Let the straight line x-b divide the area enclosed by y-1-x2- y-0 and x-0 into two parts R10- x- b and R20- x- 1 such that R1-R2-14 then b equals

Let the straight line x=b divide the area enclosed by y= (1-x)², y=0 and x=0 into two parts R₁(0≤x≤b) and R₂ (0≤x≤1) such that R₁-R₂ = 1/4 then b equals

Q. 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 \leq 1)

such that

R_{1} - R_{2} = \frac{1}{4} .

Then

b

equals

Q. The line x = 0 divides the area enclosed by the curves

| x - 1 | - y = 0, | x | + y - 3 = 0

into two area

R_{1}

and

R_{2}

where

R_{1} < R_{2}

. Then the ratio of

R_{1}

and

R_{2}

Q. If the line

x = α

divides the area of region

R = {(x, y) \in R^{2} : x^{3} \leq y \leq x, 0 \leq x \leq 1}

into two equal parts, then

Join BYJU'S Learning Program

Let the straight line x=b divide the area enclosed by y=(1−x)2, y=0 and x=0 into two parts R1(0≤x≤b) and R2(0≤x≤1) such that R1−R2=14 then b equals

The correct option is B 12 R1−R2=14b∫0(1−x2)dx−∫1b(1−x2)dx=1413−2(1−b)33=14(1−b)3=18b=12

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 $R_{2} (0 \leq x \leq 1)$ such that $R_{1} - R_{2} = \frac{1}{4}$ then $b$ equals

The correct option is B
$\frac{1}{2}$

$R_{1} - R_{2} = \frac{1}{4} b \int 0 (1 - x^{2}) d x - \int_{b}^{1} (1 - x^{2}) d x = \frac{1}{4} \frac{1}{3} - \frac{2 (1 - b)^{3}}{3} = \frac{1}{4} (1 - b)^{3} = \frac{1}{8} b = \frac{1}{2}$