Let the straight line x = b divide the area enclosed by y = (1 – x)2, y = 0 and x = 0 into two parts R1 (0 ≤ x ≤ b) and R2 (b ≤ x ≤ 1) such that R_{1}

Engineering

Mathematics

Introduction to Area

Question

Let the straight line x = b divide the area enclosed by y = (1 – x)², y = 0 and x = 0 into two parts R₁ (0 ≤ x ≤ b) and R₂ (b ≤ x ≤ 1) such that $R_{1} - R_{2} = \frac{1}{4}$ . Then b equals

$\frac{1}{2}$

$\frac{1}{4}$

$\frac{1}{3}$

$\frac{3}{4}$

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$R_{1} - R_{2} = \frac{1}{4}$

$\int_{0}^{b} {(x - 1)}^{2} dx - \int_{b}^{1} {(x - 1)}^{2} dx = \frac{1}{4}$

${[\frac{{(x - 1)}^{3}}{3}]}_{0}^{b} - {[\frac{{(x - 1)}^{3}}{3}]}_{b}^{1} = \frac{1}{4}$

$\frac{{(b - 1)}^{3}}{3} + \frac{1}{3} - 0 + \frac{{(b - 1)}^{3}}{3} = \frac{1}{4}$

$\frac{2 {(b - 1)}^{3}}{3} = \frac{1}{4} - \frac{1}{3} = \frac{- 1}{12}$

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

$b - 1 = - \frac{1}{8} \Rightarrow b = \frac{1}{2}$

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