Let g(x)= cos x^2, f(x)=√x, and α ,β (α <β ) be the roots of the quadratic equation 18x^2-9π x+π ^2=0. Then the area (in sq. units) bounded by the…

Given, $g(x)=\mathrm{cos}{x}^{2}, f(x)=\sqrt{x}$g(x)=cosx2,f(x)=x​

Given quadratic is $18{x}^{2}-9\pi x+{\pi }^{2}=0$18x2−9πx+π2=0

The roots are $x=\frac{9\pi \pm \sqrt{81{\pi }^{2}-72{\pi }^{2}}}{2\times 18}$x=2×189π±81π2−72π2​​

$$\Rightarrow x=\frac{9\pi \pm \sqrt{9{\pi }^{2}}}{36}$$⇒x=369π±9π2​​

$$\Rightarrow x=\frac{(9\pm 3)\pi }{36}$$⇒x=36(9±3)π​

$$\Rightarrow x=\frac{\pi }{3},\frac{\pi }{6}$$⇒x=3π​,6π​

$$y=gof(x)=\mathrm{cos}{(\sqrt{x})}^{2}=\mathrm{cos}x$$y=gof(x)=cos(x​)2=cosx

Required Area $\displaystyle ={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\mathrm{cos}⁡x dx$=∫6π​3π​​cos⁡xdx$={[\mathrm{sin}x]}_{\frac{\pi }{6}}^{\frac{\pi }{3}}$=[sinx]6π​3π​​

$$\Rightarrow \frac{\sqrt{3}}{2}-\frac{1}{2}$$⇒23​​−21​