Let g(x) = cos(x^2),f(x) = sqrt{x},and alpha, | vedclass

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(D) Given the quadratic equation $18x^2 - 9\pi x + \pi^2 = 0$. Factoring the quadratic: $(3x - \pi)(6x - \pi) = 0$. Thus,the roots are $\alpha = \frac

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

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(D) Given the quadratic equation $18x^2 - 9\pi x + \pi^2 = 0$. Factoring the quadratic: $(3x - \pi)(6x - \pi) = 0$. Thus,the roots are $\alpha = \frac{\pi}{6}$ and $\beta = \frac{\pi}{3}$ (since $\alpha Next,we find the composite function $(g \circ f)(x) = g(f(x)) = \cos((\sqrt{x})^2) = \cos(x)$. The area $A$ bounded by $y = \cos(x)$,$x = \frac{\pi}{6}$,$x = \frac{\pi}{3}$,and $y = 0$ is given by the integral: $A = \int_{\pi/6}^{\pi/3} \cos(x) \, dx$. Evaluating the integral: $A = [\sin(x)]_{\pi/6}^{\pi/3} = \sin(\frac{\pi}{3}) - \sin(\frac{\pi}{6})$. Substituting the values: $A = \frac{\sqrt{3}}{2} - \frac{1}{2} = \frac{\sqrt{3} - 1}{2}$ sq. units.

Let $g(x) = \cos(x^2)$,$f(x) = \sqrt{x}$,and $\alpha, \beta$ (where $\alpha < \beta$) 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:

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