Division of Area Enclosed by Parabola into Two Parts

To find the value of $b$, we first determine the total area enclosed by the curve $y = (1-x)^2$, the $x$-axis ($y = 0$), and the line $x = 0$.

The curve $y = (1-x)^2$ intersects the $x$-axis at $x = 1$. Thus, the total area $R$ under the curve from $x = 0$ to $x = 1$ is given by:
$R = \int_{0}^{1} (1-x)^2 \, dx$

Using the substitution method or direct integration:
$R = \left[ -\frac{(1-x)^3}{3} \right]_{0}^{1} = 0 - \left( -\frac{1}{3} \right) = \frac{1}{3}$

The line $x = b$ (where $0 \le b \le 1$) divides this region into two parts, $R_1$ and $R_2$, defined as:
$R_1 = \int_{0}^{b} (1-x)^2 \, dx = \left[ -\frac{(1-x)^3}{3} \right]_{0}^{b} = \frac{1 - (1-b)^3}{3}$

$R_2 = \int_{b}^{1} (1-x)^2 \, dx = \left[ -\frac{(1-x)^3}{3} \right]_{b}^{1} = \frac{(1-b)^3}{3}$

We are given that $R_1 - R_2 = \frac{1}{4}$. Substituting the expressions for $R_1$ and $R_2$:
$\frac{1 - (1-b)^3}{3} - \frac{(1-b)^3}{3} = \frac{1}{4}$

$\frac{1 - 2(1-b)^3}{3} = \frac{1}{4}$

Multiplying both sides by $3$:
$1 - 2(1-b)^3 = \frac{3}{4}$

$2(1-b)^3 = 1 - \frac{3}{4} = \frac{1}{4}$

$(1-b)^3 = \frac{1}{8}$

Taking the cube root on both sides:
$1 - b = \frac{1}{2}$

$b = 1 - \frac{1}{2} = \frac{1}{2}$

Thus, the value of $b$ is $\frac{1}{2}$, which corresponds to Option (B).