| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2022 |
| Session | June |
| Marks | 3 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Standard Integrals and Reverse Chain Rule |
| Type | Definite integral with logarithmic form |
| Difficulty | Moderate -0.8 This is a straightforward question testing the definition of a definite integral and basic integration of 1/x. Part (a) requires recognizing the Riemann sum notation, and part (b) involves routine integration of 2/x to get 2ln(x) and evaluating limits. The calculation is direct with no problem-solving insight needed, making it easier than average but not trivial since it requires understanding the limit-sum-integral connection. |
| Spec | 1.07l Derivative of ln(x): and related functions1.08d Evaluate definite integrals: between limits1.08g Integration as limit of sum: Riemann sums |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\lim_{\delta x \to 0} \sum_{x=2.1}^{6.3} \dfrac{2}{x}\,\delta x = \displaystyle\int_{2.1}^{6.3} \dfrac{2}{x}\,\mathrm{d}x\) | B1 | States \(\displaystyle\int_{2.1}^{6.3}\frac{2}{x}\,\mathrm{d}x\) or equivalent such as \(2\displaystyle\int_{2.1}^{6.3}x^{-1}\,\mathrm{d}x\) but must include limits and the \(\mathrm{d}x\). Condone \(\mathrm{d}x \leftrightarrow \delta x\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(= \big[2\ln x\big]_{2.1}^{6.3} = 2\ln 6.3 - 2\ln 2.1\) | M1 | Know that \(\int\frac{1}{x}\,\mathrm{d}x=\ln x\) and attempts to apply limits (either way). Condone \(\int\frac{2}{x}\,\mathrm{d}x = p\ln x\) (including \(p=1\)) or \(p\ln qx\) as long as limits applied. Also accept \(\ln x^2\), \(2\ln |
| \(= \ln 9\) (CSO) | A1 | Also accept \(\ln 3^2\) so \(k=9\) is fine. Condone \(\ln |
## Question 4(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\lim_{\delta x \to 0} \sum_{x=2.1}^{6.3} \dfrac{2}{x}\,\delta x = \displaystyle\int_{2.1}^{6.3} \dfrac{2}{x}\,\mathrm{d}x$ | B1 | States $\displaystyle\int_{2.1}^{6.3}\frac{2}{x}\,\mathrm{d}x$ or equivalent such as $2\displaystyle\int_{2.1}^{6.3}x^{-1}\,\mathrm{d}x$ but must include limits and the $\mathrm{d}x$. Condone $\mathrm{d}x \leftrightarrow \delta x$ |
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## Question 4(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $= \big[2\ln x\big]_{2.1}^{6.3} = 2\ln 6.3 - 2\ln 2.1$ | M1 | Know that $\int\frac{1}{x}\,\mathrm{d}x=\ln x$ and attempts to apply limits (either way). Condone $\int\frac{2}{x}\,\mathrm{d}x = p\ln x$ (including $p=1$) or $p\ln qx$ as long as limits applied. Also accept $\ln x^2$, $2\ln|x|+c$, $2\ln cx$. $[p\ln x]_{2.1}^{6.3}=p\ln6.3-p\ln2.1$ is sufficient |
| $= \ln 9$ (CSO) | A1 | Also accept $\ln 3^2$ so $k=9$ is fine. Condone $\ln|9|$. Do not accept $\ln\frac{39.69}{4.41}$. Solutions from rounded decimals should not score final mark — it is a "show that" question |
---\begin{enumerate}
\item (a) Express $\lim _ { \delta x \rightarrow 0 } \sum _ { x = 2.1 } ^ { 6.3 } \frac { 2 } { x } \delta x$ as an integral.\\
(b) Hence show that
\end{enumerate}
$$\lim _ { \delta x \rightarrow 0 } \sum _ { x = 2.1 } ^ { 6.3 } \frac { 2 } { x } \delta x = \ln k$$
where $k$ is a constant to be found.
\hfill \mbox{\textit{Edexcel Paper 1 2022 Q4 [3]}}