6.
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{89d4a7a5-3f4f-4d16-b14e-a27243cedd78-11_666_993_244_392}
\captionsetup{labelformat=empty}
\caption{Diagram not to scale}
\end{figure}
Figure 2
Figure 2 shows a sketch of the curve with equation \(y = \sqrt { ( 3 - x ) ( x + 1 ) } , 0 \leqslant x \leqslant 3\)
The finite region \(R\), shown shaded in Figure 2, is bounded by the curve, the \(x\)-axis, and the \(y\)-axis.
- Use the substitution \(x = 1 + 2 \sin \theta\) to show that
$$\int _ { 0 } ^ { 3 } \sqrt { ( 3 - x ) ( x + 1 ) } d x = k \int _ { - \frac { \pi } { 6 } } ^ { \frac { \pi } { 2 } } \cos ^ { 2 } \theta d \theta$$
where \(k\) is a constant to be determined.
- Hence find, by integration, the exact area of \(R\).