Let f : 0 ,   1 → 0 ,   1 be the function defined by f x = x 3 3 - x 2 + 5 9 x + 17 36 .…

Given,

Function $f\left(x\right)=\frac{x^3}{3}-x^2+\frac{5}{9}x+\frac{17}{36}$

Square region $S=\left[0, 1\right]\times \left[0, 1\right]$

Green region given by $G=\left\{\left(x, y\right)\in S:y>f\left(x\right)\right\}$

Red region is given by $R=\left\{\left(x, y\right)\in S:y<f\left(x\right)\right\}$

And $L_h=\left\{\left(x, h\right)\in S:x\in \left[0, 1\right]\right\}$ be the horizontal line drawn at a height $h\in \left[0, 1\right]$

Now plotting the diagram of the above function we get,

Now differentiating the function,

$f\left(x\right)=\frac{x^3}{3}-x^2+\frac{5}{9}x+\frac{17}{36}$

$\Rightarrow f^{\text{'}}\left(x\right)=x^2-2x+\frac{5}{9}$

For maxima/minima, $f^{\text{'}}\left(x\right)=0\Rightarrow x=\frac{1}{3}$

Now finding the area of red region we get,

$A_R={\int }_0^{1}f\left(x\right) dx=\frac{1}{2}$

So, area of the green region will be,

$A_G=\frac{1}{2}$, as total area is $1$ sq.unit

Now solving option $\left(\mathrm{A}\right)$ we get,

$1-h=h-\frac{1}{2}\Rightarrow h=\frac{3}{4}, \frac{3}{4}>\frac{2}{3}$

So, option $\left(\mathrm{A}\right)$ is incorrect

For option $\left(\mathrm{B}\right)$ $h=\frac{1}{2}-h\Rightarrow h=\frac{1}{4}$

So, option $\left(\mathrm{B}\right)$ is correct as $h\in \left[\frac{1}{4},\frac{2}{3}\right]$

Now solving option $\left(\mathrm{C}\right)$ we get,

${\int }_0^{1}f\left(x\right) dx=\frac{1}{2},{\int }_0^1\frac{1}{2}dx=\frac{1}{2}\Rightarrow {\int }_0^1\left(f\left(x\right)-\frac{1}{2}\right)dx=0$

$\Rightarrow h=\frac{1}{2}$

So, option $\left(\mathrm{C}\right)$ is correct.

$\left(\mathrm{D}\right)$ $\because$ Option $\left(\mathrm{C}\right)$ is correct $\Rightarrow$ option $\left(\mathrm{D}\right)$ is also correct.