JEE Advanced 2024 · Paper 2, Mathematics question 17

$y=\sqrt{x\left(\frac\pi2-x\right)}$ is the upper half of the circle with diameter $\left[0,\frac\pi2\right]$ on the $x$-axis, radius $\frac\pi4$. Its integral is the area of that semicircle:
$$\int_0^{\pi/2}g\,dx=\frac12\pi\left(\frac\pi4\right)^2=\frac{\pi^3}{32}.$$

By the previous question, $\int_0^{\pi/2}fg\,dx=\frac12\cdot\frac{\pi^3}{32}=\frac{\pi^3}{64}$. Therefore
$$\frac{16}{\pi^3}\cdot\frac{\pi^3}{64}=\frac14=0.25.$$