Discussion Forum

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 $R_{2}(b\leq x \leq1)$   such that $R_{1}-R_{2}=\frac{1}{4}$

Answer:

Explanation:

Here area between 0 to b is $R_{1}$ and b to 1 is $R_{2}$

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

$\Rightarrow$  $\left(\frac{(1-x)^{3}}{-3}\right)_{0}^{b}-\left(\frac{(1-x)^{3}}{-3}\right)^{1}_{b}=\frac{1}{4}$

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

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

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

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