| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2017 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Indefinite & Definite Integrals |
| Type | Integration by substitution |
| Difficulty | Standard +0.3 This is a standard integration by substitution question with a guided approach. Part (a) provides the substitution and asks students to verify a given result (easier than finding it independently). Part (b) applies the volume of revolution formula using the result from (a). While it requires careful algebraic manipulation and understanding of the relationship between the integrand in (a) and the y² term needed for (b), the question is structured with clear guidance and uses routine techniques expected at C3/C4 level. |
| Spec | 1.08d Evaluate definite integrals: between limits1.08h Integration by substitution4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{du}{dx} = 2\) | B1 | States or uses \(\frac{du}{dx} = 2\) or equivalent such as \(\frac{dx}{du} = 0.5\) or \(dx = \frac{1}{2}du\) |
| \(\int \frac{x}{(2x+3)^2}dx = \int \frac{u-3}{4u^2}du\) | M1 | Expression of the form \(k\int\frac{u-3}{u^2}du\), allow missing \(du\) and/or missing integral sign |
| \(= \int \frac{1}{4}u^{-1} - \frac{3}{4}u^{-2}\,du\) | dM1 | Splits into form \(\ldots u^{-1} \pm \ldots u^{-2}\), allow missing \(du\) and/or integral sign |
| \(= \frac{1}{4}\ln u + \frac{3}{4}u^{-1}\) | ddM1 A1 | For \(\ldots\ln u \pm \ldots u^{-1}\). A1: \(\frac{1}{4}\ln u + \frac{3}{4}u^{-1}\) or equivalent |
| \(\left[\frac{1}{4}\ln u + \frac{3}{4}u^{-1}\right]_3^{27} = \frac{1}{4}\ln 27 + \frac{3}{4}\times\frac{1}{27} - \left(\frac{1}{4}\ln 3 + \frac{3}{4}\times\frac{1}{3}\right) = \frac{1}{4}\ln 9 - \frac{8}{36}\) | M1 | Applies limits of 27 and 3 (in \(u\)), subtracts correct way, combines ln terms correctly. Or using \(u = 2x+3\), limits \(x=12\) and \(x=0\) |
| \(= \frac{1}{2}\ln 3 - \frac{2}{9}\) * | A1* | Given answer achieved correctly without errors. Omission of dM1 line could be allowed if implied. Need intermediate step with correct ln work before final answer |
| (7) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\int x(2x+3)^{-2}dx = \frac{x(2x+3)^{-1}}{-2} + \int\frac{(2x+3)^{-1}}{2}dx\) | M1 | |
| \(= \frac{x(2x+3)^{-1}}{-2} + \frac{1}{4}\ln(2x+3)\) | dM1 | |
| \(\left[\frac{x(2x+3)^{-1}}{-2} + \frac{1}{4}\ln(2x+3)\right]_0^{12}\) | ddM1 | |
| \(= -\frac{2}{9} + \frac{1}{4}\ln 27 - \frac{1}{4}\ln 3\) | A1 | Correct unsimplified answer |
| \(= -\frac{2}{9} + \frac{1}{4}\ln\left(\frac{27}{3}\right)\) | M1 | Collects log terms |
| \(= \frac{1}{2}\ln 3 - \frac{2}{9}\) | A0* | Scores A0* (max 5/7 without substitution method) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\int x(2x+3)^{-2}dx = \int\frac{\frac{1}{2}}{(2x+3)} + \frac{-\frac{3}{2}}{(2x+3)^2}dx\) | B0 | |
| One term of \(= \frac{1}{4}\ln(2x+3) + \frac{3(2x+3)^{-1}}{4}\) | M1 | |
| Both terms of \(= \frac{1}{4}\ln(2x+3) + \frac{3(2x+3)^{-1}}{4}\) | dM1 | |
| \(\left[\frac{1}{4}\ln(2x+3) + \frac{3(2x+3)^{-1}}{4}\right]_0^{12}\) | ddM1 | |
| \(= \frac{1}{4}\ln 27 + \frac{1}{36} - \frac{1}{4}\ln 3 - \frac{1}{4}\) | A1 | |
| \(= -\frac{2}{9} + \frac{1}{4}\ln\left(\frac{27}{3}\right)\) | M1 | |
| \(= \frac{1}{2}\ln 3 - \frac{2}{9}\) | A0* | Max 5/7 without substitution |
| (7) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(V = \pi \times \int_0^{12}\left(\frac{9\sqrt{x}}{2x+3}\right)^2 dx\) | M1 | Attempts to use part (a) to find exact volume. Condone omission of \(\pi\) or 81 or bracket |
| \(= 81\pi\left(\frac{1}{2}\ln 3 - \frac{2}{9}\right)\) | A1 | Any correct exact equivalent in terms of \(\ln 3\) and \(\pi\). Accept e.g. \(81\pi(\ln\sqrt{3}) - 18\pi\). Correct answer implies both marks |
| (2) |
# Question 9:
## Part (a) - Main scheme (substitution):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{du}{dx} = 2$ | B1 | States or uses $\frac{du}{dx} = 2$ or equivalent such as $\frac{dx}{du} = 0.5$ or $dx = \frac{1}{2}du$ |
| $\int \frac{x}{(2x+3)^2}dx = \int \frac{u-3}{4u^2}du$ | M1 | Expression of the form $k\int\frac{u-3}{u^2}du$, allow missing $du$ and/or missing integral sign |
| $= \int \frac{1}{4}u^{-1} - \frac{3}{4}u^{-2}\,du$ | dM1 | Splits into form $\ldots u^{-1} \pm \ldots u^{-2}$, allow missing $du$ and/or integral sign |
| $= \frac{1}{4}\ln u + \frac{3}{4}u^{-1}$ | ddM1 A1 | For $\ldots\ln u \pm \ldots u^{-1}$. A1: $\frac{1}{4}\ln u + \frac{3}{4}u^{-1}$ or equivalent |
| $\left[\frac{1}{4}\ln u + \frac{3}{4}u^{-1}\right]_3^{27} = \frac{1}{4}\ln 27 + \frac{3}{4}\times\frac{1}{27} - \left(\frac{1}{4}\ln 3 + \frac{3}{4}\times\frac{1}{3}\right) = \frac{1}{4}\ln 9 - \frac{8}{36}$ | M1 | Applies limits of 27 and 3 (in $u$), subtracts correct way, combines ln terms correctly. Or using $u = 2x+3$, limits $x=12$ and $x=0$ |
| $= \frac{1}{2}\ln 3 - \frac{2}{9}$ * | A1* | Given answer achieved correctly without errors. Omission of dM1 line could be allowed if implied. Need intermediate step with correct ln work before final answer |
| | **(7)** | |
## Part (a) - Alternative: By Parts:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\int x(2x+3)^{-2}dx = \frac{x(2x+3)^{-1}}{-2} + \int\frac{(2x+3)^{-1}}{2}dx$ | M1 | |
| $= \frac{x(2x+3)^{-1}}{-2} + \frac{1}{4}\ln(2x+3)$ | dM1 | |
| $\left[\frac{x(2x+3)^{-1}}{-2} + \frac{1}{4}\ln(2x+3)\right]_0^{12}$ | ddM1 | |
| $= -\frac{2}{9} + \frac{1}{4}\ln 27 - \frac{1}{4}\ln 3$ | A1 | Correct unsimplified answer |
| $= -\frac{2}{9} + \frac{1}{4}\ln\left(\frac{27}{3}\right)$ | M1 | Collects log terms |
| $= \frac{1}{2}\ln 3 - \frac{2}{9}$ | A0* | Scores A0* (max 5/7 without substitution method) |
## Part (a) - Alternative: By Partial Fractions:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\int x(2x+3)^{-2}dx = \int\frac{\frac{1}{2}}{(2x+3)} + \frac{-\frac{3}{2}}{(2x+3)^2}dx$ | B0 | |
| One term of $= \frac{1}{4}\ln(2x+3) + \frac{3(2x+3)^{-1}}{4}$ | M1 | |
| Both terms of $= \frac{1}{4}\ln(2x+3) + \frac{3(2x+3)^{-1}}{4}$ | dM1 | |
| $\left[\frac{1}{4}\ln(2x+3) + \frac{3(2x+3)^{-1}}{4}\right]_0^{12}$ | ddM1 | |
| $= \frac{1}{4}\ln 27 + \frac{1}{36} - \frac{1}{4}\ln 3 - \frac{1}{4}$ | A1 | |
| $= -\frac{2}{9} + \frac{1}{4}\ln\left(\frac{27}{3}\right)$ | M1 | |
| $= \frac{1}{2}\ln 3 - \frac{2}{9}$ | A0* | Max 5/7 without substitution |
| | **(7)** | |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $V = \pi \times \int_0^{12}\left(\frac{9\sqrt{x}}{2x+3}\right)^2 dx$ | M1 | Attempts to use part (a) to find exact volume. Condone omission of $\pi$ or 81 or bracket |
| $= 81\pi\left(\frac{1}{2}\ln 3 - \frac{2}{9}\right)$ | A1 | Any correct exact equivalent in terms of $\ln 3$ and $\pi$. Accept e.g. $81\pi(\ln\sqrt{3}) - 18\pi$. Correct answer implies both marks |
| | **(2)** | |9.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{e30f0c28-1695-40a1-8e9a-6ea7e29042bf-16_727_1491_258_239}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
\begin{enumerate}[label=(\alph*)]
\item By using the substitution $u = 2 x + 3$, show that
$$\int _ { 0 } ^ { 12 } \frac { x } { ( 2 x + 3 ) ^ { 2 } } \mathrm {~d} x = \frac { 1 } { 2 } \ln 3 - \frac { 2 } { 9 }$$
The curve $C$ has equation
$$y = \frac { 9 \sqrt { x } } { ( 2 x + 3 ) } , \quad x > 0$$
The finite region $R$, shown shaded in Figure 3, is bounded by the curve $C$, the $x$-axis and the line with equation $x = 12$. The region $R$ is rotated through $2 \pi$ radians about the $x$-axis to form a solid of revolution.
\item Use the result of part (a) to find the exact value of the volume of the solid generated.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C34 2017 Q9 [9]}}