4.
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{aa75f1c1-ee97-4fee-af98-957e6a3fbba1-05_476_1338_251_360}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{figure}
Figure 1 shows a sketch of part of the curve with equation \(y = \sqrt { x + 2 } , x \geqslant - 2\)
The finite region \(R\), shown shaded in Figure 1, is bounded by the curve, the \(x\)-axis and the line \(x = 6\)
The table below shows corresponding values of \(x\) and \(y\) for \(y = \sqrt { x + 2 }\)
| \(x\) | - 2 | 0 | 2 | 4 | 6 |
| \(y\) | 0 | 1.4142 | 2 | | 2.8284 |
- Complete the table above, giving the missing value of \(y\) to 4 decimal places.
- Use the trapezium rule, with all of the values of \(y\) in the completed table, to find an approximate value for the area of \(R\), giving your answer to 3 decimal places.
Use your answer to part (b) to find approximate values of
- \(\int _ { - 2 } ^ { 6 } \frac { \sqrt { x + 2 } } { 2 } \mathrm {~d} x\)
- \(\int _ { - 2 } ^ { 6 } ( 2 + \sqrt { x + 2 } ) \mathrm { d } x\)