Let $f:[0,\frac{\pi }{2}]\to [0,1]$ be the function defined by $f\left(x\right)={\sin }^{2}x$ and let $g:[0,\frac{\pi }{2}]\to [0,\infty )$ be the function defined by $g\left(x\right)=\sqrt{\frac{\pi x}{2}-x^2}$.

$\text{<br>}$

$\text{The value of }\frac{16}{{\mathrm{ \pi }}^3}{\int }_0^{\frac{\pi }{2}}f\left(x\right)g\left(x\right)dx\text{ is}\text{\_\_\_\_\_}$

The correct answer is 0.25.

We are given two functions defined on the interval $[0,\frac{\pi }{2}]$:
$f\left(x\right)={\sin }^{2}x$
$g\left(x\right)=\sqrt{\frac{\pi x}{2}-x^2}$

We need to evaluate the expression:
$I=\frac{16}{{\pi }^3}{\int }_0^{\frac{\pi }{2}}f\left(x\right)g\left(x\right)dx$

Step 1: Simplify the Integral using King's Property
Let $J={\int }_0^{\frac{\pi }{2}}{\sin }^{2}x·g\left(x\right)dx$  ... (Equation 1)

Using the definite integral property ${\int }_0^{a}h\left(x\right)dx={\int }_0^{a}h(a-x)dx$ with $a=\frac{\pi }{2}$, we replace $x$ with $\frac{\pi }{2}-x$:

First, evaluate $f(\frac{\pi }{2}-x)={\sin }^2(\frac{\pi }{2}-x)={\cos }^{2}x$.

Next, evaluate $g(\frac{\pi }{2}-x)$:
$g(\frac{\pi }{2}-x)=\sqrt{\frac{\pi }{2}(\frac{\pi }{2}-x)-{(\frac{\pi }{2}-x)}^2}$
$=\sqrt{\frac{{\pi }^2}{4}-\frac{\pi x}{2}-(\frac{{\pi }^2}{4}-\pi x+x^2)}$
$=\sqrt{\frac{\pi x}{2}-x^2}=g\left(x\right)$

Therefore, applying King's Property gives:
$J={\int }_0^{\frac{\pi }{2}}{\cos }^{2}x·g\left(x\right)dx$  ... (Equation 2)

Adding Equation 1 and Equation 2:
$2J={\int }_0^{\frac{\pi }{2}}({\sin }^{2}x+{\cos }^{2}x)g\left(x\right)dx$
Since ${\sin }^{2}x+{\cos }^{2}x=1$, we get:
$2J={\int }_0^{\frac{\pi }{2}}g\left(x\right)dx$
$J=\frac{1}{2}{\int }_0^{\frac{\pi }{2}}\sqrt{\frac{\pi x}{2}-x^2}dx$

Step 2: Evaluate the integral geometrically
We complete the square for the quadratic term inside the square root:
$\frac{\pi x}{2}-x^2=-(x^2-\frac{\pi x}{2})={\left(\frac{\pi }{4}\right)}^2-{(x-\frac{\pi }{4})}^2$

Thus, the integral ${\int }_0^{\frac{\pi }{2}}\sqrt{{\left(\frac{\pi }{4}\right)}^2-{(x-\frac{\pi }{4})}^2}dx$ represents the area of a upper semicircle centered at $(\frac{\pi }{4},0)$ with radius $R=\frac{\pi }{4}$.

The area of this semicircle is:
$\text{Area}=\frac{1}{2}\pi R^2=\frac{1}{2}\pi {\left(\frac{\pi }{4}\right)}^2=\frac{{\pi }^3}{32}$

Substituting this area into our expression for $J$:
$J=\frac{1}{2}·\frac{{\pi }^3}{32}=\frac{{\pi }^3}{64}$

Step 3: Calculate the final value
Now, substitute $J$ back into the required expression $I$:
$I=\frac{16}{{\pi }^3}·J=\frac{16}{{\pi }^3}·\frac{{\pi }^3}{64}=\frac{16}{64}=\frac{1}{4}=0.25$

Hence, the value of the given expression is 0.25.