| Exam Board | Edexcel |
|---|---|
| Module | CP1 (Core Pure 1) |
| Session | Specimen |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Standard Integrals and Reverse Chain Rule |
| Type | Integration with substitution given |
| Difficulty | Standard +0.3 This is a structured multi-part question requiring splitting a fraction into two standard integrals (ln and arctan forms), then applying the mean value formula. While it involves multiple techniques, each step is guided and uses standard Core Pure 1 methods with no novel insight required—slightly easier than average. |
| Spec | 4.08e Mean value of function: using integral4.08f Integrate using partial fractions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f(x) = \frac{x+2}{x^2+9} = \frac{x}{x^2+9} + \frac{2}{x^2+9}\) | B1 | Splits the fraction into two correct separate expressions |
| \(\int \frac{x}{x^2+9}\, dx = k\ln(x^2+9)(+c)\) | M1 | Recognises the required form for the first integration |
| \(\int \frac{2}{x^2+9}\, dx = k\arctan\left(\frac{x}{3}\right)(+c)\) | M1 | Recognises the required form for the second integration |
| \(\int \frac{x+2}{x^2+9}\, dx = \frac{1}{2}\ln(x^2+9) + \frac{2}{3}\arctan\left(\frac{x}{3}\right) + c\) | A1 | Both expressions integrated correctly, added together with constant of integration included |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int_0^3 f(x)\,dx = \left[\frac{1}{2}\ln(x^2+9)+\frac{2}{3}\arctan\left(\frac{x}{3}\right)\right]_0^3\) \(= \frac{1}{2}\ln 18 + \frac{2}{3}\arctan\left(\frac{3}{3}\right) - \left(\frac{1}{2}\ln 9 + \frac{2}{3}\arctan(0)\right)\) \(= \frac{1}{2}\ln\frac{18}{9} + \frac{2}{3}\arctan\left(\frac{3}{3}\right)\) | M1 | Uses limits correctly and combines logarithmic terms |
| Mean value \(= \frac{1}{3-0}\left(\frac{1}{2}\ln 2 + \frac{\pi}{6}\right)\) | M1 | Correctly applies the method for the mean value using their integration |
| \(\frac{1}{6}\ln 2 + \frac{1}{18}\pi\) | A1* | Correct work leading to the given answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{1}{6}\ln 2 + \frac{1}{18}\pi + \ln k\) | M1 | Realises that the effect of the transformation is to increase the mean value by \(\ln k\) |
| \(\frac{1}{6}\ln 2k^6 + \frac{1}{18}\pi\) | A1 | Combines \(\ln\)'s correctly to obtain the correct expression |
## Question 6:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(x) = \frac{x+2}{x^2+9} = \frac{x}{x^2+9} + \frac{2}{x^2+9}$ | B1 | Splits the fraction into two correct separate expressions |
| $\int \frac{x}{x^2+9}\, dx = k\ln(x^2+9)(+c)$ | M1 | Recognises the required form for the first integration |
| $\int \frac{2}{x^2+9}\, dx = k\arctan\left(\frac{x}{3}\right)(+c)$ | M1 | Recognises the required form for the second integration |
| $\int \frac{x+2}{x^2+9}\, dx = \frac{1}{2}\ln(x^2+9) + \frac{2}{3}\arctan\left(\frac{x}{3}\right) + c$ | A1 | Both expressions integrated correctly, added together with constant of integration included |
**(4 marks)**
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int_0^3 f(x)\,dx = \left[\frac{1}{2}\ln(x^2+9)+\frac{2}{3}\arctan\left(\frac{x}{3}\right)\right]_0^3$ $= \frac{1}{2}\ln 18 + \frac{2}{3}\arctan\left(\frac{3}{3}\right) - \left(\frac{1}{2}\ln 9 + \frac{2}{3}\arctan(0)\right)$ $= \frac{1}{2}\ln\frac{18}{9} + \frac{2}{3}\arctan\left(\frac{3}{3}\right)$ | M1 | Uses limits correctly and combines logarithmic terms |
| Mean value $= \frac{1}{3-0}\left(\frac{1}{2}\ln 2 + \frac{\pi}{6}\right)$ | M1 | Correctly applies the method for the mean value using their integration |
| $\frac{1}{6}\ln 2 + \frac{1}{18}\pi$ | A1* | Correct work leading to the given answer |
**(3 marks)**
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{6}\ln 2 + \frac{1}{18}\pi + \ln k$ | M1 | Realises that the effect of the transformation is to increase the mean value by $\ln k$ |
| $\frac{1}{6}\ln 2k^6 + \frac{1}{18}\pi$ | A1 | Combines $\ln$'s correctly to obtain the correct expression |
**(2 marks)**
---6.
$$\mathrm { f } ( x ) = \frac { x + 2 } { x ^ { 2 } + 9 }$$
\begin{enumerate}[label=(\alph*)]
\item Show that
$$\int \mathrm { f } ( x ) \mathrm { d } x = A \ln \left( x ^ { 2 } + 9 \right) + B \arctan \left( \frac { x } { 3 } \right) + c$$
where $c$ is an arbitrary constant and $A$ and $B$ are constants to be found.
\item Hence show that the mean value of $\mathrm { f } ( x )$ over the interval $[ 0,3 ]$ is
$$\frac { 1 } { 6 } \ln 2 + \frac { 1 } { 18 } \pi$$
\item Use the answer to part (b) to find the mean value, over the interval $[ 0,3 ]$, of
$$\mathrm { f } ( x ) + \ln k$$
where $k$ is a positive constant, giving your answer in the form $p + \frac { 1 } { 6 } \ln q$, where $p$ and $q$ are constants and $q$ is in terms of $k$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel CP1 Q6 [9]}}