Let the straight line x=b divide the area enc | vedclass

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(B) The total area $A$ is given by $\int_0^1 (1-x)^2 dx = \left[ -\frac{(1-x)^3}{3} \right]_0^1 = 0 - (-\frac{1}{3}) = \frac{1}{3}$.The area $R_1$ is

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$\frac{1}{2}$

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

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

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