| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Session | Specimen |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Numerical integration |
| Type | Trapezium rule accuracy improvement explanation |
| Difficulty | Standard +0.3 This is a standard trapezium rule question with routine integration. Part (a) is straightforward application of the formula with given values. Part (b) tests basic understanding of numerical methods (use more strips). Part (c) requires integration of x²ln(x) using integration by parts, which is a standard A-level technique. While multi-part, each component is textbook-standard with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.09f Trapezium rule: numerical integration |
| \(x\) | 1 | 1.5 | 2 | 2.5 | 3 |
| \(y\) | 3 | 2.3041 | 1.9242 | 1.9089 | 2.2958 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Uses or implies \(h = 0.5\) | B1 | |
| Correct form of trapezium rule | M1 | |
| \(\frac{0.5}{2}\{3 + 2.2958 + 2(2.3041 + 1.9242 + 1.9089)\} = 4.393\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Any valid statement, e.g. increase number of strips / decrease width of strips / use more trapezia | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Integration by parts on \(\int x^2 \ln x \, dx\) | M1 | |
| \(= \frac{x^3}{3}\ln x - \int \frac{x^2}{3} \, dx\) | A1 | |
| \(\int -2x + 5 \, dx = -x^2 + 5x \quad (+c)\) | B1 | |
| All integration attempted and limits used: \(\int_1^3 \frac{x^2 \ln x}{3} - 2x + 5 \, dx = \left[\frac{x^3}{9}\ln x - \frac{x^3}{27} - x^2 + 5x\right]_{x=1}^{x=3}\) | M1 | |
| Uses correct ln laws, simplifies and writes in required form | M1 | |
| Area of \(S = \frac{28}{27} + \ln 27\) \((a = 28, b = 27, c = 27)\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Strip width \(h = 0.5\) | B1 | Can be implied by sight of \(\frac{0.5}{2}\{...\}\) in trapezium rule |
| Correct bracket form: \(\{\text{first } y + \text{last } y + 2\times(\text{sum of other } y \text{ values})\}\) | M1 | Must use \(y\) values not \(x\) values |
| \(4.393\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| See scheme | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\int x^2 \ln x \, dx = Ax^3 \ln x - \int Bx^2 \, dx\) | M1 | Integration by parts the correct way around |
| \(\frac{x^3}{3}\ln x - \int \frac{x^2}{3} \, dx\) | A1 | |
| \(-2x+5\) integrates to \(-x^2+5x\) | B1 | |
| All integration completed and limits used | M1 | |
| Simplifies using ln laws to form \(\frac{a}{b} + \ln c\) | M1 | |
| \(\frac{28}{27} + \ln 27\) | A1 | Correct answer only |
## Question 14:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Uses or implies $h = 0.5$ | B1 | |
| Correct form of trapezium rule | M1 | |
| $\frac{0.5}{2}\{3 + 2.2958 + 2(2.3041 + 1.9242 + 1.9089)\} = 4.393$ | A1 | |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Any valid statement, e.g. increase number of strips / decrease width of strips / use more trapezia | B1 | |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Integration by parts on $\int x^2 \ln x \, dx$ | M1 | |
| $= \frac{x^3}{3}\ln x - \int \frac{x^2}{3} \, dx$ | A1 | |
| $\int -2x + 5 \, dx = -x^2 + 5x \quad (+c)$ | B1 | |
| All integration attempted and limits used: $\int_1^3 \frac{x^2 \ln x}{3} - 2x + 5 \, dx = \left[\frac{x^3}{9}\ln x - \frac{x^3}{27} - x^2 + 5x\right]_{x=1}^{x=3}$ | M1 | |
| Uses correct ln laws, simplifies and writes in required form | M1 | |
| Area of $S = \frac{28}{27} + \ln 27$ $(a = 28, b = 27, c = 27)$ | A1 | |
## Question 14 (continued):
### Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Strip width $h = 0.5$ | B1 | Can be implied by sight of $\frac{0.5}{2}\{...\}$ in trapezium rule |
| Correct bracket form: $\{\text{first } y + \text{last } y + 2\times(\text{sum of other } y \text{ values})\}$ | M1 | Must use $y$ values not $x$ values |
| $4.393$ | A1 | |
### Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| See scheme | B1 | |
### Part (c):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\int x^2 \ln x \, dx = Ax^3 \ln x - \int Bx^2 \, dx$ | M1 | Integration by parts the correct way around |
| $\frac{x^3}{3}\ln x - \int \frac{x^2}{3} \, dx$ | A1 | |
| $-2x+5$ integrates to $-x^2+5x$ | B1 | |
| All integration completed and limits used | M1 | |
| Simplifies using ln laws to form $\frac{a}{b} + \ln c$ | M1 | |
| $\frac{28}{27} + \ln 27$ | A1 | Correct answer only |
---14.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{f7994129-07ee-4f6d-9531-08a15a38b794-26_567_412_212_824}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
Figure 4 shows a sketch of part of the curve $C$ with equation
$$y = \frac { x ^ { 2 } \ln x } { 3 } - 2 x + 5 , \quad x > 0$$
The finite region $S$, shown shaded in Figure 4, is bounded by the curve $C$, the line with equation $x = 1$, the $x$-axis and the line with equation $x = 3$
The table below shows corresponding values of $x$ and $y$ with the values of $y$ given to 4 decimal places as appropriate.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | }
\hline
$x$ & 1 & 1.5 & 2 & 2.5 & 3 \\
\hline
$y$ & 3 & 2.3041 & 1.9242 & 1.9089 & 2.2958 \\
\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 $S$, giving your answer to 3 decimal places.
\item Explain how the trapezium rule could be used to obtain a more accurate estimate for the area of $S$.
\item Show that the exact area of $S$ can be written in the form $\frac { a } { b } + \ln c$, where $a , b$ and $c$ are integers to be found.\\
(In part c, solutions based entirely on graphical or numerical methods are not acceptable.)
\end{enumerate}
\hfill \mbox{\textit{Edexcel Paper 1 Q14 [10]}}