4.
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{245bbe52-3a14-4494-af17-7711caf79b22-10_752_1182_226_395}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{figure}
Figure 1 shows a sketch of part of the curve with equation \(y = \frac { 4 \mathrm { e } ^ { - x } } { 3 \sqrt { } \left( 1 + 3 \mathrm { e } ^ { - x } \right) }\)
The finite region \(R\), shown shaded in Figure 1, is bounded by the curve, the \(x\)-axis, the line \(x = - 3 \ln 2\) and the \(y\)-axis.
The table below shows corresponding values of \(x\) and \(y\) for \(y = \frac { 4 \mathrm { e } ^ { - x } } { 3 \sqrt { } \left( 1 + 3 \mathrm { e } ^ { - x } \right) }\)
| \(x\) | \(- 3 \ln 2\) | \(- 2 \ln 2\) | \(- \ln 2\) | 0 |
| \(y\) | 2.1333 | | 1.0079 | 0.6667 |
- Complete the table above by giving the missing value 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.
- Using the substitution \(u = 1 + 3 \mathrm { e } ^ { - x }\), or otherwise, find
$$\int \frac { 4 \mathrm { e } ^ { - x } } { 3 \sqrt { } \left( 1 + 3 \mathrm { e } ^ { - x } \right) } \mathrm { d } x$$
- Hence find the value of the area of \(R\).