| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2016 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Volumes of Revolution |
| Type | Rotation about y-axis, standard curve |
| Difficulty | Standard +0.3 This is a straightforward C4 volumes of revolution question with standard techniques: (a) requires substitution u=2x-1 to integrate a power, (b) is simple algebraic manipulation to find k, and (c) applies the standard formula V=π∫y²dx. All steps are routine textbook exercises with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.08b Integrate x^n: where n != -1 and sums4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\int (2x-1)^{\frac{3}{2}}\,dx = \frac{1}{5}(2x-1)^{\frac{5}{2}} \{+c\}\) | M1 | \((2x\pm1)^{\frac{3}{2}} \rightarrow \pm\lambda(2x\pm1)^{\frac{5}{2}}\) or \(\pm\lambda u^{\frac{5}{2}}\) where \(u=2x\pm1\); \(\lambda\neq0\) |
| \(\frac{1}{5}(2x-1)^{\frac{5}{2}}\) with or without \(+c\) | A1 | Must be simplified |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(8=(2k-1)^{\frac{3}{4}} \Rightarrow k=\frac{8^{\frac{4}{3}}+1}{2}\) | M1 | Sets \(8=(2k-1)^{\frac{3}{4}}\) or \(8=(2x-1)^{\frac{3}{4}}\) and rearranges to give \(k\) (or \(x\)) a numerical value |
| \(k=\frac{17}{2}\) | A1 | \(k\) (or \(x\))\(=\frac{17}{2}\) or \(8.5\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\pi\int\left((2x-1)^{\frac{3}{4}}\right)^2\,dx\) | B1 | For \(\pi\int\left((2x-1)^{\frac{3}{4}}\right)^2\) or \(\pi\int(2x-1)^{\frac{3}{2}}\). Ignore limits and \(dx\). Can be implied. |
| \(\left\{\int_{\frac{1}{2}}^{\frac{17}{2}}y^2\,dx\right\} = \left[\frac{(2x-1)^{\frac{5}{2}}}{5}\right]_{\frac{1}{2}}^{\frac{17}{2}} = \left(\frac{16^2}{5}\right)-(0) = \frac{1024}{5}\) | M1 | Applies \(x\)-limits of "8.5" (their answer to part (b)) and \(0.5\) to an expression of the form \(\pm\beta(2x-1)^{\frac{5}{2}}\); \(\beta\neq0\) and subtracts the correct way round |
| \(\{V_{\text{cylinder}}\} = \pi(8)^2\left(\frac{17}{2}\right) \{=544\pi\}\) | B1 ft | \(\pi(8)^2(\text{their answer to part }(b))\). \(V_{\text{cylinder}}=544\pi\) implies this mark |
| \(\left\{\text{Vol}(S)=544\pi - \frac{1024}{5}\pi\right\} \Rightarrow \text{Vol}(S)=\frac{1696}{5}\pi\) | A1 | An exact correct answer in the form \(k\pi\). E.g. \(\frac{1696}{5}\pi\), \(\frac{3392}{10}\pi\) or \(339.2\pi\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\text{Vol}(S)=\pi(8)^2\left(\frac{1}{2}\right)+\pi\int_{0.5}^{8.5}\left(8^2-(2x-1)^{\frac{3}{2}}\right)dx\) | B1 | For \(\pi\int\ldots(2x-1)^{\frac{3}{2}}\). Ignore limits and \(dx\) |
| \(=\pi(8)^2\left(\frac{1}{2}\right)+\pi\left[64x-\frac{1}{5}(2x-1)^{\frac{5}{2}}\right]_{0.5}^{8.5}\) | — | — |
| Substitution of limits | M1, B1 | As above |
| \(\left\{=32\pi+\pi\left(\left(544-\frac{1024}{5}\right)-(32-0)\right)\right\}\Rightarrow\text{Vol}(S)=\frac{1696}{5}\pi\) | A1 | — |
# Question 7:
## Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\int (2x-1)^{\frac{3}{2}}\,dx = \frac{1}{5}(2x-1)^{\frac{5}{2}} \{+c\}$ | M1 | $(2x\pm1)^{\frac{3}{2}} \rightarrow \pm\lambda(2x\pm1)^{\frac{5}{2}}$ **or** $\pm\lambda u^{\frac{5}{2}}$ where $u=2x\pm1$; $\lambda\neq0$ |
| $\frac{1}{5}(2x-1)^{\frac{5}{2}}$ with or without $+c$ | A1 | Must be simplified |
## Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $8=(2k-1)^{\frac{3}{4}} \Rightarrow k=\frac{8^{\frac{4}{3}}+1}{2}$ | M1 | Sets $8=(2k-1)^{\frac{3}{4}}$ or $8=(2x-1)^{\frac{3}{4}}$ and rearranges to give $k$ (or $x$) a numerical value |
| $k=\frac{17}{2}$ | A1 | $k$ (or $x$)$=\frac{17}{2}$ or $8.5$ |
## Part (c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\pi\int\left((2x-1)^{\frac{3}{4}}\right)^2\,dx$ | B1 | For $\pi\int\left((2x-1)^{\frac{3}{4}}\right)^2$ or $\pi\int(2x-1)^{\frac{3}{2}}$. Ignore limits and $dx$. Can be implied. |
| $\left\{\int_{\frac{1}{2}}^{\frac{17}{2}}y^2\,dx\right\} = \left[\frac{(2x-1)^{\frac{5}{2}}}{5}\right]_{\frac{1}{2}}^{\frac{17}{2}} = \left(\frac{16^2}{5}\right)-(0) = \frac{1024}{5}$ | M1 | Applies $x$-limits of "8.5" (their answer to part (b)) and $0.5$ to an expression of the form $\pm\beta(2x-1)^{\frac{5}{2}}$; $\beta\neq0$ and subtracts the correct way round |
| $\{V_{\text{cylinder}}\} = \pi(8)^2\left(\frac{17}{2}\right) \{=544\pi\}$ | B1 ft | $\pi(8)^2(\text{their answer to part }(b))$. $V_{\text{cylinder}}=544\pi$ implies this mark |
| $\left\{\text{Vol}(S)=544\pi - \frac{1024}{5}\pi\right\} \Rightarrow \text{Vol}(S)=\frac{1696}{5}\pi$ | A1 | An exact correct answer in the form $k\pi$. E.g. $\frac{1696}{5}\pi$, $\frac{3392}{10}\pi$ or $339.2\pi$ |
## Alt. Part (c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\text{Vol}(S)=\pi(8)^2\left(\frac{1}{2}\right)+\pi\int_{0.5}^{8.5}\left(8^2-(2x-1)^{\frac{3}{2}}\right)dx$ | B1 | For $\pi\int\ldots(2x-1)^{\frac{3}{2}}$. Ignore limits and $dx$ |
| $=\pi(8)^2\left(\frac{1}{2}\right)+\pi\left[64x-\frac{1}{5}(2x-1)^{\frac{5}{2}}\right]_{0.5}^{8.5}$ | — | — |
| Substitution of limits | M1, B1 | As above |
| $\left\{=32\pi+\pi\left(\left(544-\frac{1024}{5}\right)-(32-0)\right)\right\}\Rightarrow\text{Vol}(S)=\frac{1696}{5}\pi$ | A1 | — |
---7. (a) Find
$$\int ( 2 x - 1 ) ^ { \frac { 3 } { 2 } } d x$$
giving your answer in its simplest form.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{cbfbb690-bc85-46e5-a97f-35df4b6f1c84-13_727_1177_596_370}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
Figure 3 shows a sketch of part of the curve $C$ with equation
$$y = ( 2 x - 1 ) ^ { \frac { 3 } { 4 } } , \quad x \geqslant \frac { 1 } { 2 }$$
The curve $C$ cuts the line $y = 8$ at the point $P$ with coordinates $( k , 8 )$, where $k$ is a constant.\\
(b) Find the value of $k$.
The finite region $S$, shown shaded in Figure 3, is bounded by the curve $C$, the $x$-axis, the $y$-axis and the line $y = 8$. This region is rotated through $2 \pi$ radians about the $x$-axis to form a solid of revolution.\\
(c) Find the exact value of the volume of the solid generated.
\hfill \mbox{\textit{Edexcel C4 2016 Q7 [8]}}