Let f : [0, π/2] → [0, 1] be the function defined by f(x) = sin² x and let g : [0, π/2] → [0, ∞) be the function defined by g(x) = √ πx/2 − x² .

The value of 2∫₀^π/2 f(x)g(x) dx − ∫₀^π/2 g(x) dx is __________.

The correct answer is 0.

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

Expression to Evaluate:
We need to evaluate the integral expression:
$I=2{\int }_0^{\pi /2}f\left(x\right)g\left(x\right)dx-{\int }_0^{\pi /2}g\left(x\right)dx$

Let us consider the first definite integral:
$I_1={\int }_0^{\pi /2}f\left(x\right)g\left(x\right)dx={\int }_0^{\pi /2}{sin}^{2}x·<!--\; ·\; -->g\left(x\right)dx$

Applying King's Property of Definite Integrals:
Using the property ${\int }_a^{b}h\left(x\right)dx={\int }_a^{b}h(a+b-x)dx$, we replace $x$ with $\frac{\pi }{2}-x$.

First, let us examine $g\left(\frac{\pi }{2}-x\right)$:
$g\left(\frac{\pi }{2}-x\right)=\sqrt{\frac{\pi }{2}\left(\frac{\pi }{2}-x\right)-{\left(\frac{\pi }{2}-x\right)}^2}$
$=\sqrt{\frac{{\pi }^2}{4}-\frac{\pi x}{2}-\left(\frac{{\pi }^2}{4}-\pi x+x^2\right)}$
$=\sqrt{\frac{\pi x}{2}-x^2}=g\left(x\right)$

Thus, $g\left(x\right)$ is symmetric about $x=\frac{\pi }{4}$, i.e., $g\left(\frac{\pi }{2}-x\right)=g\left(x\right)$.

Now apply King's Property to $I_1$:
$I_1={\int }_0^{\pi /2}{sin}^2\left(\frac{\pi }{2}-x\right)·<!--\; ·\; -->g\left(\frac{\pi }{2}-x\right)dx$
$I_1={\int }_0^{\pi /2}{cos}^{2}x·<!--\; ·\; -->g\left(x\right)dx$

Adding the two forms of $I_1$:
$2I_1={\int }_0^{\pi /2}{sin}^{2}x·<!--\; ·\; -->g\left(x\right)dx+{\int }_0^{\pi /2}{cos}^{2}x·<!--\; ·\; -->g\left(x\right)dx$
$2I_1={\int }_0^{\pi /2}\left({sin}^{2}x+{cos}^{2}x\right)g\left(x\right)dx$
Since ${sin}^{2}x+{cos}^{2}x=1$, we have:
$2I_1={\int }_0^{\pi /2}g\left(x\right)dx$

Final Calculation:
Substitute $2I_1=2{\int }_0^{\pi /2}f\left(x\right)g\left(x\right)dx$ back into our target expression $I$:
$I=2{\int }_0^{\pi /2}f\left(x\right)g\left(x\right)dx-{\int }_0^{\pi /2}g\left(x\right)dx$
$I={\int }_0^{\pi /2}g\left(x\right)dx-{\int }_0^{\pi /2}g\left(x\right)dx=0$