqrt{5}$ | B1√ |
| (iii) Equate attempt at first derivative to $-1$ and attempt simplification | *M1 |
| Obtain $x^4 - 5x^2 + 100 = 0$ | A1 |
| Attempt evaluation of discriminant or equiv | M1 |
| Obtain $-375$ or equiv and conclude appropriately | A1 |

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# 4724 Core Mathematics 4
9 (i) Prove that $\int _ { 0 } ^ { N } \ln ( 1 + x ) \mathrm { d } x = ( N + 1 ) \ln ( N + 1 ) - N$, where $N$ is a positive constant.\\
(ii)\\
\includegraphics[max width=\textwidth, alt={}, center]{63a316f6-1c18-4224-930f-0b58112c9f71-4_616_1261_406_482}

The diagram shows the curve $y = \ln ( 1 + x )$, for $0 \leqslant x \leqslant 70$, together with a set of rectangles of unit width.
\begin{enumerate}[label=(\alph*)]
\item By considering the areas of these rectangles, explain why

$$\ln 2 + \ln 3 + \ln 4 + \ldots + \ln 70 < \int _ { 0 } ^ { 70 } \ln ( 1 + x ) d x$$
\item By considering the areas of another set of rectangles, show that

$$\ln 2 + \ln 3 + \ln 4 + \ldots + \ln 70 > \int _ { 0 } ^ { 69 } \ln ( 1 + x ) d x$$
\item Hence find bounds between which $\ln ( 70 ! )$ lies. Give the answers correct to 1 decimal place.
\end{enumerate}

\hfill \mbox{\textit{OCR FP2 2008 Q9 [12]}}