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

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

We have, $f(x) =\sqrt{x}$ and $g(x) = \cos x^2$. Then,

$gof (x) = g(f(x)) = g(\sqrt{x}) =\cos x$

Now,

$18x^2-9πx+π^2=0$

$⇒18x^2-6πx-3πx+π^2 = 0$

$⇒бx (3x-π)-π(x-3π) = 0$

$⇒(6x-π)(x-3π)=0⇒x=\frac{π}{6},\frac{π}{3}⇒α=\frac{π}{6},β=\frac{π}{3}$

Required area = $\int\limits_α^β(gof) (x)dx =\int\limits_{π/6}^{π/3}\cos x\,dx$

$=\left[\sin x\right]_{π/6}^{π/3}=\left(\frac{\sqrt{3}}{2}-\frac{1}{2}\right)$