| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Volumes of Revolution |
| Type | Volume requiring substitution or integration by parts |
| Difficulty | Standard +0.3 This is a standard C4 volume of revolution question requiring expansion of a squared binomial and integration of terms including x^(-1/2). While it involves multiple steps (squaring, expanding, integrating term-by-term, and simplifying logarithms), these are all routine techniques for C4 students with no novel problem-solving required. The 8 marks reflect the algebraic work rather than conceptual difficulty, making it slightly easier than average. |
| Spec | 4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| \(V = \pi\int_1^4 y^2\, dx = \pi\int_1^4 \left(1 + \frac{1}{2\sqrt{x}}\right)^2 dx\) | M1 | Correct volume formula |
| \(= \pi\int_1^4 \left(1 + \frac{1}{\sqrt{x}} + \frac{1}{4x}\right) dx\) | A1 | Correct expansion |
| \(= \pi\left[x + 2\sqrt{x} + \frac{1}{4}\ln x\right]_1^4\) | M1 A1 | Integrating each term correctly |
| \(= \pi\left[\left(4 + 4 + \frac{1}{4}\ln 4\right) - \left(1 + 2 + 0\right)\right]\) | M1 | Applying limits |
| \(= \pi\left[5 + \frac{1}{4}\ln 4\right] = \pi\left(5 + \frac{1}{2}\ln 2\right)\) | A1 | Using \(\frac{1}{4}\ln 4 = \frac{1}{2}\ln 2\), conclusion |
# Question 5:
| $V = \pi\int_1^4 y^2\, dx = \pi\int_1^4 \left(1 + \frac{1}{2\sqrt{x}}\right)^2 dx$ | M1 | Correct volume formula |
| $= \pi\int_1^4 \left(1 + \frac{1}{\sqrt{x}} + \frac{1}{4x}\right) dx$ | A1 | Correct expansion |
| $= \pi\left[x + 2\sqrt{x} + \frac{1}{4}\ln x\right]_1^4$ | M1 A1 | Integrating each term correctly |
| $= \pi\left[\left(4 + 4 + \frac{1}{4}\ln 4\right) - \left(1 + 2 + 0\right)\right]$ | M1 | Applying limits |
| $= \pi\left[5 + \frac{1}{4}\ln 4\right] = \pi\left(5 + \frac{1}{2}\ln 2\right)$ | A1 | Using $\frac{1}{4}\ln 4 = \frac{1}{2}\ln 2$, conclusion |
---5.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\includegraphics[alt={},max width=\textwidth]{964070ca-a2c0-4935-8a5b-f1f656495f2e-3_771_1049_251_477}
\end{center}
\end{figure}
Figure 1 shows part of the curve with equation $y = 1 + \frac { 1 } { 2 \sqrt { x } }$. The shaded region $R$, bounded by the curve, that $x$-axis and the lines $x = 1$ and $x = 4$, is rotated through $360 ^ { \circ }$ about the $x$-axis. Using integration, show that the volume of the solid generated is $\pi \left( 5 + \frac { 1 } { 2 } \ln 2 \right)$.\\
(8)\\
\hfill \mbox{\textit{Edexcel C4 Q5 [8]}}