Official solution
Given quadratic equation,
\[18{x^2} - 9\pi x + {\pi ^2} = 0\]
\[ \Rightarrow \,\,\,18{x^2} - 6\pi x - 3\pi x + {\pi ^2} = 0\]
\[ \Rightarrow \,\,\,\,6x\left( {3x - \pi } \right) - \pi \left( {3x - \pi } \right) = 0\]
\[ \Rightarrow \,\,\,\,\left( {3x - \pi } \right)\left( {6x - \pi } \right) = 0\]
\[\therefore\] \[\,\,\,\,x = {\pi \over 3},{\pi \over 6}\]
as \[\,\,\,\, \propto < B\]
\[\therefore\] \[\,\,\,\, \propto = {\pi \over 6}\] \[\,\,\,\,\] and \[\,\,\,\,\] \[\beta = {\pi \over 3}\]
Given, \[g\left( x \right) = \cos {x^2}\] and \[ + \left( x \right) = \sqrt x \]
\[y = \left( {gof} \right)x\]
\[ = \,\,\,\,\,g\left( {f\left( x \right)} \right)\]
\[ = \,\,\,\,\cos \left( {f{{\left( x \right)}^2}} \right)\]
\[ = \,\,\,\,\cos {\left( {\sqrt x} \right)^2}\]
\[ = \,\,\,\,\cos x\]
So, the required area in the curve is
Area \[ = \int\limits_{{\pi \over 6}}^{{\pi \over 3}} {\cos \,\,dx} \]
\[ = \left[ {\sin x} \right]_{{\pi \over 6}}^{{\pi \over 3}}\]
\[ = \sin {\pi \over 3} - \sin {\pi \over 6}\]
\[ = {{\sqrt 3} \over 2} - {1 \over 2}\]
\[ = {{\sqrt 3 - 1} \over 2}\]