| Exam Board | AQA |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2016 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Volumes of Revolution |
| Type | Volume requiring substitution or integration by parts |
| Difficulty | Standard +0.8 This question combines integration by parts with a non-standard function (ln(3x)/x²) and then applies it to a volume of revolution problem requiring careful setup and manipulation. The integration by parts itself is moderately challenging, and part (b) requires recognizing that the result from (a) can be adapted, plus handling the logarithmic terms at the boundaries correctly. This is above-average difficulty for C3, requiring strong technique and careful algebraic manipulation across multiple steps. |
| Spec | 1.08d Evaluate definite integrals: between limits1.08i Integration by parts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(u = \ln 3x\), \(\quad \dfrac{du}{dx} = \dfrac{1}{x}\) | B1 | PI by further work |
| \(\dfrac{dv}{dx} = \dfrac{1}{x^2}\), \(\quad v = -x^{-1}\) | B1 | PI by further work |
| \(\int = -\dfrac{1}{x}\ln 3x - \int -x^{-1} \times \dfrac{1}{x}\ dx\) | M1 | Correct substitution of their terms into the parts formula |
| \(= -\dfrac{1}{x}\ln 3x - \dfrac{1}{x}\ (+c)\) | A1 | Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((V =)\ \pi\int_{\frac{1}{3}}^{1}\left(\dfrac{\ln 3x}{x}\right)^2 dx\) | B1 | Must include \(\pi\) (not \(2\pi\)), limits and \(dx\) (each seen at some stage) |
| \([u = (\ln 3x)^2]\), \(\quad \dfrac{du}{dx} = 2\ln 3x \times \dfrac{1}{x}\) | M1 | \(\dfrac{du}{dx} = k\ln 3x \times \dfrac{1}{x}\) |
| A1 | \(k=2\) | |
| \(\dfrac{dv}{dx} = x^{-2}\), \(\quad v = -x^{-1}\) | ||
| \(\int\left(\dfrac{\ln 3x}{x}\right)^2 dx = -\dfrac{1}{x}(\ln 3x)^2 - \int -x^{-1} \times 2\ln 3x \times \dfrac{1}{x}\ dx\) | M1 | Correct substitution of their terms into the parts formula |
| \(= -\dfrac{1}{x}(\ln 3x)^2 - \dfrac{2}{x}\ln 3x - \dfrac{2}{x}\) | A1 | |
| \(= [-(\ln 3)^2 - 2\ln 3 - 2] - [-3(\ln 1)^2 - 6\ln 1 - 6]\) | M1 | Correct subst into expression of the form \(\dfrac{k}{x}(\ln 3x)^2 + \dfrac{l}{x}\ln 3x - \dfrac{m}{x}\) and \(F(1) - F(\frac{1}{3})\) |
| \([V =]\ \pi(4 - (\ln 3)^2 - 2\ln 3)\) | A1 | Total: 7 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(u = \ln 3x\), \(\quad \dfrac{dv}{dx} = \dfrac{\ln 3x}{x^2}\) | (M1) | 'splitting' in this way |
| \(\dfrac{du}{dx} = \dfrac{1}{x}\), \(\quad v = -\dfrac{1}{x}\ln 3x - \dfrac{1}{x}\) | (A1) | |
| \(\int = \ln 3x\left(-\dfrac{1}{x}\ln 3x - \dfrac{1}{x}\right) - \int\left(-\dfrac{1}{x}\ln 3x - \dfrac{1}{x}\right)\dfrac{1}{x}\ dx\) | (M1) | Correct substitution of their terms into the parts formula |
| \(= -\dfrac{1}{x}(\ln 3x)^2 - \dfrac{2}{x}\ln 3x - \dfrac{2}{x}\) | (A1) | First B1 and final 2 marks as first method |
## Question 6:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $u = \ln 3x$, $\quad \dfrac{du}{dx} = \dfrac{1}{x}$ | B1 | PI by further work |
| $\dfrac{dv}{dx} = \dfrac{1}{x^2}$, $\quad v = -x^{-1}$ | B1 | PI by further work |
| $\int = -\dfrac{1}{x}\ln 3x - \int -x^{-1} \times \dfrac{1}{x}\ dx$ | M1 | Correct substitution of **their** terms into the parts formula |
| $= -\dfrac{1}{x}\ln 3x - \dfrac{1}{x}\ (+c)$ | A1 | **Total: 4** |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(V =)\ \pi\int_{\frac{1}{3}}^{1}\left(\dfrac{\ln 3x}{x}\right)^2 dx$ | B1 | Must include $\pi$ (not $2\pi$), limits and $dx$ (each seen at some stage) |
| $[u = (\ln 3x)^2]$, $\quad \dfrac{du}{dx} = 2\ln 3x \times \dfrac{1}{x}$ | M1 | $\dfrac{du}{dx} = k\ln 3x \times \dfrac{1}{x}$ |
| | A1 | $k=2$ |
| $\dfrac{dv}{dx} = x^{-2}$, $\quad v = -x^{-1}$ | | |
| $\int\left(\dfrac{\ln 3x}{x}\right)^2 dx = -\dfrac{1}{x}(\ln 3x)^2 - \int -x^{-1} \times 2\ln 3x \times \dfrac{1}{x}\ dx$ | M1 | Correct substitution of **their** terms into the parts formula |
| $= -\dfrac{1}{x}(\ln 3x)^2 - \dfrac{2}{x}\ln 3x - \dfrac{2}{x}$ | A1 | |
| $= [-(\ln 3)^2 - 2\ln 3 - 2] - [-3(\ln 1)^2 - 6\ln 1 - 6]$ | M1 | Correct subst into expression of the form $\dfrac{k}{x}(\ln 3x)^2 + \dfrac{l}{x}\ln 3x - \dfrac{m}{x}$ and $F(1) - F(\frac{1}{3})$ |
| $[V =]\ \pi(4 - (\ln 3)^2 - 2\ln 3)$ | A1 | **Total: 7** |
**OR method:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $u = \ln 3x$, $\quad \dfrac{dv}{dx} = \dfrac{\ln 3x}{x^2}$ | (M1) | 'splitting' in this way |
| $\dfrac{du}{dx} = \dfrac{1}{x}$, $\quad v = -\dfrac{1}{x}\ln 3x - \dfrac{1}{x}$ | (A1) | |
| $\int = \ln 3x\left(-\dfrac{1}{x}\ln 3x - \dfrac{1}{x}\right) - \int\left(-\dfrac{1}{x}\ln 3x - \dfrac{1}{x}\right)\dfrac{1}{x}\ dx$ | (M1) | Correct substitution of their terms into the parts formula |
| $= -\dfrac{1}{x}(\ln 3x)^2 - \dfrac{2}{x}\ln 3x - \dfrac{2}{x}$ | (A1) | First **B1** and final 2 marks as first method |
---6
\begin{enumerate}[label=(\alph*)]
\item Use integration by parts to find $\int \frac { \ln ( 3 x ) } { x ^ { 2 } } \mathrm {~d} x$.
\item The region bounded by the curve $y = \frac { \ln ( 3 x ) } { x }$, the $x$-axis from $\frac { 1 } { 3 }$ to 1 , and the line $x = 1$ is rotated through $2 \pi$ radians about the $x$-axis to form a solid.
Find the exact value of the volume of the solid generated.\\[0pt]
[7 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA C3 2016 Q6 [11]}}