Let the straight line x -b divide the area enclosed by y-1-x-2- y-0- and x-0 into two parts R-1-0leq xleq b- and R-2-bleq xleq 1- such that R-1-R-2-displaystyle frac-1-4- Then b equalsdisplaystyle frac-3-4-displaystyle frac-1-2-displaystyle frac-1-3-displaystyle frac-1-4-

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Toppr Admin · Accepted Answer

R1-x222B-b0-x-x2212-1-2dx-xA0- -xA0- -xA0- -x-x2212-1-33-b0-b-x2212-1-3-13R2-x222B-1b-x-x2212-1-2dx-xA0- -xA0- -xA0- -x-x2212-1-33-1b-x2212-b-x2212-1-33

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
$\frac{3}{4}$
$\frac{1}{2}$
$\frac{1}{3}$
$\frac{1}{4}$

$R_{1} = \int_{0}^{b} {(x - 1)}^{2} d x$
$= {[\frac{{(x - 1)}^{3}}{3}]}_{0}^{b} = \frac{{(b - 1)}^{3} + 1}{3}$
$R_{2} = \int_{b}^{1} {(x - 1)}^{2} d x$
$= {[\frac{{(x - 1)}^{3}}{3}]}_{b}^{1} = - \frac{{(b - 1)}^{3}}{3}$

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 R1−R2=14. Then b equals34121314

R1=∫b0(x−1)2dx =[(x−1)33]b0=(b−1)3+13R2=∫1b(x−1)2dx =[(x−1)33]1b=−(b−1)33

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
$\frac{3}{4}$
$\frac{1}{2}$
$\frac{1}{3}$
$\frac{1}{4}$

$R_{1} = \int_{0}^{b} {(x - 1)}^{2} d x$
$= {[\frac{{(x - 1)}^{3}}{3}]}_{0}^{b} = \frac{{(b - 1)}^{3} + 1}{3}$
$R_{2} = \int_{b}^{1} {(x - 1)}^{2} d x$
$= {[\frac{{(x - 1)}^{3}}{3}]}_{b}^{1} = - \frac{{(b - 1)}^{3}}{3}$