| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2017 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Numerical integration |
| Type | Complete table then apply trapezium rule |
| Difficulty | Standard +0.3 This is a standard C4 integration question combining numerical methods (trapezium rule) with algebraic integration via substitution and partial fractions. Part (a) is simple substitution, (b) is routine trapezium rule application, (c) is a guided substitution, and (d) requires partial fractions—all well-practiced techniques with clear scaffolding. Slightly easier than average due to the step-by-step guidance. |
| Spec | 1.02y Partial fractions: decompose rational functions1.08h Integration by substitution1.08j Integration using partial fractions1.09f Trapezium rule: numerical integration |
| \(x\) | 0 | 0.2 | 0.4 | 0.6 | 0.8 | 1 |
| \(y\) | 2 | 1.71830 | 1.56981 | 1.41994 | 1.27165 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| At \(x = 0.2\), \(y = 1.86254\) (5 dp) | B1 cao | Look for this value in table or working |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{1}{2}(0.2)\Big[2 + 1.27165 + 2(\text{their } 1.86254 + 1.71830 + 1.56981 + 1.41994)\Big]\) | B1 o.e. | Outside brackets \(\frac{1}{2} \times (0.2)\) or \(\frac{1}{10}\) or \(\frac{1}{2} \times \frac{1}{5}\) |
| For structure of \([\ldots]\) | M1 | First and last ordinates once, middle ordinates twice |
| \(= \frac{1}{10}(16.41283) = 1.641283 = 1.6413\) (4 dp) | A1 | Anything that rounds to 1.6413 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(u = e^x\) or \(x = \ln u\); stating \(\frac{du}{dx} = e^x\) or \(\frac{du}{dx} = u\) or \(\frac{dx}{du} = \frac{1}{u}\) or \(du = u\,dx\); and \(\int \frac{6}{(e^x+2)}dx = \int \frac{6}{(u+2)u}du\) | B1* | Must start correctly and end at correct transformed integral with no incorrect working |
| \(\{x=0\} \Rightarrow a = e^0 \Rightarrow a = 1\); \(\{x=1\} \Rightarrow b = e^1 \Rightarrow b = e\) | B1 | \(a=1\) and \(b=e\) or evidence of \(0 \to 1\) and \(1 \to e\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{6}{u(u+2)} \equiv \frac{A}{u} + \frac{B}{(u+2)}\); \(\Rightarrow 6 \equiv A(u+2) + Bu\) | M1 | Writing \(\frac{6}{u(u+2)} \equiv \frac{A}{u} + \frac{B}{(u+2)}\) and complete method for at least one of \(A\) or \(B\) |
| \(u=0 \Rightarrow A=3\); \(u=-2 \Rightarrow B=-3\) | A1 | Both \(A=3\) and \(B=-3\) |
| \(\int \frac{6}{u(u+2)}du = \int\left(\frac{3}{u} - \frac{3}{(u+2)}\right)du\) | M1 | Integrates \(\frac{M}{u} \pm \frac{N}{u \pm k}\), \(M,N,k \neq 0\) (two-term partial fraction) giving \(\pm\lambda\ln(\alpha u)\) or \(\pm\mu\ln(\beta(u \pm k))\) |
| \(= 3\ln u - 3\ln(u+2)\) or \(= 3\ln 2u - 3\ln(2u+4)\) | A1 ft | Integration of both terms correctly followed through from \(M\) and \(N\) |
| \(\Big[3\ln u - 3\ln(u+2)\Big]_1^e\) | dM1 | Applies limits of \(e\) and \(1\); dependent on 2nd M mark |
| \(= 3 - 3\ln(e+2) + 3\ln 3\) or equivalents e.g. \(3\ln\!\left(\frac{e}{e+2}\right) - 3\ln\!\left(\frac{1}{3}\right)\) or \(\ln\!\left(\frac{27e^3}{(e+2)^3}\right)\) | A1 cso | See notes for full list of equivalents |
# Question 3:
## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| At $x = 0.2$, $y = 1.86254$ (5 dp) | B1 cao | Look for this value in table or working |
## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{1}{2}(0.2)\Big[2 + 1.27165 + 2(\text{their } 1.86254 + 1.71830 + 1.56981 + 1.41994)\Big]$ | B1 o.e. | Outside brackets $\frac{1}{2} \times (0.2)$ or $\frac{1}{10}$ or $\frac{1}{2} \times \frac{1}{5}$ |
| For structure of $[\ldots]$ | M1 | First and last ordinates once, middle ordinates twice |
| $= \frac{1}{10}(16.41283) = 1.641283 = 1.6413$ (4 dp) | A1 | Anything that rounds to 1.6413 |
## Part (c)
| Answer | Mark | Guidance |
|--------|------|----------|
| $u = e^x$ or $x = \ln u$; stating $\frac{du}{dx} = e^x$ or $\frac{du}{dx} = u$ or $\frac{dx}{du} = \frac{1}{u}$ or $du = u\,dx$; **and** $\int \frac{6}{(e^x+2)}dx = \int \frac{6}{(u+2)u}du$ | B1* | Must start correctly and end at correct transformed integral with no incorrect working |
| $\{x=0\} \Rightarrow a = e^0 \Rightarrow a = 1$; $\{x=1\} \Rightarrow b = e^1 \Rightarrow b = e$ | B1 | $a=1$ and $b=e$ or evidence of $0 \to 1$ and $1 \to e$ |
## Part (d) Way 1
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{6}{u(u+2)} \equiv \frac{A}{u} + \frac{B}{(u+2)}$; $\Rightarrow 6 \equiv A(u+2) + Bu$ | M1 | Writing $\frac{6}{u(u+2)} \equiv \frac{A}{u} + \frac{B}{(u+2)}$ and complete method for at least one of $A$ or $B$ |
| $u=0 \Rightarrow A=3$; $u=-2 \Rightarrow B=-3$ | A1 | Both $A=3$ **and** $B=-3$ |
| $\int \frac{6}{u(u+2)}du = \int\left(\frac{3}{u} - \frac{3}{(u+2)}\right)du$ | M1 | Integrates $\frac{M}{u} \pm \frac{N}{u \pm k}$, $M,N,k \neq 0$ (two-term partial fraction) giving $\pm\lambda\ln(\alpha u)$ or $\pm\mu\ln(\beta(u \pm k))$ |
| $= 3\ln u - 3\ln(u+2)$ or $= 3\ln 2u - 3\ln(2u+4)$ | A1 ft | Integration of both terms correctly followed through from $M$ and $N$ |
| $\Big[3\ln u - 3\ln(u+2)\Big]_1^e$ | dM1 | Applies limits of $e$ and $1$; dependent on 2nd M mark |
| $= 3 - 3\ln(e+2) + 3\ln 3$ or equivalents e.g. $3\ln\!\left(\frac{e}{e+2}\right) - 3\ln\!\left(\frac{1}{3}\right)$ or $\ln\!\left(\frac{27e^3}{(e+2)^3}\right)$ | A1 cso | See notes for full list of equivalents |
---3.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{cd958ff3-ed4e-4bd7-aa4b-339da6d618a6-08_560_1082_242_438}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows a sketch of part of the curve with equation $y = \frac { 6 } { \left( \mathrm { e } ^ { x } + 2 \right) } , x \in \mathbb { R }$\\
The finite region $R$, shown shaded in Figure 1, is bounded by the curve, the $y$-axis, the $x$-axis and the line with equation $x = 1$
The table below shows corresponding values of $x$ and $y$ for $y = \frac { 6 } { \left( \mathrm { e } ^ { x } + 2 \right) }$
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | }
\hline
$x$ & 0 & 0.2 & 0.4 & 0.6 & 0.8 & 1 \\
\hline
$y$ & 2 & & 1.71830 & 1.56981 & 1.41994 & 1.27165 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Complete the table above by giving the missing value of $y$ to 5 decimal places.
\item Use the trapezium rule, with all the values of $y$ in the completed table, to find an estimate for the area of $R$, giving your answer to 4 decimal places.
\item Use the substitution $u = \mathrm { e } ^ { x }$ to show that the area of $R$ can be given by
$$\int _ { a } ^ { b } \frac { 6 } { u ( u + 2 ) } \mathrm { d } u$$
where $a$ and $b$ are constants to be determined.
\item Hence use calculus to find the exact area of $R$. [Solutions based entirely on graphical or numerical methods are not acceptable.]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 2017 Q3 [12]}}