A2 June 2023 Paper 1 Q16
16
(a) Show that\[\int_{0.5}^{4} \frac{1}{t}\ln t\,\mathrm{d}t = a(\ln 2)^2\]
where \(a\) is a rational number to be found. [4 marks]
(b) A curve \(C\) is defined parametrically for \(t \gt 0\) by\[x = 2t \qquad y = \frac{1}{2}t^2 - \ln t\]
The arc formed by the graph of \(C\) from \(t = 0.5\) to \(t = 4\) is rotated through \(2\pi\) radians about the \(x\)-axis to generate a surface with area \(S\)
Find the exact value of \(S\), giving your answer in the form
\[S = \pi\left(b + c\ln 2 + d(\ln 2)^2\right)\]where \(b\), \(c\) and \(d\) are rational numbers to be found. [7 marks]
| Scheme | Marks | AO |
|---|---|---|
| Selects a suitable method to find the required result by integration by parts or making an appropriate substitution/inspection | M1 | 3.1a |
| Applies their correct integration method to obtain \(2\displaystyle\int \frac{1}{t}\ln t\,\mathrm{d}t = \left[(\ln t)^2\right]\) or \(\displaystyle\int \frac{1}{t}\ln t\,\mathrm{d}t = \left[\frac{u^2}{2}\right]\) OE | A1 | 1.1b |
| Substitutes limits and uses the laws of logs correctly to simplify their result of integration in terms of \(\ln 2\) | M1 | 1.1a |
| Completes a rigorous argument to show the required result. NMS = 0/4 | R1 | 2.1 |
| (4) |
Typical solution
\[u = \ln t \qquad \frac{\mathrm{d}v}{\mathrm{d}t} = \frac{1}{t}\]\[\frac{\mathrm{d}u}{\mathrm{d}t} = \frac{1}{t} \qquad v = \ln t\]\[\int_{0.5}^{4} \frac{1}{t}\ln t\,\mathrm{d}t = \Big[(\ln t)^2\Big]_{0.5}^{4} - \int_{0.5}^{4} \frac{1}{t}\ln t\,\mathrm{d}t\]\[\begin{aligned} \int_{0.5}^{4} \frac{1}{t}\ln t\,\mathrm{d}t &= \frac{1}{2}\Big[(\ln t)^2\Big]_{0.5}^{4} \\ &= \frac{1}{2}\left[(\ln 4)^2 - (\ln 0.5)^2\right] \\ &= \frac{1}{2}\left[4(\ln 2)^2 - (-\ln 2)^2\right] \end{aligned}\]\[\int_{0.5}^{4} \frac{1}{t}\ln t\,\mathrm{d}t = \frac{3}{2}(\ln 2)^2\]or
let \(u = \ln t\)
\[\ldots\mathrm{d}u = \ldots\frac{1}{t}\mathrm{d}t\]\[\begin{aligned} \int_{0.5}^{4} \frac{1}{t}\ln t\,\mathrm{d}t &= \int_{\ln 0.5}^{\ln 4} u\,\mathrm{d}u \\ &= \left[\frac{u^2}{2}\right]_{\ln 0.5}^{\ln 4} \\ &= \frac{(\ln 4)^2}{2} - \frac{(\ln 0.5)^2}{2} \\ &= \frac{(2\ln 2)^2}{2} - \frac{(-\ln 2)^2}{2} \\ &= \frac{3}{2}(\ln 2)^2 \end{aligned}\]| Scheme | Marks | AO |
|---|---|---|
| Correctly obtains \(\left(\dfrac{\mathrm{d}x}{\mathrm{d}t}\right)^2 + \left(\dfrac{\mathrm{d}y}{\mathrm{d}t}\right)^2\) in any form. | B1 | 1.1b |
| Substitutes \(y\) and their \(\left(\dfrac{\mathrm{d}x}{\mathrm{d}t}\right)^2 + \left(\dfrac{\mathrm{d}y}{\mathrm{d}t}\right)^2\) into the integrand of the formula for Surface Area of Revolution. | M1 | 1.2 |
| Obtains correctly expanded integrand | A1 | 1.1b |
| Selects integration by parts to calculate an integral of form \(k\displaystyle\int t\ln t\,\mathrm{d}t\) | M1 | 3.1a |
| Obtains correct result of integration by parts for their \(k\displaystyle\int t\ln t\,\mathrm{d}t\). No limits needed at this stage. | A1 | 1.1b |
| Substitutes limits into their expression of the form \(at^4 + bt^2 + ct^2\ln t\) and use of their answer to part (a). | M1 | 1.1a |
| Obtains \(\pi\left\{\dfrac{5103}{64} - \dfrac{129}{4}\ln 2 - 3(\ln 2)^2\right\}\) | R1 | 2.1 |
| (7) | ||
| (11 marks) |