| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2023 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Numerical integration |
| Type | Show trapezium rule gives specific value |
| Difficulty | Moderate -0.8 This is a straightforward trapezium rule question requiring only algebraic manipulation. Part (a) involves writing out the standard trapezium rule formula and simplifying to show the given result. Part (b) is a simple simultaneous equations problem (solving a+2b=51 and a+b+sum=97.2). No conceptual difficulty or problem-solving insight required—purely procedural application of a standard formula. |
| Spec | 1.09f Trapezium rule: numerical integration |
| \(x\) | 3 | 3.2 | 3.4 | 3.6 | 3.8 | 4 |
| \(y\) | \(a\) | 16.8 | \(b\) | 20.2 | 18.7 | 13.5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(h = 0.2\) | B1 | States or uses \(h = 0.2\) |
| \(\frac{1}{2} \times \text{"0.2"} \times \{a + 13.5 + 2(16.8 + b + 20.2 + 18.7)\} = 17.59\) | M1 | Forms the equation as shown. Condone copying slips. Minimum accept \(\text{"0.1"}\times\{a + 13.5 + 2(55.7 + b)\} = 17.59\). Allow adding individual trapezia areas. \(h\) must be numerical but condone \(h=1\) |
| \(a + 13.5 + 2b + 111.4 = 175.9 \Rightarrow a + 2b = 51\) | A1* | Rigorous argument leading to \(a + 2b = 51\) from correct working with no errors including brackets (missing trailing bracket condoned) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(a + 16.8 + b + 20.2 + 18.7 + 13.5 = 97.2 \Rightarrow a + b = 28\) | M1 | Attempts to form the equation \(a+16.8+b+20.2+18.7+13.5=97.2\), condoning copying errors, and attempts to solve simultaneously with given equation from (a). Score if values for \(a\) or \(b\) reached from a pair of simultaneous equations. |
| \(a = 5\) or \(b = 23\) | A1 | For either value |
| \(a = 5\) and \(b = 23\) | A1 | For both values |
## Question 5:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $h = 0.2$ | B1 | States or uses $h = 0.2$ |
| $\frac{1}{2} \times \text{"0.2"} \times \{a + 13.5 + 2(16.8 + b + 20.2 + 18.7)\} = 17.59$ | M1 | Forms the equation as shown. Condone copying slips. Minimum accept $\text{"0.1"}\times\{a + 13.5 + 2(55.7 + b)\} = 17.59$. Allow adding individual trapezia areas. $h$ must be numerical but condone $h=1$ |
| $a + 13.5 + 2b + 111.4 = 175.9 \Rightarrow a + 2b = 51$ | A1* | Rigorous argument leading to $a + 2b = 51$ from correct working with no errors including brackets (missing trailing bracket condoned) |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $a + 16.8 + b + 20.2 + 18.7 + 13.5 = 97.2 \Rightarrow a + b = 28$ | M1 | Attempts to form the equation $a+16.8+b+20.2+18.7+13.5=97.2$, condoning copying errors, and attempts to solve simultaneously with given equation from (a). Score if values for $a$ or $b$ reached from a pair of simultaneous equations. |
| $a = 5$ or $b = 23$ | A1 | For either value |
| $a = 5$ and $b = 23$ | A1 | For both values |
---\begin{enumerate}
\item A continuous curve has equation $y = \mathrm { f } ( x )$.
\end{enumerate}
The table shows corresponding values of $x$ and $y$ for this curve, where $a$ and $b$ are constants.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | }
\hline
$x$ & 3 & 3.2 & 3.4 & 3.6 & 3.8 & 4 \\
\hline
$y$ & $a$ & 16.8 & $b$ & 20.2 & 18.7 & 13.5 \\
\hline
\end{tabular}
\end{center}
The trapezium rule is used, with all the $y$ values in the table, to find an approximate area under the curve between $x = 3$ and $x = 4$
Given that this area is 17.59\\
(a) show that $a + 2 b = 51$
Given also that the sum of all the $y$ values in the table is 97.2\\
(b) find the value of $a$ and the value of $b$
\hfill \mbox{\textit{Edexcel Paper 1 2023 Q5 [6]}}