| Exam Board | Edexcel |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2021 |
| Session | January |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Numerical integration |
| Type | Complete table then apply trapezium rule |
| Difficulty | Moderate -0.3 This is a straightforward application of the trapezium rule with 5 ordinates. Part (a) requires simple substitution into a given formula, and part (b) is a standard trapezium rule calculation with all values provided or easily computed. The context is applied but the mathematics is routine P2 content with no conceptual challenges. |
| Spec | 1.09f Trapezium rule: numerical integration |
| 10 | 10.5 | 11 | 11.5 | 12 | ||
| 1.882 | 2.45 | 2.95 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((t=10, P=)\ 1.2\) or \(\frac{6}{5}\) | B1 | Allow 1.200. May be seen in table or working for part (b) |
| \((t=11.5, P=)\ 2.821\) | B1 | awrt 2.821. May be seen in table or working for part (b) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(h = 0.5\) | B1 | Seen or implied |
| \(E \approx \left(\frac{1}{2} \times \frac{1}{2}\right)\left[1.2 + 2.95 + 2(1.882 + 2.45 + 2.821)\right]\) | M1 | Correct bracket structure: first \(P\) value plus last \(P\) value, inner bracket multiplied by 2 containing remaining \(P\) values. Allow recovery of invisible brackets. M0 if \(t\) values used instead of \(P\) values |
| A1ft | Correct bracket \([\ldots]\) following through their values from (a) | |
| \(E \approx 4.61\) (kWh) | A1 | cao, must be to 2 d.p. (exact answer via integration is 4.643439737) |
## Question 7:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(t=10, P=)\ 1.2$ or $\frac{6}{5}$ | B1 | Allow 1.200. May be seen in table or working for part (b) |
| $(t=11.5, P=)\ 2.821$ | B1 | awrt 2.821. May be seen in table or working for part (b) |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $h = 0.5$ | B1 | Seen or implied |
| $E \approx \left(\frac{1}{2} \times \frac{1}{2}\right)\left[1.2 + 2.95 + 2(1.882 + 2.45 + 2.821)\right]$ | M1 | Correct bracket structure: first $P$ value plus last $P$ value, inner bracket multiplied by 2 containing remaining $P$ values. Allow recovery of invisible brackets. M0 if $t$ values used instead of $P$ values |
| | A1ft | Correct bracket $[\ldots]$ following through their values from (a) |
| $E \approx 4.61$ (kWh) | A1 | cao, must be to 2 d.p. (exact answer via integration is 4.643439737) |
---7.
Figure 1
Solar panels are installed on the roof of a building.
The power, $P$, produced on a particular day, in kW , can be modelled by the equation
$$P = 0.95 + 2 ^ { t - 12 } + 2 ^ { 12 - t } - ( t - 12 ) ^ { 2 } \quad 8.5 \leqslant t \leqslant 15.2$$
where $t$ is the time in hours after midnight. The graph of $P$ against $t$ is shown in Figure 1.
A table of values of $t$ and $P$ is shown below, with the values of $P$ given to 4 significant figures where appropriate.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | }
\hline
\begin{tabular}{ l }
Time, $t$ \\
(hours) \\
\end{tabular} & 10 & 10.5 & 11 & 11.5 & 12 \\
\hline
\begin{tabular}{ l }
Power, $P$ \\
(kW) \\
\end{tabular} & & 1.882 & 2.45 & & 2.95 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Use the given equation to complete the table, giving the values of $P$ to 4 significant figures where appropriate.
The amount of energy, in kWh , produced between 10:00 and 12:00 can be found by calculating the area of region $R$, shown shaded in Figure 1.
\item Use the trapezium rule, with all the values of $P$ in the completed table, to find an estimate for the amount of energy produced between 10:00 and 12:00. Give your answer to 2 decimal places.\\
7.\\
\includegraphics[max width=\textwidth, alt={}, center]{52c90d0e-a5e4-45fa-95a4-9523287e7588-20_769_1038_116_450}
\end{enumerate}
\hfill \mbox{\textit{Edexcel P2 2021 Q7 [6]}}