Question
Let the straight line $$x = b$$ divide the area enclosed by $$y = {\left( {1 - x} \right)^2},\,y = 0$$ and $$x=0$$ into two parts $${R_1}\left( {0 \leqslant x \leqslant b} \right)$$ and $${R_2}\left( {b \leqslant x \leqslant 1} \right)$$ such that $${R_1} - {R_2} = \frac{1}{4}.$$ Then $$b$$ equals-
A.
$$\frac{3}{4}$$
B.
$$\frac{1}{2}$$
C.
$$\frac{1}{3}$$
D.
$$\frac{1}{4}$$
Answer :
$$\frac{1}{2}$$
Solution :
$${R_1} = \int_a^b {{{\left( {x - 1} \right)}^2}dx} = \left[ {\frac{{{{\left( {x - 1} \right)}^3}}}{3}} \right]_0^b = \frac{{{{\left( {b - 1} \right)}^3} + 1}}{3}$$
$$\eqalign{ & {R_2} = \int_b^1 {{{\left( {x - 1} \right)}^2}dx} = \left[ {\frac{{{{\left( {x - 1} \right)}^3}}}{3}} \right]_b^1 = - \frac{{{{\left( {b - 1} \right)}^3}}}{3} \cr & {\text{As }}{R_1} - {R_2} = \frac{1}{4}\,\, \Rightarrow \frac{{2{{\left( {b - 1} \right)}^3}}}{3} + \frac{1}{3} = \frac{1}{4} \cr & {\text{or }}{\left( {b - 1} \right)^3} = - \frac{1}{8} \cr & {\text{or }}b - 1 = \frac{{ - 1}}{2} \cr & {\text{or }}b = \frac{1}{2} \cr} $$
$${R_1} = \int_a^b {{{\left( {x - 1} \right)}^2}dx} = \left[ {\frac{{{{\left( {x - 1} \right)}^3}}}{3}} \right]_0^b = \frac{{{{\left( {b - 1} \right)}^3} + 1}}{3}$$
$$\eqalign{ & {R_2} = \int_b^1 {{{\left( {x - 1} \right)}^2}dx} = \left[ {\frac{{{{\left( {x - 1} \right)}^3}}}{3}} \right]_b^1 = - \frac{{{{\left( {b - 1} \right)}^3}}}{3} \cr & {\text{As }}{R_1} - {R_2} = \frac{1}{4}\,\, \Rightarrow \frac{{2{{\left( {b - 1} \right)}^3}}}{3} + \frac{1}{3} = \frac{1}{4} \cr & {\text{or }}{\left( {b - 1} \right)^3} = - \frac{1}{8} \cr & {\text{or }}b - 1 = \frac{{ - 1}}{2} \cr & {\text{or }}b = \frac{1}{2} \cr} $$