Show that
$$\int _ { 1 } ^ { a } \left( 6 - \frac { 12 } { \sqrt { x } } \right) \mathrm { d } x = 6 a - 24 \sqrt { a } + 18$$
8
The curve \(y = 6 - \frac { 12 } { \sqrt { x } }\), the line \(x = 1\) and the line \(x = a\) are shown in the diagram below.
The shaded region \(R _ { 1 }\) is bounded by the curve, the line \(x = 1\) and the \(x\)-axis.
The shaded region \(R _ { 2 }\) is bounded by the curve, the line \(x = a\) and the \(x\)-axis.
\includegraphics[max width=\textwidth, alt={}, center]{9cd7f38d-a2a1-4fd3-9ed9-cb389e8ee3b6-09_705_931_632_648}
It is given that the areas of \(R _ { 1 }\) and \(R _ { 2 }\) are equal.
Find the value of \(a\)
Fully justify your answer.