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 $R_1 - R_2 = 1/4$. Then $b$ equals Options: (A) 3/4 (B) 1/2 (C) 1/3 (D) 1/4

### Analyzing the Setup We are examining the curve $y = (1-x)^2$. This parabola intersects the $x$-axis at $x=1$ and starts at $y=1$ when $x=0$. The region is bounded by the $y$-axis ($x=0$), the $x$-axis ($y=0$), and the curve. We introduce a vertical cut at $x=b$, dividing the region into two parts: $R_1$ (left) and $R_2$ (right). Our objective is to determine the value of $b$ such that the difference between these two areas is exactly $\frac{1}{4}$, specifically $R_1 - R_2 = \frac{1}{4}$. ### The Calculus of Areas To find the area of the left region $R_1$, we integrate the function from $x=0$ to $x=b$: $$R_1 = \int_{0}^{b} (1-x)^2 dx$$ Using the power rule for integration, the antiderivative of $(1-x)^2$ is $-\frac{(1-x)^3}{3}$. Evaluating this from $0$ to $b$: $$R_1 = \left[ -\frac{(1-x)^3}{3} \right]_{0}^{b} = -\frac{(1-b)^3}{3} - \left( -\frac{(1-0)^3}{3} \right) = \frac{1 - (1-b)^3}{3}$$ ### The Symmetry of the Right Side The right region $R_2$ spans from the cut $x=b$ to the boundary $x=1$. We set up the integral as follows: $$R_2 = \int_{b}^{1} (1-x)^2 dx$$ Evaluating the antiderivative at the given limits: $$R_2 = \left[ -\frac{(1-x)^3}{3} \right]_{b}^{1} = -\frac{(1-1)^3}{3} - \left( -\frac{(1-b)^3}{3} \right) = \frac{(1-b)^3}{3}$$ ### The Algebraic Dance We apply the condition $R_1 - R_2 = \frac{1}{4}$ by substituting our derived expressions: $$\frac{1 - (1-b)^3}{3} - \frac{(1-b)^3}{3} = \frac{1}{4}$$ Combining the terms over the common denominator, we obtain: $$\frac{1 - 2(1-b)^3}{3} = \frac{1}{4}$$ Cross-multiplying to solve for the variable term: $$4(1 - 2(1-b)^3) = 3$$ $$4 - 8(1-b)^3 = 3$$ $$8(1-b)^3 = 1$$ ### The Final Revelation We isolate the cubic term: $$(1-b)^3 = \frac{1}{8}$$ Taking the cube root of both sides yields: $$1 - b = \frac{1}{2}$$ Solving for $b$, we find the final position of the vertical cut: **$b = \frac{1}{2}$**

JEE Advanced 2011

LEVELJEE Main

Animated Solution for Mathematics - Definite Integration: 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 $R_{1} - R_{2} = 1/4$ . Then $b$ equals

Q: I keep getting the sign wrong when integrating $(1-x)^2$. Why is there a negative sign?

Ah, the classic trap! When you integrate $(1-x)^2$, the inner function is $u = 1-x$, so $du = -dx$. The negative sign arises from the derivative of the inner function. Always remember: $\int (ax+b)^n dx = \frac{(ax+b)^{n+1}}{a(n+1)}$. Here $a = -1$, so the negative sign is mandatory. Don't let it vanish!

Q: Is $b$ always between $0$ and $1$?

Yes, geometrically, the region only exists between $x=0$ and $x=1$. If your algebra yields a $b$ outside this range, re-check your integration limits. The line $x=b$ must slice through the existing area, so $0 < b < 1$ is a physical necessity.

Select Answer:

Visualized Solution

Visualizing the Curve

Consider the curve $y = (1 - x)^{2}$ .
The area is bounded by $y = (1 - x)^{2}$ , the x-axis ( $y = 0$ ), and the y-axis ( $x = 0$ ).
The curve intersects the x-axis at $x = 1$ .

Defining the Regions $R_{1}$ and $R_{2}$

A vertical line $x = b$ divides the area into two regions.
Region $R_{1}$ : from $x = 0$ to $x = b$ .
Region $R_{2}$ : from $x = b$ to $x = 1$ .

The Given Condition

We are given the condition: $R_{1} - R_{2} = \frac{1}{4}$ .
To use this, we need to find the areas of $R_{1}$ and $R_{2}$ using definite integrals.

Setting up the Integral for $R_{1}$

The area $R_{1}$ is the integral of the curve from $0$ to $b$ .
$R_{1} = \int_{0}^{b} (1 - x)^{2} d x$

Evaluating $R_{1}$

Using the power rule: $\int (1 - x)^{2} d x = - \frac{( 1 - x ) ^{3}}{3}$
$R_{1} = [- \frac{( 1 - x ) ^{3}}{3}]_{0}^{b}$
$R_{1} = - \frac{( 1 - b ) ^{3}}{3} - (- \frac{1}{3})$
$R_{1} = \frac{1 - ( 1 - b ) ^{3}}{3}$

Setting up the Integral for $R_{2}$

The area $R_{2}$ is the integral of the curve from $b$ to $1$ .
$R_{2} = \int_{b}^{1} (1 - x)^{2} d x$

Evaluating $R_{2}$

Using the same antiderivative: $R_{2} = [- \frac{( 1 - x ) ^{3}}{3}]_{b}^{1}$
Evaluating at upper limit $1$ : $- \frac{( 1 - 1 ) ^{3}}{3} = 0$
$R_{2} = 0 - (- \frac{( 1 - b ) ^{3}}{3})$
$R_{2} = \frac{( 1 - b ) ^{3}}{3}$

Substituting into the Condition

Given: $R_{1} - R_{2} = \frac{1}{4}$
Substitute the expressions we found:
$\frac{1 - ( 1 - b ) ^{3}}{3} - \frac{( 1 - b ) ^{3}}{3} = \frac{1}{4}$

Combining the Terms

Since the denominators are the same, combine the numerators:
$\frac{1 - ( 1 - b ) ^{3} - ( 1 - b ) ^{3}}{3} = \frac{1}{4}$
$\frac{1 - 2 ( 1 - b ) ^{3}}{3} = \frac{1}{4}$

Cross-Multiplying

Cross-multiply to eliminate fractions:
$4 \cdot (1 - 2 (1 - b)^{3}) = 3 \cdot 1$
$4 - 8 (1 - b)^{3} = 3$

Solving for the Cube Term

Rearrange the equation to isolate the cube term:
$4 - 3 = 8 (1 - b)^{3}$
$1 = 8 (1 - b)^{3}$
$(1 - b)^{3} = \frac{1}{8}$

Finding the Final Value of $b$

Take the cube root of both sides:
$1 - b = 3 \frac{1}{8}$
$1 - b = \frac{1}{2}$
$b = 1 - \frac{1}{2} = \frac{1}{2}$
The line $x = \frac{1}{2}$ divides the area as required.

00:00 / 00:00

The Sigma Insight: Area Bounded by Curves

Solution Diagram

Analyzing the Setup

We are examining the curve

y = (1 - x)^{2}

. This parabola intersects the

x

-axis at

x = 1

and starts at

y = 1

when

x = 0

The region is bounded by the

y

-axis (

x = 0

), the

x

-axis (

y = 0

), and the curve. We introduce a vertical cut at

x = b

, dividing the region into two parts:

R_{1}

(left) and

R_{2}

(right).

Our objective is to determine the value of

b

such that the difference between these two areas is exactly

\frac{1}{4}

, specifically

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

The Calculus of Areas

To find the area of the left region

R_{1}

, we integrate the function from

x = 0

x = b

R_{1} = \int_{0}^{b} (1 - x)^{2} d x

Using the power rule for integration, the antiderivative of

(1 - x)^{2}

- \frac{( 1 - x ) ^{3}}{3}

. Evaluating this from

0

b

R_{1} = [- \frac{( 1 - x ) ^{3}}{3}]_{0}^{b} = - \frac{( 1 - b ) ^{3}}{3} - (- \frac{( 1 - 0 ) ^{3}}{3}) = \frac{1 - ( 1 - b ) ^{3}}{3}

The Symmetry of the Right Side

The right region

R_{2}

spans from the cut

x = b

to the boundary

x = 1

. We set up the integral as follows:

R_{2} = \int_{b}^{1} (1 - x)^{2} d x

Evaluating the antiderivative at the given limits:

R_{2} = [- \frac{( 1 - x ) ^{3}}{3}]_{b}^{1} = - \frac{( 1 - 1 ) ^{3}}{3} - (- \frac{( 1 - b ) ^{3}}{3}) = \frac{( 1 - b ) ^{3}}{3}

The Algebraic Dance

We apply the condition

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

by substituting our derived expressions:

\frac{1 - ( 1 - b ) ^{3}}{3} - \frac{( 1 - b ) ^{3}}{3} = \frac{1}{4}

Combining the terms over the common denominator, we obtain:

\frac{1 - 2 ( 1 - b ) ^{3}}{3} = \frac{1}{4}

Cross-multiplying to solve for the variable term:

4 (1 - 2 (1 - b)^{3}) = 3

4 - 8 (1 - b)^{3} = 3

8 (1 - b)^{3} = 1

The Final Revelation

We isolate the cubic term:

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

Taking the cube root of both sides yields:

1 - b = \frac{1}{2}

Solving for

b

, we find the final position of the vertical cut:

$b = \frac{1}{2}$

Common Traps & FAQs

Why can't we just use the total area to solve this?

You absolutely can, and that is a brilliant observation! The total area

A

\int_{0}^{1} (1 - x)^{2} d x = \frac{1}{3}

. Since

R_{1} + R_{2} = \frac{1}{3}

and

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

, you have a system of two linear equations. Adding them gives

2 R_{1} = \frac{1}{3} + \frac{1}{4} = \frac{7}{12}

, so

R_{1} = \frac{7}{24}

. This is a fantastic shortcut that saves time in the exam hall!

I keep getting the sign wrong when integrating

(1 - x)^{2}

. Why is there a negative sign?

Ah, the classic trap! When you integrate

(1 - x)^{2}

, the inner function is

u = 1 - x

, so

d u = - d x

. The negative sign arises from the derivative of the inner function. Always remember:

\int (a x + b)^{n} d x = \frac{( a x + b ) ^{n + 1}}{a ( n + 1 )}

. Here

a = - 1

, so the negative sign is mandatory. Don't let it vanish!

b

always between

0

and

1

Yes, geometrically, the region only exists between

x = 0

and

x = 1

. If your algebra yields a

b

outside this range, re-check your integration limits. The line

x = b

must slice through the existing area, so

0 < b < 1

is a physical necessity.

Animated Solution for Mathematics - Definite Integration: 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 $R_{1} - R_{2} = 1/4$ . Then $b$ equals

Select Answer:

Visualized Solution

Visualizing the Curve

Defining the Regions $R_{1}$ and $R_{2}$

The Given Condition

Setting up the Integral for $R_{1}$

Evaluating $R_{1}$

Setting up the Integral for $R_{2}$

Evaluating $R_{2}$

Substituting into the Condition

Combining the Terms

Cross-Multiplying

Solving for the Cube Term

Finding the Final Value of $b$

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

The Calculus of Areas

The Symmetry of the Right Side

The Algebraic Dance

The Final Revelation

Similar Questions

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If the area enclosed between the curves $y = k x^{2}$ and $x = k y^{2}, (k > 0)$ , is 1 square unit. Then k is:

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Select Answer:

Visualized Solution

Visualizing the Curve

Defining the Regions .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R1​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R2​

The Given Condition

Setting up the Integral for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R1​

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R1​

Setting up the Integral for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R2​

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R2​

Substituting into the Condition

Combining the Terms

Cross-Multiplying

Solving for the Cube Term

Finding the Final Value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

The Calculus of Areas

The Symmetry of the Right Side

The Algebraic Dance

The Final Revelation

Similar Questions

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Defining the Regions $R_{1}$ and $R_{2}$

Setting up the Integral for $R_{1}$

Evaluating $R_{1}$

Setting up the Integral for $R_{2}$

Evaluating $R_{2}$

Finding the Final Value of $b$

If the line $x = a$ 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

If the line $y = m x$ bisects the area enclosed by the lines $x = 0, y = 0, x = \frac{3}{2}$ and the curve $y = 1 + 4 x - x^{2}$ , then $12 m$ is equal to .

The area enclosed between the curves $y = a x^{2}$ and $x = a y^{2}$ ( $a > 0$ ) is 1 sq. unit, then the value of $a$ is

If the area (in sq. units) of the region ${(x, y) : y^{2} \leq 4 x, x + y \leq 1, x \geq 0, y \geq 0}$ is $a 2 + b$ , then $a - b$ is equal to :

Let $S$ be the area of the region enclosed by $y = e^{- x^{2}}, y = 0, x = 0$ and $x = 1$ ; then

The area bounded by the parabolas $y = (x + 1)^{2}$ and $y = (x - 1)^{2}$ and the line $y = 1/4$ is

If the area enclosed between the curves $y = k x^{2}$ and $x = k y^{2}, (k > 0)$ , is 1 square unit. Then k is:

The area bounded by the curves $y = ∣ x^{2} - 1∣$ and $y = 1$ is

If the area (in sq. units) bounded by the parabola $y^{2} = 4 λ x$ and the line $y = λ x, λ > 0$ , is $1/9$ , then $λ$ is equal to :

If the area of the region bounded by the curves, $y = x^{2}, y = \frac{1}{x}$ and the lines $y = 0$ and $x = t (t > 1)$ is 1 sq. unit, then t is equal to :