| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2018 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Standard Integrals and Reverse Chain Rule |
| Type | Definite integral with logarithmic form |
| Difficulty | Standard +0.3 This question requires standard integration techniques (logarithmic and power rule with reverse chain rule) applied to two definite integrals. Part (a) involves recognizing ln|3x-k| and showing the result simplifies to ln(8), while part (b) requires integrating (2x-k)^{-2} and showing the result equals 1/k. Both parts are straightforward applications of A-level integration with algebraic simplification, slightly above average due to the parameter k and the 'show that' proof structure requiring clear working. |
| Spec | 1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)1.08d Evaluate definite integrals: between limits |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int \frac{2}{(3x-k)}\,dx = A\ln(3x-k)\) | M1 | Condone missing bracket; allow recovery if \(A\ln 9k-k \to A\ln 8k\) in subsequent work |
| \(\int \frac{2}{(3x-k)}\,dx = \frac{2}{3}\ln(3x-k)\) | A1 | |
| \(\int_k^{3k} \frac{2}{(3x-k)}\,dx = \frac{2}{3}\ln(9k-k)-\frac{2}{3}\ln(3k-k)\) | dM1 | Substituting \(k\) and \(3k\) and subtracting |
| \(= \frac{2}{3}\ln\!\left(\frac{8k}{2k}\right) = \frac{2}{3}\ln 4\) | A1 | Uses correct ln work to show result independent of \(k\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int \frac{2}{(2x-k)^2}\,dx = \frac{C}{(2x-k)}\) | M1 | |
| \(\int_k^{2k} \frac{2}{(2x-k)^2}\,dx = -\frac{1}{(4k-k)}+\frac{1}{(2k-k)}\) | dM1 | Substituting \(k\) and \(2k\) and subtracting |
| \(= \frac{2}{3k}\) \(\left(\propto \frac{1}{k}\right)\) | A1 | Shows inversely proportional to \(k\); answer of form \(\frac{A}{k}\) with \(A=\frac{2}{3}\); accept from \(-\frac{1}{(3k)}+\frac{1}{(k)}=\left(-\frac{1}{3}+1\right)\times\frac{1}{k} \propto \frac{1}{k}\) |
# Question 7:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int \frac{2}{(3x-k)}\,dx = A\ln(3x-k)$ | M1 | Condone missing bracket; allow recovery if $A\ln 9k-k \to A\ln 8k$ in subsequent work |
| $\int \frac{2}{(3x-k)}\,dx = \frac{2}{3}\ln(3x-k)$ | A1 | |
| $\int_k^{3k} \frac{2}{(3x-k)}\,dx = \frac{2}{3}\ln(9k-k)-\frac{2}{3}\ln(3k-k)$ | dM1 | Substituting $k$ and $3k$ and subtracting |
| $= \frac{2}{3}\ln\!\left(\frac{8k}{2k}\right) = \frac{2}{3}\ln 4$ | A1 | Uses correct ln work to show result independent of $k$ |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int \frac{2}{(2x-k)^2}\,dx = \frac{C}{(2x-k)}$ | M1 | |
| $\int_k^{2k} \frac{2}{(2x-k)^2}\,dx = -\frac{1}{(4k-k)}+\frac{1}{(2k-k)}$ | dM1 | Substituting $k$ and $2k$ and subtracting |
| $= \frac{2}{3k}$ $\left(\propto \frac{1}{k}\right)$ | A1 | Shows inversely proportional to $k$; answer of form $\frac{A}{k}$ with $A=\frac{2}{3}$; accept from $-\frac{1}{(3k)}+\frac{1}{(k)}=\left(-\frac{1}{3}+1\right)\times\frac{1}{k} \propto \frac{1}{k}$ |
---\begin{enumerate}
\item Given that $k \in \mathbb { Z } ^ { + }$\\
(a) show that $\int _ { k } ^ { 3 k } \frac { 2 } { ( 3 x - k ) } \mathrm { d } x$ is independent of $k$,\\
(b) show that $\int _ { k } ^ { 2 k } \frac { 2 } { ( 2 x - k ) ^ { 2 } } \mathrm {~d} x$ is inversely proportional to $k$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel Paper 1 2018 Q7 [7]}}