5.
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{fe07afad-9cfc-48c0-84f1-5717f81977d4-14_688_691_251_630}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{figure}
Figure 2 shows a sketch of the curve with parametric equations
$$x = \sqrt { 9 - 4 t } \quad y = \frac { t ^ { 3 } } { \sqrt { 9 + 4 t } } \quad 0 \leqslant t \leqslant \frac { 9 } { 4 }$$
The curve touches the \(x\)-axis when \(t = 0\) and meets the \(y\)-axis when \(t = \frac { 9 } { 4 }\)
The region \(R\), shown shaded in Figure 2, is bounded by the curve, the \(x\)-axis and the \(y\)-axis.
- Show that the area of \(R\) is given by
$$K \int _ { 0 } ^ { \frac { 9 } { 4 } } \frac { t ^ { 3 } } { \sqrt { 81 - 16 t ^ { 2 } } } \mathrm {~d} t$$
where \(K\) is a constant to be found.
- Using the substitution \(u = 81 - 16 t ^ { 2 }\), or otherwise, find the exact area of \(R\).
(Solutions relying on calculator technology are not acceptable.)