| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/2 (Pre-U Mathematics Paper 2) |
| Year | 2010 |
| Session | June |
| Marks | 15 |
| Topic | Integration by Parts |
| Type | Sequential multi-part (building on previous) |
| Difficulty | Challenging +1.2 This is a structured multi-part integration question requiring integration by parts and careful algebraic manipulation. Part (i) is a standard 'show that' proof using integration by parts. Part (ii) involves finding intersection points (straightforward algebra) and computing area between curves using the result from (i). Part (iii) applies volume of revolution with a substitution. While requiring multiple techniques and careful work across 15 marks total, each component uses standard A-level methods without requiring novel insight or particularly sophisticated problem-solving—moderately above average difficulty due to length and algebraic complexity. |
| Spec | 1.08e Area between curve and x-axis: using definite integrals1.08f Area between two curves: using integration1.08i Integration by parts1.09f Trapezium rule: numerical integration4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| Using integration by parts, with \(u = \ln x\) and \(\frac{dv}{dx} = x^n\) | M1 A1 | |
| The indefinite integral is: \(\frac{1}{n+1}x^{n+1}\ln x - \int\frac{1}{n+1}x^{n+1}\frac{1}{x} dx\) | A1 | |
| \(\frac{1}{n+1}x^{n+1}\ln x - \frac{x^{n+1}}{(n+1)^2} = \frac{x^{n+1}}{(n+1)^2}((n+1)\ln x - 1)\) | A1 | |
| Correctly substituting the limits in their integrated expressions | M1 | |
| Obtain: \(\frac{a^{n+1}}{(n+1)^2}((n+1)\ln a - 1) + \frac{1}{(n+1)^2}\) | A1 | [6] |
| Answer | Marks | Guidance |
|---|---|---|
| (a) The curves intersect when \(x^2 \ln x = x \ln 2^x\) \(\left(= x^2 \ln 2\right)\) | M1 | |
| The solution is \(x = 2\) | A1 | [2] |
| (b) The required area is given by: \(\int_1^2 x^2 \ln 2 \, dx - \int_1^2 x^2 \ln x \, dx\) (condone reverse order) | M1 | |
| Attempt at using the result from part (i) and correctly substituting the limits for the first integral, the value of this area is: \(\left(\frac{8}{3} - \frac{1}{3}\right)\ln 2 - \left(\frac{8}{9}(3\ln 2 - 1) + \frac{1}{9}\right)\) | M1 A1 | |
| This simplifies to: \(\frac{7}{9} - \frac{1}{3}\ln 2\) | A1 | [4] |
| Answer | Marks | Guidance |
|---|---|---|
| The volume of revolution is: \((\pi)\int_{(i)}^{(e)} y^2(dx) = (\pi)\int_{(i)}^{(e)} x^2 \ln x \, (dx)\) | B1 | |
| Using the result from part (i), the exact value of the volume is: \(\pi\left(\frac{e^4}{16}(4\ln e - 1) + \frac{1}{16}\right) = 32.36\) | M1 A1 | [3] |
### (i)
Using integration by parts, with $u = \ln x$ and $\frac{dv}{dx} = x^n$ | M1 A1 |
The indefinite integral is: $\frac{1}{n+1}x^{n+1}\ln x - \int\frac{1}{n+1}x^{n+1}\frac{1}{x} dx$ | A1 |
$\frac{1}{n+1}x^{n+1}\ln x - \frac{x^{n+1}}{(n+1)^2} = \frac{x^{n+1}}{(n+1)^2}((n+1)\ln x - 1)$ | A1 |
Correctly substituting the limits in their integrated expressions | M1 |
Obtain: $\frac{a^{n+1}}{(n+1)^2}((n+1)\ln a - 1) + \frac{1}{(n+1)^2}$ | A1 | [6]
### (ii)
**(a)** The curves intersect when $x^2 \ln x = x \ln 2^x$ $\left(= x^2 \ln 2\right)$ | M1 |
The solution is $x = 2$ | A1 | [2]
**(b)** The required area is given by: $\int_1^2 x^2 \ln 2 \, dx - \int_1^2 x^2 \ln x \, dx$ (condone reverse order) | M1 |
Attempt at using the result from part (i) and correctly substituting the limits for the first integral, the value of this area is: $\left(\frac{8}{3} - \frac{1}{3}\right)\ln 2 - \left(\frac{8}{9}(3\ln 2 - 1) + \frac{1}{9}\right)$ | M1 A1 |
This simplifies to: $\frac{7}{9} - \frac{1}{3}\ln 2$ | A1 | [4]
### (iii)
The volume of revolution is: $(\pi)\int_{(i)}^{(e)} y^2(dx) = (\pi)\int_{(i)}^{(e)} x^2 \ln x \, (dx)$ | B1 |
Using the result from part (i), the exact value of the volume is: $\pi\left(\frac{e^4}{16}(4\ln e - 1) + \frac{1}{16}\right) = 32.36$ | M1 A1 | [3]\begin{enumerate}[label=(\roman*)]
\item Show that
$$\int x^n \ln x \, dx = \frac{x^{n+1}}{(n+1)^2}\left((n+1)\ln a - 1\right) + \frac{1}{(n+1)^2},$$
where $n \neq -1$ and $a > 1$. [6]
\item \begin{enumerate}[label=(\alph*)]
\item Determine the $x$-coordinate of the point of intersection of the curves $y = x^3 \ln x$ and $y = x \ln 2^x$, where $x > 0$. [2]
\item Find the exact value of the area of the region enclosed between these two curves, the line $x = 1$ and their point of intersection. Express your answer in the form $b + c \ln 2$, where $b$ and $c$ are rational. [4]
\end{enumerate}
\item The curve $y = (x^3 \ln x)^{0.5}$, for $1 < x < e$, is rotated through $2\pi$ radians about the $x$-axis. Determine the value of the resulting volume of revolution, giving your answer correct to 4 significant figures. [3]
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9794/2 2010 Q9 [15]}}