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If the straight-line $x=b$ divides the area enclosed by $y=(1-x)_{-}^2$ $y=0$ and $x=0$ in 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 $\mathrm{b}$ equals
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$\frac{1}{2}$
$\begin{aligned} & \int_0^b(1-x)^2 \mathrm{~d} x-\int_b^1(1-x)^2 \mathrm{~d} x=\frac{1}{4} \\ & \Rightarrow\left[\frac{(1-x)^3}{-3}\right]_0^b-\left[\frac{(1-x)^3}{-3}\right]_b^1=\frac{1}{4}\end{aligned}$
$\begin{aligned} & \Rightarrow \frac{(1-b)^3-(1-0)^3-(1-1)^3+(1-b)^3}{-3}=\frac{1}{4} \\ & 2(1-b)^3-1=\frac{-3}{4} \\ & \Rightarrow 2(1-b)^3-1=\frac{-3}{4} \\ & \Rightarrow(1-b)^3=\frac{1}{8} \\ & \Rightarrow b=\frac{1}{2}\end{aligned}$
$\begin{aligned} & \Rightarrow \frac{(1-b)^3-(1-0)^3-(1-1)^3+(1-b)^3}{-3}=\frac{1}{4} \\ & 2(1-b)^3-1=\frac{-3}{4} \\ & \Rightarrow 2(1-b)^3-1=\frac{-3}{4} \\ & \Rightarrow(1-b)^3=\frac{1}{8} \\ & \Rightarrow b=\frac{1}{2}\end{aligned}$
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