JEE Main PAPER - 1

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