| Exam Board | OCR MEI |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2019 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Numerical integration |
| Type | Over/underestimate justification with graph |
| Difficulty | Moderate -0.3 This is a straightforward trapezium rule application with standard interval width, requiring only routine calculation and basic understanding that the trapezium rule over-estimates for concave functions. The justification in part (b) requires recognizing the curve's shape but is a standard textbook concept, making this slightly easier than average. |
| Spec | 1.09f Trapezium rule: numerical integration |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{0.5}{2}\left[\sqrt{1+(-1)^3} + 2\sqrt{1+(-0.5)^3} + \sqrt{1+0^3}\right]\) oe | M1 | Condone omission of brackets; must be three terms in the bracket |
| \(\sqrt{1+(-0.5)^3}\) soi | B1 | NB \(\frac{\sqrt{14}}{4} = 0.9(35414346693)\) |
| \(0.717707\) cao | A1 [3] | NB \(\frac{2+\sqrt{14}}{8}\) unsupported implies M1B1; if unsupported allow SC3 for 0.717707 and SC2 for 0.717707173347 unsupported to 7 or more dp |
| Answer | Marks | Guidance |
|---|---|---|
| Under-estimate since curve is concave down/convex up oe | B1 [1] | Or eg the slant lines of both trapezia are entirely below the curve; allow annotated diagram with at least one trapezium; condone eg trapezium below curve; allow integral is 0.841309 BC so my answer is an underestimate |
## Question 4:
### Part (a):
$\frac{0.5}{2}\left[\sqrt{1+(-1)^3} + 2\sqrt{1+(-0.5)^3} + \sqrt{1+0^3}\right]$ oe | **M1** | Condone omission of brackets; must be three terms in the bracket
$\sqrt{1+(-0.5)^3}$ soi | **B1** | NB $\frac{\sqrt{14}}{4} = 0.9(35414346693)$
$0.717707$ cao | **A1 [3]** | NB $\frac{2+\sqrt{14}}{8}$ unsupported implies **M1B1**; if unsupported allow **SC3** for 0.717707 and **SC2** for 0.717707173347 unsupported to 7 or more dp
### Part (b):
Under-estimate since curve is concave down/convex up oe | **B1 [1]** | Or eg the slant lines of both trapezia are entirely below the curve; allow annotated diagram with at least one trapezium; condone eg trapezium below curve; allow integral is 0.841309 BC so my answer is an underestimate
---4 Fig. 4 shows the graph of $y = \sqrt { 1 + x ^ { 3 } }$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{95eb3bcc-6d3c-4f7e-9b27-5e046ab57ec5-05_544_639_338_248}
\captionsetup{labelformat=empty}
\caption{Fig. 4}
\end{center}
\end{figure}
\begin{enumerate}[label=(\alph*)]
\item Use the trapezium rule with $h = 0.5$ to find an estimate of $\int _ { - 1 } ^ { 0 } \sqrt { 1 + x ^ { 3 } } \mathrm {~d} x$, giving your answer correct to 6 decimal places.
\item State whether your answer to part (a) is an under-estimate or an over-estimate, justifying your answer.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 2 2019 Q4 [4]}}