Let alpha and beta (alpha

Question

(D) Given $f(x) = x^2$ and $g(x) = \cos x$,we have $g(f(x)) = \cos(x^2)$.For the quadratic equation $18x^2 - 9\pi x + \pi^2 = 0$,the roots are given b

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$\frac{1}{2} (\sin \frac{\pi^2}{9} - \sin \frac{\pi^2}{36})$

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(D) Given $f(x) = x^2$ and $g(x) = \cos x$,we have $g(f(x)) = \cos(x^2)$.For the quadratic equation $18x^2 - 9\pi x + \pi^2 = 0$,the roots are given by the quadratic formula:$x = \frac{9\pi \pm \sqrt{(9\pi)^2 - 4(18)(\pi^2)}}{2(18)} = \frac{9\pi \pm \sqrt{81\pi^2 - 72\pi^2}}{36} = \frac{9\pi \pm 3\pi}{36}$.Thus,the roots are $x = \frac{12\pi}{36} = \frac{\pi}{3}$ and $x = \frac{6\pi}{36} = \frac{\pi}{6}$.Since $\alpha We need to evaluate $I = \int_{\pi/6}^{\pi/3} x \cos(x^2) dx$.Let $t = x^2$,then $dt = 2x dx$,which implies $x dx = \frac{dt}{2}$.When $x = \frac{\pi}{6}$,$t = \frac{\pi^2}{36}$. When $x = \frac{\pi}{3}$,$t = \frac{\pi^2}{9}$.Substituting these into the integral:$I = \int_{\pi^2/36}^{\pi^2/9} \cos(t) \frac{dt}{2} = \frac{1}{2} [\sin(t)]_{\pi^2/36}^{\pi^2/9} = \frac{1}{2} (\sin \frac{\pi^2}{9} - \sin \frac{\pi^2}{36})$.

Let $\alpha$ and $\beta$ $(\alpha < \beta)$ be the roots of $18x^2 - 9\pi x + \pi^2 = 0$,$f(x) = x^2$,and $g(x) = \cos x$. Then $\int_{\alpha}^{\beta} x (g \circ f(x)) dx =$

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