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

$\frac{1}{2}$

Clearly,

$R_1=\int\limits_0^b(1-x)^2dx$ and $R_2=\int\limits_0^1(1-x^2)dx$

$⇒R_1=\left[\frac{(x-1)^3}{3}\right]_0^b$ and, $R_2=\left[\frac{(x-1)^3}{3}\right]_b^1$

$⇒R_1=\frac{(b-1)^3}{3}+\frac{1}{3}$ and, $R_2=-\frac{(b-1)^3}{3}$

$∴R_1-R_2=\frac{1}{4}$

$⇒\frac{2}{3}(b-1)^3=-\frac{1}{12}$

$⇒(b-1)^3=-\frac{1}{8}⇒b-1=-\frac{1}{2}⇒b=\frac{1}{2}$