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 | \(\frac { \sqrt { } 2 } { 4 }\) | \(\frac { \sqrt { } 2 } { 2 }\) | \(\frac { 3 \sqrt { } 2 } { 4 }\) | \(\sqrt { } 2\) |
| \(y\) | 0 | | 0.3240 | | 3.9210 |
- Complete the table above giving the missing values of \(y\) to 4 decimal places.
- 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.
- Use the substitution \(u = x ^ { 2 } + 2\) to show that the area of \(R\) is
$$\frac { 1 } { 2 } \int _ { 2 } ^ { 4 } ( u - 2 ) \ln u \mathrm {~d} u$$
- Hence, or otherwise, find the exact area of \(R\).