OCR Further Pure Core 2 2018 December — Question 8 7 marks

Exam BoardOCR
ModuleFurther Pure Core 2 (Further Pure Core 2)
Year2018
SessionDecember
Marks7
TopicVolumes of Revolution
TypeVolume with logarithmic functions
DifficultyChallenging +1.8 This is a Further Maths question requiring volume of revolution with a non-standard integrand involving √(ln x). Students must identify correct limits (where y=0), set up π∫y² dx, expand (x-3)²ln x, and integrate term-by-term using integration by parts twice. The multi-step nature, Further Maths context, and need for careful algebraic manipulation with logarithms make this significantly harder than average, though the technique itself is standard for FM students.
Spec4.08d Volumes of revolution: about x and y axes
\includegraphics{figure_8} The figure shows part of the graph of \(y = (x - 3)\sqrt{\ln x}\). The portion of the graph below the \(x\)-axis is rotated by \(2\pi\) radians around the \(x\)-axis to form a solid of revolution, S. Determine the exact volume of S. [7]
AnswerMarks Guidance
Limits 1 & 3 seenB1
\(V = \pi\int_1^3 [(x - 3)\ln x] \, dx = \pi\int_1^3 (x - 3)^2 \ln xdx\)M1 Correct substitution into formula (ignore limits) and simplification to integrable (by parts) form
\(V = \pi\left[\frac{1}{3}(x - 3)^3 \ln x - \frac{1}{3}\int_1^3 \frac{(x - 3)^3}{x} \, dx\right]\)*M1 Integration by parts with \((x - 3)^2\) (may be expanded) being integrated.
\(\frac{1}{x}(x - 3)^3 = x^2 - 9x + 27 - \frac{27}{x}\) soiA1 May come implicitly from previously expanded form
\(V = \frac{\pi}{3}\left[\frac{(x - 3)^3 \ln x - \frac{x^3}{3} + 9x^2 - 27 \ln x}{9x^2 - 2 + 27x - 27 \ln x}\right]\)A1 Completing the integral. NB \(\int(x - 3)^3 \ln x \cdot 1\right]^3_1 = 0\) so may be omitted provided it is seen earlier
dep1.1a Correctly dealing with limits
*M1
\(\frac{\pi}{3}[(3 - 3)^3 \ln 3 - \frac{3^3}{3} + \frac{9 \times 3^2}{2} - 27 \times 3 + 27\ln 3 - (-(1-3)^3 \ln 1 + \frac{1^3}{3} - \frac{9 \times 1^2}{2} + 27 \times 1 - 27\ln 1)]\)A1
So volume of S is \(\frac{\pi}{9}(81\ln 3 - 80)\) oeA1 
Limits 1 & 3 seen | B1 |

$V = \pi\int_1^3 [(x - 3)\ln x] \, dx = \pi\int_1^3 (x - 3)^2 \ln xdx$ | M1 | Correct substitution into formula (ignore limits) and simplification to integrable (by parts) form

$V = \pi\left[\frac{1}{3}(x - 3)^3 \ln x - \frac{1}{3}\int_1^3 \frac{(x - 3)^3}{x} \, dx\right]$ | *M1 | Integration by parts with $(x - 3)^2$ (may be expanded) being integrated.

$\frac{1}{x}(x - 3)^3 = x^2 - 9x + 27 - \frac{27}{x}$ soi | A1 | May come implicitly from previously expanded form

$V = \frac{\pi}{3}\left[\frac{(x - 3)^3 \ln x - \frac{x^3}{3} + 9x^2 - 27 \ln x}{9x^2 - 2 + 27x - 27 \ln x}\right]$ | A1 | Completing the integral. NB $\int(x - 3)^3 \ln x \cdot 1\right]^3_1 = 0$ so may be omitted provided it is seen earlier

dep | 1.1a | Correctly dealing with limits
*M1 |

$\frac{\pi}{3}[(3 - 3)^3 \ln 3 - \frac{3^3}{3} + \frac{9 \times 3^2}{2} - 27 \times 3 + 27\ln 3 - (-(1-3)^3 \ln 1 + \frac{1^3}{3} - \frac{9 \times 1^2}{2} + 27 \times 1 - 27\ln 1)]$ | A1 |

So volume of S is $\frac{\pi}{9}(81\ln 3 - 80)$ oe | A1 |

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\includegraphics{figure_8}

The figure shows part of the graph of $y = (x - 3)\sqrt{\ln x}$. The portion of the graph below the $x$-axis is rotated by $2\pi$ radians around the $x$-axis to form a solid of revolution, S.

Determine the exact volume of S. [7]

\hfill \mbox{\textit{OCR Further Pure Core 2 2018 Q8 [7]}}
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