3.
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{cd958ff3-ed4e-4bd7-aa4b-339da6d618a6-08_560_1082_242_438}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\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 \(y = \frac { 6 } { \left( \mathrm { e } ^ { x } + 2 \right) }\)
| \(x\) | 0 | 0.2 | 0.4 | 0.6 | 0.8 | 1 |
| \(y\) | 2 | | 1.71830 | 1.56981 | 1.41994 | 1.27165 |
- Complete the table above by giving the missing value of \(y\) to 5 decimal places.
- 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.
- 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.
- Hence use calculus to find the exact area of \(R\). [Solutions based entirely on graphical or numerical methods are not acceptable.]