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

Given, curve $y=(1-x)^{2}$ is a parabola, which open upward

$\therefore R_{1}=\int_{0}^{b} y \,dx $
$=\int_{0}^{b} \left(1-x\right)^{2} dx=\left[-\frac{\left(1-x\right)^{3}}{3}\right]_{0}^{b}$
$=\frac{-1}{3}\left[\left(1-b\right)^{3}-1\right]$ and $R_{2}=\int_{1}^{b} y' dx$
$=\int^{b}_{1}\left(1-x\right)^{2} dx =\left[-\frac{\left(1-x\right)^{3}}{3}\right]_{1}^{b}$
$=-\frac{1}{3}\left[-\frac{\left(1-b\right)^{3}}{3}-0\right]$
$=+\frac{1}{3}\left[\left(1-b\right)^{3}\right]$
But it is given, $R_{1}-R_{2}=\frac{1}{4}$
$\therefore -\frac{1}{3}\left[\left(1-b\right)^{3}-1\right]-\frac{1}{3}\left[\left(1-b\right)^{3}\right]=\frac{1}{4}$
$\Rightarrow -\frac{1}{3}\left[2\left(1-b\right)^{3}-1\right]=\frac{1}{4}$
$\Rightarrow 2\left(1-b\right)^{3}=-\frac{3}{4}+1$
$\Rightarrow \left(1-b\right)^{3}=1 /8$
$\Rightarrow 1-b=1/ 2$
$\Rightarrow b=1-\frac{1}{2}$
$=\frac{1}{2}$

Answer 2

Given that;
Straight line x=b
area enclosed by \(y=(1-x)^2,y=0,x=0\)
\(R_1(0≤x≤b)\) and \(R_2(b≤x≤1)\) such that \(R_1-R_2=\frac{1}{2}\)
\(\int^b_0(1-x)^2dx-\int^1_b(1-x)^2dx=\frac{1}{4}\)
⇒\([\frac{(x-1)^3}{3}]^6_0-[\frac{(x-1)^3}{3}]^1_b=\frac{1}{4}\)
⇒\(\frac{2(b-1)^3}{3}=-\frac{1}{12}\)
⇒\((b-1)^3=-\frac{1}{8}\)
⇒\(b=\frac{1}{2}\)