Mathematics - Area Under Curves Question with Solution | TestHub

Given, $g\left(x\right)=\cos x^2, f\left(x\right)=\sqrt{x}$

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

The roots are $x=\frac{9\pi \pm \sqrt{81{\pi }^2-72{\pi }^2}}{2\times 18}$

$\Rightarrow x=\frac{9\pi \pm \sqrt{9{\pi }^2}}{36}$

$\Rightarrow x=\frac{\left(9\pm 3\right)\pi }{36}$

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

$y=gof\left(x\right)=\cos {\left(\sqrt{x}\right)}^2=\cos x$

Required Area $={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\cos x dx$$={\left[\sin x\right]}_{\frac{\pi }{6}}^{\frac{\pi }{3}}$
$\Rightarrow \frac{\sqrt{3}}{2}-\frac{1}{2}$