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(A) The area $R_1$ is given by $\int_0^b (1-x)^2 \, dx$ and the area $R_2$ is given by $\int_b^1 (1-x)^2 \, dx$.Given $R_1 - R_2 = \frac{1}{4}$,we hav

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$\frac{1}{2}$

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(A) The area $R_1$ is given by $\int_0^b (1-x)^2 \, dx$ and the area $R_2$ is given by $\int_b^1 (1-x)^2 \, dx$.Given $R_1 - R_2 = \frac{1}{4}$,we have:$\int_0^b (1-x)^2 \, dx - \int_b^1 (1-x)^2 \, dx = \frac{1}{4}$Evaluating the integrals:$\left[ \frac{-(1-x)^3}{3} \right]_0^b - \left[ \frac{-(1-x)^3}{3} \right]_b^1 = \frac{1}{4}$$\left( \frac{-(1-b)^3}{3} - \frac{-(1-0)^3}{3} \right) - \left( \frac{-(1-1)^3}{3} - \frac{-(1-b)^3}{3} \right) = \frac{1}{4}$$\left( \frac{1 - (1-b)^3}{3} \right) - \left( \frac{(1-b)^3}{3} \right) = \frac{1}{4}$$\frac{1 - 2(1-b)^3}{3} = \frac{1}{4}$$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}$$1-b = \frac{1}{2}$$b = 1 - \frac{1}{2} = \frac{1}{2}$

If the straight line $x=b$ divides 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:

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