Edexcel C4 2011 June — Question 4 15 marks

Exam BoardEdexcel
ModuleC4 (Core Mathematics 4)
Year2011
SessionJune
Marks15
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicNumerical integration
TypeComplete table then apply trapezium rule
DifficultyStandard +0.3 This is a standard C4 integration question combining numerical methods (trapezium rule) with integration by parts. Part (a) requires calculator work, (b) is routine trapezium rule application, (c) is a guided substitution, and (d) requires integration by parts twice but with clear structure. The question is slightly easier than average due to its highly scaffolded nature and standard techniques.
Spec1.06d Natural logarithm: ln(x) function and properties1.08i Integration by parts1.09f Trapezium rule: numerical integration4.08e Mean value of function: using integral
4. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{9d513d77-b8f9-4223-832f-f566c5f50457-05_673_1058_264_443} \captionsetup{labelformat=empty} \caption{Figure 2}
\end{figure} Figure 2 shows a sketch of the curve with equation \(y = x ^ { 3 } \ln \left( x ^ { 2 } + 2 \right) , x \geqslant 0\).
The finite region \(R\), shown shaded in Figure 2, is bounded by the curve, the \(x\)-axis and the line \(x = \sqrt { } 2\). The table below shows corresponding values of \(x\) and \(y\) for \(y = x ^ { 3 } \ln \left( x ^ { 2 } + 2 \right)\).
\(x\)0\(\frac { \sqrt { } 2 } { 4 }\)\(\frac { \sqrt { } 2 } { 2 }\)\(\frac { 3 \sqrt { } 2 } { 4 }\)\(\sqrt { } 2\)
\(y\)00.32403.9210
  1. Complete the table above giving the missing values of \(y\) to 4 decimal places.
  2. Use the trapezium rule, with all the values of \(y\) in the completed table, to obtain an estimate for the area of \(R\), giving your answer to 2 decimal places.
  3. Use the substitution \(u = x ^ { 2 } + 2\) to show that the area of \(R\) is $$\frac { 1 } { 2 } \int _ { 2 } ^ { 4 } ( u - 2 ) \ln u \mathrm {~d} u$$
  4. Hence, or otherwise, find the exact area of \(R\).
AnswerMarks Guidance
(a) awrt 0.0333, 1.3596B1 B1 (2)
(b) Area\((R) \approx \frac{1}{2} \times \frac{\sqrt{2}}{4}[\ldots]\)B1
\(\approx \ldots [0 + 2(0.0333 + 0.3240 + 1.3596) + 3.9210]\)M1
\(\approx 1.30\) AcceptA1 (3)
(c) \(u = x^2 + 2 \Rightarrow \frac{du}{dx} = 2x\)B1
Area\((R) = \int_0^{\sqrt{2}} x^3 \ln(x^2 + 2) dx\)B1
\(\int x^3 \ln(x^2 + 2) dx = \int x^2 \ln(x^2 + 2) x dx = \int(u-2)(\ln u)\frac{1}{2} du\)M1
Hence Area\((R) = \frac{1}{2}\left[\int_2^4 (u-2)\ln u \, du\right]\) *A1 cso
(d) \(\int(u-2)\ln u \, du = \left(\frac{u^2}{2} - 2u\right)\ln u - \int\left(\frac{u^2}{2} - 2u\right)\frac{1}{u} du\)M1 A1
\(= \left(\frac{u^2}{2} - 2u\right)\ln u - \int\left(\frac{u}{2} - 2\right) du\)M1 A1
\(= \left(\frac{u^2}{2} - 2u\right)\ln u - \left(\frac{u^2}{4} - 2u\right) (+C)\)M1 A1
Area\((R) = \frac{1}{2}\left[\left(\frac{u^2}{2} - 2u\right)\ln u - \left(\frac{u^2}{4} - 2u\right)\right]_2^4\)M1
\(= \frac{1}{2}\left[(8-8)\ln 4 - 4 + 8 - ((2-4)\ln 2 - 1 + 4)\right]\)M1
\(= \frac{1}{2}(2\ln 2 + 1)\)A1 \(\ln 2 + \frac{1}{2}\) 
**(a)** awrt 0.0333, 1.3596 | B1 B1 | (2)

**(b)** Area$(R) \approx \frac{1}{2} \times \frac{\sqrt{2}}{4}[\ldots]$ | B1 |

$\approx \ldots [0 + 2(0.0333 + 0.3240 + 1.3596) + 3.9210]$ | M1 |

$\approx 1.30$ Accept | A1 | (3)

**(c)** $u = x^2 + 2 \Rightarrow \frac{du}{dx} = 2x$ | B1 |

Area$(R) = \int_0^{\sqrt{2}} x^3 \ln(x^2 + 2) dx$ | B1 |

$\int x^3 \ln(x^2 + 2) dx = \int x^2 \ln(x^2 + 2) x dx = \int(u-2)(\ln u)\frac{1}{2} du$ | M1 |

Hence Area$(R) = \frac{1}{2}\left[\int_2^4 (u-2)\ln u \, du\right]$ * | A1 | cso | (4)

**(d)** $\int(u-2)\ln u \, du = \left(\frac{u^2}{2} - 2u\right)\ln u - \int\left(\frac{u^2}{2} - 2u\right)\frac{1}{u} du$ | M1 A1 |

$= \left(\frac{u^2}{2} - 2u\right)\ln u - \int\left(\frac{u}{2} - 2\right) du$ | M1 A1 |

$= \left(\frac{u^2}{2} - 2u\right)\ln u - \left(\frac{u^2}{4} - 2u\right) (+C)$ | M1 A1 |

Area$(R) = \frac{1}{2}\left[\left(\frac{u^2}{2} - 2u\right)\ln u - \left(\frac{u^2}{4} - 2u\right)\right]_2^4$ | M1 |

$= \frac{1}{2}\left[(8-8)\ln 4 - 4 + 8 - ((2-4)\ln 2 - 1 + 4)\right]$ | M1 |

$= \frac{1}{2}(2\ln 2 + 1)$ | A1 | $\ln 2 + \frac{1}{2}$ | (6) [15] |

---
4.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{9d513d77-b8f9-4223-832f-f566c5f50457-05_673_1058_264_443}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}

Figure 2 shows a sketch of the curve with equation $y = x ^ { 3 } \ln \left( x ^ { 2 } + 2 \right) , x \geqslant 0$.\\
The finite region $R$, shown shaded in Figure 2, is bounded by the curve, the $x$-axis and the line $x = \sqrt { } 2$.

The table below shows corresponding values of $x$ and $y$ for $y = x ^ { 3 } \ln \left( x ^ { 2 } + 2 \right)$.

\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | }
\hline
$x$ & 0 & $\frac { \sqrt { } 2 } { 4 }$ & $\frac { \sqrt { } 2 } { 2 }$ & $\frac { 3 \sqrt { } 2 } { 4 }$ & $\sqrt { } 2$ \\
\hline
$y$ & 0 &  & 0.3240 &  & 3.9210 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Complete the table above giving the missing values of $y$ to 4 decimal places.
\item Use the trapezium rule, with all the values of $y$ in the completed table, to obtain an estimate for the area of $R$, giving your answer to 2 decimal places.
\item Use the substitution $u = x ^ { 2 } + 2$ to show that the area of $R$ is

$$\frac { 1 } { 2 } \int _ { 2 } ^ { 4 } ( u - 2 ) \ln u \mathrm {~d} u$$
\item Hence, or otherwise, find the exact area of $R$.
\end{enumerate}

\hfill \mbox{\textit{Edexcel C4 2011 Q4 [15]}}
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