Area Question 22
Question 22
- If 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 (a) $\frac{3}{4}$ (b) $\frac{1}{2}$ (c) $\frac{1}{3}$ (d) $\frac{1}{4}$
(2011)
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Solution:
- Here, area between 0 to $b$ is $R_{1}$ and $b$ to 1 is $R_{2}$.
$\therefore \quad \int_{0}^{b}(1-x)^{2} d x-\int_{b}^{1}(1-x)^{2} d x=\frac{1}{4}$
$$ \Rightarrow \quad \frac{(1-x)^{3}}{-3}{ }{0}^{b}-{\frac{(1-x)^{3}}{-3}}{b}^{1}=\frac{1}{4} $$
$\Rightarrow-\frac{1}{3}\left[(1-b)^{3}-1\right]+\frac{1}{3}\left[0-(1-b)^{3}\right]=\frac{1}{4}$
$\Rightarrow \quad-\frac{2}{3}(1-b)^{3}=-\frac{1}{3}+\frac{1}{4}=-\frac{1}{12} \Rightarrow(1-b)^{3}=\frac{1}{8}$
$\Rightarrow \quad(1-b)=\frac{1}{2} \quad \Rightarrow \quad b=\frac{1}{2}$
