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)(0lexleb) and R_(2)(blexle1)` such that `R_(1)-R_(2)=(1)/(4).` Then b equals

Options

A. `3//4`
B. `1//2`
C. `1//3`
D. `1//4`

Explanation

To solve the problem, we need to find the value of \( b \) such that the area \( R_1 \) (from \( x = 0 \) to \( x = b \)) minus the area \( R_2 \) (from \( x = b \) to \( x = 1 \)) equals \( \frac{1}{4} \). The area is enclosed by the curve \( y = (1 - x)^2 \) and the x-axis.

### Step 1: Determine the total area under the curve
The total area \( A \) under the curve from \( x = 0 \) to \( x = 1 \) can be calculated using the integral:

\[ A = \int_0^1 (1 - x)^2 \, dx \]

### Step 2: Calculate the integral
To compute the integral:

\[ A = \int_0^1 (1 - 2x + x^2) \, dx \]

Calculating this step-by-step:

1. Integrate each term:
\[ = \left[ x - x^2 + \frac{x^3}{3} \right]_0^1 \]

2. Evaluate at the limits:
\[ = \left( 1 - 1 + \frac{1}{3} \right) - \left( 0 - 0 + 0 \right) = \frac{1}{3} \]

### Step 3: Set up the areas \( R_1 \) and \( R_2 \)
The areas \( R_1 \) and \( R_2 \) can be expressed as:

\[ R_1 = \int_0^b (1 - x)^2 \, dx \]
\[ R_2 = \int_b^1 (1 - x)^2 \, dx \]

### Step 4: Express \( R_2 \) in terms of \( R_1 \)
From the total area, we know:

\[ R_2 = A - R_1 = \frac{1}{3} - R_1 \]

### Step 5: Set up the equation based on the condition
Given that \( R_1 - R_2 = \frac{1}{4} \), we can substitute \( R_2 \):

\[ R_1 - \left(\frac{1}{3} - R_1\right) = \frac{1}{4} \]

This simplifies to:

\[ 2R_1 - \frac{1}{3} = \frac{1}{4} \]

### Step 6: Solve for \( R_1 \)
Rearranging gives:

\[ 2R_1 = \frac{1}{4} + \frac{1}{3} \]

Finding a common denominator (12):

\[ 2R_1 = \frac{3}{12} + \frac{4}{12} = \frac{7}{12} \]

Thus,

\[ R_1 = \frac{7}{24} \]

### Step 7: Calculate \( b \) using \( R_1 \)
Now we need to find \( b \) such that:

\[ \int_0^b (1 - x)^2 \, dx = \frac{7}{24} \]

Following the same integration process:

\[ \int_0^b (1 - 2x + x^2) \, dx = \left[ x - x^2 + \frac{x^3}{3} \right]_0^b = b - b^2 + \frac{b^3}{3} \]

Setting this equal to \( \frac{7}{24} \):

\[ b - b^2 + \frac{b^3}{3} = \frac{7}{24} \]

### Step 8: Clear the fraction
Multiply through by 24 to eliminate the fraction:

\[ 24b - 24b^2 + 8b^3 = 7 \]

Rearranging gives:

\[ 8b^3 - 24b^2 + 24b - 7 = 0 \]

### Step 9: Solve the cubic equation
This cubic equation can be solved using numerical methods or graphing techniques. By testing possible rational roots, we find that \( b = \frac{1}{2} \) satisfies the equation.

### Final Answer
Thus, the value of \( b \) is:

\[ \boxed{\frac{1}{2}} \]