Let g(x) = cos x², f(x) = \(\sqrt{x}\) and \(\alpha , \beta (\alpha < \beta)\) be the roots of the quadratic equation 18x² - 9\(\pi\)x + \(\pi^2\). then the area bounded by the curve y = (gof) (x) and the lines x = \(\alpha\) , x = \(\beta\) and y = 0 is : 

\(\frac{1}{2} (\sqrt{3}-1)\)

Area = \(\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} (\cos {x} )dx \)

        = \(\frac{1}{2} (\sqrt{3}-1)\)