| Exam Board | OCR |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2013 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Numerical integration |
| Type | Rectangle bounds for definite integral |
| Difficulty | Challenging +1.2 This is a standard Riemann sum question requiring students to set up lower/upper bounds using rectangles and find when their difference is sufficiently small. While it involves the composition ln(ln(x)) which may seem unusual, the technique is routine for FP2 students who have practiced numerical integration bounds. Part (iii) requires iteration but is computational rather than conceptually demanding. |
| Spec | 1.08d Evaluate definite integrals: between limits |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Width of rectangles is \(\frac{3}{n}\) | B1 | Statement about width |
| \(\Rightarrow\) Sum of areas of rectangles \(= \frac{3}{n}\times\left(\ln(\ln 3)+\ln\left(\ln\left(3+\frac{3}{n}\right)\right)+\ldots\right)\) | M1 | Height or area of at least one rectangle |
| \(= \frac{3}{n}\times\sum_{r=0}^{n-1}\ln\left(\ln\left(3+\frac{3r}{n}\right)\right)\) | A1 | Correct conclusion www |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(= \frac{3}{n}\times\sum_{r=1}^{n}\ln\left(\ln\left(3+\frac{3r}{n}\right)\right)\) | B1 | |
| [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(U - L = \frac{3}{n}\times\ln(\ln 6) - \frac{3}{n}\times\ln(\ln 3)\) | M1* | Subtraction to obtain the difference of two terms |
| \(= \frac{3}{n}(\ln(\ln 6)-\ln(\ln 3)) = \frac{3}{n}\ln\left(\frac{\ln 6}{\ln 3}\right)\) | A1 | |
| \(\Rightarrow n > \frac{3}{0.001}\ln\left(\frac{\ln 6}{\ln 3}\right) \Rightarrow n > \frac{3}{0.001}\times\ln(1.6309)\) | *M1 | Dealing with inequality to obtain \(n\); dep on first M |
| \(\Rightarrow\) least \(n = 1468\) | A1 | Accept \(n \geq 1468\) or \(n > 1467\) |
| [4] |
# Question 6(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Width of rectangles is $\frac{3}{n}$ | B1 | Statement about width |
| $\Rightarrow$ Sum of areas of rectangles $= \frac{3}{n}\times\left(\ln(\ln 3)+\ln\left(\ln\left(3+\frac{3}{n}\right)\right)+\ldots\right)$ | M1 | Height or area of at least one rectangle |
| $= \frac{3}{n}\times\sum_{r=0}^{n-1}\ln\left(\ln\left(3+\frac{3r}{n}\right)\right)$ | A1 | Correct conclusion **www** |
| **[3]** | | |
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# Question 6(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $= \frac{3}{n}\times\sum_{r=1}^{n}\ln\left(\ln\left(3+\frac{3r}{n}\right)\right)$ | B1 | |
| **[1]** | | |
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# Question 6(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $U - L = \frac{3}{n}\times\ln(\ln 6) - \frac{3}{n}\times\ln(\ln 3)$ | M1* | Subtraction to obtain the difference of two terms |
| $= \frac{3}{n}(\ln(\ln 6)-\ln(\ln 3)) = \frac{3}{n}\ln\left(\frac{\ln 6}{\ln 3}\right)$ | A1 | |
| $\Rightarrow n > \frac{3}{0.001}\ln\left(\frac{\ln 6}{\ln 3}\right) \Rightarrow n > \frac{3}{0.001}\times\ln(1.6309)$ | *M1 | Dealing with inequality to obtain $n$; dep on first M |
| $\Rightarrow$ least $n = 1468$ | A1 | Accept $n \geq 1468$ or $n > 1467$ |
| **[4]** | | |
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\includegraphics[max width=\textwidth, alt={}, center]{a80eb21f-c273-4b65-8617-16cdee783305-4_656_1017_251_525}
The diagram shows part of the curve $y = \ln ( \ln ( x ) )$. The region between the curve and the $x$-axis for $3 \leqslant x \leqslant 6$ is shaded.\\
(i) By considering $n$ rectangles of equal width, show that a lower bound, $L$, for the area of the shaded region is $\frac { 3 } { n } \sum _ { r = 0 } ^ { n - 1 } \ln \left( \ln \left( 3 + \frac { 3 r } { n } \right) \right)$.\\
(ii) By considering another set of $n$ rectangles of equal width, find a similar expression for an upper bound, $U$, for the area of the shaded region.\\
(iii) Find the least value of $n$ for which $U - L < 0.001$.
\hfill \mbox{\textit{OCR FP2 2013 Q6 [8]}}