Let g(x) = cos x ^ 2 , f(x) = x , and α, β(α < β) be the roots of...

Given, $g\left(x\right) = cos{x}^{2}, f\left(x\right) = \sqrt{x}$

Given quadratic is $18{x}^{2} - 9\pix + {\pi}^{2} = 0$

The roots are $x = \dfrac{9\pi \pm \sqrt{81{\pi}^{2} - 72{\pi}^{2}}}{2 \times 18}$

$⇒ x = \dfrac{9\pi \pm \sqrt{9{\pi}^{2}}}{36}$

$⇒ x = \dfrac{\left(9 \pm 3\right)\pi}{36}$

$⇒ x = \dfrac{\pi}{3}, \dfrac{\pi}{6}$

$y = gof\left(x\right) = cos{\left(\sqrt{x}\right)}^{2} = \cos x$

Required Area $= {∫}_{\dfrac{\pi}{6}}^{\dfrac{\pi}{3}}cos xdx$$= {\left[\sin x\right]}_{\dfrac{\pi}{6}}^{\dfrac{\pi}{3}}$
$⇒ \dfrac{\sqrt{3}}{2} - \dfrac{1}{2}$