Standard +0.3 This question tests standard integration of 1/u and 1/u² forms with linear substitutions. Part (a) requires recognizing the logarithmic integral and showing the result simplifies to a constant (ln(3/5) or similar). Part (b) involves integrating to get -1/u form and showing the result is proportional to 1/k. Both parts are straightforward applications of reverse chain rule with algebraic manipulation, requiring more steps than basic integration but no novel insight—slightly easier than average A-level.
7. Given that \(k \in \mathbb { Z } ^ { + }\)
a. show that \(\int _ { 2 k } ^ { 3 k } \frac { 6 } { ( 7 k - 2 x ) } \mathrm { d } x\) is independent of \(k\),
b. show that \(\int _ { k } ^ { 2 k } \frac { 2 } { 3 ( 2 x - k ) ^ { 2 } } \mathrm {~d} x\) is inversely proportional to \(k\).
### Part a:
$-3\ln\left(\frac{1}{3}\right)$ or $\ln 27$ | A1 | Uses correct ln work and notation to show that $I = -3\ln\left(\frac{1}{3}\right)$ or $\ln 27$ i.e. independent of $k$
| M1 | For substituting $3k$ and $2k$ into their $A\ln(7k - 2x)$ and subtracting either way around.
| A1 | $\int \frac{6}{(7k-2x)} dx = -3\ln(7k - 2x)$
| M1 | $\int \frac{6}{(7k-2x)} dx = A\ln(7k - 2x)$
### Part b:
Shows that it is inversely proportional to $k$ | A1 |
| M1 | For substituting $2k$ and $k$ into their $\frac{B}{(2x-k)}$ and subtracting
| M1 | $\int \frac{2}{3(2x-k)^2} dx = \frac{B}{(2x-k)}$
7. Given that $k \in \mathbb { Z } ^ { + }$\\
a. show that $\int _ { 2 k } ^ { 3 k } \frac { 6 } { ( 7 k - 2 x ) } \mathrm { d } x$ is independent of $k$,\\
b. show that $\int _ { k } ^ { 2 k } \frac { 2 } { 3 ( 2 x - k ) ^ { 2 } } \mathrm {~d} x$ is inversely proportional to $k$.\\
\hfill \mbox{\textit{Edexcel PMT Mocks Q7 [7]}}