| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2021 |
| Session | October |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Numerical integration |
| Type | Trapezium rule then exact integration comparison |
| Difficulty | Standard +0.3 This is a standard two-part integration question combining numerical and algebraic methods. Part (a) is routine trapezium rule application with given values. Part (b) requires integration by parts twice for (ln x)², which is a well-practiced A-level technique, though the algebraic manipulation to reach the required form adds minor complexity. Overall slightly easier than average due to clear structure and standard methods. |
| Spec | 1.02z Models in context: use functions in modelling |
| \(x\) | 2 | 2.5 | 3 | 3.5 | 4 |
| \(y\) | 0.4805 | 0.8396 | 1.2069 | 1.5694 | 1.9218 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(h = 0.5\) | B1 | Correct strip width; may be implied by \(\frac{1}{2} \times \frac{1}{2}\{...\}\) or \(\frac{1}{4}\{...\}\) |
| \(A \approx \frac{1}{2} \times \frac{1}{2}\{0.4805 + 1.9218 + 2(0.8396 + 1.2069 + 1.5694)\}\) | M1 | Correct application of trapezium rule; look for \(\frac{1}{2} \times "h"\{0.4805 + 1.9218 + 2(0.8396 + 1.2069 + 1.5694)\}\) condoning slips in digits |
| \(= 2.41\) | A1 | 2.41 only, not awrt; bracketing must be correct but implied by awrt 2.41 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int (\ln x)^2\, dx = x(\ln x)^2 - \int x \times \frac{2\ln x}{x}\, dx\) | M1 | Attempts parts the correct way round to achieve \(\alpha x(\ln x)^2 - \beta \int \ln x\, dx\) |
| A1 | Correct expression, may be unsimplified | |
| \(= x(\ln x)^2 - 2\int \ln x\, dx = x(\ln x)^2 - 2(x\ln x - \int dx)\) \(= x(\ln x)^2 - 2\int \ln x\, dx = x(\ln x)^2 - 2x\ln x + 2x\) | dM1 | Attempts parts again (condone coefficient errors) to achieve \(\alpha x(\ln x)^2 - \beta x\ln x \pm \gamma x\) |
| \(\int_2^4 (\ln x)^2\, dx = \left[x(\ln x)^2 - 2x\ln x + 2x\right]_2^4\) \(= 4(\ln 4)^2 - 2\times 4\ln 4 + 2\times 4 - (2(\ln 2)^2 - 2\times 2\ln 2 + 2\times 2)\) \(= 4(2\ln 2)^2 - 16\ln 2 + 8 - 2(\ln 2)^2 + 4\ln 2 - 4\) | ddM1 | Applies limits 4 and 2 to expression of form \(\pm\alpha x(\ln x)^2 \pm \beta x\ln x \pm \gamma x\); uses \(\ln 4 = 2\ln 2\) at least once; both M's must have been awarded |
| \(= 14(\ln 2)^2 - 12\ln 2 + 4\) | A1 | Correct answer |
# Question 11:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $h = 0.5$ | B1 | Correct strip width; may be implied by $\frac{1}{2} \times \frac{1}{2}\{...\}$ or $\frac{1}{4}\{...\}$ |
| $A \approx \frac{1}{2} \times \frac{1}{2}\{0.4805 + 1.9218 + 2(0.8396 + 1.2069 + 1.5694)\}$ | M1 | Correct application of trapezium rule; look for $\frac{1}{2} \times "h"\{0.4805 + 1.9218 + 2(0.8396 + 1.2069 + 1.5694)\}$ condoning slips in digits |
| $= 2.41$ | A1 | 2.41 only, not awrt; bracketing must be correct but implied by awrt 2.41 |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int (\ln x)^2\, dx = x(\ln x)^2 - \int x \times \frac{2\ln x}{x}\, dx$ | M1 | Attempts parts the correct way round to achieve $\alpha x(\ln x)^2 - \beta \int \ln x\, dx$ |
| | A1 | Correct expression, may be unsimplified |
| $= x(\ln x)^2 - 2\int \ln x\, dx = x(\ln x)^2 - 2(x\ln x - \int dx)$ $= x(\ln x)^2 - 2\int \ln x\, dx = x(\ln x)^2 - 2x\ln x + 2x$ | dM1 | Attempts parts again (condone coefficient errors) to achieve $\alpha x(\ln x)^2 - \beta x\ln x \pm \gamma x$ |
| $\int_2^4 (\ln x)^2\, dx = \left[x(\ln x)^2 - 2x\ln x + 2x\right]_2^4$ $= 4(\ln 4)^2 - 2\times 4\ln 4 + 2\times 4 - (2(\ln 2)^2 - 2\times 2\ln 2 + 2\times 2)$ $= 4(2\ln 2)^2 - 16\ln 2 + 8 - 2(\ln 2)^2 + 4\ln 2 - 4$ | ddM1 | Applies limits 4 and 2 to expression of form $\pm\alpha x(\ln x)^2 \pm \beta x\ln x \pm \gamma x$; uses $\ln 4 = 2\ln 2$ at least once; both M's must have been awarded |
| $= 14(\ln 2)^2 - 12\ln 2 + 4$ | A1 | Correct answer |
---11.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{08ede5ea-85e9-44eb-be6a-5878096734e2-34_705_837_248_614}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows a sketch of part of the curve with equation
$$y = ( \ln x ) ^ { 2 } \quad x > 0$$
The finite region $R$, shown shaded in Figure 2, is bounded by the curve, the line with equation $x = 2$, the $x$-axis and the line with equation $x = 4$
The table below shows corresponding values of $x$ and $y$, with the values of $y$ given to 4 decimal places.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | }
\hline
$x$ & 2 & 2.5 & 3 & 3.5 & 4 \\
\hline
$y$ & 0.4805 & 0.8396 & 1.2069 & 1.5694 & 1.9218 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Use the trapezium rule, with all the values of $y$ in the table, to obtain an estimate for the area of $R$, giving your answer to 3 significant figures.
\item Use algebraic integration to find the exact area of $R$, giving your answer in the form
$$y = a ( \ln 2 ) ^ { 2 } + b \ln 2 + c$$
where $a$, $b$ and $c$ are integers to be found.
\end{enumerate}
\hfill \mbox{\textit{Edexcel Paper 1 2021 Q11 [8]}}