| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2013 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Numerical integration |
| Type | Complete table then apply trapezium rule |
| Difficulty | Moderate -0.3 This is a straightforward multi-part question testing standard C4 techniques: substitution into a parametric equation, trapezium rule application, and integration by parts. While part (c) requires integration by parts (a key C4 skill), the structure is highly scaffolded with clear steps and no novel problem-solving required. Slightly easier than average due to the routine nature of all parts. |
| Spec | 1.08d Evaluate definite integrals: between limits1.09f Trapezium rule: numerical integration |
| \(t\) | 0 | 2 | 4 | 6 | 8 |
| \(x\) | 3 | 7.107 | 7.218 | 5.223 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(6.248046798... = 6.248\) (3dp) | B1 | 6.248 or awrt 6.248 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Area \(\approx \frac{1}{2} \times 2 \times \left[3 + 2(7.107 + 7.218 + 6.248) + 5.223\right]\) | B1; M1 | Outside brackets \(\frac{1}{2}\times 2\); correct trapezium rule structure. Allow one miscopy of values |
| \(= 49.369 = 49.37\) (2 dp) | A1 | 49.37 or awrt 49.37 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int 4te^{-\frac{1}{3}t} \rightarrow \pm Ate^{-\frac{1}{3}t} \pm B\int e^{-\frac{1}{3}t}\{dt\}\), \(A\neq 0\), \(B\neq 0\) | M1 | |
| \(te^{-\frac{1}{3}t} \rightarrow -3te^{-\frac{1}{3}t} - \int -3e^{-\frac{1}{3}t}\) | A1 | |
| \(+ 3t\) | B1 | \(3 \rightarrow 3t\) |
| \(= -12te^{-\frac{1}{3}t} - 36e^{-\frac{1}{3}t}\ \{+3t\}\) | A1 | \(-12te^{-\frac{1}{3}t} - 36e^{-\frac{1}{3}t}\) |
| Substitutes limits 8 and 0 into integrated function of form \(\pm\lambda te^{-\frac{1}{3}t} \pm \mu e^{-\frac{1}{3}t}\) or \(\pm\lambda te^{-\frac{1}{3}t} \pm \mu e^{-\frac{1}{3}t} + Bt\), subtracts correct way | dM1 | |
| \(= \left(-96e^{-\frac{8}{3}} - 36e^{-\frac{8}{3}} + 24\right) - (0 - 36 + 0)\) | ||
| \(= 60 - 132e^{-\frac{8}{3}}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Difference \(= \left\ | 60 - 132e^{-\frac{8}{3}} - 49.37\right\ | = 1.458184439... = 1.46\) (2 dp) |
# Question 5:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $6.248046798... = 6.248$ (3dp) | B1 | 6.248 or awrt 6.248 |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Area $\approx \frac{1}{2} \times 2 \times \left[3 + 2(7.107 + 7.218 + 6.248) + 5.223\right]$ | B1; M1 | Outside brackets $\frac{1}{2}\times 2$; correct trapezium rule structure. Allow one miscopy of values |
| $= 49.369 = 49.37$ (2 dp) | A1 | 49.37 or awrt 49.37 |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int 4te^{-\frac{1}{3}t} \rightarrow \pm Ate^{-\frac{1}{3}t} \pm B\int e^{-\frac{1}{3}t}\{dt\}$, $A\neq 0$, $B\neq 0$ | M1 | |
| $te^{-\frac{1}{3}t} \rightarrow -3te^{-\frac{1}{3}t} - \int -3e^{-\frac{1}{3}t}$ | A1 | |
| $+ 3t$ | B1 | $3 \rightarrow 3t$ |
| $= -12te^{-\frac{1}{3}t} - 36e^{-\frac{1}{3}t}\ \{+3t\}$ | A1 | $-12te^{-\frac{1}{3}t} - 36e^{-\frac{1}{3}t}$ |
| Substitutes limits 8 and 0 into integrated function of form $\pm\lambda te^{-\frac{1}{3}t} \pm \mu e^{-\frac{1}{3}t}$ or $\pm\lambda te^{-\frac{1}{3}t} \pm \mu e^{-\frac{1}{3}t} + Bt$, subtracts correct way | dM1 | |
| $= \left(-96e^{-\frac{8}{3}} - 36e^{-\frac{8}{3}} + 24\right) - (0 - 36 + 0)$ | | |
| $= 60 - 132e^{-\frac{8}{3}}$ | A1 | |
## Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Difference $= \left\|60 - 132e^{-\frac{8}{3}} - 49.37\right\| = 1.458184439... = 1.46$ (2 dp) | B1 | 1.46 or awrt 1.46 |
---5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{08f62966-2e63-4542-a10a-c6453a3215e7-06_689_992_118_484}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows part of the curve with equation $x = 4 t \mathrm { e } ^ { - \frac { 1 } { 3 } t } + 3$. The finite region $R$ shown shaded in Figure 1 is bounded by the curve, the $x$-axis, the $t$-axis and the line $t = 8$.
\begin{enumerate}[label=(\alph*)]
\item Complete the table with the value of $x$ corresponding to $t = 6$, giving your answer to 3 decimal places.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | }
\hline
$t$ & 0 & 2 & 4 & 6 & 8 \\
\hline
$x$ & 3 & 7.107 & 7.218 & & 5.223 \\
\hline
\end{tabular}
\end{center}
\item Use the trapezium rule with all the values of $x$ in the completed table to obtain an estimate for the area of the region $R$, giving your answer to 2 decimal places.
\item Use calculus to find the exact value for the area of $R$.
\item Find the difference between the values obtained in part (b) and part (c), giving your answer to 2 decimal places.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 2013 Q5 [11]}}