Standard +0.3 This is a straightforward volumes of revolution question requiring integration of y² with respect to x. The algebraic manipulation involves expanding (9² - [9/(2x-3)^1.25]²), substituting u=2x-3, and integrating a power function. While it requires careful algebra and the substitution technique, it follows a standard template with no conceptual surprises—slightly easier than average for a P4/Further Maths question.
2.
\includegraphics[max width=\textwidth, alt={}, center]{960fe82f-c180-422c-b409-a5cdc5fae924-06_974_1088_116_548}
\section*{Figure 1}
Figure 1 shows a sketch of part of the curve with equation
$$y = \frac { 9 } { ( 2 x - 3 ) ^ { 1.25 } } \quad x > \frac { 3 } { 2 }$$
The finite region \(R\), shown shaded in Figure 1, is bounded by the curve, the line with equation \(y = 9\) and the line with equation \(x = 6\)
This region is rotated through \(2 \pi\) radians about the \(x\)-axis to form a solid of revolution. Find, by algebraic integration, the exact volume of the solid generated.
M1: Integrates form \(\int\frac{...}{(2x-3)^{2.5}}dx\) achieving \(\frac{...}{(2x-3)^{1.5}}\). A1: Correct integration, following through on their \(\alpha\). Typical values of \(\alpha\): 1, \(\pi\) or \(2\pi\)
Attempts volume of solid under curve. Dependent on previous M1. Top limit 6, bottom limit from \(\frac{9}{(2x-3)^{1.25}}=9\)
Volume \(= 26\pi\)
A1
Correct exact volume under curve. May be part of larger expression e.g. \(...-26\pi\)
Volume of solid \(= \pi\times 9^2\times(6-"2")-"26\pi" = 298\pi\)
ddM1, A1
ddM1: Correct method for volume above curve. Dependent on both previous M's. A1: \(298\pi\)
# Question 2:
| Answer/Working | Marks | Guidance |
|---|---|---|
| Achieves a lower limit of 2 | B1 | May be awarded anywhere, accept on diagram or in limits of integral |
| $\alpha\int y^2\,dx = \alpha\int\frac{81}{(2x-3)^{2.5}}\,dx = \alpha\times\frac{-27}{(2x-3)^{1.5}}$ | M1, A1 | M1: Integrates form $\int\frac{...}{(2x-3)^{2.5}}dx$ achieving $\frac{...}{(2x-3)^{1.5}}$. A1: Correct integration, following through on their $\alpha$. Typical values of $\alpha$: 1, $\pi$ or $2\pi$ |
| $= \pi\int_{``2''}^{6} y^2\,dx = \left[\frac{k}{(2x-3)^{1.5}}\right]_{``2''}^{6} = \left(\left(-\frac{27}{27}\right)-\left(-\frac{27}{1}\right)\right)$ | dM1 | Attempts volume of solid under curve. Dependent on previous M1. Top limit 6, bottom limit from $\frac{9}{(2x-3)^{1.25}}=9$ |
| Volume $= 26\pi$ | A1 | Correct exact volume under curve. May be part of larger expression e.g. $...-26\pi$ |
| Volume of solid $= \pi\times 9^2\times(6-"2")-"26\pi" = 298\pi$ | ddM1, A1 | ddM1: Correct method for volume above curve. Dependent on both previous M's. A1: $298\pi$ |
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2.\\
\includegraphics[max width=\textwidth, alt={}, center]{960fe82f-c180-422c-b409-a5cdc5fae924-06_974_1088_116_548}
\section*{Figure 1}
Figure 1 shows a sketch of part of the curve with equation
$$y = \frac { 9 } { ( 2 x - 3 ) ^ { 1.25 } } \quad x > \frac { 3 } { 2 }$$
The finite region $R$, shown shaded in Figure 1, is bounded by the curve, the line with equation $y = 9$ and the line with equation $x = 6$
This region is rotated through $2 \pi$ radians about the $x$-axis to form a solid of revolution. Find, by algebraic integration, the exact volume of the solid generated.\\
\hfill \mbox{\textit{Edexcel P4 2021 Q2 [7]}}