OCR MEI C4 — Question 2 4 marks

Exam BoardOCR MEI
ModuleC4 (Core Mathematics 4)
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVolumes of Revolution
TypeRotation about x-axis: polynomial or root function
DifficultyModerate -0.8 This is a straightforward volumes of revolution question requiring direct application of the standard formula V = π∫y²dx with a simple polynomial function. The integrand (x²+1)² expands easily and integrates routinely with clear bounds 0 to 2. No problem-solving or conceptual insight needed—pure mechanical application of a standard technique.
Spec4.08d Volumes of revolution: about x and y axes
2 The graph shows part of the curve \(y = x ^ { 2 } + 1\). \includegraphics[max width=\textwidth, alt={}, center]{62dbc58e-f498-483f-a9aa-05cb5aa44881-2_380_876_715_575} Find the volume when the area between this curve, the axes and the line \(x = 2\) is rotated through \(360 ^ { 0 }\) about the \(x\)-axis.
Question 2:
AnswerMarks Guidance
AnswerMarks Guidance
\(V=\pi\int_0^2(x^2+1)^2\,dx\)M1 Integral
\(=\pi\int_0^2(x^4+2x^2+1)\,dx\)M1 Multiply out
\(=\pi\left[\frac{x^5}{5}+\frac{2x^3}{3}+x\right]_0^2=\pi\left(\frac{2^5}{5}+\frac{2\cdot2^3}{3}+2\right)-0\)A1
\(=\frac{206\pi}{15}\)A1
Total: 4 marks
## Question 2:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $V=\pi\int_0^2(x^2+1)^2\,dx$ | M1 | Integral |
| $=\pi\int_0^2(x^4+2x^2+1)\,dx$ | M1 | Multiply out |
| $=\pi\left[\frac{x^5}{5}+\frac{2x^3}{3}+x\right]_0^2=\pi\left(\frac{2^5}{5}+\frac{2\cdot2^3}{3}+2\right)-0$ | A1 | |
| $=\frac{206\pi}{15}$ | A1 | |

**Total: 4 marks**

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2 The graph shows part of the curve $y = x ^ { 2 } + 1$.\\
\includegraphics[max width=\textwidth, alt={}, center]{62dbc58e-f498-483f-a9aa-05cb5aa44881-2_380_876_715_575}

Find the volume when the area between this curve, the axes and the line $x = 2$ is rotated through $360 ^ { 0 }$ about the $x$-axis.

\hfill \mbox{\textit{OCR MEI C4  Q2 [4]}}
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Q1 5 Q2 4 Q3 5 Q4 6 Q5 8 Q6 8 Q7 18 Q8 18