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 $\frac{1}{2}$

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

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