Standard +0.3 This question requires recognizing the reverse chain rule pattern, performing a straightforward integration to get -1/(2x+k), evaluating at limits, and then algebraically simplifying to show the result is proportional to 1/k. While it involves multiple steps and requires some algebraic manipulation, the integration technique is standard and the 'show that' format provides clear direction, making it slightly easier than average.
11 Given that \(k\) is a positive constant, show that \(\int _ { k } ^ { 2 k } \frac { 2 } { ( 2 x + k ) ^ { 2 } } d x\) is inversely proportional to \(k\).
Substituting \(u = 2x + k\). Allow for a different substitution giving an integral in \(u\) only. Allow \(\int \frac{a}{u^2}\,du\) for any constant seen
Substituting correct new limits into their integrated expression, or substituting in terms of \(x\) and using original limits
Alternatively, by inspection: \(\int_k^{2k} \frac{2}{(2x+k)^2}\,dx = \left[-(2x+k)^{-1}\right]_k^{2k}\)
M1
Integrating by inspection to obtain any multiple of \((2x+k)^{-1}\)
(fully correct indefinite integral)
A2
Fully correct indefinite integral – need not be simplified
\(-\frac{1}{5k} + \frac{1}{3k}\)
M1
Substituting limits into their integrated expression
\(= \frac{2}{15k}\)
A1
Allow \(\left(-\frac{1}{5} + \frac{1}{3}\right)\frac{1}{k}\) seen
This is inversely proportional to \(k\) [with constant of proportionality \(\frac{2}{15}\)]
E1
FT their definite integral. Must use phrase "inversely proportional" to \(k\) or indicate \(\propto \frac{1}{k}\). Allow if \(\frac{a}{k}\) required at the start of the argument
[6]
## Question 11:
| Answer | Marks | Guidance |
|--------|-------|----------|
| Let $u = 2x + k$, $2\,dx = du$ | M1 | Substituting $u = 2x + k$. Allow for a different substitution giving an integral in $u$ only. Allow $\int \frac{a}{u^2}\,du$ for any constant seen |
| $\int \frac{2}{(2x+k)^2}\,dx = \int \frac{1}{u^2}\,du$ | A1 | Correct integrand in terms of $u$. Ignore limits |
| $= -\frac{1}{u}\ [+c]$ | A1 | Correct indefinite integral; constant need not be seen |
| $\int_k^{2k} \frac{2}{(2x+k)^2}\,dx = \int_{3k}^{5k}\left(\frac{1}{u^2}\right)du = -\frac{1}{5k} + \frac{1}{3k}$ | M1 | Substituting correct new limits into their integrated expression, or substituting in terms of $x$ and using original limits |
| **Alternatively, by inspection:** $\int_k^{2k} \frac{2}{(2x+k)^2}\,dx = \left[-(2x+k)^{-1}\right]_k^{2k}$ | M1 | Integrating by inspection to obtain any multiple of $(2x+k)^{-1}$ |
| (fully correct indefinite integral) | A2 | Fully correct indefinite integral – need not be simplified |
| $-\frac{1}{5k} + \frac{1}{3k}$ | M1 | Substituting limits into their integrated expression |
| $= \frac{2}{15k}$ | A1 | Allow $\left(-\frac{1}{5} + \frac{1}{3}\right)\frac{1}{k}$ seen |
| This is inversely proportional to $k$ [with constant of proportionality $\frac{2}{15}$] | E1 | FT their definite integral. Must use phrase "inversely proportional" to $k$ or indicate $\propto \frac{1}{k}$. Allow if $\frac{a}{k}$ required at the start of the argument |
| **[6]** | | |
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11 Given that $k$ is a positive constant, show that $\int _ { k } ^ { 2 k } \frac { 2 } { ( 2 x + k ) ^ { 2 } } d x$ is inversely proportional to $k$.
\hfill \mbox{\textit{OCR MEI Paper 1 2022 Q11 [6]}}