Let the straight line $x = b$ divide the area enclosed by $y=(1-x)^2$, $y=0$, and $x=0$ into two parts $R_1\,(0 \leq x \leq b)$ and $R_2\,(b \leq x \leq 1)$ such that $R_1 - R_2 = \frac{1}{4}$. Then $b$ equals

$y = (1-x)^2, \quad y = 0, \quad x = 0; \qquad \text{line } x = b$ $R_1 = \int_0^b (1-x)^2\,dx, \qquad R_2 = \int_b^1 (1-x)^2\,dx$ $R_1 - R_2 = \dfrac{1}{4}$ $\text{To find: } b$ $1.\ A = \int_0^1 (1-x)^2\,dx$ $2.\ R_1 + R_2 = A, \ R_1 - R_2 = \dfrac{1}{4}$ $3.\ R_1 = \int_0^b (1-x)^2\,dx, \text{ solve for } b$ $A = \int_0^1 (1-x)^2\,dx = \left[-\dfrac{(1-x)^3}{3}\right]_0^1 = \dfrac{1}{3}$ $R_1 + R_2 = A = \dfrac{1}{3}, \qquad R_1 - R_2 = \dfrac{1}{4}$ $2R_1 = \dfrac{1}{3} + \dfrac{1}{4} = \dfrac{7}{12} \Rightarrow R_1 = \dfrac{7}{24}$ $R_1 = \int_0^b (1-x)^2\,dx = \dfrac{1}{3} - \dfrac{(1-b)^3}{3} = \dfrac{7}{24}$ $\Rightarrow (1-b)^3 = \dfrac{1}{8}$ $1-b = \dfrac{1}{2} \Rightarrow b = \dfrac{1}{2} \ \Rightarrow \text{option } (2)$