AQA Further AS Paper 1 Specimen — Question 5 5 marks

Exam BoardAQA
ModuleFurther AS Paper 1 (Further AS Paper 1)
SessionSpecimen
Marks5
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TopicVolumes of Revolution
TypeRotation about x-axis: polynomial or root function
DifficultyModerate -0.3 This is a straightforward volume of revolution question requiring the standard formula V = π∫y²dx. The integrand (3 + √x)² expands to simple terms (9 + 6√x + x) that integrate routinely using basic power rules. While it's a 5-mark Further Maths question, it requires no problem-solving insight—just careful algebraic expansion and standard integration technique, making it slightly easier than average overall.
Spec4.08d Volumes of revolution: about x and y axes
5 The region bounded by the curve with equation \(y = 3 + \sqrt { x }\), the \(x\)-axis and the lines \(x = 1\) and \(x = 4\) is rotated through \(2 \pi\) radians about the \(x\)-axis. Use integration to show that the volume generated is \(\frac { 125 \pi } { 2 }\) [0pt] [5 marks]
Question 5:
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(y^2 = 9 + 6\sqrt{x} + x\)B1 Obtains \(y^2\)
Volume \(= \pi\int_1^4 (9 + 6\sqrt{x} + x)\,dx = \pi\left[9x + 4x^{\frac{3}{2}} + \frac{x^2}{2}\right]_1^4\)M1 Integrates 'their' \(y^2\) within integral to find volume of revolution with at least two terms correct (condone missing \(\pi\))
All terms correctly integratedA1F FT 'their' \(y^2\), provided M1 awarded
\(\left[9\times4 + 4\times4^{\frac{3}{2}} + \frac{4^2}{2}\right] - \left[9\times1 + 4\times1^2 + \frac{1^2}{2}\right]\)M1 Substitutes correct limits into 'their' volume expression; FT provided previous M1 awarded
\(= 76 - 13\frac{1}{2} = 62\frac{1}{2} = \frac{125}{2}\)
Hence volume \(= \frac{125\pi}{2}\) AGA1 Completes fully correct argument; AG 
## Question 5:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $y^2 = 9 + 6\sqrt{x} + x$ | B1 | Obtains $y^2$ |
| Volume $= \pi\int_1^4 (9 + 6\sqrt{x} + x)\,dx = \pi\left[9x + 4x^{\frac{3}{2}} + \frac{x^2}{2}\right]_1^4$ | M1 | Integrates 'their' $y^2$ within integral to find volume of revolution with at least two terms correct (condone missing $\pi$) |
| All terms correctly integrated | A1F | FT 'their' $y^2$, provided M1 awarded |
| $\left[9\times4 + 4\times4^{\frac{3}{2}} + \frac{4^2}{2}\right] - \left[9\times1 + 4\times1^2 + \frac{1^2}{2}\right]$ | M1 | Substitutes correct limits into 'their' volume expression; FT provided previous M1 awarded |
| $= 76 - 13\frac{1}{2} = 62\frac{1}{2} = \frac{125}{2}$ | | |
| Hence volume $= \frac{125\pi}{2}$ **AG** | A1 | Completes fully correct argument; AG |

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5 The region bounded by the curve with equation $y = 3 + \sqrt { x }$, the $x$-axis and the lines $x = 1$ and $x = 4$ is rotated through $2 \pi$ radians about the $x$-axis.

Use integration to show that the volume generated is $\frac { 125 \pi } { 2 }$\\[0pt]
[5 marks]\\

\hfill \mbox{\textit{AQA Further AS Paper 1  Q5 [5]}}
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Q1 1 Q2 1 Q3 1 Q4 8 Q5 5 Q6 12 Q7 4 Q8 8 Q9 3 Q10 8 Q11 5 Q12 12