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Question

Mathematics Question on applications of integrals

Let the straight line x=bx = b divide the area enclosed by y=(1x)2y = (1 - x)^2 , y=0y = 0 and x=0x = 0 into two parts R1(0xb)R_1(0 \le x \le b) and R2(bx1)R_2(b \le x \le 1) such that R1R2=14.R_1-R_2 = \frac {1}{4}. Then, bb equals

A

34\frac{3}{4}

B

12\frac{1}{2}

C

13\frac{1}{3}

D

14\frac{1}{4}

Answer

12\frac{1}{2}

Explanation

Solution

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


R1=0bydx\therefore R_{1}=\int_{0}^{b} y \,dx
=0b(1x)2dx=[(1x)33]0b=\int_{0}^{b} \left(1-x\right)^{2} dx=\left[-\frac{\left(1-x\right)^{3}}{3}\right]_{0}^{b}
=13[(1b)31]=\frac{-1}{3}\left[\left(1-b\right)^{3}-1\right] and R2=1bydxR_{2}=\int_{1}^{b} y' dx
=1b(1x)2dx=[(1x)33]1b=\int^{b}_{1}\left(1-x\right)^{2} dx =\left[-\frac{\left(1-x\right)^{3}}{3}\right]_{1}^{b}
=13[(1b)330]=-\frac{1}{3}\left[-\frac{\left(1-b\right)^{3}}{3}-0\right]
=+13[(1b)3]=+\frac{1}{3}\left[\left(1-b\right)^{3}\right]
But it is given, R1R2=14R_{1}-R_{2}=\frac{1}{4}
13[(1b)31]13[(1b)3]=14\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}
13[2(1b)31]=14\Rightarrow -\frac{1}{3}\left[2\left(1-b\right)^{3}-1\right]=\frac{1}{4}
2(1b)3=34+1\Rightarrow 2\left(1-b\right)^{3}=-\frac{3}{4}+1
(1b)3=1/8\Rightarrow \left(1-b\right)^{3}=1 /8
1b=1/2\Rightarrow 1-b=1/ 2
b=112\Rightarrow b=1-\frac{1}{2}
=12=\frac{1}{2}