Question

Let the straight line $x=b$ divide the area enclosed by $y=(1-x{)}^2,y=0$ and $x=0$ into two parts $R_1(0\le x\le b)$ and $R_2(b\le x\le 1)$ 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 1

The correct option is B $\frac{1}{2}$

$R_1=\underset{0}{\overset{b}{\int }}(x-1{)}^{2}dx={\left[\frac{(x-1{)}^3}{3}\right]}_0^b=\frac{(b-1{)}^3+1}{3}$
$R_2=\underset{b}{\overset{1}{\int }}(x-1{)}^{2}dx={\left[\frac{(x-1{)}^3}{3}\right]}_b^1=-\frac{(b-1{)}^3}{3}$
As $R_1-R_2=\frac{1}{4}$
$\Rightarrow \frac{2(b-1{)}^3}{3}+\frac{1}{3}=\frac{1}{4}$
$\Rightarrow (b-1{)}^3=-\frac{1}{8}$
$\Rightarrow b-1=-\frac{1}{2}\Rightarrow b=\frac{1}{2}$