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 $y = (gof)(x)$ and the lines $x = \alpha, x = \beta$ and $y = 0$, is

The roots of the quadratic equation 18x^2 - 9(pi)x + pi^2 = 0 are alpha = pi/6 and beta = pi/3. We evaluate the composite function (gof)(x) = cos((sqrt(x))^2) = cos(x). The required area is the definite integral from alpha to beta of cos(x) dx, which is sin(x) evaluated from pi/6 to pi/3. Calculating this gives sin(pi/3) - sin(pi/6) = sqrt(3)/2 - 1/2 = (sqrt(3) - 1)/2. The result is 1/2(sqrt(3) - 1).

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